# A minimax method in shape and topological derivatives and homogenization:
the case of Helmholtz equation

###### Abstract

In this paper, we perform a rigourous version of shape and topological
derivatives for optimizations problems under constraint Helmoltz problems. A
shape and topological optimization problem is formulated by introducing cost
functional. We derive first by considering the lagradian method the shape
derivative of the functional. It is also proven a topological derivative with
the same approach. An application to several unconstrained shape functions
arising from differential geometry are also given.

Mame Gor Ngom<EMAIL_ADDRESS>
Université Alioune Diop de Bambey,

Ecole Doctorale des Sciences et Techniques et Sciences de la Société.

Laboratoire de Mathématiques de la Décision et d’Analyse Numérique

(L.M.D.A.N) F.A.S.E.G)/F.S.T.

Ibrahima Faye<EMAIL_ADDRESS>Diaraf Seck
<EMAIL_ADDRESS>
Université Alioune Diop de Bambey, UFR S.A.T.I.C, BP 30 Bambey (Sénégal),

Ecole Doctorale des Sciences et Techniques et Sciences de la Société.

Laboratoire de Mathématiques de la Décision et d’Analyse Numérique

(L.M.D.A.N).

Keywords: Minimax method, shape, topological derivatives, averaged Lagrangian,
homogenization, two scale convergence, Helmholtz equation.

##  1 Introduction

This paper deals with the notion of shape and topological derivative of
functional under Helmholtz equation as constraint. The shape derivative is a
differential on some appropriate metric spaces [9, 16] and references there
in. The topological derivative introduced by Sokolowski and Zochowski [30] is
only a semi differential, this arises from the fact that the tangent space to
the underlying metric space of geometries is only a cone. This type of
derivative is obtained by expansion methods. In [13], Delfour introduced the
notion of differentials and semi-differentials for metric spaces and
geometries. The semi differential of the objective function is obtained by
introducing a Lagrangian and the adjoint state of the initial problem. This
problem is equivalent to the derivative of the minimax of the Lagrangian with
respect to a positive parameter $s$ as it goes to 0. We have inspired by the
results of Sturm in his thesis [32] and Delfour and Sturm [12, 14], to
introduce the new approach for shape and topological derivative. For this, we
introduce first the basic notion in differentials calculus for geometries and
some methods to derive shape and topological derivative. Nowadays, some
interesting contributions have already been brought in this topic, a lof of
question are still open. We derive the minimax of a Lagrangian with respect to
a positive parameter $t.$ We improve the minimax approach [8] introduced by K.
Sturm [32] to find optimal conditions.

Let $\Omega$ be a bounded lipschitzien open domain in
$\mathbb{R}^{N},\,\,N\geq 2$ with a regular boundary and $\eta$ be the
solution of the following Helmhottz equation

$-\Delta\eta-k^{2}\eta=f\,\,\text{in}\,\,\Omega,\,\,\eta=0\,\,\text{on}\,\,\partial\Omega.$
(1.1)

Let us consider the functional defined by

$J(\Omega)=\int_{\Omega}|\nabla\eta-A|^{2}dx+\int_{\Omega}|\eta-\eta_{d}|^{2}dx$
(1.2)

where $\eta$ is solution to (1.1), $A$ is a given vector and $\eta_{d}$ a
given target. Our objective, in this paper, is to compute the shape and
topological derivative of functional $J(\Omega_{s})=j(\Omega_{s},\eta_{s}),$
where the perturbed domain $\Omega_{s}$ of $\Omega$ is defined by
$\Omega_{s}=T_{s}(\Omega)$ or $\Omega_{s}=\Omega\backslash E_{r}$ depending on
the derivative to be calculated. $\eta_{s}$ is solution to (1.1) posed in
$\Omega_{s}.$ The idea is to use variational techniques due to Sturm [32] and
Delfour [8]. Under useful hypotheses, Delfour [8] and Sturm [32] consider that
the topological derivative of the functional is equal to the s-derivative of
the Lagrangian. The calculation of the shape derivative or the topological
derivative is possible only if the limit of a certain quantity $R(s)$ exists.
In this work, still inspired by their work, we give a sufficient condition for
the existence of this limit and therefore the calculation of the shape or
topological derivatives of the functional $J(\Omega)=j(\Omega,\eta_{\Omega}),$
where $\eta_{\Omega}$ is solution to the constraint equation. This
necessitates the addition of a new assumption on the solution
$\eta_{\Omega_{s}}$ in the perturbed domain $\Omega_{s}.$ The rest of the
paper is organized as follows: in section 2, we present the main problem and
the result obtained from the Lagrangian method. This first result is an
adaptation of result due to Delfour [8] and Sturm [32]. In section 3, we
derive the topological gradient by using also the minimax Lagrangian method.
Some useful hypotheses and various cases are also given. We give also some
examples where it is not possible to obtain the topological derivative. In the
last section, we give our main result based on asymptotic analysis tools, to
ensure topological derivatives by using the two scales convergence in
homogenization.

##  2 Problem formulation and shape derivative

Let $\Omega\subset\mathbb{R}^{N},\,\,N\geq 2$ be a bounded lipschitzian open
with a regular boundary $\partial\Omega$. Let $\eta=\eta(\Omega)$ be the
solution of the following Helmholtz problem

$\left\\{\begin{array}[]{ccc}-\Delta\eta-k^{2}\eta=f\,\,\text{in}\;\;\Omega,\\\
\eta=0\,\,\text{on}\,\,\partial\Omega.\end{array}\right.$ (2.1)

We consider the functional $J$ defined by

$\displaystyle
J(\Omega)=\int_{\Omega}\lvert\nabla\eta-A\rvert^{2}dx+\int_{\Omega}\lvert\eta-\eta_{d}\rvert^{2}dx$
(2.2)

where $\eta$ is the solution of (2.1), $A\in(L^{2}(\Omega))^{N}$ a given
vector of $\mathbb{R}^{N}$ and $\eta_{d}\in L^{2}(\Omega)$ a given function.
We want to calculate the shape derivative of the functional $J$.
To compute the shape derivative of $J(\Omega)$ by the velocity method, we
perturb the bounded open domain $\Omega$ by a family of $T_{s}$
diffeomorphisms of $\mathbb{R}^{N}$, $s\geq 0$ generated by a sufficiently
regular velocity field $V$. The perturbed domain $\Omega_{s}$ is then defined
by $\Omega_{s}=T_{s}(\Omega)$, $0\leq s\leq\tau$, with $\tau>0$. In most cases
[27, 15] the shape derivative of $J$ in $\Omega$ the direction $V$ direction
is defined as

$\displaystyle dJ(\Omega;V)=\lim_{s\searrow
0}\;\;\frac{J(\Omega_{s})-J(\Omega)}{s}$

when the limit exists where $\Omega_{s}=T_{s}(\Omega)$. The idea here is to
use some techniques due to [8, 12] to compute the shape derivative of the
functional $J,$ defined in the perturbed domain $\Omega_{s}$ by

$\displaystyle
J(\Omega_{s})=\int_{\Omega_{s}}\lvert\nabla\eta_{s}-A\rvert^{2}dx+\int_{\Omega_{s}}\lvert\eta_{s}-\eta_{d}\rvert^{2}dx$
(2.3)

where $\eta_{s}$ is the solution to the problem

$\left\\{\begin{array}[]{ccc}-\Delta\eta_{s}-k^{2}\eta_{s}=f\;\;\text{in}\,\,\Omega_{s}\\\
\qquad\qquad\quad\eta_{s}=0\;\;\text{on}\,\,\partial\Omega_{s},\end{array}\right.$
(2.4)

with $\partial\Omega_{s}$ is the boundary of $\Omega_{s}$.
The variational formulation of (2.4) is to find $\eta_{s}\in
H^{1}_{0}(\Omega_{s})$ such that

$\displaystyle\int_{\Omega_{s}}\nabla\eta_{s}\nabla
v^{\prime}-k^{2}\eta_{s}v^{\prime}dx=\int_{\Omega_{s}}fv^{\prime}dx,\;\text{for
all}\;v^{\prime}\in H^{1}_{0}(\Omega_{s}).$ (2.5)

We see clearly that the space of functions $H^{1}_{0}(\Omega_{s})$ depends on
the parameter $s$. To obtain a space independent of $s$, we introduce the
following parametrization

$\displaystyle H^{1}_{0}(\Omega_{s})=\\{\varphi\circ T^{-1}_{s},\;\;\varphi\in
H^{1}_{0}(\Omega)\\}.$ (2.6)

Therefore to work in the fixed domain $H^{1}_{0}(\Omega)$, we introduce the
compositions $\eta^{s}=\eta_{s}\circ T_{s}$ and $v=v^{\prime}\circ T_{s}$. The
variational formulation of (2.5) then amounts to finding $\eta^{s}\in
H^{1}_{0}(\Omega)$ such that

$\displaystyle\int_{\Omega_{s}}\nabla(\eta^{s}\circ T_{s}^{-1})\nabla(v\circ
T_{s}^{-1})-k^{2}(\eta^{s}\circ T_{s}^{-1})(v\circ
T_{s}^{-1})dx=\int_{\Omega_{s}}f(v\circ T_{s}^{-1})dx,\;\;\forall\;v\in
H^{1}_{0}(\Omega).$ (2.7)

Applying the change of variables formula, we get

$\displaystyle\int_{\Omega}\left\\{\nabla(\eta^{s}\circ
T_{s}^{-1})\nabla(v\circ T_{s}^{-1})-k^{2}(\eta^{s}\circ T_{s}^{-1})(v\circ
T_{s}^{-1})\right\\}\circ T_{s}J_{s}dx=\int_{\Omega}\left\\{f(v\circ
T_{s}^{-1})\right\\}\circ T_{s}J_{s}dx.$

Furthermore, we have

$\displaystyle\int_{\Omega}\left\\{k^{2}(\eta^{s}\circ T_{s}^{-1})(v\circ
T_{s}^{-1})\right\\}\circ T_{s}J_{s}dx$ $\displaystyle=$
$\displaystyle\int_{\Omega}k^{2}(\eta^{s}\circ T_{s}^{-1}\circ T_{s})(v\circ
T_{s}^{-1}\circ T_{s})J_{s}dx$ $\displaystyle=$
$\displaystyle\int_{\Omega}k^{2}\eta^{s}vJ_{s}dx$
$\displaystyle\int_{\Omega}\left\\{f(v\circ T_{s}^{-1})\right\\}\circ
T_{s}J_{s}dx=\int_{\Omega}(f\circ T_{s})(v\circ T_{s}^{-1}\circ
T_{s})J_{s}dx=\int_{\Omega}f\circ T_{s}vJ_{s}dx$

Let $DT_{s}(X)$ be the Jacobian matrix of $T_{s}$ evaluated in $X$,
${}^{*}DT_{s}(X)$ the transpose matrix of $DT_{s}(X)$, then we have the
following proposition

We are therefore able to define the $s$-dependent Lagrangian:

$\displaystyle L(s,\varphi,\phi)$ $\displaystyle=$
$\displaystyle\int_{\Omega}\left\\{B(s)(\nabla\varphi-\nabla\tilde{\eta}^{s})\right\\}\cdot(\nabla\varphi-\nabla\tilde{\eta}^{s})+\lvert\varphi-\eta_{d}\circ
T_{s}\rvert^{2}J_{s}dx+$ (2.8) $\displaystyle+$
$\displaystyle\int_{\Omega}B(s)\nabla\varphi\cdot\nabla\phi-k^{2}\varphi\phi
J_{s}-f\circ T_{s}\phi J_{s}dx$ (2.9)

and defined $j(s)$ as follows

$\displaystyle g(s)=\inf_{\varphi\in H^{1}_{0}(\Omega)}\sup_{\phi\in
H^{1}_{0}(\Omega)}L(s,\varphi,\phi);\;\;dg(0)=\lim_{s\searrow
0}\;(g(s)-g(0))/s=dJ(\Omega;V(0)).$

In (2.8)-(2.9), $B(s)$ designate the following formula
$B(s)=J_{s}DT^{-1}_{s}(DT^{-1}_{s})^{*}$ where $J_{s}=detDT_{s}.$ For $V\in
C^{0}([0,\tau];C_{0}^{1}(\mathbb{R}^{N},\mathbb{R}^{N}))$ and the
diffeomorphism $T_{s}(V)=T_{s},$ we have

$\displaystyle\frac{d}{ds}T_{s}(X)=V(s,T_{s}(X)),\,\;\;T_{0}(X)=X,\,\;\;\frac{dT_{s}}{ds}=V(s)\circ
T_{s},\,\;\;T_{0}=I.$

where $V(s)(X)=V(s,X)$ and $I$ the identity matrix in $\mathbb{R}^{N}$.
However

$\displaystyle\frac{d}{ds}DT_{s}=DV(s)\circ
T_{s}DT_{s},\,\;\;DT_{0}=I,\,\;\;\frac{d}{ds}J_{s}=divV(s)\circ
T_{s}J_{s},\,\;\;J_{0}=1$

where $DV(s)$ and $DT_{s}$ are the Jacobian matrices of $V(s)$ and $T(s)$.
For $k\geq 1$, $C_{0}^{k}(\mathbb{R}^{N},\mathbb{R}^{N})$ is the space of $k$
times continuously differentiable functions from $\mathbb{R}^{N}$ to
$\mathbb{R}^{N}$ tending to zero at infinity; for $k=0$,
$C_{0}^{0}(\mathbb{R}^{N},\mathbb{R}^{N})$ is the space of continuous
functions from $\mathbb{R}^{N}$ to $\mathbb{R}^{N}$ tending to zero at
infinity.

We will use the following notations

$\displaystyle V_{s}=V(s)\circ
T_{s},\;\;V_{s}(X)=V(s,T_{s}(X)),\;\;j(s)=T_{s}-I,\;\;l(s)(x)=T_{s}(X)-X.$

In the following, we give the following lemma, which can found in [15].

###### Lemma 2.1

Assume that $V\in C^{0}([0,\tau];C_{0}^{1}(\mathbb{R}^{N},\mathbb{R}^{N}))$,
$l\in C^{0}([0,\tau];C_{0}^{1}(\mathbb{R}^{N},\mathbb{R}^{N}))$.
For $\tau>0$ sufficiently small $J_{s}=\det DT_{s}=\lvert\det
DT_{s}\rvert=\lvert J_{s}\rvert$, $0\leq s\leq\tau$ and there are constants
$0<\alpha<\beta$ such that

$\displaystyle\forall\;\xi\in\mathbb{R}^{N},\;\alpha\lvert\xi\rvert^{2}\leq
A(s)\xi\cdot\xi\leq\beta\lvert\xi\rvert^{2}\;\;\text{and}\;\;\alpha\leq
J_{s}\leq\beta.$ (2.10)

  * (i)

As $s$ goes to zero, $V_{s}\rightarrow V(0)$ in
$C_{0}^{1}(\mathbb{R}^{N},\mathbb{R}^{N})$, $DT_{s}\rightarrow I$ in
$C_{0}^{0}(\mathbb{R}^{N},\mathbb{R}^{N})$, $J_{s}\rightarrow 1$ in
$C_{0}^{0}(\mathbb{R}^{N},\mathbb{R})$

$\displaystyle\frac{DT_{s}-I}{s}\;\;\text{is bounded
in}\;C_{0}^{0}(\mathbb{R}^{N},\mathbb{R}^{N}),\;\;\frac{J_{s}-1}{s}\;\;\text{is
bounded in}\;C_{0}^{0}(\mathbb{R}^{N}).$

  * (ii)

As $s$ goes to zero

$\displaystyle B(s)\rightarrow
I\;\;\text{in}\;C_{0}^{0}(\mathbb{R}^{N},\mathbb{R}^{N}),\;\;\;\frac{B(s)-I}{s}\;\;\text{is
bounded in}\;C_{0}^{0}(\mathbb{R}^{N};\mathbb{R}^{N})$

$B^{\prime}(s)=divV_{s}I-DV_{s}-DV_{s}^{*}\rightarrow
B^{\prime}(0)=divV(0)-DV(0)-DV(0)^{*}$ in
$C_{0}^{0}(\mathbb{R}^{N};\mathbb{R}^{N})$,
where $DV_{s}$ is the Jacobian matrix of $V_{s}$ and $DV_{s}^{*}$ the
transpose matrix of $DV_{s}$.

  * (iii)

Given $h\in H^{1}(\mathbb{R}^{N})$, as $s$ goes to zero

$\displaystyle h\circ T_{s}\rightarrow
h\;\;\text{in}\;\;L^{2}(\Omega),\;\;\frac{h\circ T_{s}-h}{s}\;\;\text{is
bounded in}\;\;L^{2}(\Omega)$ $\displaystyle\nabla h\cdot
V_{s}\rightarrow\nabla h\cdot V(0)\;\;\text{in}\;\;L^{2}(\Omega).$

###### Proof.

See [15]. ∎

In the following, we need to calculate the derivatives of the Lagrangian with
respect to the variables $s,\,\varphi$ and $\phi.$ when the limits exist, in
oder to get the shape derivative of the funtionnal. In this case we have

$\displaystyle d_{s}L(0,x,y)=\lim_{s\searrow 0}\frac{L(s,x,y)-L(0,x,y)}{s}$
$\displaystyle\varphi\in X,\;\;d_{x}L(s,x,y;\varphi)=\lim_{\theta\searrow
0}\frac{L(s,x+\theta\varphi,y)-L(s,x,y)}{\theta}$ $\displaystyle\phi\in
Y\;\;d_{y}L(s,x,y;\phi)=\lim_{\theta\searrow
0}\frac{L(s,x,y+\theta\phi)-L(s,x,y)}{\theta}$

$s\searrow 0$ and $\theta\searrow 0$ mean that $s$ and $\theta$ go to $0$ by
strictly positive values.
Thus, we have, since $L(s,x,y)$ is affine en $y$, for all
$(s,x)\in[0,\tau]\times X$,

$\displaystyle\forall\;y,\psi\in
H_{0}^{1}(\Omega)\;\;d_{y}L(s,x,y;\psi)=L(s,x,\psi)-L(s,x,0)=d_{y}G(s,x,0,\psi).$
(2.11)

Recall that the $s$ dependent lagragian is given by

$\displaystyle L(s,\varphi,\phi)$ $\displaystyle=$
$\displaystyle\int_{\Omega}\left\\{B(s)(\nabla\varphi-\nabla\tilde{\eta}^{s})\right\\}\cdot(\nabla\varphi-\nabla\tilde{\eta}^{s})+\lvert\varphi-\eta_{d}\circ
T_{s}\rvert^{2}J_{s}dx+$ (2.12) $\displaystyle+$
$\displaystyle\int_{\Omega}B(s)\nabla\varphi\cdot\nabla\phi-k^{2}\varphi\phi
J_{s}-f\circ T_{s}\phi J_{s}dx$ (2.13)

The set of solutions (states) $x^{s}$ at $s\geq 0$ is denoted

$\displaystyle E(s)=\left\\{x^{s}\in H_{0}^{1}(\Omega),\;\;\forall\;\psi\in
H_{0}^{1}(\Omega),\;\;d_{y}L(s,x^{s},0;\psi)=0\right\\}.$ (2.14)

The set of solutions $p^{s}$ at $s\geq 0$ is denoted

$\displaystyle Y(s,x^{s})=\left\\{p^{s}\in
H_{0}^{1}(\Omega),\;\;\forall\;\varphi\in
H_{0}^{1}(\Omega),\;\;d_{x}L(s,x^{s},p^{s};\varphi)=0\right\\}.$ (2.15)

Finally the set of minimisers for the minimax is given by

$\displaystyle X(s)=\left\\{x^{s}\in H_{0}^{1}(\Omega),\;\,g(s)=\inf_{x\in
H_{0}^{1}(\Omega)}\sup_{y\in H_{0}^{1}(\Omega)}L(s,x,y)=\sup_{y\in
H_{0}^{1}(\Omega)}L(s,x^{s},y)\right\\}.$ (2.16)

Following the work of Delfour [8], we have the following lemma giving the way
to get the infimum of the Lagragian.

