Shape Sensitivity of Eigenvalue Functionals for Scalar Problems: Computing the Semi-derivative of a Minimum

Abstract

This paper is devoted to the computation of certain directional semi-derivatives of eigenvalue functionals of self-adjoint elliptic operators involving a variety of boundary conditions. A uniform treatment of these problems is possible by considering them as a problem of calculating the semi-derivative of a minimum with respect to a parameter. The applicability of this approach, which can be traced back to the works of Danskin [8, 9] and Zolésio [28], to the treatment of eigenvalue problems (where the full shape derivative may not exist, due to multiplicity issues), has been illustrated by Zolésio in [29] (see also [10, Chap. 10] and included references). Despite this, some of the recent literature (see, for example, [1] or [7]) on the shape sensitivity of eigenvalue problems still continue to employ methods such as the material derivative method or Lagrangian methods which seem less adapted to this class of problems. The Delfour–Zolésio approach does not seem to be fully exploited in the existing literature: we aim to recall the importance and the simplicity of the ideas from [8, 28], by applying it to the analysis of the shape sensitivity for eigenvalue functionals for a class of elliptic operators in the scalar setting (Laplacian or diffusion in heterogeneous media), thus recovering known results in the case of Dirichlet or Neumann boundary conditions and obtaining new results in the case of Steklov or Wentzell boundary conditions.

Appendix

1.1 Technical Results

The purpose of this subsection is to recall some auxiliary results or notions used in the calculations of the shape sensitivity. We recall here some versions in relation to the perturbation of identity from [20]. For the corresponding versions involving velocity fields we refer to [10, Chap. 9].

Classical shape derivative formulæ.

Lemma A.1

Let $\delta >0$. Let a vector field ${\varvec{V}}\in \varvec{\mathrm {W}}^{1,\infty } ({\mathbb {R}}^d)$ and let

$$\begin{aligned} \Psi \; : \; t \in [0,\delta ) \mapsto \Psi _{t} = \varvec{\mathrm {I}}+ t {\varvec{V}}\in \varvec{\mathrm {W}}^{1,\infty }({\mathbb {R}}^d). \end{aligned}$$

Let a bounded Lipschitz open set $\Omega$ in ${\mathbb {R}}^d$ and let $\Omega _t:=\Psi _t(\Omega )$ for all $t \in [0,\delta )$. We consider a function f such that $t \in [0,\delta ) \mapsto f(t) \in \mathrm {L}^{1}({\mathbb {R}}^d)$ is differentiable at 0 with $f(0) \in \mathrm {W}^{1,1}({\mathbb {R}}^d)$. Then the function

$$\begin{aligned} t \in [0, \delta ) \mapsto {\mathcal {F}}(t)= \int _{\Omega _t} f(t,x) \ \mathrm {d}x \end{aligned}$$

is differentiable at 0 (we say that ${\mathcal {F}}$ admits a semi-derivative) and we have

$$\begin{aligned} {\mathcal {F}}'(0)= \int _{\partial \Omega } f(0,x) V_\mathrm{n}\ \mathrm {d}\varsigma (x) + \int _{\Omega } f'(0,x) \ \mathrm {d}x , \end{aligned}$$

where $V_\mathrm{n}= {\varvec{V}}\cdot \varvec{\mathrm {n}}$.

Lemma A.2

Let $\delta >0$. Let a vector field ${\varvec{V}}\in {\varvec{C}}^{1,\infty } ({\mathbb {R}}^d)$ and let

$$\begin{aligned} \Psi \; : \; t \in [0,\delta ) \mapsto \Psi _{t} = \varvec{\mathrm {I}}+ t {\varvec{V}}\in {\varvec{C}}^{1,\infty }({\mathbb {R}}^d). \end{aligned}$$

Let a bounded open set $\Omega$ in ${\mathbb {R}}^d$ of classe ${\mathcal {C}}^2$ and let $\Omega _t:=\Psi _t(\Omega )$ for all $t \in [0,\delta )$. We consider a function g such that $t \in [0,\delta ) \mapsto g(t) \circ \Psi _{t} \in \mathrm {W}^{1,1}(\Omega )$ is differentiable at 0 with $g(0) \in \mathrm {W}^{2,1}(\Omega )$. Then the function

$$\begin{aligned} t \in [0, \delta ) \mapsto {\mathcal {G}}(t)= \int _{\partial \Omega _t} g(t,x) \ \mathrm {d}x \end{aligned}$$

is differentiable at 0 (we say that ${\mathcal {G}}$ admits a semi-derivative), the function is differentiable at 0 for all open set $\omega \subset \overline{\omega } \subset \Omega$ and the derivative $g'(0)$ belongs to $W^{1,1}(\Omega )$ and we have

$$\begin{aligned} {\mathcal {G}}'(0)= \int _{\partial \Omega } \left( g'(0,x) + \left( \mathrm {H}\, g(0,x) + \partial _{\mathrm {n}} g \right) V_\mathrm{n}\right) \mathrm {d}\varsigma (x) , \end{aligned}$$

where $V_\mathrm{n}= {\varvec{V}}\cdot \varvec{\mathrm {n}}$ and where $\mathrm {H}$ is the mean curvature function on $\partial \Omega$.

