About the rst- and second-order shape derivatives of the probability to nd a xed number of electrons chemically characterized by a Slater multi-determinant wave function Dalphin Jérémy October 19, 2016

Abstract In Quantum Chemistry, researchers are interested in nding new ways to characterize the electronic structures of molecules and their interactions. The model of Maximal Probability Domains (MPDs) is a developing method based on probabilities that allows such a geometrical and spatial characterization of the electronic structures of chemical systems. In this article, we consider a quantum system of n electrons. We derive formulas for the rst-, second-, and higher-order shape derivatives of the functional pν : Ω 7→ pν (Ω), with pν (Ω) the probability to nd exactly a xed number ν of electrons in a given spatial region Ω ⊆ R3 , where exactly means that the n − ν remaining ones are located in the complement R3 \Ω. First, we study the case of chemical systems characterized by general wave functions, with precise general regularity results concerning the shape dierentiability of pν at any order. Then, by restricting our analysis to the rst- and second-order shape derivatives, we consider the case of wave functions given as a sum of Slater determinants, and nally we present the specic case of (Hartree-Fock) wave functions determined by a single Slater determinant. In the case of single-Slater-determinant wave functions, we derive a new recursive formula for the second-order shape derivative of pν . We also give a new proof of the recursive formula obtained in [8] for the rst-order shape derivative of pν , which does not assume the simplicity of the eigenvalues associated with the overlap matrix, as it is the case in [8].

1 Introduction On the one hand, the traditional chemical intuition i.e. the way chemists understand how molecules interact together have been deeply inuenced by a localized vision of electrons around the cores. On the other hand, Quantum Mechanics allows the electrons to be delocalized over the whole space. Indeed, a chemical system of n electrons is entirely characterized by its wave function, which is a priori dened everywhere. Hence, there is a loss of chemical informations that Quantum Chemistry tries to recover in several manners. One way to reconnect the usual expectations of chemists with the results of accurate quantum mechanical calculations consists in removing the problematical high-dimensionality of the wave function by averaging it correctly over the positions of all electrons. More precisely, computing the probability pν (Ω) to nd exactly a xed number ν of electrons in a given region Ω of the real three-dimensional space R3 , where exactly means that the n − ν remaining ones are located in the complement R3 \Ω, we can try to solve the following shape optimization problem: sup pν (Ω) . Ω⊆R3

(1)

The model of Maximal Probability Domains i.e. searching for the local/global maximizers and the critical points of (1) is a developing method based on probabilities that allows such a geometrical and spatial characterization of the electronic structures of molecules and their interactions in chemical systems. The goal of this article is to properly derive formulas for the rst- and second-order shape derivatives of the functional pν : Ω 7→ pν (Ω) in the case of general and Slater multi/single-determinant wave functions. These expressions are fundamental in the implementation of numerical algorithms trying to solve (1).

1

In this article, our rst main contribution has been to study the dierentiability properties of the map pν,Ω : θ 7→ pν [(I + θ)(Ω)] associated to the shape functional pν : Ω 7→ pν (Ω) for general wave functions Ψ. Under the W k,2 -regularity of Ψ, we get that pν,Ω is of class C k around the origin for any integer k > 1 and for any measurable subset Ω of R3 . In particular, we obtain a formula for the shape derivative of pν at any order. The results with their precise references in the text are sum up in Table 1. Ω ⊆ R3 Measurable Measurable Measurable Measurable Lipschitz Measurable Lipschitz C 1,1 -domain

Ψ L2 L2 L2 W 1,2 W 1,2 W 2,2 W 2,2 W 2,2

θ : R3 → R3

Measurable

W k,2

W 1,∞

C 0,1 W 1,∞ W 1,∞ W 1,∞ W 1,∞ W 1,∞ ∩ C 1 W 1,∞ ∩ C 1

Regularity of

pν,Ω : θ 7→ pν [(I + θ) (Ω)] The map pν : Ω 7→ pν (Ω) is well dened. pν,Ω is well dened on B0,1 . pν,Ω is C 0 on B0,1 ∩ W 1,∞ . pν,Ω is C 1 on B0,1 ∩ W 1,∞ . pν has a well-dened shape gradient. pν,Ω is C 2 on B0,1 ∩ W 1,∞ . pν,Ω is C 2 on B0,1 ∩ W 1,∞ ∩ C 1 . pν has a well-dened shape Hessian. pν,Ω is C k on B0,1 ∩ W 1,∞ .

Proof

Denition 2.2 Proposition 2.5 Proposition 2.5 Corollary 2.7 Theorem 2.6 Corollary 2.9 Corollary 2.9 Theorem 2.8 Theorem 2.10

Table 1: Summary of the dierent regularity results concerning the functional pν,Ω : θ 7→ pν [(I + θ)(Ω)]. The counterpart of such a general result is the poor structure we obtain for the shape derivative of pν . Hence, restricting our study to the rst- and second-order shape dierentiability properties, our second main contribution is to give various formulas for the probability pν (Ω) and for the rst/second-order shape derivatives of the functional pν : Ω 7→ pν (Ω) in the case of general and Slater-multi/single-determinant wave functions. The results with their precise references in the text are sum up in Table 2.