###### Lemma 2.2

(Constrained infimum and minimax)
We have the following assertions

  * (i)

$\displaystyle\inf_{x\in H_{0}^{1}(\Omega)}\sup_{y\in
H_{0}^{1}(\Omega)}L(s,x,y)=\inf_{x\in E(s)}L(s,x,0)$

  * (ii)

The minimax $g(s)=+\infty$ if and only if $E(s)=\emptyset$. And in this case
we have $X(s)=X$.

  * (iii)

If $E(s)\neq\emptyset$, then

$\displaystyle X(s)=\left\\{x^{s}\in E(s):\;\;G(s,x^{s},0)=\inf_{x\in
E(s)}G(s,x,0)\right\\}\subset E(s)$

and $g(s)<+\infty$.

###### Proof.

(i) Since the function $y\mapsto G(s,x,y)$ is affine, then for all $(s,x,y)$

$\displaystyle G(s,x,y)=G(s,x,0)+d_{y}G(s,x,0;y).$

In this case we have

$\displaystyle\sup_{y\in Y}G(s,x,y)=G(s,x,0)+\sup_{y\in
Y}d_{y}G(s,x,0;y)=\begin{cases}G(s,x,0)\;\;\text{if}\;\;x\in E(s)\\\
+\infty\;\;\text{if}\;\;x\in X\backslash E(s)\end{cases}$

In this case we have

$\displaystyle\inf_{x\in X}\sup_{y\in Y}G(s,x,y)=\inf_{x\in E(s)}G(s,x,0)$

(ii) If $g(s)=+\infty$, then for all $x\in X$, $\sup_{y\in Y}G(s,x,y)=+\infty$
or quite simply

$\displaystyle G(s,x,0)+\sup_{y\in Y}d_{y}G(s,x,0;y)=+\infty$

This leads to $\sup_{y\in Y}d_{y}G(s,x,0;y)=+\infty$ and then there exists
$y\in Y$ such that $d_{y}G(s,x,0;y)>0$ and then we have $\forall\,x\in X$,
$x\notin E(s)$ and $E(s)=E(s)\cap X=\emptyset$.
Conversely if $E(s)=\emptyset$ we have directly by definition of the infinimum

$\displaystyle g(s)=\inf_{x\in E(s)}G(s,x,0)=+\infty$

(iii) If $E(s)\neq\emptyset$ and $x^{s}\in X(s)$ and $x^{s}\in E(s)$ from the
definition of $(i)$

$\displaystyle G(s,x^{s},0)\leq\sup_{y\in Y}G(s,x^{s},y)=g(s)=\inf_{x\in
E(s)}G(s,x,0)\leq+\infty$

if $x^{s}\in E(s)$, $\sup_{y\in Y}d_{y}G(s,x^{s},0;y)=+\infty$ and

$\displaystyle g(s)=\sup_{y\in Y}d_{y}G(s,x^{s},y)=G(s,x,0)+\sup_{y\in
Y}d_{y}G(s,x,0;y)=+\infty$

This contradicts the fact that $g(s)$ is finite. Therefore $X(s)\subset E(s)$
and

$\displaystyle G(s,x^{s},0)\leq\sup_{y\in Y}G(s,x^{s},y)=g(s)=\inf_{x\in
E(s)}G(s,x,0)\leq G(s,x^{s},0)$

implies that

$\displaystyle x^{s}\in X(s),\;\;g(s)=\inf_{x\in E(s)}G(s,x,0).$

Conversely, if there is $x^{s}\in E(s)$ such that

$\displaystyle G(s,x^{s},0)=\inf_{x\in E(s)}G(s,x,0).$

So for all $y\in Y$, $d_{y}G(s,x^{s},0,y)=0$, $G(s,x^{s},y)=G(s,x^{s},0)$ and
from the definition (i)

$\displaystyle\sup_{y\in Y}G(s,x^{s},0)=\inf_{x\in E(s)}G(s,x,0)=\inf_{x\in
X}\sup_{y\in Y}G(s,x,y)$

and $x^{s}\in X(s)$. ∎

We also need the following assumptions

Hypothesis (H0)

  * (i)

: For all $s\in[0,\tau]$, $x^{0}\in X(0)$, $x^{s}\in X(s)$ and $y\in
H_{0}^{1}(\Omega)$, the function $\theta\mapsto
L(s,x^{0}+\theta(x^{s}-x^{0}),y):[0,1]\rightarrow\mathbb{R}$ is absolutely
continuous. This implies that for almost all $\theta$ the derivative exists
and is equal to $d_{x}L(s,x^{0}+\theta(x^{s}-x^{0}),y;x^{s}-x^{0})$ and it is
the integral of its derivative. In particular

$\displaystyle
L(s,x^{s},y)=L(s,x^{0},y)+\int_{0}^{1}d_{x}L(s,x^{0}+\theta(x^{s}-x^{0}),y;x^{s}-x^{0})d\theta.$

  * ii)

: For all $s\in[0,\tau]$, $x^{0}\in X(0)$, $x^{s}\in X(s)$ and $y\in
H_{0}^{1}(\Omega)$, $\phi\in H_{0}^{1}(\Omega)$ and for almost all
$\theta\in[0,1]$, $d_{x}L(s,x^{0}+\theta(x^{s}-x^{0}),y;\phi)$ exist et the
functions $\theta\mapsto L(s,x^{0}+\theta(x^{s}-x^{0}),y;\phi)$ belong to
$L^{1}[0,1];$

Hypothesis (H1) for all $s\in[0,\tau]$, $g(s)$ is finite, $X(s)=\\{x^{s}\\}$
and $Y(s,x^{0},x^{s})=\\{y^{s}\\}$ are singletons
Hypothesis (H2) $d_{s}L(0,x^{0},y^{0})$ exists.
Hypothesis(H3) The following limit exists

$\displaystyle R(x^{0},y^{0})=\lim_{s\searrow
0}\int_{0}^{1}d_{x}G\left(s,x^{0}+\theta(x^{s}-x^{0}),p^{0};\frac{x^{s}-x^{0}}{s}\right)d\theta.$

We have the following result

###### Theorem 2.3

([32, 33]).

Under assymptions (H0), (H1), (H2), (H3), $dg(0)$ exists and

$dg(0)=d_{s}G(0,x^{0},p^{0})+R(x^{0},p^{0}).$ (2.17)

###### Proof.

Recall that $g(s)=G(s,x^{s},y)$ and $g(0)=G(0,x^{0},y)$ for each $y\in Y$,
then for a standard adjoint state $p^{0}$ at $s=0$

$\displaystyle
g(s)-g(0)=G(s,x^{s},p^{0})-G(s,x^{0},p^{0})+(G(s,x^{0},p^{0})-G(0,x^{0},p^{0}))$

Dividing by $s>0$, we obtain

$\displaystyle\frac{g(s)-g(0)}{s}$ $\displaystyle=$
$\displaystyle\frac{G(s,x^{s},p^{0})-G(s,x^{0},p^{0})}{s}+\frac{G(s,x^{0},p^{0})-G(0,x^{0},p^{0})}{s}$
$\displaystyle=$
$\displaystyle\int_{0}^{1}d_{x}G\left(s,x^{0}+\theta(x^{s}-x^{0}),p^{0};\frac{x^{s}-x^{0}}{s}\right)d\theta$
$\displaystyle+$ $\displaystyle\frac{G(s,x^{0},p^{0})-G(0,x^{0},p^{0})}{s}$

Going to the limit when $s$ goes to $0$, we obtain

$\displaystyle dg(0)$ $\displaystyle=$ $\displaystyle\lim_{s\searrow
0}\int_{0}^{1}d_{x}G\left(s,x^{0}+\theta(x^{s}-x^{0}),p^{0};\frac{x^{s}-x^{0}}{s}\right)d\theta+d_{s}G(0,x^{0},p^{0})$
$\displaystyle=$ $\displaystyle d_{s}G(0,x^{0},p^{0})+R(x^{0},p^{0}).$

To prove the formula (2.17), it is sufficient to verify the hypotheses of the
theorem of Delfour [8], Sturm [32, 33]. For that, let us recall the Lagrangian

$\displaystyle L(s,\varphi,\phi)$ $\displaystyle=$
$\displaystyle\int_{\Omega}\left\\{B(s)(\nabla\varphi-\nabla\tilde{\eta}^{s})\right\\}\cdot(\nabla\varphi-\nabla\tilde{\eta}^{s})+\lvert\varphi-\eta_{d}\circ
T_{s}\rvert^{2}J_{s}dx+$ (2.18) $\displaystyle+$
$\displaystyle\int_{\Omega}B(s)\nabla\varphi\cdot\nabla\phi-k^{2}\varphi\phi
J_{s}-f\circ T_{s}\phi J_{s}dx$ (2.19)

Then the state equation is given by

$\text{find }\,\,x^{s}\in H_{0}^{1}(\Omega)\,\,\text{such that for
all}\,\,\psi\in H_{0}^{1}(\Omega),\,\,\int_{\Omega}B(s)\nabla
x^{s}\nabla\varphi-k^{2}x_{s}\varphi=\int_{\Omega}f(T_{s})\varphi J_{s}dx$
(2.20)

The adjoint state $p^{s}$ is solution to

$\text{Find}\,p^{s}\in H_{0}^{1}(\Omega)\,\,\text{such
that}\,\forall\,\varphi\in
H_{0}^{1}(\Omega),\,\,2\int_{\Omega}|\eta^{0}-\eta_{d}(Ts)|\varphi-\Delta\eta^{0}\varphi+\int_{\Omega}B(s)\nabla
p^{s}\nabla\varphi-k^{2}p^{s}\varphi=0.$ (2.21)

The bilinear forms (2.20) and (2.21) are coercives, then there exists a unique
solution $\eta^{s}$ and $(\eta^{0},p^{0})$ solution (2.20) and (2.21). This
allows us to write that, $\forall s\in[0,\tau]$ the spaces

$X(s)=\\{\eta^{s}\\}\,\,\text{and}\,\,Y(0,u^{0})=\\{p^{0}\\}$

are singletons, then Hypothesis (H1) is verified.
For the Hypothesis (H2), the $s$-derivative of the lagragian
$L(s,\varphi,\phi)$ is given by

$\displaystyle d_{s}L(s,\varphi,\phi)$ $\displaystyle=$
$\displaystyle\int_{\Omega}\left\\{B^{\prime}(s)(\nabla\varphi-\nabla\tilde{\eta}^{s})\right\\}\cdot(\nabla\varphi-\nabla\tilde{\eta}^{s})-2B(s)(\nabla\varphi-\nabla\tilde{\eta}^{s})\cdot\nabla\left([\nabla\tilde{\eta}\cdot
V(s)]\circ T_{s}\right)dx$ $\displaystyle+$
$\displaystyle\int_{\Omega}\lvert\varphi-\eta_{d}\circ
T_{s}\rvert^{2}divV_{s}J_{s}-2(\varphi-\eta_{d}\circ T_{s})\nabla\eta_{d}\cdot
V_{s}J_{s}dx$ $\displaystyle+$
$\displaystyle\int_{\Omega}B(s)\nabla\varphi\cdot\nabla\phi-k^{2}\varphi\phi
divV_{s}J_{s}-\left(f\circ T_{s}divV_{s}J_{s}+\nabla f\cdot
V_{s}J_{s}\right)\phi dx.$

Taking $\varphi=\eta^{0}$ and $\phi=p^{0}$, we have

$\displaystyle d_{s}L(s,\eta^{0},p^{0})$ $\displaystyle=$
$\displaystyle\int_{\Omega}\left\\{B^{\prime}(s)(\nabla\eta^{0}-\nabla\tilde{\eta}^{s})\right\\}\cdot(\nabla\eta^{0}-\nabla\tilde{\eta}^{s})-2B(s)(\nabla\eta^{0}-\nabla\tilde{\eta}^{s})\cdot\nabla\left([\nabla\tilde{\eta}\cdot
V(s)]\circ T_{s}\right)dx$ $\displaystyle+$
$\displaystyle\int_{\Omega}\lvert\eta^{0}-\eta_{d}\circ
T_{s}\rvert^{2}divV_{s}J_{s}-2(\eta^{0}-\eta_{d}\circ
T_{s})\nabla\eta_{d}\cdot V_{s}J_{s}dx$ $\displaystyle+$
$\displaystyle\int_{\Omega}B^{\prime}(s)\nabla\eta^{0}\cdot\nabla
p^{0}-k^{2}\eta^{0}p^{0}divV_{s}J_{s}-\left(f\circ T_{s}divV_{s}J_{s}+\nabla
f\cdot V_{s}J_{s}\right)p^{0}dx.$

Using the lemma 2.1, we get

$\displaystyle d_{s}L(0,\eta^{0},p^{0})$ $\displaystyle=$
$\displaystyle\int_{\Omega}\left\\{B^{\prime}(0)(\nabla\eta^{0}-\nabla\tilde{\eta})\right\\}\cdot(\nabla\eta^{0}-\nabla\tilde{\eta})-2\left(\nabla\eta^{0}-\nabla\tilde{\eta}\right)\cdot\nabla\left([\nabla\tilde{\eta}\cdot
V(0)]\right)dx$ $\displaystyle+$
$\displaystyle\int_{\Omega}\lvert\eta^{0}-\eta_{d}\rvert^{2}divV(0)-2(\eta^{0}-\eta_{d})\nabla\eta_{d}\cdot
V(0)dx$ $\displaystyle+$
$\displaystyle\int_{\Omega}B^{\prime}(0)\nabla\eta^{0}\cdot\nabla
p^{0}-k^{2}\eta^{0}p^{0}divV(0)-\left(fdivV(0)+\nabla f\cdot
V(0)\right)p^{0}dx,$

then Hypothesis (H2) is satisfied.
To prove the last hypothesis, we give $x$-derivative of the lagrangian
$L(s,\varphi,\psi)$ with respect to $\varphi$ in the direction
$\varphi^{\prime}$:

$\displaystyle
d_{\varphi}L(s,\varphi,\phi;\varphi^{\prime})=\int_{\Omega}2B(s)(\nabla\varphi-\nabla\tilde{\eta}^{s})\cdot\nabla\varphi^{\prime}+2(\varphi-\eta_{d}\circ
T_{s})\varphi^{\prime}J_{s}+B(s)\nabla\varphi^{\prime}\cdot\nabla\phi-\lambda\varphi^{\prime}\phi
J_{s}.$ (2.22)

The derivative of the $s$-dependent Lagrangian with respect to $\phi$ in the
direction $\phi$

$\displaystyle
d_{\phi}L(s,\varphi,\phi;\phi^{\prime})=\int_{\Omega}B(s)\nabla\varphi\cdot\nabla\phi^{\prime}-k^{2}\varphi\phi^{\prime}J_{s}-f\circ
T_{s}\phi^{\prime}J_{s}dx.$ (2.23)

The state equation at $s\geq 0$ and the adjoint state equation at $s=0$ are

$\displaystyle\eta^{s}\in H^{1}_{0}(\Omega),\;\;\forall\;\phi^{\prime}\in
H^{1}_{0}(\Omega),\;\;\int_{\Omega}B(s)\nabla\eta^{s}\cdot\nabla\phi^{\prime}-k^{2}\eta^{s}\phi^{\prime}J_{s}-f\circ
T_{s}\phi^{\prime}J_{s}dx=0$ (2.24) $\displaystyle p^{0}\in
H^{1}_{0}(\Omega),\;\forall\;\varphi^{\prime}\in
H^{1}_{0}(\Omega),\;\int_{\Omega}2(\nabla\eta^{0}-\nabla\tilde{\eta})\cdot\nabla\varphi^{\prime}+2(\eta^{0}-\eta_{d})\varphi^{\prime}+\nabla
p^{0}\cdot\nabla\varphi^{\prime}-k^{2}p^{0}\varphi^{\prime}dx=0.$ (2.25)
$\displaystyle R(s)$ $\displaystyle=$
$\displaystyle\int_{0}^{1}d_{\varphi}L\left(s,\eta^{0}+\theta(\eta^{s}-\eta^{0}),p^{0};\frac{\eta^{s}-\eta^{0}}{s}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{\Omega}2B(s)\left\\{\nabla\left(\frac{\eta^{s}+\eta^{0}}{2}\right)-\nabla\tilde{\eta}^{s}\right\\}\cdot\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)+2\left(\frac{\eta^{s}+\eta^{0}}{2}-\eta_{d}\circ
T_{s}\right)\left(\frac{\eta^{s}-\eta^{0}}{s}\right)J_{s}dx$ $\displaystyle+$
$\displaystyle\int_{\Omega}B(s)\nabla
p^{0}\cdot\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)-k^{2}p^{0}\left(\frac{\eta^{s}-\eta^{0}}{s}\right)J_{s}dx.$

Taking $\varphi^{\prime}=\frac{\eta^{s}-\eta^{0}}{s}$ in the adjoint equation
(2.25), we have

$\displaystyle\int_{\Omega}2\nabla(\eta^{0}-\tilde{\eta})\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)+2(\eta^{0}-\eta_{d})\left(\frac{\eta^{s}-\eta^{0}}{s}\right)+\nabla
p^{0}\cdot\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)-k^{2}p^{0}\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx=0.$

This means that

$\displaystyle\int_{\Omega}2\left\\{\nabla\left(\frac{\eta^{s}+\eta^{0}}{2}\right)-\nabla\left(\frac{\eta^{s}-\eta^{0}}{2}\right)-\nabla\tilde{\eta}^{s}+\nabla\tilde{\eta}^{s}-\nabla\tilde{\eta}\right\\}\cdot\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx+$
$\displaystyle\int_{\Omega}2\left(\frac{\eta^{s}+\eta^{0}}{2}-\frac{\eta^{s}-\eta^{0}}{2}-\eta_{d}\circ
T_{s}+\eta_{d}\circ
T_{s}-\eta_{d}\right)\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx+$
$\displaystyle\int_{\Omega}\nabla
p^{0}\cdot\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)-k^{2}p^{0}\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx=0.$