Some tangential calculus. The following lemma are formulated in terms of an intrinsic tangential calculus developed by Delfour and Zolésio [12, 13] (see also [10, Chap. 9, Sect. 5]) for modelling thin shells.

An important role is played by the signed distance to the boundary $\partial \Omega$ defined by

$$\begin{aligned} b(x) := \left\{ \begin{array}{rl} \mathrm {d}(x,\partial \Omega ) , &{} \text{ if } x \in \Omega , \\ - \mathrm {d}(x,\partial \Omega ) , &{} \text{ if } x \in {\mathbb {R}}^d\setminus \overline{\Omega }. \\ \end{array} \right. \end{aligned}$$

Indeed, if $\partial \Omega$ is smooth then we have $Db = \varvec{\mathrm {n}}$ is the outward unit normal, $D^2b$ is second fundamental form on the boundary. If $\Pi _d = \mathrm {I}_{d} - \varvec{\mathrm {n}}\otimes \varvec{\mathrm {n}}$ is the projection on the tangent plane, then $\nabla _\Gamma u = \Pi _d \nabla u$ is the tangential gradient of a scalar function in $\mathrm {H}^1(\Omega )$. For any smooth vector field V on $\Gamma = \partial \Omega$, the tangential divergence is given by $\mathrm {div}_\Gamma V := \mathrm {tr}(DV \, \Pi _d )$ and the mean curvature at any point on $\partial \Omega$ is given by

$$\begin{aligned} \mathrm {H}:= \mathrm {div}_{\Gamma }\, \varvec{\mathrm {n}}. \end{aligned}$$

The Laplace-Beltrami operator is defined by

$$\begin{aligned} \Delta _\Gamma u = \mathrm {div}_\Gamma (\nabla _\Gamma u) \end{aligned}$$

for every sufficiently smooth function u on $\Gamma$. Keep in mind also that $\mathrm {D}^2 b \, \varvec{\mathrm {n}}= 0$ and the tangential Stoke formula

$$\begin{aligned} \int _\Gamma \mathrm {div}_\Gamma V \ dS = \int _\Gamma H \ V \cdot \varvec{\mathrm {n}}\ dS\, \end{aligned}$$

while working with the intrinsic tangential calculus.

Lemma A.3

Given a bounded open set $\Omega$ in ${\mathbb {R}}^d$ of class ${\mathcal {C}}^2$ and $u \in \mathrm {H}^2({\mathbb {R}}^d)$, we have

$$\begin{aligned} \partial _{\mathrm {n}} |\nabla _\Gamma u|^2 = 2 \nabla _\Gamma u \cdot ( \nabla _\Gamma (\partial _{\mathrm {n}} u )- \mathrm {D}^2 b \, \nabla _\Gamma u ), \end{aligned}$$

where b is the signed distance to the boundary $\partial \Omega$.

Proof

From that $\mathrm {D}^2 b n = 0$ it follows that $\partial _{\mathrm {n}}\Pi _{d} = 0$ and that $\Pi _{d}\mathrm {D}^2 b =\mathrm {D}^2 b \Pi _{d} = \mathrm {D}^2 b$. After recalling that $\nabla _{\Gamma } u = \Pi _{d} \nabla u$, we have

$$\begin{aligned} \partial _{\mathrm {n}}(\nabla _{\Gamma } u) = \partial _{\mathrm {n}} (\Pi _{d} \nabla u) = \Pi _{d} \partial _{\mathrm {n}} (\nabla u) \qquad \text{ and } \qquad \partial _{\mathrm {n}} (\nabla u ) = \mathrm {D}^2 u \, \varvec{\mathrm {n}}. \end{aligned}$$

Thus $\nabla (\partial _{\mathrm {n}} u ) = \nabla (\nabla u \cdot \varvec{\mathrm {n}}) = \mathrm {D}^2 u \, \varvec{\mathrm {n}}+ \nabla \varvec{\mathrm {n}}\, \nabla u = \partial _{\mathrm {n}} (\nabla u ) + \mathrm {D}^2 b \, \nabla u$. Hence we obtain

$$\begin{aligned} \nabla _{\Gamma }(\partial _{\mathrm {n}} u) = \Pi _{d} \nabla (\partial _{\mathrm {n}} u) = \Pi _{d} \partial _{\mathrm {n}}(\nabla u ) + \Pi _{d} \mathrm {D}^2 b \nabla {\varvec{u}}= \partial _{\mathrm {n}}(\nabla _{\Gamma } u ) + \mathrm {D}^2 b \nabla u . \end{aligned}$$

Therefore, we obtain the result since $\partial _{\mathrm {n}} \left( \left| \nabla _{\Gamma }u \right| ^2\right) = 2 \partial _{\mathrm {n}} \left( \nabla _{\Gamma }u\right) \cdot \nabla _{\Gamma }u$. $\square$

Lemma A.4

Given a bounded open set $\Omega$ in ${\mathbb {R}}^d$ of class ${\mathcal {C}}^2$, ${\varvec{V}}$ in $\varvec{{\mathcal {C}}}^1({\mathbb {R}}^d;{\mathbb {R}}^d)$ and $u \in \mathrm {H}^2({\mathbb {R}}^d)$, we have

where $V_{\mathrm {n}}$ is the normal component ${\varvec{V}}\cdot \varvec{\mathrm {n}}$ of the vector field ${\varvec{V}}$.