Expressions for

General wave functions Sum of Slater determinants Single Slater determinant

pν (Ω) Denition 2.2 Proposition 3.2 Corollary 4.4

ν D0 pν,Ω (θ) and ∂p (Ω) ∂Ω Theorem 2.6 Proposition 3.4 Proposition 4.5

˜ and D02 pν,Ω (θ, θ) Theorem 2.8 Proposition 3.5 Theorem 4.7

∂ 2 pν ∂Ω2

(Ω)

Table 2: Summary of the dierent shape dierentiability results concerning the functional pν : Ω 7→ pν (Ω). Finally, our third main contribution is the recursive formula we get for the second-order shape derivative of pν : Ω 7→ pν (Ω) in the case of single-Slater-determinant wave functions. It extends the results of [8] concerning the rst-order shape derivative of pν . Moreover, our approach is dierent from [8] in the sense that our proof does not rely on the shape derivative of the eigenvalues associated with the overlap matrix, a property that holds true only if we assume the simplicity of these eigenvalues. Hence, a new proof of the results in [8] are given, which does not require such a restrictive hypothesis. The results state as follows.

Proposition 1.1 (2004 [8, Section 2.2]).

Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any open bounded subset of R3 with a Lipschitz boundary as in Denition 5.46. We consider 2n well-dened maps ±1/2 ±1/2 φ1 , . . . , φn ∈ W 1,2 (R3 , C) such that S(R3 ) = In , where S(Ω) is the overlap matrix dened by (70). Then, the wave function Ψ given by (69) satises Assumption 2.1 and the map pν : Ω 7−→ pν (Ω) of Denition 2.2 is well dened. We also have c0 = 1, where c0 is the normalizing constant dened by (7), and the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is well dened around the origin. Moreover, it is dierentiable at the origin and its (rst-order) dierential at the origin is given by the following continuous linear form dened for any θ ∈ W 1,∞ (R3 , R3 ) by:    n X n  X Y Y  X    Uil (Ω)Ujl (Ω) ε (l) λ (Ω) [1 − λ (Ω)] Iν k k   ∂Ω  Iν ⊆J1,nK  k∈Iν k∈J1,nK σ∈{− 1 , 1 } i,j=1 l=1

Z D0 pν,Ω (θ) =

X

2 2

k6=l

card Iν =ν

k∈I / ν k6=l

φσi (x) φσj (x) hθ (x) | nΩ (x)i dA (x) ,

(2) where the integration is done with respect to the two-dimensional Hausdor measure referred to as A(•), where nΩ (x) denotes the unit vector normal to the boundary ∂Ω at the point x pointing outwards Ω, where εIν (l) = 1 if l ∈ Iν otherwise εIν (l) = −1, where the boundary values of φσi φσj ∈ W 1,1 (R3 , C) have to be understood in the sense of trace, where (λk (Ω))16k6n are the real eigenvalues of the Hermitian matrix S(Ω) given by (70), and where U (Ω) is a unitary matrix such that [U (Ω)]T S(Ω)U (Ω) is a diagonal matrix whose k-th diagonal coecient is λk (Ω) for any k ∈ J1, nK.

2

Theorem 1.2.

Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any open bounded subset of R3 with a C 1,1 ±1/2 boundary as in Denition 5.47. We consider 2n well-dened maps φ±1/2 , . . . , φn ∈ W 2,2 (R3 , C) such 1 3 that S(R ) = In , where S(Ω) is the overlap matrix (70). Then, the wave function Ψ given by (69) satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) of Denition 2.2 is well dened. We also have c0 = 1, where c0 is the normalizing constant (7), and the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is well dened and dierentiable around the origin. Moreover, it is twice dierentiable at the origin and its second-order dierential at the origin is given by the following continuous symmetric bilinear form dened ˜ ∈ (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) × (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) by: for any (θ, θ)   ˜ = D02 pν,Ω (θ, θ)

 n X n  X Y Y  X    Uil (Ω)Ujl (Ω) ε (l) λ (Ω) [1 − λ (Ω)] I k k ν   ∂Ω   k∈Iν Iν ⊆J1,nK k∈J1,nK σ∈{− 1 , 1 } i,j=1 l=1

Z

X

2 2

k6=l

card Iν =ν



k∈I / ν k6=l




 HΩ (x)φσi (x)φσj (x) + ∇ φσi φσj (x) | nΩ (x) θn (x) θ˜n (x) dA (x) 