In other words,

$\displaystyle\int_{\Omega}2\left\\{\nabla\left(\frac{\eta^{s}+\eta^{0}}{2}\right)-\nabla\tilde{\eta}^{s}\right\\}\cdot\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx-s\int_{\Omega}\left\lvert\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)\right\rvert^{2}dx+$
$\displaystyle\int_{\Omega}2(\eta_{d}\circ
T_{s}-\eta_{d})\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx+\int_{\Omega}2(\nabla\tilde{\eta}^{s}-\nabla\tilde{\eta})\cdot\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx+$
$\displaystyle\int_{\Omega}2\left(\frac{\eta^{s}+\eta^{0}}{2}-\eta_{d}\circ
T_{s}\right)\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx-s\int_{\Omega}\left\lvert\frac{\eta^{s}-\eta^{0}}{s}\right\rvert^{2}dx+$
$\displaystyle\int_{\Omega}\nabla
p^{0}\cdot\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)-k^{2}p^{0}\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx=0.$

We can therefore rewrite the expression for $R(s)$ as follows

$\displaystyle R(s)$ $\displaystyle=$
$\displaystyle\int_{\Omega}2\left(\frac{B(s)-I}{s}\right)\left\\{\nabla\left(\frac{\eta^{s}+\eta^{0}}{2}\right)-\nabla\tilde{\eta}^{s}\right\\}\nabla\left(\eta^{s}-\eta^{0}\right)dx+s\int_{\Omega}\left\lvert\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)\right\rvert^{2}dx+$
(2.26)
$\displaystyle\int_{\Omega}2\left(\frac{J_{s}-1}{s}\right)\left(\frac{\eta^{s}+\eta^{0}}{2}-\eta_{d}\circ
T_{s}\right)\left(\eta^{s}-\eta^{0}\right)dx-\int_{\Omega}2(\nabla\tilde{\eta}^{s}-\nabla\tilde{\eta})\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx-$
(2.27) $\displaystyle\int_{\Omega}2\left(\eta_{d}\circ
T_{s}-\eta_{d}\right)\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx+s\int_{\Omega}\left\lvert\frac{\eta^{s}-\eta^{0}}{s}\right\rvert^{2}dx+$
(2.28) $\displaystyle\int_{\Omega}\frac{B(s)-I}{s}\nabla
p^{0}\nabla\left(\eta^{s}-\eta^{0}\right)-k^{2}\left(\frac{J_{s}-1}{s}\right)p^{0}\left(\eta^{s}-\eta^{0}\right)dx.$
(2.29)

Thus the following inequality is verified

$\displaystyle R(s)$ $\displaystyle\leq$ $\displaystyle
2\left\lVert\frac{B(s)-I}{s}\right\rVert\left\lVert\nabla\left(\frac{\eta^{s}+\eta^{0}}{2}\right)-\nabla\tilde{\eta}^{s}\right\rVert\left\lVert\nabla\left(\eta^{s}-\eta^{0}\right)\right\rVert+s\left\lVert\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)\right\rVert^{2}+$
$\displaystyle
2\left\lVert\frac{J_{s}-1}{s}\right\rVert\left\lVert\frac{\eta^{s}+\eta^{0}}{2}-\eta_{d}\circ
T_{s}\right\rVert\left\lVert\eta^{s}-\eta^{0}\right\rVert+2\left\lVert\nabla\tilde{\eta}^{s}-\nabla\tilde{\eta}\right\rVert\left\lVert\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)\right\rVert+$
$\displaystyle 2\left\lVert\frac{\eta_{d}\circ
T_{s}-\eta_{d}}{s}\right\rVert\left\lVert\eta^{s}-\eta^{0}\right\rVert+s\left\lVert\frac{\eta^{s}-\eta^{0}}{s}\right\rVert^{2}+$
$\displaystyle\;\;\,\left\lVert\frac{B(s)-I}{s}\right\rVert\lVert\nabla
p^{0}\rVert\left\lVert\nabla(\eta^{s}-\eta^{0})\right\rVert+k^{2}\left\lVert\frac{J_{s}-1}{s}\right\rVert\left\lVert
p^{0}\right\rVert\left\lVert\eta^{s}-\eta^{0}\right\rVert.$

We need to show that the limit of $R(s)$ exists and is zero. To do this, we
first note that by the lemma 2.1, the terms $(B(s)-I)/s$ and $(J(s)-1)/s$ are
uniformly bounded. Thus it suffices only to show that
$\eta^{s}\rightarrow\eta^{0}$ in $H^{1}_{0}(\Omega)$-strong and that the norm
$H_{0}^{1}(\Omega)$ of $(\eta^{s}-\eta^{0})/s$ is bounded. From (2.24) of
$\eta^{s}$ and $\eta^{0}$, we see that for any $\phi^{\prime}\in
H^{1}_{0}(\Omega)$,

$\displaystyle\int_{\Omega}B(s)\nabla\eta^{s}\nabla\phi^{\prime}-k^{2}\eta^{s}\phi^{\prime}J_{s}-f\circ
T_{s}\phi^{\prime}J_{s}dx=\int_{\Omega}\nabla\eta^{0}\nabla\phi^{\prime}-k^{2}\eta^{0}\phi^{\prime}-f\phi^{\prime}dx$
$\displaystyle\int_{\Omega}\nabla\eta^{s}\nabla\phi^{\prime}dx+\int_{\Omega}[B(s)-I]\nabla\eta^{s}\nabla\phi^{\prime}-k^{2}\eta^{s}\phi^{\prime}J_{s}-f\circ
T_{s}\phi^{\prime}J_{s}dx=\int_{\Omega}\nabla\eta^{0}\nabla\phi^{\prime}-k^{2}\eta^{0}\phi^{\prime}-f\phi^{\prime}dx$
$\displaystyle\int_{\Omega}\nabla(\eta^{s}-\eta^{0})\nabla\phi^{\prime}-k^{2}(\eta^{s}J_{s}-\eta^{0})\phi^{\prime}dx=-\int_{\Omega}[B(s)-I]\nabla\eta^{s}\nabla\phi^{\prime}dx+\int_{\Omega}(f\circ
T_{s}J_{s}-f)\phi^{\prime}dx$
$\displaystyle\int_{\Omega}\nabla(\eta^{s}-\eta^{0})\nabla\phi^{\prime}-k^{2}(\eta^{s}J_{s}-\eta^{0}J_{s}+\eta^{0}J_{s}-\eta^{0})\phi^{\prime}dx=-\int_{\Omega}[B(s)-I]\nabla\eta^{s}\nabla\phi^{\prime}dx+\int_{\Omega}(f\circ
T_{s}J_{s}-f)\phi^{\prime}dx$

$\displaystyle\int_{\Omega}\nabla(\eta^{s}-\eta^{0})\nabla\phi^{\prime}-k^{2}(\eta^{s}-\eta^{0})J_{s}\phi^{\prime}dx$
$\displaystyle=$
$\displaystyle-\int_{\Omega}[B(s)-I]\nabla\eta^{s}\nabla\phi^{\prime}+f\circ
T_{s}(J_{s}-1)\phi^{\prime}dx$ $\displaystyle+\int_{\Omega}(f\circ
T_{s}-f)\phi^{\prime}+\eta^{0}(J_{s}-1)\phi^{\prime}dx.$

Taking $\phi^{\prime}=\eta^{s}-\eta^{0}$, we get

$\displaystyle\lVert\nabla(\eta^{s}-\eta^{0})\rVert^{2}-k^{2}J_{s}\lVert\eta^{s}-\eta^{0}\rVert^{2}$
$\displaystyle=$
$\displaystyle-\int_{\Omega}[B(s)-I]\nabla\eta^{s}\nabla(\eta^{s}-\eta^{0})+f\circ
T_{s}(J_{s}-1)(\eta^{s}-\eta^{0})dx$ $\displaystyle+\int_{\Omega}(f\circ
T_{s}-f)(\eta^{s}-\eta^{0})+\eta^{0}(J_{s}-1)(\eta^{s}-\eta^{0})dx.$

According to the lemma 2.1, we have $\alpha\leq J_{s}\leq\beta$. Using the
Cauchy Schwarz inequality

$\displaystyle\lVert\nabla(\eta^{s}-\eta^{0})\rVert^{2}-k^{2}\beta\lVert\eta^{s}-\eta^{0}\rVert^{2}$
$\displaystyle\leq$ $\displaystyle\lVert
B(s)-I\rVert\lVert\nabla\eta^{s}\rVert\lVert\nabla(\eta^{s}-\eta^{0})\rVert$
$\displaystyle+\lVert f\circ T_{s}\rVert\lVert
J_{s}-1\rVert\lVert\eta^{s}-\eta^{0}\rVert$ $\displaystyle+\lVert f\circ
T_{s}-f\rVert\lVert\eta^{s}-\eta^{0}\rVert+\lVert\eta^{0}\rVert\lVert
J_{s}-1\rVert\lVert\eta^{s}-\eta^{0}\rVert.$

Due to the fact that $\Omega$ is a bounded open lipschitzian domain, there
exists a positive constant $c(\Omega)>0$ such that
$\lVert\eta^{s}-\eta^{0}\rVert\leq
c(\Omega)\lVert\nabla(\eta^{s}-\eta^{0})\rVert$. This will allow us to write

$\displaystyle\left(1-k^{2}c(\Omega)\beta\right)\lVert\nabla(\eta^{s}-\eta^{0})\rVert^{2}$
$\displaystyle\leq$ $\displaystyle\lVert
B(s)-I\rVert\lVert\nabla\eta^{s}\rVert\lVert\nabla(\eta^{s}-\eta^{0})\rVert$
$\displaystyle+\lVert f\circ T_{s}\rVert\lVert
J_{s}-1\rVert\lVert\nabla(\eta^{s}-\eta^{0})\rVert c(\Omega)$
$\displaystyle+\lVert f\circ
T_{s}-f\rVert\lVert\nabla(\eta^{s}-\eta^{0})\rVert c(\Omega)$
$\displaystyle+\lVert\eta^{0}\rVert\lVert
J_{s}-1\rVert\lVert\nabla(\eta^{s}-\eta^{0})\rVert c(\Omega).$

So we have

$\displaystyle\left(1-k^{2}c(\Omega)\beta\right)\lVert\nabla(\eta^{s}-\eta^{0})\rVert$
$\displaystyle\leq$ $\displaystyle\lVert
B(s)-I\rVert\lVert\nabla\eta^{s}\rVert+\lVert f\circ T_{s}\rVert\lVert
J_{s}-1\rVert c(\Omega)$ $\displaystyle+\lVert f\circ T_{s}-f\rVert
c(\Omega)+\lVert\eta^{0}\rVert\lVert J_{s}-1\rVert c(\Omega).$

The right hand side of this inequality tends to zero when $s$ goes to zero.
But this does not mean directly that $\lVert\nabla(\eta^{s}-\eta^{0})\rVert$
goes to zero. In fact it will depend on the sign of $1-k^{2}c(\Omega)\beta$.
If $1-k^{2}c(\Omega)\beta$ is positive, then
$\lVert\nabla(\eta^{s}-\eta^{0})\rVert$ goes to zero. On the other hand if
$1-k^{2}c(\Omega)\beta$ is negative, we can always find a positive constant
$M$ such that $M+1-k^{2}c(\Omega)\beta>0$ and such that we have

$\displaystyle\left(M+1-k^{2}c(\Omega)\beta\right)\lVert\nabla(\eta^{s}-\eta^{0})\rVert$
$\displaystyle\leq$ $\displaystyle\lVert
B(s)-I\rVert\lVert\nabla\eta^{s}\rVert+\lVert f\circ T_{s}\rVert\lVert
J_{s}-1\rVert c(\Omega)$ $\displaystyle+\lVert f\circ T_{s}-f\rVert
c(\Omega)+\lVert\eta^{0}\rVert\lVert J_{s}-1\rVert c(\Omega).$

So we can conclude that $\eta^{s}\rightarrow\eta^{0}$ in
$H^{1}_{0}(\Omega)$-strong. To show that $(\nabla(\eta^{s}-\eta^{0}))/s$ is
bounded, we divide the equation by $s>0$

$\displaystyle\left(1-k^{2}c(\Omega)\beta\right)\left\lVert\frac{\nabla(\eta^{s}-\eta^{0})}{s}\right\rVert$
$\displaystyle\leq$
$\displaystyle\left\lVert\frac{B(s)-I}{s}\right\rVert\lVert\nabla\eta^{s}\rVert+\left\lVert\frac{J_{s}-1}{s}\right\rVert\lVert
f\circ T_{s}\rVert c(\Omega)$ $\displaystyle+\left\lVert\frac{f\circ
T_{s}-f}{s}\right\rVert
c(\Omega)+\left\lVert\frac{J_{s}-1}{s}\right\rVert\lVert\eta^{0}\rVert
c(\Omega).$

The term on the right is bounded, by playing with the sign of
$1-k^{2}c(\Omega)\beta$, we show that $(\nabla(\eta^{s}-\eta^{0}))/s$ is
bounded. This means that $(\eta^{s}-\eta^{0})/s$ is bounded in
$H^{1}_{0}(\Omega)$ and therefore $(\eta^{s}-\eta^{0})/s$ is bounded in
$L^{2}(\Omega)$.

Then $R(s)\rightarrow 0$ when $s\rightarrow 0,$ i.e. the term $R(x^{0},p^{0})$
is zero and the shape derivative is given by $d_{s}L(0,\eta^{0},p^{0}).$

∎

The use of the averaged adjoint revealed the possible occurrence of an extra
term and provided a simpler expression of the former hypothesis (H3). It turns
out that the extra term can also be obtained by using the standard adjoint at
t = 0 significantly simplifying the checking of that condition.

###### Corollary 2.4

Consider the Lagrangian functional

$\displaystyle(s,x,y)\mapsto G(s,x,y):[0,\tau]\times X\times
Y\rightarrow\mathbb{R},\;\;\tau>0$

where $X$ and $Y$ are vector spaces and the function $y\mapsto G(s,x,y)$ is
affine. Assume that H(0) and the following assumptions are satisfied:
H(1) for all $s\in[0,\tau]$, $X(s)\neq\emptyset$, $g(s)$ is finite, and for
each $x\in X(0)$, $Y(0,x)\neq\emptyset$,
H(2) for all $x\in X(0)$ and $p\in Y(0,x)$ $d_{s}G(0,x,p)$ exists,
H(3”) there exist $x^{0}\in X(0)$ and $p^{0}\in Y(0,x^{0})$ such that the
following limit exists

$\displaystyle R(x^{0},p^{0})=\lim_{s\searrow
0}\int_{0}^{1}d_{x}G\left(s,x^{0}+\theta(x^{s}-x^{0}),p^{0};\frac{x^{s}-x^{0}}{s}\right)d\theta.$

Then, $dg(0)$ exists and there exist $x^{0}\in X(0)$ and $p^{0}\in Y(0,x^{0})$
such that $dg(0)=d_{s}G(0,x^{0},p^{0})+R(x^{0},p^{0})$.

##  3 Topological derivative with geometrical perturbation of the domain

Consider the functional defined by

$\displaystyle
J(\Omega)=\int_{\Omega}\lvert\nabla\eta-A\rvert^{2}dx+\int_{\Omega}\lvert\eta-\eta_{d}\rvert^{2}dx$
(3.1)

where $\Omega$ is a domain of $\mathbb{R}^{N}$, $N=2,3$ with a lipschitzian
$\partial\Omega$ boundary and $\eta\in H^{1}_{0}(\Omega)$ is a solution of the
variational equation of state

$\displaystyle\exists\;\eta\in H^{1}_{0}(\Omega)\;\text{such
that}\int_{\Omega}\nabla\eta\nabla v-k^{2}\eta
vdx=\int_{\Omega}fvdx,\;\text{for all}\;v\in H^{1}_{0}(\Omega).$ (3.2)

Let $x_{0}\in\Omega\subset\mathbb{R}^{N}$, $E$ be a set $d$-rectifiable with
$0\leq d<N$ and $E_{r}$ the $r$-dilatation of $E$. We recall that

$\displaystyle d_{E}(x)=\inf_{x_{0}\in E}\lvert
x-x_{0}\rvert\;\;\text{and}\;\;E_{r}=\\{x\in\mathbb{R}^{N}:\;d_{E}(x)\leq
r\\}.$

We define the auxiliary variable $s$ as the volume of the $r$-dilatation
$E_{r}$. The perturbed domains are this time defined by
$r\rightarrow\Omega_{r}=\Omega\backslash E_{r}$.  $s$ is the volume of
$E_{r}$. It is also given by $s=\alpha_{N-d}r^{N-d}$, where $\alpha_{N-d}$ is
the volume of the ball of radius 1 in $\mathbb{R}^{N-d}$. Therefore the
perturbation takes the form
$\textbf{1}_{\Omega_{r}}=\textbf{1}_{\Omega}-\textbf{1}_{E_{r}}$. We define
the following sets

$\displaystyle
E_{r}^{o}=\\{x\in\mathbb{R}^{N}:\;\;d_{E}(x)<r\\}\;\;\text{and}\;\;\;E_{r}^{\prime}=E_{r}^{o}\backslash
E=\\{x\in\mathbb{R}^{N}:\;\;0<d_{E}(x)<r\\}.$

###### Definition 3.1

Let $E$ be a closed subset of $\mathbb{R}^{N}$.
i) The set of points of $\mathbb{R}^{N}$ which have a unique projection on
$E$.

$\displaystyle Unp(E)=\\{y\in\mathbb{R}^{N}:\;\exists\;\text{an unique}\;x\in
E\;\text{such that}\;d_{E}(y)=\lVert y-x\rVert\\}$

where $d_{E}(y)$ is the distance function from a point $y$ to $E$.
ii) The reach of a point $x\in E$ and the reach of $E$ are defined

$\displaystyle reach(E,x)=\sup\\{r>0:\;B(x,r)\subset
Unp(E)\\},\;\;reach(E)=\inf_{x\in E}reach(E,x).$

We say that $E$ is a positive reach set if $reach(E)>0$.