Proof

We first recall that, since , we have . Hence we obtain the result noticing that $\nabla _{\Gamma } u = \Pi _{d} \nabla u$. $\square$

Some reminders on $\Gamma$-convergence.

For the convenience of the reader, we recall the definition and the main property of the $\Gamma$-convergence. For further details we refer to the book of Dal Maso [5].

Definition A.5

(Sequential $\Gamma$-convergence) A family of functionals $\{F_t\}_{t > 0}$ defined on a topological space X is said to be sequentially $\Gamma$-convergent to a functional F as $t \rightarrow 0^+$ if the two following statements hold:

(i):

$\Gamma -\liminf$ inequality: for every sequence $\{x_t\}$ converging to $x \in X$, we have

$$\begin{aligned} \liminf _{t \rightarrow 0^+} F_t(x_t) \ge F(x) ; \end{aligned}$$

(ii):

$\Gamma -\limsup$ inequality: for every $x \in X$ there exists a sequence $\{x_t\}$ converging to x such that

$$\begin{aligned} \limsup _{t \rightarrow 0^+} F_t(x_t) \le F(x). \end{aligned}$$

When Properties (i) and (ii) are satisfied, we write $\displaystyle { F = \mathop {\Gamma - \mathrm {lim}}_{t \rightarrow 0^+} F_t}$.

Proposition A.6

Let $F_t: X \rightarrow {\mathbb {R}}$ be a sequence of functionals on a topological space such that:

• (i) $\displaystyle { F = \mathop {\Gamma -\mathrm {lim}}_{t \rightarrow 0^+} F_t}$;

• (ii) $\sup _t F_t(x_t) <+\infty \ \Rightarrow \ \{x_t\} \text{ is } \text{ sequentially } \text{ relatively } \text{ compact } \text{ in } X$.

Then we have the convergence: $\inf F_t \rightarrow \inf F$ as $t \rightarrow 0^+$ and, every cluster point of a minimizing sequence $\{x_t\}$ (i.e. such that $\displaystyle {F_t(x_t) = \inf _{x \in X} F_t(x)}$) achieves the minimum of F.

1.2 Second Proof of Theorem 2.5 in the Dirichlet Case Using the Material Derivative

In this section, we shall recalculate the expression for ${\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}})$ obtained in Theorem 2.5, while considering the particular case of Dirichlet boundary condition on $\partial \Omega$, based on the material derivative approach. We omit the proof of the existence of the material derivative which is a direct adaptation of the existing works (see for example [6]). It can be made out from the following calculations that those based on the classical material derivative method are much more tedious as compared to the calculations obtained in the previous sections.

1.2.1 First Characterization with the Material Derivative

We use the notations for the problem on the perturbed domain given in the beginning of Sect. 3.4.1. Let $u_t$ be a normalized eigenfunctions for ${\mathfrak {M}}_{\Omega }(\Omega _t)$. We set $u^t= u_t \circ \Psi _t$. The existence of the shape derivative and of the material derivative of u, and of the shape derivative ${\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}})$ are assumed to begin with and we will perform calculations using them.

The shape derivative ${\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}})$ can be obtained by deriving the Rayleigh quotient on $\Omega _t$ evaluated at a normalized eigenfunction $u^t$. In view of the normalization condition $\displaystyle \int _{\Omega _{t}} \rho _{t} \left| u_{t} \right| ^2 \, \mathrm {d}x = 1$, it is enough to derive $\displaystyle \int _{\Omega _t}\sigma _t(x) \left| \nabla u_t \right| ^2 \mathrm {d}x$ for which we use the Hadamard’s formula. So, arguing similarly as in Proposition 3.5, we obtain,

$$\begin{aligned} {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}})= & {} 2 \int _{\Omega } \sigma (x)\nabla u' \cdot \nabla u \, \mathrm {d}x + \int _{\Gamma } \left[ \sigma (x)\left| \nabla u \right| ^2 \right] V_\mathrm{n}\, \mathrm {d}\varsigma (x)\nonumber \\&+ \int _{\partial \Omega }\sigma _{2}\left| \nabla {u} \right| ^2V_\mathrm{n}\, \mathrm {d}\varsigma (x). \end{aligned}$$

(A.1)

This does not give a boundary expression of the shape derivative and also involves $u'$ which has to be characterized through a boundary value problem. This involves several difficulties and so, classically, one takes the route through the material derivative (cf. [6]).

We have the following variational formulation for the perturbed problem on $\Omega _t$

$$\begin{aligned} \int _{\Omega _{t}} \sigma _{t}(x) \nabla u_{t} \cdot \nabla \varphi _{t} \, \mathrm {d}x= {\mathfrak {M}}_{\Omega }(\Omega _t) \int _{\Omega _{t}} \rho _{t}u_{t} \varphi _{t}\, \mathrm {d}x , \qquad \text{ for } \text{ all } \varphi _{t} \in \mathrm {H}^1_{0}(\Omega _{t}). \end{aligned}$$

So, for any $\varphi \in \mathrm {H}^1_{0}(\Omega )$, by choosing $\varphi _{t} := \varphi \circ \Psi _{t}^{-1}$ and then making a change of variables in the variational problem $y=\Psi _{t}(x)$ we have

$$\begin{aligned} \int _{\Omega } \sigma (x) \left( A(t) \nabla u^{t} \right) \cdot \nabla \varphi \, \mathrm {d}x = {\mathfrak {M}}_{\Omega }(\Omega _t) \int _{\Omega } \rho \, u^{t} \varphi j(t) \, \mathrm {d}x, \end{aligned}$$

by noticing that $\sigma _{t}(\Psi _{t}(x))=\sigma (x)$ and $\rho _{t}(\Psi _{t}(x))=\rho (x)$. We set $A(t) := j(t) \mathrm {D}\Psi _{t}^{-1}\left( \mathrm {D}\Psi _{t}^{-1}\right) ^\top$ while recalling that $j(t)=\det (\mathrm {D}\Psi _t)$. Deriving the equation with respect to t at $t=0$ under the integral sign we obtain that,

$$\begin{aligned}&\int _{\Omega } \sigma (x) \left( \nabla {\dot{u}} + A'(0)\nabla u \right) \cdot \nabla \varphi \, \mathrm {d}x\nonumber \\&\quad = \int _{\Omega } \rho \big ( {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}}) u + {\mathfrak {M}}_{\Omega }(\Omega ) \left( j'(0)u + {\dot{u}}\right) \big ) \varphi \, \mathrm {d}x. \end{aligned}$$

(A.2)

Using the Hadamard’s formula given in Lemma A.1 on the normalization condition

$$\begin{aligned} 1 = \int _{\Omega _{t}} \rho _{t} \left| u_{t} \right| ^2 \, \mathrm {d}x = \int _{\Omega _{1,t}} \rho _{1} \left| u_{t} \right| ^2 \, \mathrm {d}x + \int _{\Omega _{2,t}} \rho _{2} \left| u_{t} \right| ^2 \, \mathrm {d}x , \end{aligned}$$

we obtain

$$\begin{aligned} 0= & {} \int _{\Omega } 2 \rho \, u' u\, \mathrm {d}x + \int _{\partial \Omega } \rho \, \left| u \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x)+ (\rho _{1}-\rho _{2}) \int _{\Gamma } \left| u \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x)\nonumber \\= & {} 2 \int _{\Omega } \rho \, u' u \, \mathrm {d}x + \int _{\Gamma } \left[ \rho \right] \left| u \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x), \end{aligned}$$

(A.3)

since $u = 0$ on $\partial \Omega$.

1.2.2 Rewriting of Some Terms

Let $\varphi \in \mathrm {H}^1_{0}(\Omega )$. Since $A'(0) = \left( \mathrm {div}\,{\varvec{V}}\right) \mathrm {I}- \left( \mathrm {D}{\varvec{V}}+ \mathrm {D}{\varvec{V}}^\top \right)$ (see, e.g., [24, Lemma 2.31]), we have in both $\Omega _{i}$, $i=1,2$, where $\sigma$ is constant,

$$\begin{aligned} \sigma (x)A'(0)\nabla u\cdot \nabla \varphi= & {} \sigma (x) \left( \mathrm {div}\,{\varvec{V}}\right) \nabla u \cdot \nabla \varphi - \sigma (x) \left( \mathrm {D}{\varvec{V}}+ \mathrm {D}{\varvec{V}}^\top \right) \nabla u\cdot \nabla \varphi \\= & {} \mathrm {div} \Big ( \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) {\varvec{V}}\Big ) - \nabla \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) \cdot {\varvec{V}}\\&- \sigma (x) \left( \mathrm {D}{\varvec{V}}+ \mathrm {D}{\varvec{V}}^\top \right) \nabla u\cdot \nabla \varphi \\= & {} \mathrm {div} \Big ( \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) {\varvec{V}}\Big ) - \sigma (x) \nabla \left( \nabla u \cdot {\varvec{V}}\right) \\&\cdot \nabla \varphi - \sigma (x) \nabla u \cdot \nabla (\nabla \varphi \cdot {\varvec{V}}). \end{aligned}$$

Moreover

$$\begin{aligned} \int _{\Omega } \mathrm {div} \Big ( \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) {\varvec{V}}\Big ) \, \mathrm {d}x= & {} \int _{\partial \Omega } \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) V_\mathrm{n}\, \mathrm {d}\varsigma (x)\nonumber \\&+ \int _{\Gamma }\left[ \sigma (x)\nabla u\cdot \nabla \varphi \right] V_\mathrm{n}\, \mathrm {d}\varsigma (x), \end{aligned}$$

and

$$\begin{aligned}&\int _{\Omega } \sigma (x) \nabla u \cdot \nabla (\nabla \varphi \cdot {\varvec{V}})\, \mathrm {d}x \\&\quad = \displaystyle - \int _{\Omega } \mathrm {div} \sigma (x)\nabla u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}x + \int _{\partial \Omega } \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}\varsigma (x) + \int _{\Gamma } \left[ \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}} \right] \, \mathrm {d}\varsigma (x)\\&\quad = \displaystyle {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Omega } \rho \, u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}x + \int _{\partial \Omega } \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}\varsigma (x) + \int _{\Gamma } \left[ \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}} \right] \, \mathrm {d}\varsigma (x). \end{aligned}$$