The definition of $Unp(E)$ obviously implies the existence of a projection on
$E$, a function $p_{E}:Unp(E)\rightarrow E$ which associates $x\in Unp(E)$ the
unique point $p_{E}(x)\in E$ such that $d_{E}(x)=\lVert x-p_{E}(x)\rVert$. For
$0<R<reach(E)$, the projection $p_{E}$ is equal to the gradient of the
following convex function

$\displaystyle f_{E}(x)=\frac{1}{2}\left(\lVert
x\rVert^{2}-d_{E}^{2}(x)\right),\;\;p_{E}(x)=\nabla
f_{E}(x)=x-\frac{1}{2}d_{E}^{2}(x).$

Hypothesis 1
Let $E$ be a closed connected subset of $\mathbb{R}^{N}$ and $R>0$ such that
$E_{2R}=\\{x\in\mathbb{R}^{N};\;d_{E}(x)\leq 2R\\}\subset\Omega$. Suppose that
$E$ has a positive reach greater than $2R$ $(reach(E)>2R)$ and that
$0<H^{d}(E)<\infty$ for an integer $d$, $0\leq d<N$.
We also need the following theorem which extends the Lebesgue differentiation
theorem from $d=0$ to $0\leq d<N$.

###### Theorem 3.1

Let $E$ be a compact subset of $\mathbb{R}^{N}$ and $0\leq d<N$ an integer.
Let $\alpha_{k}$ denote the volume of the unit ball in $\mathbb{R}^{k}$.
Suppose that

  * (1)

$E$ is a $d$-rectifiable subset of $\mathbb{R}^{N}$ such that $\partial E=E$
and $0<H^{d}(E)<\infty$,

  * (2)

$E$ has a positive reach, i.e. there is $R>0$ such that $d^{2}_{E}\in
C^{1,1}(E_{R})$

  * (3)

and $f$ is continuous in $E_{R}$. Then

$\displaystyle\lim_{r\searrow
0}\frac{1}{\alpha_{N-d}r^{N-d}}\int_{E_{r}}fd\xi=\lim_{r\searrow
0}\frac{1}{\alpha_{N-d}r^{N-d}}\int_{E_{r}}fop_{E}d\xi=\int_{E}fdH^{d}.$

###  3.1 Topological derivative using the Lagrangian and minimax methods with
perturbation on the source term

We define the auxiliary variable $s$ as the volume of the $r$-dilation
$E_{r}$. The perturbed domains are then defined by

$\displaystyle s\rightarrow\Omega_{s}=\Omega\backslash E_{r}$

$s$ is the volume of $E_{r}$. It is also given by $s=\alpha_{N-d}r^{N-d}$,
where $\alpha_{N-d}$ is the volume of the ball of radius 1 in
$\mathbb{R}^{N-d}$. Therefore the perturbation takes the form
$\mathds{1}_{\Omega_{s}}=\mathds{1}_{\Omega}-\mathds{1}_{E_{r}}$. we introduce
a topological perturbation of the second member $f$, by posing for example

$f^{s}=\begin{cases}f(x)\;\;if\;\;x\in\Omega\backslash E_{r}\\\\[5.69046pt]
\gamma f(x)\;\;if\;\;x\in E_{r}\end{cases}$

$f$ is a function assumed to be continuous at the point $x_{0}$ and that
$f(x)$ is constant in a neighbourhood of $x_{0}$.

$\displaystyle\int_{\Omega}\nabla\eta^{s}\nabla
vdx-\int_{\Omega}k^{2}\eta^{s}vdx$ $\displaystyle=$
$\displaystyle\int_{\Omega}f^{s}vdx$ $\displaystyle=$
$\displaystyle\int_{\Omega\backslash
E_{r}}f^{s}vdx+\int_{E_{r}}f^{s}vdx=\int_{\Omega\backslash
E_{r}}fvdx+\gamma\int_{E_{r}}fvdx$ $\displaystyle=$
$\displaystyle\int_{\Omega}fvdx-\int_{E_{r}}fvdx+\gamma\int_{E_{r}}fvdx$
$\displaystyle=$
$\displaystyle\int_{\Omega}fvdx-(1-\gamma)\int_{E_{r}}fvdx=\int_{\Omega}[f-(1-\gamma)\chi_{E_{r}}f]vdx.$

This leads to

$\displaystyle\int_{\Omega}\nabla\eta^{s}\nabla
vdx-\int_{\Omega}k^{2}\eta^{s}vdx$ $\displaystyle=$
$\displaystyle\int_{\Omega}[f-(1-\gamma)\chi_{E_{r}}f]vdx.$ (3.3)

We can therefore define the Lagrangian depending on $s$ as

$\displaystyle
L(s,\varphi,\Phi)=\int_{\Omega}\lvert\nabla\varphi-A\rvert^{2}dx+\int_{\Omega}\lvert\varphi-\eta_{d}\rvert^{2}dx+\int_{\Omega}\nabla\varphi\nabla\Phi
dx-\int_{\Omega}k^{2}\varphi\Phi
dx-\int_{\Omega}[f-(1-\gamma)\chi_{E_{r}}f]\Phi dx.$

The derivative of the Lagrangian depending on $s$ with respect to $\varphi$
is:

$\displaystyle
d_{\varphi}L(s,\varphi,\Phi;\varphi^{\prime})=\int_{\Omega}2(\nabla\varphi-A)\nabla\varphi^{\prime}+2(\varphi-\eta_{d})\varphi^{\prime}+\int_{\Omega}\nabla\varphi^{\prime}\nabla\Phi
dx-\int_{\Omega}k^{2}\varphi^{\prime}\Phi dx.$

The initial adjoint state $p^{0}$ is a solution of
$d_{\varphi}L(s,\eta^{0},p^{0};\varphi^{\prime})=0$ for all
$\varphi^{\prime}\in H^{1}_{0}(\Omega)$ with $\eta^{0}=\eta=\eta^{s}$ for
$s=0$. So the variational formulation of the adjoint equation of state is
given by: Find $p^{0}\in H^{1}_{0}(\Omega)$ such that

$\displaystyle\int_{\Omega}2(\nabla\eta^{0}-A)\nabla\varphi^{\prime}+2(\eta^{0}-\eta_{d})\varphi^{\prime}+\int_{\Omega}\nabla\varphi^{\prime}\nabla
p^{0}dx-\int_{\Omega}k^{2}\varphi^{\prime}p^{0}dx=0.$ (3.4)

The derivative of the Lagrangian depending on $s$ with respect to $\Phi$ is:

$\displaystyle
d_{\Phi}L(s,\varphi,\Phi;\Phi^{\prime})=\int_{\Omega}\nabla\varphi\nabla\Phi^{\prime}dx-\int_{\Omega}k^{2}\varphi\Phi^{\prime}dx-\int_{\Omega}[f-(1-\gamma)\chi_{E_{r}}f]\Phi^{\prime}dx.$

The initial state $\eta=\eta^{0}$ is solution of
$d_{\Phi}L(0,\eta^{0},0;\Phi^{\prime})=0$ for all $\Phi^{\prime}\in
H^{1}_{0}(\Omega)$ and we find obviously

$\displaystyle\int_{\Omega}\nabla\eta^{0}\nabla\Phi^{\prime}dx-\int_{\Omega}k^{2}\eta^{0}\Phi
dx=\int_{\Omega}f\Phi^{\prime}dx\;\;\text{pour tout}\;\Phi^{\prime}\in
H^{1}_{0}(\Omega)$

and the state $\eta^{s}$ for $s\geq 0$ verifies

$\displaystyle\int_{\Omega}\nabla\eta^{s}\nabla\Phi^{\prime}dx-\int_{\Omega}k^{2}\eta^{s}\Phi^{\prime}dx=\int_{\Omega}[f-(1-\gamma)\chi_{E_{r}}f]\Phi^{\prime}dx\;\;\text{pour
tout}\;\Phi^{\prime}\in H^{1}_{0}(\Omega).$ (3.5)

Now we want to calculate the derivative of the Lagrangian depending on $s$
with respect to $s$. To do this we first calculate the ratio
$\frac{L(s,\varphi,\Phi)-L(0,\varphi,\Phi)}{s}$

$\displaystyle\frac{L(s,\varphi,\Phi)-L(0,\varphi,\Phi)}{s}$ $\displaystyle=$
$\displaystyle\frac{1-\gamma}{s}\int_{E_{r}}f(x)\Phi(x)dx$ $\displaystyle=$
$\displaystyle\frac{1-\gamma}{|E_{r}\rvert}\int_{E_{r}}f(x)\Phi(x)dx.$

For $d=0$, $E=\\{x_{0}\\}$, $E_{r}=\\{x\in\mathbb{R}^{N}:\;\lvert
x-x_{0}\rvert\leq r\\}=\overline{B(x_{0},r)}$; the closed ball of centre
$x_{0}$ and radius $r$. So by applying Lebesgue’s differentiation theorem we
get $d_{s}L(0,\varphi,\Phi)=(1-\gamma)f(x_{0})\Phi(x_{0})$. So at the point
$\eta^{0}$ and $p^{0}$

$\displaystyle d_{s}L(0,\eta^{0},p^{0})=(1-\gamma)f(x_{0})p^{0}(x_{0}).$ (3.6)

In cases where $d=1,2$, we have

$\displaystyle\frac{L(s,\varphi,\Phi)-L(0,\varphi,\Phi)}{s}$ $\displaystyle=$
$\displaystyle\frac{1-\gamma}{|E_{r}\rvert}\int_{E_{r}}f(x)\Phi(x)dx$
$\displaystyle=$
$\displaystyle\frac{1-\gamma}{\alpha_{N-d}r^{N-d}}\int_{E_{r}}f(x)\Phi(x)dx\rightarrow(1-\gamma)\int_{E}f(x)\Phi(x)dH^{d}.$

Therfore at the point $\eta^{0}$ et $p^{0}$

$\displaystyle d_{s}L(0,\eta^{0},p^{0})=(1-\gamma)\int_{E}f(x)\Phi(x)dH^{d}.$
(3.7)

We now need to calculate $R(\eta^{0},p^{0})$ which is defined by

$\displaystyle R(\eta^{0},p^{0})=\lim_{s\searrow
0}\int_{0}^{1}d_{x}L\left(s,\eta^{0}+\theta(\eta^{s}-\eta^{0}),p^{0};\frac{\eta^{s}-\eta^{0}}{s}\right)=\lim_{s\searrow
0}\;R(s)$ (3.8) $\displaystyle R(s)$ $\displaystyle=$
$\displaystyle\int_{\Omega}2\left\\{\nabla\left(\frac{\eta^{s}+\eta^{0}}{2}\right)-A\right\\}\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)+2\left(\frac{\eta^{s}+\eta^{0}}{2}-\eta_{d}\right)\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx$
$\displaystyle+$
$\displaystyle\int_{\Omega}\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)\nabla
p^{0}-k^{2}\left(\frac{\eta^{s}-\eta^{0}}{s}\right)p^{0}dx.$

Taking $\varphi^{\prime}=\frac{\eta^{s}-\eta^{0}}{s}$ in the adjoint equation
of state verified by $p^{0}$ (3.4), we obtain

$\displaystyle\int_{\Omega}2(\nabla\eta^{0}-A)\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)+2(\eta^{0}-\eta_{d})\left(\frac{\eta^{s}-\eta^{0}}{s}\right)dx+$
$\displaystyle\int_{\Omega}\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)\nabla
p^{0}dx-\int_{\Omega}k^{2}\left(\frac{\eta^{s}-\eta^{0}}{s}\right)p^{0}dx=0.$

So injecting this equality into the expression of $R(s)$ gives

$\displaystyle R(s)$ $\displaystyle=$
$\displaystyle\int_{\Omega}\nabla\left(\eta^{s}-\eta^{0}\right)\nabla\left(\frac{\eta^{s}-\eta^{0}}{s}\right)+\left(\eta^{s}-\eta^{0}\right)\left(\frac{\eta^{s}-\eta^{0}}{s}\right)$
$\displaystyle=$
$\displaystyle\int_{\Omega}\left\lvert\nabla\left(\frac{\eta^{s}-\eta^{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega}\left\lvert\frac{\eta^{s}-\eta^{0}}{s^{1/2}}\right\rvert^{2}dx=\left\lVert\frac{\eta^{s}-\eta^{0}}{s^{1/2}}\right\rVert^{2}_{H^{1}(\Omega)}$

To show that $R(s)$ tends to $0$, we can show that the norm of
$\eta^{s}-\eta^{0}$ in $H^{1}(\Omega)$ is an $O(s)$. In this case we need the
following lemma

###### Lemma 3.2

Let $\eta^{0}$ and $\eta^{s}$ be the solutions of (3.2) and (3.3)
respectively. Then, we have the following estimate:

$\displaystyle\lVert\eta^{s}-\eta^{0}\rVert_{H^{1}(\Omega)}\leq Cs$

with $C$ a positive constant not depending of $s.$

###### Proof.

Let $a(\eta,v)$ be the bilinear formula defined by

$\displaystyle a(\eta,v)=\int_{\Omega}\nabla\eta\nabla
vdx-k^{2}\int_{\Omega}\eta vdx$

it is clear that there are two constants $\alpha\in\mathbb{R}$ and $\beta>0$
such that

$\displaystyle\lvert
a(\eta,\eta)\rvert\geq\beta\lVert\eta\rVert^{2}_{H^{1}_{0}(\Omega)}-\alpha\lVert\eta\rVert^{2}_{L^{2}(\Omega)}$
(3.9)

indeed, we have

$\displaystyle a(\eta,v)$ $\displaystyle=$
$\displaystyle\int_{\Omega}\lvert\nabla\eta\rvert^{2}dx-\int_{\Omega}k^{2}\lvert\eta\rvert^{2}dx=\lVert\nabla\eta\rVert^{2}_{L^{2}(\Omega)}-k^{2}\lVert\eta\rVert^{2}_{L^{2}(\Omega)}$
$\displaystyle=$
$\displaystyle\lVert\eta\rVert^{2}_{H^{1}_{0}(\Omega)}-k^{2}\lVert\eta\rVert^{2}_{L^{2}(\Omega)}$
$\displaystyle\geq$
$\displaystyle\beta\lVert\eta\rVert^{2}_{H^{1}_{0}(\Omega)}-k^{2}\lVert\eta\rVert^{2}_{L^{2}(\Omega)}.$

Therefore $a(\eta,v)$ is $H^{1}_{0}(\Omega)$ elliptical for $\alpha=k^{2}$.
Subtracting the two problems:

$\displaystyle\int_{\Omega}\nabla(\eta^{s}-\eta^{0})\nabla
vdx-\int_{\Omega}k^{2}(\eta^{s}-\eta^{0})vdx$ $\displaystyle=$
$\displaystyle\int_{\Omega}f^{s}vdx-\int_{\Omega}fvdx$ $\displaystyle=$
$\displaystyle\int_{\Omega\backslash
E_{r}}fvdx+\gamma\int_{E_{r}}fvdx-\int_{\Omega}fvdx$ $\displaystyle=$
$\displaystyle\int_{\Omega}fvdx-(1-\gamma)\int_{E_{r}}fvdx-\int_{\Omega}fv$
$\displaystyle=$ $\displaystyle-(1-\gamma)\int_{E_{r}}fvdx.$

This leads to

$\displaystyle\int_{\Omega}\nabla(\eta^{s}-\eta^{0})\nabla
vdx-\int_{\Omega}k^{2}(\eta^{s}-\eta^{0})vdx=-(1-\gamma)\int_{E_{r}}fvdx.$
(3.10)

Since $\eta^{s}-\eta^{0}\in H^{1}_{0}(\Omega)$, we can take
$v=\eta^{s}-\eta^{0}$. This gives

$\displaystyle\int_{\Omega}\lvert\nabla(\eta^{s}-\eta^{0})\rvert^{2}dx-\int_{\Omega}k^{2}(\eta^{s}-\eta^{0})\rvert^{2}dx=-(1-\gamma)\int_{E_{r}}f(\eta^{s}-\eta^{0})dx.$

Using the inequality verified by $a(\eta,\eta)$

$\displaystyle\beta\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}_{0}(\Omega)}-k^{2}\lVert\eta^{s}-\eta^{0}\rVert^{2}_{L^{2}(\Omega)}\leq\lvert
a(\eta^{s}-\eta^{0},\eta^{s}-\eta^{0})\rvert\leq\lvert
1-\gamma\rvert\left\lvert\int_{E_{r}}f(\eta^{s}-\eta^{0})dx\right\rvert.$

The Cauchy Schwarz inequality leads to:

$\displaystyle\beta\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}_{0}(\Omega)}-k^{2}\lVert\eta^{s}-\eta^{0}\rVert^{2}_{L^{2}(\Omega)}$
$\displaystyle\leq$ $\displaystyle\lvert 1-\gamma\rvert\lVert
f\rVert_{L^{2}(E_{r})}\lVert\eta^{s}-\eta^{0}\rVert_{L^{2}(E_{r})}$
$\displaystyle\leq$ $\displaystyle
C_{1}s\lVert\eta^{s}-\eta^{0}\rVert_{L^{2}(E_{r})}$ $\displaystyle\leq$
$\displaystyle C_{2}s\lVert\eta^{s}-\eta^{0}\rVert_{H^{1}(\Omega)}.$

We can also write

$\displaystyle\beta\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}_{0}(\Omega)}-k^{2}\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}(\Omega)}\leq
C_{2}s\lVert\eta^{s}-\eta^{0}\rVert_{H^{1}(\Omega)}.$

As the norms $\lVert\cdot\rVert_{H^{1}_{0}(\Omega)}$ and
$\lVert\cdot\rVert_{H^{1}(\Omega)}$ are equivalent, we can find $\beta_{1}>0$
such that

$\displaystyle\beta_{1}\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}(\Omega)}\leq\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}_{0}(\Omega)}.$

Therefore, taking $\beta_{2}$ as the product of $\beta$ and $\beta_{1}$

$\displaystyle\beta_{2}\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}(\Omega)}-k^{2}\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}(\Omega)}\leq
C_{2}s\lVert\eta^{s}-\eta^{0}\rVert_{H^{1}(\Omega)}.$
$\displaystyle(\beta_{2}-k^{2})\lVert\eta^{s}-\eta^{0}\rVert_{H^{1}(\Omega)}\leq
C_{2}s.$

If $\beta_{2}-k^{2}>0$, then we have, for $C=\frac{C_{2}}{\beta_{2}-k^{2}}$

$\displaystyle\lVert\eta^{s}-\eta^{0}\rVert^{2}_{H^{1}(\Omega)}\leq Cs.$

If $\beta_{2}-k^{2}\leq 0$, we can find $M_{1}>0$ such that
$M_{1}+\beta_{2}-k^{2}>0$ and such that

$\displaystyle(M_{1}+\beta_{2}-k^{2})\lVert\eta^{s}-\eta^{0}\rVert_{H^{1}(\Omega)}\leq
C_{2}s$

and so taking $C=\frac{C_{2}}{M_{1}+\beta_{2}-k^{2}}$, the desired result is
obtained. ∎

Because $\lVert\eta^{s}-\eta^{0}\rVert=O(s)$, by definition, they exist a rank
$N\in\mathbb{N}$ and a bounded sequence $b_{s}$ such that for all $s\geq N$,
$\lVert\eta^{s}-\eta^{0}\rVert_{H^{1}(\Omega)}=sb_{s}$. This implies that