Thus

$$\begin{aligned} \int _{\Omega }\sigma (x)A'(0)\nabla u\cdot \nabla \varphi \, \mathrm {d}x= & {} \int _{\partial \Omega } \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) V_\mathrm{n}\, \mathrm {d}\varsigma (x) - \int _{\partial \Omega } \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}\varsigma (x)\\&+ \int _{\Gamma }\left[ \sigma (x)\nabla u\cdot \nabla \varphi \right] V_\mathrm{n}\, \mathrm {d}\varsigma (x) - \int _{\Gamma } \left[ \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}} \right] \, \mathrm {d}\varsigma (x)\\&- \int _{\Omega } \sigma (x) \nabla \left( \nabla u \cdot {\varvec{V}}\right) \cdot \nabla \varphi \, \mathrm {d}x - {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Omega } \rho \, u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}x . \end{aligned}$$

Using this equality in (A.2) and since $j'(0) = \mathrm {div}\,{\varvec{V}}$ (see, e.g., [24, Lemma 2.31]), we obtain

$$\begin{aligned}&\int _{\Omega } \sigma (x) \nabla {\dot{u}} \cdot \nabla \varphi \, \mathrm {d}x + \int _{\partial \Omega } \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) V_\mathrm{n}\, \mathrm {d}\varsigma (x)\\&\qquad - \int _{\partial \Omega } \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}\varsigma (x) + \int _{\Gamma }\left[ \sigma (x)\nabla u\cdot \nabla \varphi \right] V_\mathrm{n}\, \mathrm {d}\varsigma (x) \\&\qquad - \int _{\Gamma } \left[ \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}} \right] \, \mathrm {d}\varsigma (x)\\&\qquad - \int _{\Omega } \sigma (x) \nabla \left( \nabla u \cdot {\varvec{V}}\right) \cdot \nabla \varphi \, \mathrm {d}x - {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Omega } \rho \, u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}x \\&\quad = \int _{\Omega }\rho \left( {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}}) u \varphi + {\mathfrak {M}}_{\Omega } (\Omega )\left( (\mathrm {div} {\varvec{V}}) u + {\dot{u}}\right) \varphi \right) \, \mathrm {d}x. \end{aligned}$$

Hence

$$\begin{aligned}&\int _{\Omega } \sigma (x) \left( \nabla {\dot{u}} - \nabla \left( \nabla u \cdot {\varvec{V}}\right) \right) \cdot \nabla \varphi \, \mathrm {d}x + \int _{\partial \Omega } \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) V_\mathrm{n}\, \mathrm {d}\varsigma (x)\\&\qquad - \int _{\partial \Omega } \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}}\, \mathrm {d}\varsigma (x)\\&\quad = - \int _{\Gamma }\left[ \sigma (x)\nabla u\cdot \nabla \varphi \right] V_\mathrm{n}\, \mathrm {d}\varsigma (x) + \int _{\Gamma } \left[ \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}} \right] \, \mathrm {d}\varsigma (x)\\&\qquad +{\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}}) \int _{\Omega } \rho \, u \varphi \, \mathrm {d}x + {\mathfrak {M}}_{\Omega } (\Omega )\int _{\Omega } \rho \, {\dot{u}} \varphi \, \mathrm {d}x \\&\qquad + {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Omega } \rho \left( (\mathrm {div} {\varvec{V}}) \varphi + \nabla \varphi \cdot {\varvec{V}}\right) u \, \mathrm {d}x. \end{aligned}$$

Moreover, using the fact that $u=0$ and $\varphi =0$ on $\partial \Omega$,

$$\begin{aligned}&\int _{\Omega } \rho \left( (\mathrm {div} {\varvec{V}}) \varphi + \nabla \varphi \cdot {\varvec{V}}\right) u \, \mathrm {d}x\\&\quad = \int _{\Omega } \rho \, \mathrm {div}\left( \varphi \cdot {\varvec{V}}\right) u \, \mathrm {d}x\\&\quad = \int _{\partial \Omega } \rho \varphi V_\mathrm{n}u \, \mathrm {d}\varsigma (x) + (\rho _{1}-\rho _{2}) \int _{\Gamma } \varphi V_\mathrm{n}u \, \mathrm {d}\varsigma (x) - \int _{\Omega } \rho \varphi {\varvec{V}}\cdot \nabla u \, \mathrm {d}x \\&\quad = (\rho _{1}-\rho _{2}) \int _{\Gamma } \left( \varphi \otimes {\varvec{V}}\right) \varvec{\mathrm {n}}\cdot u \, \mathrm {d}x - \int _{\Omega } \left( \varphi \otimes {\varvec{V}}\right) \cdot \nabla u \, \mathrm {d}x \\&\quad = \int _{\Gamma } \left[ \rho \right] \varphi V_\mathrm{n}u \, \mathrm {d}\varsigma (x) - \int _{\Omega } \left( \nabla u \cdot {\varvec{V}}\right) \varphi \, \mathrm {d}x, \end{aligned}$$

and since ${\varvec{V}}=V_\mathrm{n}\varvec{\mathrm {n}}$, we have, on $\Gamma$,

$$\begin{aligned} - \left[ \left( \sigma (x)\nabla u\cdot \nabla \varphi \right) \right] V_\mathrm{n}+ \left[ \sigma (x)\partial _{\mathrm {n}}u \nabla \varphi \cdot {\varvec{V}} \right] = - \left[ \sigma (x) \right] \nabla _{\Gamma }(u) \cdot \nabla _{\Gamma }(\varphi ) V_\mathrm{n}. \end{aligned}$$

Notice that we have used here the fact that u has no jump on $\Gamma$ and so do $\nabla _{\Gamma }u$. Then

$$\begin{aligned}&\int _{\Omega } \sigma (x) \left( \nabla {\dot{u}} - \nabla \left( \nabla u \cdot {\varvec{V}}\right) \right) \cdot \nabla \varphi \, \mathrm {d}x \\&\quad = - \int _{\Gamma } \left[ \sigma (x) \right] \nabla _{\Gamma }(u) \cdot \nabla _{\Gamma }(\varphi ) V_\mathrm{n}\, \mathrm {d}\varsigma (x) +{\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}})\int _{\Omega } \rho \, u \varphi \, \mathrm {d}x \\&\qquad + {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Omega } \rho \, {\dot{u}} \varphi \, \mathrm {d}x + {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Gamma } \left[ \rho \right] \varphi V_\mathrm{n}u \, \mathrm {d}\varsigma (x)\\&\qquad - {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Omega } \rho \left( \nabla u \cdot {\varvec{V}}\right) \varphi \, \mathrm {d}x . \end{aligned}$$

1.2.3 Conclusion: Characterization with the Shape Derivative

Since $u'={\dot{u}}-\nabla u \cdot {\varvec{V}}$, we deduce from the above equality that for all $\varphi \in \mathrm {H}^1_{0}(\Omega )$,

$$\begin{aligned} \int _{\Omega } \sigma (x) \nabla u' \cdot \nabla \varphi \, \mathrm {d}x= & {} - \int _{\Gamma } \left[ \sigma (x) \right] \nabla _{\Gamma }(u) \cdot \nabla _{\Gamma }(\varphi ) V_\mathrm{n}\, \mathrm {d}\varsigma (x)\nonumber \\&+ {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}}) \int _{\Omega } \rho \, u \varphi \, \mathrm {d}x\nonumber \\&+ {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Omega } \rho \, u' \varphi \, \mathrm {d}x + {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Gamma } \left[ \rho \right] \varphi V_\mathrm{n}u \, \mathrm {d}\varsigma (x).\nonumber \\ \end{aligned}$$

(A.4)

Thus, taking $\varphi =u$ and using the normalization conditions $\displaystyle \int _{\Omega } \rho \, \left| u \right| ^2=1$ and (A.3),

$$\begin{aligned} \int _{\Omega } {\sigma (x) \nabla u' \cdot \nabla u }\, \mathrm {d}x= & {} - \int _{\Gamma } \left[ \sigma (x) \right] \left| \nabla _{\Gamma }(u) \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x) + {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}}) \\&- \frac{1}{2} {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Gamma } \left[ \rho \right] \left| u \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x) + {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Gamma } \left[ \rho \right] u V_\mathrm{n}u \, \mathrm {d}\varsigma (x). \end{aligned}$$

Then, using (A.1), we can eliminate the volume term $\displaystyle \int _{\Omega } \sigma (x) \nabla u' \cdot \nabla u$ to obtain

$$\begin{aligned} {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}})= & {} 2 \int _{\Gamma } \left[ \sigma (x) \right] \left| \nabla _{\Gamma }(u) \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x) - \int _{\Gamma } \left[ \sigma (x)\left| \nabla u \right| ^2 \right] V_\mathrm{n}\, \mathrm {d}\varsigma (x) \\&- {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Gamma } \left[ \rho \right] \left| u \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x) - \int _{\partial \Omega } \sigma _{2} \left| \nabla u \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x). \end{aligned}$$

Finally, we obtain

$$\begin{aligned} {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}})= & {} \int _{\Gamma } \left[ \sigma (x) \right] \left| \nabla _{\Gamma }(u) \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x) - \int _{\Gamma } \left[ \sigma (x) \left| \partial _{\mathrm {n}}u \right| ^2 \right] V_\mathrm{n}\, \mathrm {d}\varsigma (x) \\&- {\mathfrak {M}}_{\Omega } (\Omega ) \int _{\Gamma } \left[ \rho \right] \left| u \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x) - \int _{\partial \Omega } \sigma _{2} \left| \nabla u \right| ^2 V_\mathrm{n}\, \mathrm {d}\varsigma (x) , \end{aligned}$$

using the facts that $\left[ \nabla _{\Gamma }u \right] =0$ since $\left[ u \right] =0$ on $\Gamma$ and $\left| \nabla _{\Gamma }(u) \right| ^2=\left| \nabla _{\Gamma } u \right| ^2 + \left| \partial _{\mathrm {n}}u \right| ^2$. This concludes the proof of Theorem 2.5 in the case of Dirichlet boundary conditions.