$\displaystyle\left\lVert\frac{\eta^{s}-\eta^{0}}{s^{1/2}}\right\rVert^{2}_{H^{1}(\Omega)}=s(b_{s})^{2}.$
(3.11)

Thus by tending $s$ to zero, $s(b_{s})^{2}$ goes to zero. We conclude that
$\lim_{s\searrow 0}R(s)=0$. We then have the following proposition

###### Proposition 3.3

For a compact $d$-rectifiable set $E\subset\Omega$ ($d=0$) the topological
derivative for the functionnal $J$ is

$\displaystyle D_{T}J(x_{0})=(1-\gamma)f(x_{0})p^{0}(x_{0}).$

For a compact $d$-rectifiable set $E\subset\Omega$ ($d=1$ for a curve, $d=2$
for a surface) the topological derivative for the functionnal $J$ is

$\displaystyle D_{T}J=(1-\gamma)\int_{E}f(x)\Phi(x)dH^{d}$

###  3.2 Dirichlet condition on the boundary of the hole

Associate with $r$, $0<r\leq R$ the perturbed domain
$\Omega_{r}=\Omega\backslash E_{r}$, where by assumption,
$\partial\Omega_{r}=\partial\Omega\cup\partial E_{r}$,
$\partial\Omega\cap\partial E_{r}=\emptyset$ and $\partial E_{r}\in C^{1,1}$.
We first want to put a Dirichlet condition at the boundary of the hole (the
$r$-dilatation). Let $\eta_{r}$ be the solution of the following system

$\begin{cases}-\Delta\eta_{r}-k^{2}\eta_{r}=f\qquad\text{in}\;\;\Omega_{r}\\\\[5.69046pt]
\qquad\qquad\;\;\,\eta_{r}=0\qquad\text{on}\;\;\partial\Omega\\\\[5.69046pt]
\qquad\qquad\;\;\,\eta_{r}=0\qquad\text{on}\;\;\partial E_{r}\end{cases}$
(3.12)

which can be extended to $\Omega$ by introducing the solution
$\eta^{0}_{r}:E_{r}^{0}\rightarrow\mathbb{R}$ of the problem

$\displaystyle-\Delta\eta^{0}_{r}-k^{2}\eta^{0}_{r}=f\;\;\text{in}\;\;E^{0}_{r},\;\;\eta^{0}_{r}=\eta_{r}\;\text{on}\;\partial
E_{r}.$

We suppose that $\Omega_{r}$ has two components $\Omega_{r}^{m}$ and
$\Omega_{r}^{0}$. $\Omega_{r}^{m}$ is the component of $\Omega_{r}$ for which
$\partial\Omega$ is part of its boundary. $\Omega_{r}^{0}$ is the blind
component of $\Omega_{r}$ whose boundary has an empty intersection with
$\partial\Omega$. The function $\eta_{r}$ is distributed between the two
components $\Omega_{r}^{0}$ and $\Omega_{r}^{m}$ as

$\displaystyle\eta_{r}=\eta_{0}\;\;\text{in}\;\;\Omega_{r}^{0}\;\;\;\text{and}\;\;\;-\Delta\eta_{r}-k^{2}\eta_{r}=f\;\;\text{in}\;\;\Omega_{r}^{m},$
(3.13)
$\displaystyle\begin{cases}\eta_{r}=0\;\;\text{on}\;\;\partial\Omega\\\\[5.69046pt]
\eta_{r}=0\;\;\text{on}\;\;\partial\Omega_{r}^{m}\cap\partial
E_{r}.\end{cases}$

Since $\partial E_{r}$ is made up of two disjoint boundary
$\partial\Omega_{r}^{0}$ and $\partial\Omega_{r}^{m}\cap\partial E_{r}$, we
can construct an extension to $\Omega$ by defining the solution
$\eta^{0}_{r}:E_{r}^{0}\rightarrow\mathbb{R}$ of the problem

$\displaystyle-\Delta\eta^{0}_{r}-k^{2}\eta^{0}_{r}=f\;\;\text{in}\;\;E_{r}^{0},\;\;\;\;\eta_{r}^{0}$
$\displaystyle=$
$\displaystyle\eta_{r}\;\;\text{on}\;\;\partial\Omega_{r}^{m}\cap\partial
E_{r}$ $\displaystyle\eta_{r}^{0}$ $\displaystyle=$
$\displaystyle\eta_{r}\;\;\text{on}\;\;\partial\Omega_{r}^{0}.$

For simplicity, we can assume that $\Omega_{r}^{0}$ is empty.
Let us associate with the state $\eta_{r}$ the perturbed objective function

$\displaystyle
J(\Omega_{r})=\int_{\Omega_{r}}\left(\lvert\nabla\eta_{r}-A\rvert^{2}+\lvert\eta_{r}-\eta_{d}\lvert^{2}\right)dx.$

We want to find the topological derivative of the objective function
$J(\Omega_{r})$

$\displaystyle dJ=\lim_{r\searrow
0}\frac{J(\Omega_{r})-J(\Omega)}{\alpha_{N-d}r^{N-d}}.$ (3.14)

We define the following set

$\displaystyle\mathcal{V}_{r}=\left\\{\eta\in
H^{1}(\Omega_{r}):\;\;\eta=0\;\text{on}\;\partial\Omega,\;\eta=0\;\text{on}\;\partial
E_{r}\right\\}.$ (3.15)

The variational formulation of (3.12) is to find $\eta_{r}\in\mathcal{V}_{r}$
such that

$\displaystyle-\int_{\partial\Omega_{r}}\frac{\partial\eta_{r}}{\partial
n}vd\sigma+\int_{\Omega_{r}}\nabla\eta_{r}\nabla
vdx-\int_{\Omega_{r}}k^{2}\eta_{r}vdx=\int_{\Omega_{r}}fvdx,\;\;\text{for
all}\;v\in\mathcal{V}$

that is to say

$\displaystyle-\int_{\partial\Omega}\frac{\partial\eta_{r}}{\partial
n}vd\sigma-\int_{\partial E_{r}}\frac{\partial\eta_{r}}{\partial
n}vd\sigma+\int_{\Omega_{r}}\nabla\eta_{r}\nabla
vdx-\int_{\Omega_{r}}k^{2}\eta_{r}vdx=\int_{\Omega_{r}}fvdx,\;\;\text{for
all}\;v\in\mathcal{V}.$

By taking $\mathcal{V}=\mathcal{V}_{r}$, we get

$\displaystyle\int_{\Omega_{r}}\nabla\eta_{r}\nabla
vdx-\int_{\Omega_{r}}k^{2}\eta_{r}vdx=\int_{\Omega_{r}}fvdx,\;\;\text{for
all}\;v\in\mathcal{V}_{r}.$ (3.16)

We can therefore define the $r$-dependent Lagrangian as

$\displaystyle
L(r,\varphi,\phi)=\int_{\Omega_{r}}\lvert\nabla\varphi-A\rvert^{2}dx+\int_{\Omega_{r}}\lvert\varphi-\eta_{d}\rvert^{2}dx+\int_{\Omega_{r}}\nabla\varphi\nabla\phi
dx-\int_{\Omega_{r}}k^{2}\varphi\phi dx-\int_{\Omega_{r}}f\phi dx.$
$\displaystyle
J(\Omega_{r})=\inf_{\varphi\in\mathcal{V}_{r}}\sup_{\phi\in\mathcal{V}_{r}}L(r,\varphi,\phi)=\int_{\Omega_{r}}\left\\{\lvert\nabla\eta_{r}-A\rvert^{2}+\lvert\eta_{r}-\eta_{d}\rvert^{2}+\nabla\eta_{r}\nabla
p^{0}-k^{2}\eta_{r}p_{0}-fp_{0}\right\\}dx.$ $\displaystyle
J(\Omega)=\inf_{\varphi\in H^{1}_{0}(\Omega)}\sup_{\phi\in
H^{1}_{0}(\Omega)}L(0,\varphi,\phi)=\int_{\Omega}\left\\{\lvert\nabla\eta_{0}-A\rvert^{2}+\lvert\eta_{0}-\eta_{d}\rvert^{2}+\nabla\eta_{0}\nabla
p_{0}-k^{2}\eta_{0}p_{0}-fp_{0}\right\\}dx.$

We can rewrite the $r$-dependent Lagrangian as follows

$\displaystyle L(r,\varphi,\phi)$ $\displaystyle=$
$\displaystyle\int_{\Omega}\left\\{\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}+\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right\\}dx$
$\displaystyle-$
$\displaystyle\int_{E_{r}}\left\\{\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}+\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right\\}dx.$

The derivative of the $r$-dependent Lagrangian with respect to $\varphi$ is:

$\displaystyle
d_{\varphi}L(r,\varphi,\phi;\varphi^{\prime})=\int_{\Omega_{r}}\left\\{2(\nabla\varphi-A)\nabla\varphi^{\prime}+2(\varphi-\eta_{d})\varphi^{\prime}\right\\}dx+\int_{\Omega_{r}}\nabla\varphi^{\prime}\nabla\phi
dx-\int_{\Omega_{r}}k^{2}\varphi^{\prime}\phi dx.$

The initial adjoint state $p_{0}$ is a solution of
$d_{\varphi}L(0,\eta_{0},p_{0};\varphi^{\prime})=0$ for all
$\varphi^{\prime}\in H^{1}_{0}(\Omega)$ with $\eta_{0}=\eta=\eta_{r}$ for
$r=0$. Thus the variational formulation of the adjoint equation of state is
given by

$\displaystyle\text{Find}\,p_{0}\in
H^{1}_{0}(\Omega),\;\,\int_{\Omega}\left\\{2(\nabla\eta_{0}-A)\nabla\varphi^{\prime}+2(\eta_{0}-\eta_{d})\varphi^{\prime}\right\\}dx+\int_{\Omega}\left\\{\nabla\varphi^{\prime}\nabla
p_{0}-k^{2}\varphi^{\prime}p_{0}\right\\}dx=0.$ (3.17)

The derivative of the Lagrangian depending on $r$ with respect to $\phi$ is:

$\displaystyle
d_{\phi}L(r,\varphi,\phi;\phi^{\prime})=\int_{\Omega_{r}}\nabla\varphi\nabla\phi^{\prime}dx-\int_{\Omega_{r}}k^{2}\varphi\phi^{\prime}dx-\int_{\Omega_{r}}f\phi^{\prime}dx.$

The initial state $\eta=\eta_{0}$ is solution of
$d_{\phi}L(0,\eta_{0},0;\phi^{\prime})=0$ for all $\phi^{\prime}\in
H^{1}_{0}(\Omega)$ and we obviously find the initial variational equation

$\displaystyle\int_{\Omega}\nabla\eta_{0}\nabla\phi^{\prime}dx-\int_{\Omega}k^{2}\eta_{0}\phi^{\prime}dx=\int_{\Omega}f\phi^{\prime}dx,\;\;\text{for
all}\;\phi^{\prime}\in H^{1}_{0}(\Omega)$ (3.18)

and the state $\eta_{r}$ for $r\geq 0$ verifies

$\displaystyle\int_{\Omega_{r}}\nabla\eta_{r}\nabla\phi^{\prime}dx-\int_{\Omega_{r}}k^{2}\eta_{r}\phi^{\prime}dx=\int_{\Omega_{r}}f\phi
dx,\;\;\text{for all}\;\phi^{\prime}\in H^{1}_{0}(\Omega).$ (3.19)

Now we want to calculate the derivative of the Lagrangian depending on $r$
with respect to $s$. To do this we first calculate the ratio
$\frac{L(r,\varphi,\phi)-L(0,\varphi,\phi)}{s}$

$\displaystyle L(r,\varphi,\phi)-L(0,\varphi,\phi)$ $\displaystyle=$
$\displaystyle\int_{\Omega_{r}}\left\\{\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}\right\\}dx+\int_{\Omega_{r}}\left\\{\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right\\}dx$
$\displaystyle-$
$\displaystyle\int_{\Omega}\left\\{\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}\right\\}dx+\int_{\Omega}\left\\{\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right\\}dx$
$\displaystyle=$
$\displaystyle-\int_{E_{r}}\left\\{\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}+\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right\\}dx.$

For $d=0$, $E=\\{x_{0}\\}$, $E_{r}=\\{x\in\mathbb{R}^{N}:\;\lvert
x-x_{0}\rvert\leq r\\}=\overline{B(x_{0},r)}$; the closed ball centred $x_{0}$
and radius $r$. So by applying Lebesgue’s differentiation theorem we get

$\displaystyle d_{s}L(0,\varphi,\phi)$ $\displaystyle=$
$\displaystyle\lim_{s\searrow 0}\;\;-\frac{1}{\lvert
B(x_{0},r)\rvert}\int_{B(x_{0},r)}\left\\{\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}+\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right\\}dx$
$\displaystyle=$
$\displaystyle-\lvert\nabla\varphi(x_{0})-A(x_{0})\rvert^{2}-\lvert\varphi(x_{0})-\eta_{d}(x_{0})\rvert^{2}-\nabla\varphi(x_{0})\nabla\phi(x_{0})$
$\displaystyle+$ $\displaystyle
k^{2}\varphi(x_{0})\phi(x_{0})+f(x_{0})\phi(x_{0}).$

So at the point $\eta_{0}$ and $p_{0}$

$\displaystyle d_{s}L(0,\eta_{0},p_{0})$ $\displaystyle=$
$\displaystyle-\lvert\nabla\eta_{0}(x_{0})-A(x_{0})\rvert^{2}-\lvert\eta_{0}(x_{0})-\eta_{d}(x_{0})\rvert^{2}-\nabla\eta_{0}(x_{0})\nabla
p_{0}(x_{0})$ $\displaystyle+$ $\displaystyle
k^{2}\eta_{0}(x_{0})p_{0}(x_{0})+f(x_{0})p_{0}(x_{0}).$

In cases where $0<d\leq N-1$, we have

$\displaystyle\frac{L(r,\varphi,\Phi)-L(0,\varphi,\Phi)}{s}$ $\displaystyle=$
$\displaystyle-\frac{1}{|E_{r}\rvert}\int_{E_{r}}\left(\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}+\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right)dx$
$\displaystyle=$
$\displaystyle-\frac{1}{\alpha_{N-d}r^{N-d}}\int_{E_{r}}\left(\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}+\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right)dx$
$\displaystyle\rightarrow$
$\displaystyle-\int_{E}\left(\lvert\nabla\varphi-A\rvert^{2}+\lvert\varphi-\eta_{d}\rvert^{2}+\nabla\varphi\nabla\phi-k^{2}\varphi\phi-f\phi\right)dH^{d}.$

In the cases $\eta_{0}$ and $p_{0}$

$\displaystyle
d_{s}L(0,\eta_{0},p_{0})=-\int_{E}\left(\lvert\nabla\eta_{0}-A\rvert^{2}+\lvert\eta_{0}-\eta_{d}\rvert^{2}+\nabla\eta_{0}\nabla
p_{0}-k^{2}\eta_{0}p_{0}-fp_{0}\right)dH^{d}.$

We define the function $R(r)$ by

$\displaystyle
R(r)=\int_{0}^{1}d_{x}L\left(r,\eta_{0}+\theta(\eta_{r}-\eta_{0}),p_{0};\frac{\eta_{r}-\eta_{0}}{s}\right).$
(3.20) $\displaystyle R(r)$ $\displaystyle=$
$\displaystyle\int_{\Omega_{r}}2\left\\{\nabla\left(\frac{\eta_{r}+\eta_{0}}{2}\right)-A\right\\}\nabla\left(\frac{\eta_{r}-\eta_{0}}{s}\right)+2\left(\frac{\eta_{r}+\eta_{0}}{2}-\eta_{d}\right)\left(\frac{\eta_{r}-\eta_{0}}{s}\right)dx$
$\displaystyle+$
$\displaystyle\int_{\Omega_{r}}\nabla\left(\frac{\eta_{r}-\eta_{0}}{s}\right)\nabla
p_{0}-k^{2}\left(\frac{\eta_{r}-\eta_{0}}{s}\right)p_{0}dx$ $\displaystyle=$
$\displaystyle\frac{1}{s}\int_{\Omega_{r}}2\left\\{\nabla\left(\frac{\eta_{r}+\eta_{0}}{2}\right)-A\right\\}\nabla\left(\eta_{r}-\eta_{0}\right)+2\left(\frac{\eta_{r}+\eta_{0}}{2}-\eta_{d}\right)\left(\eta_{r}-\eta_{0}\right)dx$
$\displaystyle+$
$\displaystyle\frac{1}{s}\int_{\Omega_{r}}\nabla\left(\eta_{r}-\eta_{0}\right)\nabla
p_{0}-k^{2}\left(\eta_{r}-\eta_{0}\right)p_{0}dx$ $\displaystyle=$
$\displaystyle\frac{1}{s}\int_{\Omega_{r}}2\left\\{\nabla\left(\frac{\eta_{r}+\eta_{0}}{2}\right)-\nabla\eta_{0}+\nabla\eta_{0}-A\right\\}\nabla\left(\eta_{r}-\eta_{0}\right)dx$
$\displaystyle+$
$\displaystyle\frac{1}{s}\int_{\Omega_{r}}2\left(\frac{\eta_{r}+\eta_{0}}{2}-\eta_{0}+\eta_{0}-\eta_{d}\right)\left(\eta_{r}-\eta_{0}\right)+\nabla\left(\eta_{r}-\eta_{0}\right)\nabla
p_{0}-k^{2}\left(\eta_{r}-\eta_{0}\right)p_{0}dx$ $\displaystyle=$
$\displaystyle\frac{1}{s}\int_{\Omega_{r}}2\left\\{\nabla\left(\frac{\eta_{r}-\eta_{0}}{2}\right)+\nabla\eta_{0}-A\right\\}\nabla\left(\eta_{r}-\eta_{0}\right)+2\left(\frac{\eta_{r}-\eta_{0}}{2}+\eta_{0}-\eta_{d}\right)\left(\eta_{r}-\eta_{0}\right)dx$
$\displaystyle+$
$\displaystyle\frac{1}{s}\int_{\Omega_{r}}\nabla\left(\eta_{r}-\eta_{0}\right)\nabla
p_{0}-k^{2}\left(\eta_{r}-\eta_{0}\right)p_{0}dx.$

Then we get

$R(r)=\int_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega_{r}}\left\lvert\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right\rvert^{2}dx+\frac{1}{s}\int_{\Omega_{r}}\nabla\left(\eta_{r}-\eta_{0}\right)\nabla
p_{0}-k^{2}\left(\eta_{r}-\eta_{0}\right)p_{0}dx\\\ $
$\frac{1}{s}\int_{\Omega_{r}}2(\nabla\eta_{0}-A)\nabla(\eta_{r}-\eta_{0})+2(\eta_{0}-\eta_{d})(\eta_{r}-\eta_{0})dx.$
(3.21)

An other writing of (3.2) shows that for $\eta\in H_{0}^{1}(\Omega)$,

$\displaystyle\int_{\Omega_{r}}\left(\nabla\eta\nabla v-k^{2}\eta vdx-
fv\right)dx=-\int_{E_{r}}\left(\nabla\eta\nabla v-k^{2}\eta vdx-
fv\right)dx=\int_{\partial E_{r}}\frac{\partial\eta}{\partial
n}vdH^{N-1},\;\;\forall\;v\in H_{0}^{1}(\Omega).$

Thanks to the assumption on $E$, $\partial E_{r}\in C^{1,1}$

$\displaystyle\frac{\partial\eta}{\partial n}=\nabla\eta\cdot
n_{\Omega_{r}}=-\nabla\eta\cdot n_{\partial
E_{r}}=-\nabla\eta\cdot\nabla_{d_{E}}\;\text{sur}\;E_{r}.$

This allows us to write that

$\displaystyle\int_{\Omega_{r}}\left(\nabla\eta\nabla v-k^{2}\eta vdx-
fv\right)dx=-\int_{\partial E_{r}}\nabla\eta\cdot\nabla_{d_{E}}vdH^{N-1}.$
(3.22)

Between (3.16) and (3.22), we obtain

$\displaystyle\int_{\Omega_{r}}\left\\{\nabla(\eta_{r}-\eta_{0})\nabla
v-k^{2}(\eta_{r}-\eta_{0})v\right\\}dx=\int_{\partial
E_{r}}\nabla\eta_{0}\cdot\nabla_{d_{E}}vdH^{N-1}.$

At this level, we introduce the adjoint equation for $r\geq 0$

$\displaystyle\int_{\Omega_{r}}\left\\{2(\nabla\eta_{0}-A)\nabla\varphi^{\prime}+2(\eta_{0}-\eta_{d})\varphi^{\prime}\right\\}dx+\int_{\Omega_{r}}\nabla\varphi^{\prime}\nabla
p_{r}dx-\int_{\Omega_{r}}k^{2}\varphi^{\prime}p_{r}dx=0.$ (3.23)

By taking $\varphi^{\prime}=\eta_{r}-\eta_{0}$ in (3.23), it comes

$\displaystyle\int_{\Omega_{r}}\left\\{2(\nabla\eta_{0}-A)\nabla(\eta_{r}-\eta_{0})+2(\eta_{0}-\eta_{d})(\eta_{r}-\eta_{0})\right\\}dx=-\int_{\Omega_{r}}\left\\{\nabla(\eta_{r}-\eta_{0})\nabla
p_{r}-k^{2}(\eta_{r}-\eta_{0})p_{r}\right\\}dx.$

We can therefore rewrite it as follows

$\displaystyle R(r)$ $\displaystyle=$
$\displaystyle\int_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega_{r}}\left\lvert\frac{\eta_{s}-\eta_{0}}{s^{1/2}}\right\rvert^{2}dx+\frac{1}{s}\int_{\Omega_{r}}\left\\{\nabla\left(\eta_{r}-\eta_{0}\right)\nabla
p_{0}-k^{2}\left(\eta_{r}-\eta_{0}\right)p_{0}\right\\}dx$ (3.24)
$\displaystyle-$
$\displaystyle\frac{1}{s}\int_{\Omega_{r}}\left\\{\nabla(\eta_{r}-\eta_{0})\nabla
p_{r}-k^{2}(\eta_{r}-\eta_{0})p_{r}\right\\}dx$ (3.25) $\displaystyle=$
$\displaystyle\int_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega_{r}}\left\lvert\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right\rvert^{2}dx$
(3.26) $\displaystyle-$
$\displaystyle\frac{1}{s}\int_{\Omega_{r}}\left\\{\nabla(\eta_{r}-\eta_{0})\nabla(p_{r}-p_{0})dx-k^{2}(\eta_{r}-\eta_{0})(p_{r}-p_{0})\right\\}dx$
(3.27) $\displaystyle=$
$\displaystyle\int_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega_{r}}\left\lvert\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right\rvert^{2}dx-\frac{1}{s^{1/2}}\int_{\partial
E_{r}\cap\partial\Omega_{r}^{m}}\nabla p_{0}\cdot\nabla
d_{E}\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)dx.$ (3.28)

We therefore have the following theorem

###### Theorem 3.4

Let $E$ verify Hypothesis1 and $s=\alpha_{N-d}r^{N-d}$.
The topological derivative (3.14) exists if and only if the following limit
exists

$\displaystyle l=\lim_{r\searrow 0}\;(l_{0}(r)+l_{1}(r));\qquad l_{0}(r)$
$\displaystyle=$
$\displaystyle\int_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega_{r}}\left\lvert\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right\rvert^{2}dx$
$\displaystyle l_{1}(r)$ $\displaystyle=$
$\displaystyle\frac{1}{s^{1/2}}\int_{\partial
E_{r}\cap\partial\Omega_{r}^{m}}\nabla p_{0}\cdot\nabla
d_{E}\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)dx.$

The topological derivative is given by the expression

$\displaystyle
dJl-\int_{E}\left(\lvert\nabla\eta_{0}-A\rvert^{2}+\lvert\eta_{0}-\eta_{d}\rvert^{2}+\nabla\eta_{0}\nabla
p_{0}-k^{2}\eta_{0}p_{0}-fp_{0}\right)dH^{d}$

where $(\eta_{0},p_{0})$ is the solution of the coupled system (3.18)-(3.17).
In particular for $d=0$;

$\displaystyle
dJ=l-\lvert\nabla\eta_{0}(x_{0})-A(x_{0})\rvert^{2}-\lvert\eta_{0}(x_{0})-\eta_{d}(x_{0})\rvert^{2}-\nabla\eta_{0}(x_{0})\nabla
p_{0}(x_{0})+k^{2}\eta_{0}(x_{0})p_{0}(x_{0})+f(x_{0})p_{0}(x_{0}).$

###  3.3 Neumann condition on the boundary of the hole

In the case where we put a Neumann condition on the boundary of the hole, the
perturbed problem becomes

$\begin{cases}-\Delta\eta_{r}-k^{2}\eta_{r}=f\qquad\text{in}\;\;\Omega_{r}\\\\[5.69046pt]
\qquad\qquad\;\;\,\eta_{r}=0\qquad\text{on}\;\;\partial\Omega\\\\[5.69046pt]
\qquad\qquad\frac{\partial\eta_{r}}{\partial n}=0\qquad\text{on}\;\;\partial
E_{r}\end{cases}$ (3.29)

First of all, we note that the calculations remain the same. The small changes
that exist are simply in the spaces considered. The techniques for calculating
the derivative with respect to the $r$-Lagrangian variables remain unchanged.
So in the following the details of the calculations will be omitted.
We construct an extension to $\Omega$ by always introducing the solution
$\eta^{0}_{r}:E_{r}^{0}\rightarrow\mathbb{R}$ of the problem

$\displaystyle-\Delta\eta^{0}_{r}-k^{2}\eta^{0}_{r}=f\;\;\text{in}\;\;E^{0}_{r},\;\;\eta^{0}_{r}=\eta_{r}\;\text{on}\;\partial
E_{r}.$

Always the hypothesis $\Omega_{r}$ has two components $\Omega_{r}^{m}$ and
$\Omega_{r}^{0}$ is assumed. $\Omega_{r}^{m}$ is the component of $\Omega_{r}$
for which $\partial\Omega$ is part of its boundary. $\Omega_{r}^{0}$ is the
blind component of $\Omega_{r}$ whose boundary has an empty intersection with
$\partial\Omega$. The function $\eta_{r}$ is distributed between the two
components $\Omega_{r}^{0}$ and $\Omega_{r}^{m}$ as

$\displaystyle\eta_{r}=\eta_{0}\;\;\text{in}\;\;\Omega_{r}^{0}\;\;\;\text{and}\;\;\;-\Delta\eta_{r}-k^{2}\eta_{r}=f\;\;\text{in}\;\;\Omega_{r}^{m},$
$\displaystyle\begin{cases}\eta_{r}=0\;\;\text{on}\;\;\partial\Omega\\\\[5.69046pt]
\eta_{r}=0\;\;\text{on}\;\;\partial\Omega_{r}^{m}\cap\partial
E_{r}.\end{cases}$

Since $\partial E_{r}$ is made up of two disjoint edges
$\partial\Omega_{r}^{0}$ and $\partial\Omega_{r}^{m}\cap\partial E_{r}$, we
can construct an extension to $\Omega$ by defining the solution
$\eta^{0}_{r}:E_{r}^{0}\rightarrow\mathbb{R}$ of the problem

$\displaystyle-\Delta\eta^{0}_{r}-k^{2}\eta^{0}_{r}=f\;\;\text{in}\;\;E_{r}^{0},\;\;\;\;\eta_{r}^{0}$
$\displaystyle=$
$\displaystyle\eta_{r}\;\;\text{on}\;\;\partial\Omega_{r}^{m}\cap\partial
E_{r}$ $\displaystyle\eta_{r}^{0}$ $\displaystyle=$
$\displaystyle\eta_{r}\;\;\text{on}\;\;\partial\Omega_{r}^{0}.$

For simplicity, $\Omega_{r}^{0}$ is always assumed to be empty. The solution
space of the perturbed problem to consider is

$\displaystyle\mathcal{U}_{r}=\left\\{\eta\in
H^{1}(\Omega_{r}):\;\;\eta=0\;\text{on}\;\partial\Omega\right\\}.$

The variational formulation of (3.29) consists in finding
$\eta_{r}\in\mathcal{U}_{r}$ such that

$\displaystyle-\int_{\partial\Omega}\frac{\partial\eta_{r}}{\partial
n}vd\sigma-\int_{\partial E_{r}}\frac{\partial\eta_{r}}{\partial
n}vd\sigma+\int_{\Omega_{r}}\nabla\eta_{r}\nabla
vdx-\int_{\Omega_{r}}k^{2}\eta_{r}vdx-\int_{\Omega_{r}}fvdx=0,\;\;\text{for
all}\;\;v\in\mathcal{\tilde{U}}_{r}.$

According to the equation (3.29), the integral on the boundary of $E_{r}$ is
zero. It will therefore suffice to take the set of test functions,

$\displaystyle\mathcal{\tilde{U}}_{r}=\left\\{v\in
H^{1}(\Omega_{r}):\;\;v=0\;\;\;\text{on}\;\partial\Omega\right\\}=\mathcal{U}_{r},$

in order to get

$\displaystyle\int_{\Omega_{r}}\nabla\eta_{r}\nabla
vdx-\int_{\Omega_{r}}k^{2}\eta_{r}vdx-\int_{\Omega_{r}}fvdx=0,\;\;\text{for
all}\;\;v\in\mathcal{\tilde{U}}_{r}.$

The $r$-dependent Lagrangian is then a function defined from
$[0,R]\times\mathcal{U}_{r}$ to values in $\mathbb{R}$ by

$\displaystyle
L(r,\varphi,\phi)=\int_{\Omega_{r}}\lvert\nabla\varphi-A\rvert^{2}dx+\int_{\Omega_{r}}\lvert\varphi-\eta_{d}\rvert^{2}dx+\int_{\Omega_{r}}\nabla\varphi\nabla\phi
dx-\int_{\Omega_{r}}k^{2}\varphi\phi dx-\int_{\Omega_{r}}f\phi dx$
$\displaystyle
J(\Omega_{r})=\inf_{\varphi\in\mathcal{U}_{r}}\sup_{\phi\in\mathcal{U}_{r}}L(s,\varphi,\phi)=\int_{\Omega_{r}}\left(\lvert\nabla\eta_{r}-A\rvert^{2}+\lvert\eta_{r}-\eta_{d}\rvert^{2}+\nabla\eta_{r}\nabla
p_{0}-k^{2}\eta_{r}p_{0}-fp_{0}\right)dx$ $\displaystyle
J(\Omega)=\inf_{\varphi\in H^{1}_{0}(\Omega)}\sup_{\phi\in
H^{1}_{0}(\Omega)}L(0,\varphi,\phi)=\int_{\Omega}\left(\lvert\nabla\eta_{0}-A\rvert^{2}+\lvert\eta_{0}-\eta_{d}\rvert^{2}+\nabla\eta_{0}\nabla
p_{0}-k^{2}\eta_{0}p_{0}-fp_{0}\right)dx$

The derivative of the $r$-dependent Lagrangian with respect to $\varphi$ is:

$\displaystyle
d_{\varphi}L(r,\varphi,\phi;\varphi^{\prime})=\int_{\Omega_{r}}\left\\{2(\nabla\varphi-A)\nabla\varphi^{\prime}+2(\varphi-\eta_{d})\varphi^{\prime}\right\\}dx+\int_{\Omega_{r}}\nabla\varphi^{\prime}\nabla\phi
dx-\int_{\Omega_{r}}k^{2}\varphi^{\prime}\phi dx$

The derivative of the $r$-dependent Lagrangian with respect to $phi$ is:

$\displaystyle
d_{\phi}L(s,\varphi,\phi;\phi^{\prime})=\int_{\Omega_{r}}\nabla\varphi\nabla\phi^{\prime}dx-\int_{\Omega_{r}}k^{2}\varphi\phi^{\prime}dx-\int_{\Omega}f\phi^{\prime}dx.$

For $d=0$, $E_{r}=\overline{B(x_{0},r)}$,

$\displaystyle
d_{s}L(0,\eta_{0},p_{0})=-\lvert\nabla\eta(x_{0})-A(x_{0})\rvert^{2}-\lvert\eta_{0}(x_{0})-\eta_{d}(x_{0})\rvert^{2}-\nabla\eta_{0}(x_{0})\nabla
p_{0}(x_{0})+k^{2}\eta_{0}(x_{0})p_{0}(x_{0}).$

For the cases where $d=1,2$,

$\displaystyle
d_{s}L(0,\eta_{0},p_{0})=-\int_{E}\left(\lvert\nabla\eta_{0}-A\rvert^{2}+\lvert\eta_{0}-\eta_{d}\rvert^{2}+\nabla\eta_{0}\nabla
p_{0}-k^{2}\eta_{0}p_{0}\right)dH^{d}.$

The state and adjoint state $\eta_{0}$ and $p_{0}$ are solutions of

$\displaystyle\text{Find}\,\eta_{0}\in
H^{1}_{0}(\Omega),\,\,\int_{\Omega}\nabla\eta_{0}\nabla\phi^{\prime}dx-\int_{\Omega}k^{2}\eta_{0}\phi^{\prime}dx=\int_{\Omega}f\phi^{\prime}dx,\;\forall\,\phi^{\prime}\in
H^{1}_{0}(\Omega).$ (3.30) $\displaystyle\text{Find}\,p_{0}\in
H^{1}_{0}(\Omega),\;\int_{\Omega}\left\\{2(\nabla\eta_{0}-A)\nabla\varphi^{\prime}+2(\eta_{0}-\eta_{d})\varphi^{\prime}\right\\}dx+\int_{\Omega}\left\\{\nabla\varphi^{\prime}\nabla
p_{0}-k^{2}\varphi^{\prime}p_{0}\right\\}dx=0,\;\forall\,\varphi^{\prime}\in
H^{1}_{0}(\Omega).$ (3.31)

The function $R(r)$ is defined by

$R(r)=\int_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega_{r}}\left\lvert\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right\rvert^{2}dx-\frac{1}{s^{1/2}}\int_{\partial
E_{r}\cap\partial\Omega_{r}^{m}}\nabla p_{0}\cdot\nabla
d_{E}\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)dx.$ (3.32)

We therefore have the following theorem

###### Theorem 3.5

Let $0\leq d<N$, $E$ verifying the hypothesis H1 and $s=\alpha_{N-d}r^{N-d}$.
The topological derivative exists if and only if the following limit exists

$\displaystyle l=\lim_{r\searrow 0}(l_{0}(r)+l_{1}(r);\qquad l_{0}(r)$
$\displaystyle=$
$\displaystyle\int_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega_{r}}\left\lvert\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right\rvert^{2}dx$
$\displaystyle l_{1}(r)$ $\displaystyle=$
$\displaystyle\frac{1}{s^{1/2}}\int_{\partial
E_{r}\cap\partial\Omega_{r}^{m}}\nabla p_{0}\cdot\nabla
d_{E}\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)dx.$

Then, in this case, the topological derivative is given by the following
expression

$\displaystyle
dJ=l-\int_{E}\left(\lvert\nabla\eta_{0}-A\rvert^{2}+\lvert\eta_{0}-\eta_{d}\rvert^{2}+\nabla\eta_{0}\nabla
p_{0}-k^{2}\eta_{0}p_{0}-fp_{0}\right)dH^{d}$

where $(\eta_{0},p_{0})$ is the solution of the coupled system (3.30)-(3.31).
In particular for $d=0$;

$\displaystyle
dJ=l-\lvert\nabla\eta_{0}(x_{0})-A(x_{0})\rvert^{2}-\lvert\eta_{0}(x_{0})-\eta_{d}(x_{0})\rvert^{2}-\nabla\eta_{0}(x_{0})\nabla
p_{0}(x_{0})+k^{2}\eta_{0}(x_{0})p_{0}(x_{0})+f(x_{0})p_{0}(x_{0}).$

###  3.4 Example of the one-dimensional Helmholtz equation

The calculation of $R(s)$ is not generally easy to obtain. And even if one
manages to calculate the limit giving $R(s),$ one is not always reassured of
the existence of its limit. Sometimes the limit does not exist, sometimes if
it is non-zero, sometimes it is zero. In the following, we give examples of
explicit calculations where we have $l=\infty.$ But unfortunately the
construction is not automatic, there are many things involved.
Let $\Omega=]-1,1[$, $E=\\{0\\}$ and

$\displaystyle\eta_{0}^{\prime\prime}+k^{2}\eta_{0}=x\;\;\text{in}\;\;]-1,1[\;\;\text{and}\;\;\eta_{0}(\pm
1)=0$ (3.33)

where $\eta_{0}$ is a function of a variable $x$ and $\eta_{0}^{\prime\prime}$
its second derivative with respect to $x$. The general solution of (3.33)
takes the form

$\displaystyle\eta_{0}(x)=A\cos(kx)+B\sin(kx)+\frac{x}{k^{2}},$

where $A$ and $B$ are constants to be determined from the initial conditions.
After the calculations we can have $A=0$ and $B=-\frac{1}{k^{2}\sin k}$, and
in this case we obtain the following solution

$\displaystyle\eta_{0}(x)=-\frac{\sin(kx)}{k^{2}\sin k}+\frac{x}{k^{2}}.$

We define for $0<r<1$ respectively $E_{r}$ and $\Omega_{r}$ by

$\displaystyle E_{r}=\\{x\in\mathbb{R},\;\;\lvert x\rvert\leq
r\\}=[-r,r]\;\;\text{and}\;\;\Omega_{r}=]-1,-r[\cup]r,1[.$