Remark A.7

The method exposed in this appendix use the classical shape derivative approach. Notice that recently, A. Laurain and K. Sturm present in [21] another approach based on a pure material derivative approach which permits to obtain the same result faster. Indeed, in this very particular case of Dirichlet boundary conditions, we can proceed as follows.

Taking ${\dot{u}} \in \mathrm {H}^1_{0}(\Omega )$ as a test function in the variational formulation of the problem in $\Omega$, we have

$$\begin{aligned} \int _{\Omega } \sigma (x) \nabla u \cdot \nabla {\dot{u}} = {\mathfrak {M}}_{\Omega }(\Omega ) \int _{\Omega } \rho u {\dot{u}} . \end{aligned}$$

Then, taking $\varphi =u$ in Eq. (A.2), using the previous equality and the fact that $\displaystyle \int _{\Omega }\rho u^2 = 1$ and that $j'(0) = \mathrm {div} {\varvec{V}}$, we obtain

$$\begin{aligned} {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}}) = \int _{\Omega } \sigma (x) A'(0) \left| \nabla u \right| ^2 \mathrm {d}x - {\mathfrak {M}}_{\Omega }(\Omega ) \int _{\Omega } \rho u^2 \mathrm {div} {\varvec{V}}\mathrm {d}x. \end{aligned}$$

Notice that

$$\begin{aligned} \mathrm {div} {\varvec{V}}= \mathrm {I}_{d} \,\mathrm {:}\,\mathrm {D}{\varvec{V}}\qquad \text{ and } \qquad \mathrm {D}{\varvec{V}}\nabla u \cdot \nabla u = \mathrm {D}{\varvec{V}}\,\mathrm {:}\,(\nabla u \otimes \nabla u ) . \end{aligned}$$

Thus, using the fact that $A'(0) = \left( \mathrm {div}\,{\varvec{V}}\right) \mathrm {I}- \left( \mathrm {D}{\varvec{V}}+ \mathrm {D}{\varvec{V}}^\top \right)$, this leads

$$\begin{aligned} {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}}) = \int _{\Omega } \left( \left( \sigma (x) \left| \nabla u \right| ^2 - {\mathfrak {M}}_{\Omega }(\Omega ) \rho u^2 \right) \mathrm {I}_{d} - 2 \sigma (x) \nabla u \otimes \nabla u \right) \,\mathrm {:}\,\mathrm {D}{\varvec{V}}\mathrm {d}x. \end{aligned}$$

and, from [21], one can then prove that

$$\begin{aligned} {\mathfrak {M}}_{\Omega }'(\Omega ;{\varvec{V}})= & {} \int _{\partial \Omega } \left( \left( \sigma (x) \left| \nabla u \right| ^2 - {\mathfrak {M}}_{\Omega }(\Omega ) \rho u^2 \right) \mathrm {I}_{d} - 2 \sigma (x) \nabla u \otimes \nabla u \right) \varvec{\mathrm {n}}\cdot \varvec{\mathrm {n}}\, V_\mathrm{n}\mathrm {d}\varsigma (x) \\&+ \int _{\Gamma } \left[ \left( \sigma (x) \left| \nabla u \right| ^2 - {\mathfrak {M}}_{\Omega }(\Omega ) \rho u^2 \right) \mathrm {I}_{d} - 2 \sigma (x) \nabla u \otimes \nabla u \right] \varvec{\mathrm {n}}\cdot \varvec{\mathrm {n}}\, V_\mathrm{n}\mathrm {d} \varsigma (x) . \end{aligned}$$

Therefore we obtain the announced result since

$$\begin{aligned}&\left( \left( \sigma (x) \left| \nabla u \right| ^2 - {\mathfrak {M}}_{\Omega }(\Omega ) \rho u^2 \right) \mathrm {I}_{d} - 2 \sigma (x) \nabla u \otimes \nabla u \right) \varvec{\mathrm {n}}\cdot \varvec{\mathrm {n}}\\&\quad = \sigma (x) \left| \nabla u \right| ^2 - {\mathfrak {M}}_{\Omega }(\Omega ) \rho u^2 - 2 \sigma (x) \left| \partial _{\mathrm {n}}u \right| ^2 = - \sigma _{2}\left| \nabla u \right| ^2 \end{aligned}$$

on $\partial \Omega$ (where we have the homogeneous Dirichlet boundary condition $u=0$), and since

$$\begin{aligned}&\left[ \left( \sigma (x) \left| \nabla u \right| ^2 - {\mathfrak {M}}_{\Omega }(\Omega ) \rho u^2 \right) \mathrm {I}_{d} - 2 \sigma (x) \nabla u \otimes \nabla u \right] \varvec{\mathrm {n}}\cdot \varvec{\mathrm {n}}\\&\quad = \left[ \sigma (x) \left| \nabla u \right| ^2 \right] - {\mathfrak {M}}_{\Omega }(\Omega ) \left[ \rho \right] u^2 - 2 \left[ \sigma (x) \left| \partial _{\mathrm {n}}u \right| ^2 \right] \\&\quad = \left[ \sigma (x) \right] \left[ \left| \nabla _{\Gamma } u \right| ^2 \right] - {\mathfrak {M}}_{\Omega }(\Omega ) \left[ \rho \right] u^2 - \left[ \sigma (x) \left| \partial _{\mathrm {n}}u \right| ^2 \right] \end{aligned}$$

on $\Gamma$.