The perturbed system (for a Dirichlet condition on the boundary of $\partial
E_{r}$) is then defined by

$\displaystyle\eta_{r}^{\prime\prime}+k^{2}\eta_{r}=x\;\;\text{in}\;\;]-1,-r[\cup]r,1[\;\;\text{et}\;\;\eta_{r}(\pm
1)=0\;\;\eta_{r}(\pm r)=0.$

In $]-1,-r[$,

$\displaystyle\eta_{r}(x)$ $\displaystyle=$
$\displaystyle\left\\{\frac{1}{k^{2}\cos k}+\frac{\tan k}{k^{2}(\tan k\cos
kr-\sin kr)}\left(r-\frac{\cos kr}{\cos k}\right)\right\\}\cos kx$
$\displaystyle+$ $\displaystyle\left\\{\frac{1}{k^{2}(\tan k\cos kr-\sin
kr)}\left(r-\frac{\cos kr}{\cos k}\right)\right\\}\sin kx+\frac{x}{k^{2}}$

and in $]r,1[$,

$\displaystyle\eta_{r}(x)$ $\displaystyle=$
$\displaystyle\left\\{-\frac{1}{k^{2}\cos k}-\frac{\tan k}{k^{2}(\sin kr-\tan
k\cos kr)}\left(\frac{\cos kr}{\cos k}-r\right)\right\\}\cos kx$
$\displaystyle+$ $\displaystyle\left\\{\frac{1}{k^{2}(\sin kr-\tan k\cos
kr)}\left(\frac{\cos kr}{\cos k}-r\right)\right\\}\sin kx+\frac{x}{k^{2}}.$

As for the extension in $[-r,r]$, we define the function $\eta_{r}^{0}$
satisfying

$\displaystyle(\eta^{0}_{r})^{\prime\prime}+k^{2}\eta^{0}_{r}=x\;\;\text{in}\;\;[-r,r]\;\;\text{and}\;\;\eta^{0}_{r}(\pm
r)=\eta^{0}_{r}(\pm r)=0$

whose solution is

$\displaystyle\eta^{0}_{r}(x)=-\frac{r}{k^{2}\sin
kr}\sin(kx)+\frac{x}{k^{2}}.$

The extended function $\eta^{r}$ in $]-1,1[$ is then defined by:
in $]-1,-r[$,

$\displaystyle\eta^{r}(x)$ $\displaystyle=$
$\displaystyle\left\\{\frac{1}{k^{2}\cos k}+\frac{\tan k}{k^{2}(\tan k\cos
kr-\sin kr)}\left(r-\frac{\cos kr}{\cos k}\right)\right\\}\cos kx$
$\displaystyle+$ $\displaystyle\left\\{\frac{1}{k^{2}(\tan k\cos kr-\sin
kr)}\left(r-\frac{\cos kr}{\cos k}\right)\right\\}\sin kx+\frac{x}{k^{2}},$

in $[-r,r]$

$\displaystyle\eta^{r}(x)=-\frac{r}{k^{2}\sin kr}\sin(kx)+\frac{x}{k^{2}}$

and in $]r,1[$,

$\displaystyle\eta^{r}(x)$ $\displaystyle=$
$\displaystyle\left\\{-\frac{1}{k^{2}\cos k}-\frac{\tan k}{k^{2}(\sin kr-\tan
k\cos kr)}\left(\frac{\cos kr}{\cos k}-r\right)\right\\}\cos kx$
$\displaystyle+$ $\displaystyle\left\\{\frac{1}{k^{2}(\sin kr-\tan k\cos
kr)}\left(\frac{\cos kr}{\cos k}-r\right)\right\\}\sin kx+\frac{x}{k^{2}}.$

Let us note by $w^{r}=\eta^{r}-\eta_{0}$, to calculate the limit of $R(r)$,
in $]-1,-r[$,

$\displaystyle w^{r}(x)$ $\displaystyle=$
$\displaystyle\left\\{\frac{1}{k^{2}\cos k}+\frac{\tan k}{k^{2}(\tan k\cos
kr-\sin kr)}\left(r-\frac{\cos kr}{\cos k}\right)\right\\}\cos kx$
$\displaystyle+$ $\displaystyle\left\\{\frac{1}{k^{2}\sin
k}+\frac{1}{k^{2}(\tan k\cos kr-\sin kr)}\left(r-\frac{\cos kr}{\cos
k}\right)\right\\}\sin kx,$

in $[-r,r]$

$\displaystyle w^{r}(x)=\left(\frac{1}{k^{2}\sin k}-\frac{r}{k^{2}\sin
kr}\right)\sin(kx)$

and in $]r,1[$,

$\displaystyle w^{r}(x)$ $\displaystyle=$
$\displaystyle\left\\{-\frac{1}{k^{2}\cos k}-\frac{\tan k}{k^{2}(\sin kr-\tan
k\cos kr)}\left(\frac{\cos kr}{\cos k}-r\right)\right\\}\cos kx$
$\displaystyle+$ $\displaystyle\left\\{\frac{1}{k^{2}\sin
k}+\frac{1}{k^{2}(\sin kr-\tan k\cos kr)}\left(\frac{\cos kr}{\cos
k}-r\right)\right\\}\sin kx$
$\displaystyle\left\lVert\frac{w^{r}}{\sqrt{2r}}\right\rVert^{2}_{L^{2}(\Omega_{r})}$
$\displaystyle=$ $\displaystyle\frac{1}{2r}\left\\{\frac{1}{k^{2}\cos
k}+\frac{\tan k}{k^{2}(\tan k\cos kr-\sin kr)}\left(r-\frac{\cos kr}{\cos
k}\right)\right\\}^{2}\left(\frac{1}{2}(1-r)+\frac{1}{4}(\sin 2k-\sin
2kr)\right)$ $\displaystyle+$
$\displaystyle\frac{1}{2r}\left\\{\frac{1}{k^{2}\sin k}+\frac{1}{k^{2}(\tan
k\cos kr-\sin kr)}\left(r-\frac{\cos kr}{\cos
k}\right)\right\\}^{2}\left(\frac{1}{2}(1-r)-\frac{1}{4}(\sin 2k-\sin
2kr)\right)$ $\displaystyle+$
$\displaystyle\frac{1}{2r}\left\\{-\frac{1}{k^{2}\cos k}-\frac{\tan
k}{k^{2}(\sin kr-\tan k\cos kr)}\left(\frac{\cos kr}{\cos
k}-r\right)\right\\}^{2}\left(\frac{1}{2}(1-r)+\frac{1}{4}(\sin 2k-\sin
2kr)\right)$ $\displaystyle+$
$\displaystyle\frac{1}{2r}\left\\{\frac{1}{k^{2}\sin k}+\frac{1}{k^{2}(\sin
kr-\tan k\cos kr)}\left(\frac{\cos kr}{\cos
k}-r\right)\right\\}^{2}\left(\frac{1}{2}(1-r)-\frac{1}{4}(\sin 2k-\sin
2kr)\right)$ $\displaystyle\rightarrow+\infty,\;\;\text{when}\;\;r\rightarrow
0$
$\displaystyle\left\lVert\frac{w^{r}}{\sqrt{2r}}\right\rVert^{2}_{L^{2}(E_{r})}=\frac{1}{2r}\left(\frac{1}{k^{2}\sin
k}-\frac{r}{k^{2}\sin kr}\right)^{2}\left(r-\frac{1}{2}\sin 2kr\right).$

The function R(s) takes the following form:

$\displaystyle R(r)$ $\displaystyle=$
$\displaystyle\int_{-1}^{-r}\left\lvert\frac{(w^{r})^{\prime}}{\sqrt{2r}}\right\rvert^{2}dx+\int_{r}^{1}\left\lvert\frac{(w^{r})^{\prime}}{\sqrt{2r}}\right\rvert^{2}dx+\int_{-1}^{-r}\left\lvert\frac{w^{r}}{\sqrt{2r}}\right\rvert^{2}dx$
$\displaystyle+$
$\displaystyle\int_{r}^{1}\left\lvert\frac{w^{r}}{\sqrt{2r}}\right\rvert^{2}dx+\frac{1}{\sqrt{2r}}\int_{-r}^{r}p_{0}^{\prime}d^{\prime}_{E}\left(\frac{w^{r}}{\sqrt{2r}}\right)dx$
$\displaystyle=$
$\displaystyle\left\lVert\frac{w^{r}}{\sqrt{2r}}\right\rVert^{2}_{L^{2}(\Omega_{r})}+\left\lVert\frac{(w^{r})^{\prime}}{\sqrt{2r}}\right\rVert^{2}_{L^{2}(\Omega_{r})}+\frac{1}{\sqrt{2r}}\int_{-r}^{r}p_{0}^{\prime}d^{\prime}_{E}\left(\frac{w^{r}}{\sqrt{2r}}\right)dx.$

Solving the adjoint equation In $\Omega=]-1,1[,$ the adjoint equation is
solution to

$\displaystyle-p^{\prime\prime}_{0}(x)-k^{2}p_{0}(x)=2(\eta^{\prime\prime}_{0}-A^{\prime})+2(\eta_{0}-\eta_{d})\;\;\text{in}\;\;]-1,1[\;\text{and}\;\;p_{0}(\pm
1)=0$

To simplify, let $A$ be a constant function and $\eta_{d}=\frac{x}{k^{2}}$.
Then by replacing $\eta^{\prime\prime}_{0}$ and $\eta_{0}$ by their
expression, the equation becomes

$\displaystyle-p^{\prime\prime}_{0}(x)-k^{2}p_{0}(x)=\frac{2(k^{2}-1)}{k^{2}\sin
k}\sin kx$ (3.34)

whose characteristic equation $r^{2}+k^{2}=0$ has a negative discriminant, so
the solutions of the equation without second member are the functions
$x\rightarrow p_{H}(x)=\alpha\cos kx+\beta\sin kx$ where
$\alpha,\beta\in\mathbb{R}$.
In the following, we are looking for a particular solution. Let’s put
$A=\frac{2(k^{2}-1)}{k^{2}\sin k}$. Note that here the second member is of the
form $AIm(e^{ikx})$, so we are looking for a solution of the form
$p_{p}(x)=Im(Bxe^{ikx})$. Deriving $p_{p}(x)$ twice and replacing it in the
equation gives

$\displaystyle-Im\left((2iBk-
Bk^{2}x)e^{ikx}\right)+k^{2}Im(Bxe^{ikx})=AIm(e^{ikx})$
$\displaystyle\Rightarrow-Im\left[(2iBk-
Bk^{2}x+Bk^{2}x)e^{ikx}\right]=AIm(e^{ikx})$ $\displaystyle\Rightarrow-
Im\left[(2iBk)e^{ikx}\right]=AIm(e^{ikx})$

And by identification $-2iBk=A$, which implies that
$B=-\frac{A}{2ki}=\frac{iA}{2k}$. So the particular solution is

$\displaystyle
p_{p}(x)=Im\left(\frac{iA}{2k}xe^{ikx}\right)=Im\left(\frac{iA}{2k}x\cos
kx-\frac{A}{2k}x\sin kx\right)=\frac{A}{2k}x\cos kx,$

and the general solution $p_{0}$ is

$\displaystyle p_{0}(x)=\alpha\cos kx+\beta\sin kx+\frac{A}{2k}x\cos kx$

$\displaystyle p_{0}(-1)=0$ $\displaystyle\Rightarrow$
$\displaystyle\alpha\cos k-\beta\sin k-\frac{A}{2k}\cos k=0$ (3.35)
$\displaystyle p_{0}(1)=0$ $\displaystyle\Rightarrow$ $\displaystyle\alpha\cos
k+\beta\sin k+\frac{A}{2k}\cos k=0$ (3.36)

According to equations (3.35) and (3.36), $\alpha=0$ and
$\beta=-\frac{A}{2k\sin k}\cos k=-\frac{A}{2k\tan k}$. The general solution
$p_{0}$ is then

$\displaystyle p_{0}(x)=-\frac{A}{2k\tan k}\sin kx+\frac{A}{2k}x\cos
kx\;\;\text{and}\;\;p^{\prime}_{0}(x)=\frac{A}{2k}\cos kx-\frac{A}{2\tan
k}\cos kx-\frac{A}{2}x\sin kx$

The derivative of the function $w_{r}$, $(w_{r})^{\prime}$ is also given as
in $]-1,-r[$,

$\displaystyle(w_{r})^{\prime}(x)$ $\displaystyle=$
$\displaystyle\left\\{-\frac{1}{k\cos k}-\frac{\tan k}{k(\tan k\cos kr-\sin
kr)}\left(r-\frac{\cos kr}{\cos k}\right)\right\\}\sin kx$ $\displaystyle+$
$\displaystyle\left\\{\frac{1}{k\sin k}+\frac{1}{k(\tan k\cos kr-\sin
kr)}\left(r-\frac{\cos kr}{\cos k}\right)\right\\}\cos kx,$

in $[-r,r]$

$\displaystyle(w_{r})^{\prime}(x)=\left(\frac{1}{k\sin k}-\frac{r}{k\sin
kr}\right)\cos(kx)$

and in $]r,1[$,

$\displaystyle(w_{r})^{\prime}(x)$ $\displaystyle=$
$\displaystyle\left\\{\frac{1}{k\cos k}+\frac{\tan k}{k(\sin kr-\tan k\cos
kr)}\left(\frac{\cos kr}{\cos k}-r\right)\right\\}\sin kx$ $\displaystyle+$
$\displaystyle\left\\{\frac{1}{k\sin k}+\frac{1}{k(\sin kr-\tan k\cos
kr)}\left(\frac{\cos kr}{\cos k}-r\right)\right\\}\cos kx$
$\displaystyle\left\lVert\frac{(w^{r})^{\prime}}{\sqrt{2r}}\right\rVert^{2}_{L^{2}(\Omega_{r})}$
$\displaystyle=$ $\displaystyle\frac{1}{2r}\left\\{-\frac{1}{k\cos
k}-\frac{\tan k}{k(\tan k\cos kr-\sin kr)}\left(r-\frac{\cos kr}{\cos
k}\right)\right\\}^{2}\left(\frac{1}{2}(1-r)-\frac{1}{4}(\sin 2k-\sin
2kr)\right)$ $\displaystyle+$ $\displaystyle\frac{1}{2r}\left\\{\frac{1}{k\sin
k}+\frac{1}{k(\tan k\cos kr-\sin kr)}\left(r-\frac{\cos kr}{\cos
k}\right)\right\\}^{2}\left(\frac{1}{2}(1-r)+\frac{1}{4}(\sin 2k-\sin
2kr)\right)$ $\displaystyle+$ $\displaystyle\frac{1}{2r}\left\\{\frac{1}{k\cos
k}+\frac{\tan k}{k(\sin kr-\tan k\cos kr)}\left(\frac{\cos kr}{\cos
k}-r\right)\right\\}^{2}\left(\frac{1}{2}(1-r)-\frac{1}{4}(\sin 2k-\sin
2kr)\right)$ $\displaystyle+$ $\displaystyle\frac{1}{2r}\left\\{\frac{1}{k\sin
k}+\frac{1}{k(\sin kr-\tan k\cos kr)}\left(\frac{\cos kr}{\cos
k}-r\right)\right\\}^{2}\left(\frac{1}{2}(1-r)+\frac{1}{4}(\sin 2k-\sin
2kr)\right)$ $\displaystyle\rightarrow 0,\;\;\text{when}\;\;r\rightarrow 0$
$\displaystyle l_{1}(r)$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2r}}\int_{\partial
E_{r}}p_{0}^{\prime}d^{\prime}_{E}\left(\frac{w^{r}(x)}{\sqrt{2r}}\right)dH^{0}=-\frac{1}{\sqrt{2r}}p^{\prime}(x)\frac{x}{\lvert
x\rvert}\frac{w^{r}(x)}{\sqrt{2r}}\Big{|}_{x=-r}^{r}$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2r}}p^{\prime}(r)\frac{r}{\lvert
r\rvert}\frac{w^{r}(r)}{\sqrt{2r}}+\frac{1}{\sqrt{2r}}p^{\prime}(-r)\frac{(-r)}{\lvert
r\rvert}\frac{w^{r}(-r)}{\sqrt{2r}}=-\frac{1}{\sqrt{2r}}p^{\prime}(r)\frac{w^{r}(r)}{\sqrt{2r}}-\frac{1}{\sqrt{2r}}p^{\prime}(-r)\frac{w^{r}(-r)}{\sqrt{2r}}$
$\displaystyle=$ $\displaystyle-\frac{\sin(kr)}{2r}\left(\frac{A}{2k}\cos
kr-\frac{A}{2\tan k}\cos kr-\frac{A}{2}r\sin kr\right)\left(\frac{1}{k^{2}\sin
k}-\frac{r}{k^{2}\sin kr}\right)$
$\displaystyle+\frac{\sin(kr)}{2r}\left(\frac{A}{2k}\cos kr-\frac{A}{2\tan
k}\cos kr-\frac{A}{2}r\sin kr\right)\left(\frac{1}{k^{2}\sin
k}-\frac{r}{k^{2}\sin kr}\right)=0$

In summary, from all the above, we have: $\displaystyle\lim_{s\rightarrow
0}R(r)=\infty.$
From this example, considering the min max approach developed by Delfour, in
order to have the existence of a topological sensitivity which is a semi-
differential, we will be led to restrict ourselves to specific cases to hope
to calculate this topological semi-differential.

##  4  Homogenization with two scale convergence and topological semi
differential

We propose in this section a method allowing to pronounce on the limit of
$R(s)$ when s tends to zero. Here we are looking for sufficient conditions to
be able to calculate the limit of $R(s).$ Since this limit may exist or not,
on is wondering if it is possible to propose a family of boundary value
problems for which it will be quite possible to get the limit of $R(s)$ when
$s$ goes to $0.$
It is important to underline that in topological optimization, the
perturbation of domain introduces singular situations on the behavior of the
state of the system, the most important object in the analysis, the so-called
polarization matrix is to be fully understood. The polarization matrices are
present, either in explicit forms or implicit in all asymptotic formulae. For
more details about this notion, see for instance [25], [2], [11] and
references therein. The method proposed here, is based on an asymptotic
analysis approach and more particularly on two-scale convergence. For this
method, we can therefore refer to the work of Allaire [1], Nguetseng [24],
Frénod et al. [21], Faye et al.[18, 19]. One can always show, depending on the
field of study, the existence of a second scale either in relation to space or
in relation to time. For this, we assume that the solution $\eta^{s}$ of the
perturbed problem admits an asymptotic expansion of the form

$\eta^{s}(x)=\eta_{0}(x,\frac{x}{s})+s\eta_{1}(x,\frac{x}{s})+s^{2}\eta_{2}(x,\frac{x}{s})+\ldots$
(4.1)

where the function $\eta_{i},\,\,i=0,1,\ldots$ are regular and periodic
functions with respect to the second variable. Many characterization results
of the $\eta_{i}(x,\frac{x}{s})$ exist in the literature. One can quote, for
example, the work of Allaire [1] and Ngetseng [24] in the case of elliptic
equations, those of Faye et al [18, 19] in the case of the parabolic equations
and the work of Frénod et al. [21] in the case of hyperbolic equations; see
also [4], where parabolic and hyperbolic equations are coupled. We are going
to prove, in this paper, that if any sequence $\eta^{s}$ verifying an
asymptotic expansion of the type (4.1) where the functions $\eta_{i}(x,y)$ are
smooth and $Y$-periodic in $y,$ and two-scale converges to the first term of
the expansion, namely, $\eta_{0}(x,y)$ in a sense that we shall precise. In
many cases, $Y$ is choseen to be equals to $[0,1]^{N},\Omega\subset R^{N}.$ In
addition, we have:

$\eta^{s}(x)-\eta_{0}(x,\frac{x}{s})-s\eta_{1}(x,\frac{x}{s})\longrightarrow
0\,\,\text{in}\,\,H^{1}(\Omega)\,\,\text{strongly}.$ (4.2)

Let $\eta_{0}$ be the two scale limit of $\eta^{s}$ and
$\eta_{0}^{s}(x)=\eta_{0}(x,\frac{x}{s}).$ The proof of the main result of
this part is based on the existence of a corrector term of the type (4.2),
i.e.

$\frac{\eta^{s}-\eta_{0}^{s}}{s}\rightarrow\eta_{1}\,\,\text{in}\,\,H^{1}(\Omega).$

So we have the following theorem

###### Theorem 4.1

Considering $\eta^{s}$ the solution to (2.4) and
$\eta_{0}^{s}=\eta_{0}^{s}(x)=\eta_{0}(x,\frac{x}{s})$ where $\eta_{0}$ is the
two scale limits of $\eta^{s},$ the following estimate holds

$||\frac{\eta^{s}-\eta_{0}^{s}}{s}||_{H^{1}(\Omega)}\leq\alpha,$

where $\alpha$ is a constant not depending on $s.$ Furthermore, the sequence
$\frac{\eta^{s}-\eta_{0}^{s}}{s}$ two scale converges to a profile
$\eta_{1}\in H^{1}(\Omega,L^{2}_{\\#}(Y)).$

###### Proof.

The sequence $\eta^{s}$ is solution to the following equation

$\Delta\eta^{s}+\eta^{s}=f\,\,\text{in}\,\,\Omega\\\ $ (4.3)

Let us first characterized the equation satisfied by the two scale limits.

Defining a test function as follows, $\psi^{s}(x)=\psi(x,\frac{x}{s})$ for any
$\psi(x,y)$, regular with compact support in $Y\times{\mathbb{T}}^{2}$ and
periodic in $y$ with period 1, multiplying (4.3) by $\psi^{s}$ and integrating
in $\Omega,$ we have

$\int_{\Omega}(\Delta\eta^{s}+\eta^{s})\psi^{s}dx=\int_{\Omega}f\psi^{s}dtdx.$
(4.4)

Then using Green formula in the first integral over $\Omega$ and the boundary
condition, we get

$\int_{\Omega}\big{(}-\nabla\eta^{s}\nabla\psi^{s}+\eta^{s}\psi^{s}\big{)}dx=\int_{\Omega}f\psi^{s}dx$
(4.5)

But

$\nabla\psi^{s}=\left(\nabla_{x}\psi\right)^{s}+\frac{1}{s}\left(\nabla_{y}\psi\right)^{s},$
(4.6)

where

$\left(\nabla_{x}\psi\right)^{s}(x)=\nabla_{x}\psi(x,\frac{x}{s})\,\,\textrm{
and }\left(\nabla_{y}\psi\right)^{s}(x)=\nabla_{y}\psi(x,\frac{x}{s}),$ (4.7)

then we have

$\int_{\Omega}\big{(}-\nabla\eta^{s}\Big{(}\left(\nabla_{x}\psi\right)^{s}+\frac{1}{s}\left(\nabla_{y}\psi\right)^{s}\Big{)}+\eta^{s}\psi^{s}\big{)}dx=\int_{\Omega}f\psi^{s}dx$
(4.8)

Using the two-scale convergence due to Nguetseng [24] and Allaire [1] (see
also Frénod Raviart and Sonnendrücker [21]), if a sequence $f^{s}$ is bounded
in $L^{2}(\Omega)$, then there exists a profile $U(x,y)$, periodic of period 1
with respect to $y$, such that for all
$\psi(x,y)\in\mathcal{D}(\Omega,\mathcal{C}^{\infty}(Y)),$ we have

$\int_{\Omega}\int_{Y}f^{s}\psi^{s}dx\longrightarrow\int_{\Omega}\int_{Y}U\psi
dydx,$ (4.9)

for a subsequence extracted from $(f^{s})$.
Multiplying (4.8) by $s$ and passing to the limit as $s\rightarrow 0$ and
using (4.9) we have

$\int_{\Omega}\int_{Y}\Big{(}-\nabla\eta_{0}\nabla_{x}\psi+\eta_{0}\psi\Big{)}dydx=\int_{\Omega}\int_{Y}f\psi
dydx,$ (4.10)

and

$\int_{\Omega}\int_{Y}div_{y}(\nabla_{x}\eta_{0})\psi dx\,dy=0.$ (4.11)

We obtain from (4.10) the equation satisfied by $\eta_{0}$:

$\left\\{\begin{array}[]{ccc}\Delta\eta_{0}+\eta_{0}=f\,\,\text{in}\,\,\Omega\times
Y\\\ \eta_{0}=0\,\,\text{on}\,\,\partial\Omega\\\
\nabla_{y}\cdot\nabla_{x}\eta_{0}=0\,\,\text{in}\,\,\Omega\times
Y\end{array}\right.$ (4.12)

For the following, let

$\eta^{s}_{0}(x)=\eta_{0}(x,\frac{x}{s}),$ (4.13)

where $\eta_{0}$ is solution to (4.12). Then we have

$\nabla\eta_{0}^{s}=\left(\nabla_{x}\eta_{0}^{s}\right)^{s}+\frac{1}{s}\left(\nabla_{y}\eta_{0}^{s}\right)^{s},$
(4.14)

where

$\left(\nabla_{x}\eta_{0}^{s}\right)^{s}(x)=\nabla_{x}\eta_{0}^{s}(x,\frac{x}{s})\,\,\textrm{
and
}\left(\nabla_{y}\eta_{0}^{s}\right)^{s}(x)=\nabla_{y}\eta_{0}^{s}(x,\frac{x}{s}),$
(4.15)

It follows from a result of Allaire [1], that the sequence $\eta^{s}$ two
scale converge to a function $\eta_{(}x)=\int_{Y}\eta_{0}(x,y)dy$ and the
sequence

$\eta^{s}(x)-\eta_{0}(x)-s\eta_{1}(x,\frac{x}{s})\rightarrow
0\,,\,\text{in}\,\,H^{1}(\Omega)\,\,\text{strongly}$

and the

$\nabla\eta^{s}(x)-\nabla\eta_{0}(x)-\nabla_{y}\eta_{1}(x,\frac{x}{s})$

converges strongly to zero in $(L^{2}(\Omega))^{N}.$ This result prove that
the sequence $\frac{\eta^{s}-\eta_{0}^{s}}{s}$ is bounded and two scale
converges to a profile $\eta_{1}\in H^{1}(\Omega,L^{2}_{\\#}(Y)).$
In an other hand, $\frac{\eta^{s}-\eta_{0}^{s}}{s}$ is solution to

$\Delta\frac{\eta^{s}-\eta_{0}^{s}}{s}+\frac{\eta^{s}-\eta_{0}^{s}}{s}=\nabla_{x}\cdot(\nabla_{y}\eta_{0})$
(4.16)

The sequence $\frac{\eta^{s}-\eta_{0}^{s}}{s}$ is bounded in $H^{1}(\Omega)$
then, from a result of Allaire [1] and Nguetseng [24], it two scale converges
to a profile $\eta_{1}$ solution to

$\left\\{\begin{array}[]{ccc}\Delta\eta_{1}+\eta_{1}=\nabla_{x}\cdot(\nabla_{y}\eta_{0})=0\,\,\text{in}\,\,\Omega\times
Y\\\ \eta_{1}=0\,\,\partial\Omega,\\\ y\rightarrow\eta_{1}\,\,\text{is 1
periodic}\end{array}\right.$ (4.17)

∎

###### Theorem 4.2

Under asymptions (4.1),(4.2) and Hypotheses (H0), (H1), (H2), (H3), If $R(s)$
is given by (3.28) and (3.32); the limits of $R(s)$ exists et tends to zero as
$s\rightarrow 0.$
In addition, the functional defined by (3.1) is differentiable and the
topoogical derivative is given by

$DJ(\Omega)=\frac{d}{ds}L(0,\eta^{0},p^{0})$ (4.18)

where $L$ designates the Lagrangian, $\eta^{0}$ and $p^{0}$ are solutions to
the direct and adjoint states.

###### Proof.

The proof is identical for the two cases considered. Without loss of
generality, we will only consider the case where $R(s)$ is given by (3.28),
i.e.

$R(r)=\int_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)\right\rvert^{2}dx+\int_{\Omega_{r}}\left\lvert\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right\rvert^{2}dx-\frac{1}{s^{1/2}}\int_{\partial
E_{r}\cap\partial\Omega_{r}^{m}}\nabla p_{0}\cdot\nabla
d_{E}\left(\frac{\eta_{r}-\eta_{0}}{s^{1/2}}\right)dx.$

To prove that $R(r)$ tends to zero, it is sufficient to show that the
expressions of $l_{1}(r)$ and $l_{2}(r)$ given in Theorem 3.5 tend to zero
when $r\rightarrow 0.$

We are therefore going to focus on the convergence of each of the terms of the
expression of $R(r).$ From (4.1) and (4.2), we get that; if
$\eta^{0}(x)=\eta_{0}(x,\frac{x}{s}),$ the sequence
$\frac{\eta^{s}-\eta^{0}}{s}$ to scale converges to a profile $\eta^{1}\in
L^{\infty}_{\\#}((\mathbb{R},L^{2}(\Omega)).$ This makes possible to get the
following result

$||\frac{\eta^{s}-\eta_{0}}{s}||_{H^{1}(\Omega)}\leq\alpha$ (4.19)

where $\alpha$ is a constant not depending on $s.$
The parameters $s$ and $r$ are proportional, when $r$ goes to zero, $s$ also
goes to zero. Instead of $\eta_{r}$ we can write $\eta_{s}$. So using the fact
that, for small
$s,\,\,||\frac{\eta^{s}-\eta_{0}}{\sqrt{s}}||_{H^{1}(\Omega)}\leq||\frac{\eta^{s}-\eta_{0}}{s}||_{H^{1}(\Omega)}\leq\alpha$
we have

$l_{1}(r)=l^{\prime}_{1}(s)=\sqrt{s}\int_{\Omega}\chi_{\Omega_{r}}\left\lvert\nabla\left(\frac{\eta_{s}-\eta_{0}}{s}\right)\right\rvert^{2}dx+\sqrt{s}\int_{\Omega}\chi_{\Omega_{r}}\left\lvert\frac{\eta_{s}-\eta_{0}}{s}\right\rvert^{2}dx$

Taking into account (4.19), we get that as $\lim_{r\rightarrow 0}l_{1}(r)=0.$
For the same reasons, the expression given by $l_{2}(r)$ will also cancel.
This means that the limit of $R(r)$ will tend to 0. For the proof of the last
result, it is only necessary to note that
$R(\eta^{0},p^{0})=\lim_{r\rightarrow 0}R(r)=0.$ By application of Sturm’s and
Delfour’s theorem, the shape derivative or the topological derivative are
nothing else than the derivative of the Lagrangian with respect to the
variable $s.$ Then (4.18) is true. ∎

### ACKNOWLEDGEMENT

This work has been supported by the Deutsche Forschungsgemeinschaft within the
Priority program SPP 1962 ”Non-smooth and Complementarity based Distributed
Parameter Systems: Simulation and Hierarchical Optimization”. The authors
would like to thank Volker Schulz ( University Trier, Trier, Germany) and Luka
Schlegel (University Trier, Trier, Germany) for helpful and interesting
discussions within the project Shape Optimization Mitigating Coastal Erosion
(SOMICE).

## References

  * [1] G. Allaire, _Homogenization and Two-Scale convergence_ , SIAM J. Math. Anal. 23 (1992), 1482–1518.
  * [2] H. Ammari and H. Kang, _Polarization and Moment Tensors With Applications to Inverse Problems and Effective Medium Theory,_ Springer Appl. Math. Sci. 162, Springer, New York, 2007.
  * [3] S. Amstutz _Analyse de sensibilité topologique et applications en optimisation de formes_. Optimisation et contrôle [Math.OC]. Université d’Avignon, 2011.
  * [4] M. A. M. T. Baldé, D. Seck, _Homogenization and corrector result for a coupled parabolic hyperbolic system,_ Journal of Mathematical Analysis and Applications vol. 484, Issue 1, 2020.
  * [5] D. Bucur, _How to prove existence in shape optimization._ Arch. Rational Mech. Anal. 122 (1993), 183-195.
  * [6] D. Bucur, G. Buttazzo, _Variational methods in some shape optimization problems_ , SNS Pisa (2002).
  * [7] D. Bucur, G. Buttazzo A. Henrot, _Existence results for some optimal partition problems._ , Adv. Math. Sci. Appl.; 8 (1998), 571-579.
  * [8] M. C. Delfour. _Control, Shape, and Topological Derivatives via Minimax Differentiability of Lagrangians_. In Numerical methods for optimal control problems. Proceedings of the workshop, Rome, Italy, June 1923, 2017, pages 137-164. Cham: Springer, 2018.
  * [9] M. C. Delfour. _Metrics spaces of shapes and geometries from set parametrized functions._ In Pratelli, A., Leugering, G.(eds.) New Trends in Shape Optimization. International Series of Numerical Mathematics, 166 57=101. Birkhauser, Basel 2015.
  * [10] M. C. Delfour. _Topological derivatives via one-sided derivative of parametrized minima and minimax_. Centre de recherches mathématiques and Département de mathématiques et de statistique, Université de Montréal, Montréal, Canada
  * [11] M. C. Delfour. _Topological derivative of state-constrained objective functions: a direct method*_. SIAM J. CONTROL OPTIM. Vol. 60, No. 1, pp. 22-47, 2022.
  * [12] M. C. Delfour, K. Sturm. _Minimax differentiability via the averaged adjoint for control/shape sensitivity_. In: Proc. of the 2nd IFAC Workshop on Control of Systems Governed by Partial Differential Equations, IFAC- Paper online, vol. 49-8, pp. 142-149, 2016.
  * [13] M. C. Delfour, _Differentials and Semidifferentials for Metric Spaces of Shapes and Geometries_ System Modeling and Optimization, International Federation for Information Processing 2016 Published by Springer International Publishing AG 2016. All Rights Reserved, 230-239, 2016.
  * [14] M. C. Delfour and K. Sturm. _Parametric semidifferentiability of minimax of lagrangians averaged adjoint approach_. Journal of Convex Analysis, 24(4): 1117-1142, 2017.
  * [15] M. Delfour and J. P. Zolesio. _Shapes and Geometries: Metrics, Analysis, Differential Calculus and Optimization_. SIAM Series on Advances in Design and Control, 2nd edn. Society for Industrial and Applied Mathematics, Philadelphia , 2011.
  * [16] M. C. Delfour, J. P. Zolésio, _Shapes and Geometries: Metrics, Analysis, Differential Calculus, and Optimization._ SIAM Series on Advances in Design and Control, 2nd edn. Society for Industrial and Applied Mathematics, Philadelphia (2011).
  * [17] L. C. Evans. _Partial Differential Equations,_ Graduate Studies in Mathematics, vol. 19, American Mathematical Society, Providence, RI, 1998.
  * [18] I. Faye, E. Frénod, D. Seck, _Singularly perturbed degenerated parabolic equations and application to seabed morphodynamics in tided environment_ ,Discrete and Continuous Dynamical Systems, Vol 29 $N^{o}3$ March 2011, pp 1001-1030.
  * [19] I. Faye, E. Frénod, D. Seck, _Long term behaviour of singularly perturbed parabolic degenerated equation, Journal of Nonlinear Analysis and Application_ , 2 (2016), 82- 105\. https://doi.org/10.5899/2016/jnaa-00297.
  * [20] H. Federer. _Geometric Measure Theory_. Springer, Berlin (1969).
  * [21] E. Frénod, Raviart P. A., and E. Sonnendrücker, _Asymptotic expansion of the Vlasov equation in a large external magnetic field_ , J. Math. Pures et Appl. 80 (2001), 815–843.
  * [22] A. Henrot and M. Pierre, _Variation et Optimisation de Formes. Une Analyse Géométrique,_ Mathématiques et Applications, 48, Springer, Berlin , 2005.
  * [23] A. Henrot et M. Pierre. _Variations et optimisation de forme: Une analyse géométrique_. Mathématiques et Applications. Springer Berlin Heidelberg, 2006.
  * [24] G. Nguetseng, _A general convergence result for a functional related to the theory of homogenization_ , SIAM J. Math. Anal. 20 (1989), 608–623.
  * [25] S.A.Nazarov and J.Sokolowski, _Asymptotic analysis of shape functionals,_ J.Math.Pures Appl., 82 (2003), pp. 125-196.
  * [26] A. A. Novotny et J. Sokolowski _Topological Derivative in Shape Optimization_. Interaction of mechanics and mathematics, Springer, Heidelberg, New York, 2013.
  * [27] J. Sokolowski and J. P. Zolezio _Introduction to Shape Optimization_. Shape Sensivity Analysis, Springer-Verlag Berlin Heidelberg GmbH, 1991.
  * [28] A. A. Novotny and J. Sokolowski Topological Derivative in Shape Optimization. Interaction of mechanics and mathematics, Springer, Heidelberg, New York, 2013.
  * [29] B. Samet. _The topological asymptotic analysis for the Maxwell equations and applications_. Mathematics [math]. Université Paul Sabatier - Toulouse III, 2004. English.
  * [30] J. Sokolowski, A. Zochowski. _On the topological derivative in shape optimization._ SIAM J. Control Optim. 37(4), 1251-1272, 1999.
  * [31] J. Sokolowski and J. P. Zolesio, _Introduction to Shape Optimization: Shape Sensitivity Analysis,_ Springer Series in Computational Mathematics, 10 Springer, Berlin , 1992.
  * [32] K. Sturm. _On shape optimization with non-linear partial differential equations_. Doctoral thesis, Technische Universität of Berlin , 2014. , 2015.
  * [33] K. Sturm. _Minimax Lagrangian approach to the differentiability of non-linear PDE constrained shape functions without saddle point assumption_. SIAM J. Control Optim. 53(4), 2017-2039, 2015