Once again, it is very important to notice that these methods are based on the existence of material or shape derivatives, which is not necessarily true for multiple eigenvalue. The approach that we present in this paper is uniform and permits to deal with several boundary conditions, without assuming the simplicity of the eigenvalue, using the notion of semi-derivative.

1.2.4 Characterization of the Shape Derivative as the Solution of a Transmission Problem

We conclude this section noticing that we can, classically, characterize the shape derivative of the initial problem as the solution of a transmission problem. Indeed, from (A.4), we obtain that, for all $\varphi \in \mathrm {H}^1_{0}(\Omega )$,

$$\begin{aligned}&- \int _{\Omega } \mathrm {div}\,(\sigma (x) \nabla u') \varphi \, \mathrm {d}x + \int _{\Gamma } \left[ \sigma \partial _{\mathrm {n}}u' \right] \varphi \, \mathrm {d}\varsigma (x)\\&\quad = \int _{\Gamma } \mathrm {div}_{\Gamma } \left( \left[ \sigma \right] \nabla _{\Gamma } u V_\mathrm{n}\right) \varphi \, \mathrm {d}\varsigma (x) + {\mathfrak {M}}_{\Omega }'(\Omega ,{\varvec{V}}) \int _{\Omega } \rho \, u \varphi \, \mathrm {d}x\\&\qquad + {\mathfrak {M}}_{\Omega }(\Omega ) \int _{\Omega } \rho \, u' \varphi \, \mathrm {d}x + {\mathfrak {M}}_{\Omega }(\Omega ) \int _{\Gamma } \left[ \rho \right] \varphi V_\mathrm{n}u \, \mathrm {d}\varsigma (x), \end{aligned}$$

and then

$$\begin{aligned} \left[ \sigma \partial _{\mathrm {n}}u' \right]= & {} \mathrm {div}_{\Gamma } \left( \left[ \sigma \right] \nabla _{\Gamma } u V_\mathrm{n}\right) + {\mathfrak {M}}_{\Omega }(\Omega ) \left[ \rho \right] u V_\mathrm{n}\\= & {} \left[ \sigma \right] \mathrm {div}_{\Gamma } \left( \nabla _{\Gamma } u V_\mathrm{n}\right) + {\mathfrak {M}}_{\Omega }(\Omega ) \left[ \rho \right] u V_\mathrm{n}. \end{aligned}$$

Moreover, since $u_{1}=u_{2}$ on $\Gamma$, we have

$$\begin{aligned} u'_{1}-u'_{2} = \left( \nabla u_{2} - \nabla u_{1}\right) \cdot {\varvec{V}}= - \left[ \partial _{\mathrm {n}}u \right] V_\mathrm{n}, \end{aligned}$$

the last equality being obtained using the fact that $\nabla _{\Gamma } u _{1} = \nabla _{\Gamma } u _{2}$. Hence

$$\begin{aligned} \left[ u \right] = - \left[ \partial _{\mathrm {n}}u \right] V_\mathrm{n}. \end{aligned}$$

Finally we classically have

$$\begin{aligned} u' = -\partial _{\mathrm {n}}u V_\mathrm{n}\, \text{ on } \partial \Omega . \end{aligned}$$

Hence we obtain that the shape derivative $u'$ is solution of

$$\begin{aligned} \left\{ \begin{array}{rclll} &{}&{}- \mathrm {div}\,\left( \sigma (x) \nabla u' \right) = {\mathfrak {M}}_{\Omega }(\Omega ) \rho u' + {\mathfrak {M}}_{\Omega }'(\Omega ,{\varvec{V}}) \rho u &{} &{} \text{ in } \left( \Omega \backslash \overline{\omega }\right) \cup \omega ,\\ &{}&{}\left[ u \right] = - \left[ \partial _{\mathrm {n}}u \right] V_\mathrm{n}&{} &{} \text{ on } \Gamma , \\ &{}&{}\left[ \sigma \partial _{\mathrm {n}}u' \right] = \left[ \sigma \right] \mathrm {div}_{\Gamma } \left( \nabla _{\Gamma } u V_\mathrm{n}\right) + {\mathfrak {M}}_{\Omega }(\Omega ) \left[ \rho \right] u V_\mathrm{n}&{} &{} \text{ on } \Gamma , \\ &{}&{}u' = -\partial _{\mathrm {n}}u V_\mathrm{n}&{} &{} \text{ on } \partial \Omega . \end{array} \right. \end{aligned}$$