Z

Z

+ ∂Ω



 n Y Y X X X   X    (q) λ (Ω) (p) ε [1 − λ (Ω)] ε m I m I ν ν   ∂Ω   m∈Iν m∈J1,nK Iν ⊆J1,nK σ∈{− 1 , 1 } i,j=1 k∈J1,nK p=1 q∈J1,nK X

2 2

n X

q6=p

l∈J1,nK k6=i l6=j

m6=p,q

card Iν =ν

m∈I / ν m6=p,q

   Uip (Ω)Ukq (Ω) − Upk (Ω)Uiq (Ω) Ujp (Ω) Ulq (Ω) − Upl (Ω) Ujq (Ω) φσi (x)φσj (x)φσk (y)φσl (y)θn (x) θ˜n (y) dA (x) dA (y) 



 Z n X n  X X Y Y  X  σ σ  − εIν (l) λk (Ω) [1 − λk (Ω)]   Uil (Ω)Ujl (Ω) φi (x)φj (x) ∂Ω   1 1 i,j=1 l=1 k∈I I ⊆J1,nK k∈J1,nK ν ν σ∈{− , } 2 2

k6=l

card Iν =ν

k∈I / ν

k6=l    D E D E IIΩ θ˜∂Ω (x) , θ∂Ω (x) + ∇∂Ω [θn ] (x) | θ˜∂Ω (x) + ∇∂Ω [θ˜n ] (x) | θ∂Ω (x) dA (x) ,

(3) where we refer to Proposition 1.1 for the notation A(•), nΩ , εIν , (λm (Ω))16m6n , and U (Ω), where the boundary values of φσi φσj ∈ W 2,1 (R3 , C) and ∇(φσi φσj ) have to be understood in the sense of trace, where (•)n := h(•) | nΩ i is the normal component of a vector eld and (•)∂Ω := (•) − (•)n nΩ its tangential component, which means that in particular ∇∂Ω (•) := ∇(•) − h∇(•) | nΩ inΩ refers to the tangential component of the gradient operator, where IIΩ (•, •) := −hD∂Ω nΩ (•) | (•)i is the second fundamental form associated with the C 1,1 -surface ∂Ω, which is a symmetric bilinear form on the tangent space, with D∂Ω (•) := D(•) − D(•)nΩ [nΩ ]T denoting the tangential component of the dierential operator on vector elds, and where HΩ := div∂Ω (nΩ ) is the scalar mean curvature associated with the C 1,1 -surface ∂Ω, with div∂Ω (•) := div(•) − hD(•)nΩ | nΩ i denoting the tangential component of the divergence operator. To conclude, we emphasize the structure of (2)(3) obtained in Proposition 1.1 and Theorem 1.2 respectively for the rst- and second-order shape derivative of pν . They are optimal in the sense that this is the form we expect from shape derivatives (see e.g. [27, Section 5.9.4] and [34]). • The rst-order shape derivative in (2) is a continuous linear form, which has a shape gradient structure of the form: Z ∂pν (Ω)θn dA. D0 pν,Ω (θ) = ∂Ω ∂Ω • The last term of (3) can be interpreted as a continuous linear form l1 as follows: observing from (2) ˜ , where Z that we can write D0 pν,Ω (θ) = l1 (θn ), then the third term in (3) has the form l1 [Z(θ, θ)] is a vector eld which is is equal to zero if θ and θ˜ are normal to the boundary ∂Ω. In this case, the last term of (3) is thus equal to zero. Moreover, if Ω is a critical shape for pν i.e. if D0 pν,Ω ≡ 0, then l1 ≡ 0 and this term is also equal to zero. • The middle term of (3) is a continuous symmetric bilinear form, which takes the form of a kernel: Z Z KΩ (x, y) θn (x)θ˜n (y)dA(x)dA(y). ∂Ω

∂Ω

• The rst term of (3) is a continuous symmetric bilinear form, which can be interpreted as the shape Hessian part of the second-order shape derivative. Indeed, it takes the following form: Z ∂ 2 pν (Ω)θn θ˜n dA. 2 ∂Ω ∂Ω

3

The paper is organized as follows. In the remainder of the introduction, Section 1.1 is concerned with the various motivations and references related to the model of Maximal Probability Domains. Then, in Sections 234, we respectively detail the case of general, Slater-multi-determinant, and Slater-singledeterminant wave functions. Each of them is divided into three subsections. The rst one gives the probability as a shape functional, the second one study the rst-order shape dierentiability properties of this map, and the third one consider its second- or higher-order shape dierentiability properties. Finally, Section 5 introduce all the material and the proofs of standard results needed throughout the article.

Contents 1 Introduction

1

2 Shape dierentiability in the case of general wave functions

7

1.1 2.1 2.2 2.3

On the various motivations concerning maximal probability domains . . . . . . . . . . . . . On the expression of the probability for general wave functions . . . . . . . . . . . . . . . . On the rst-order shape derivative of the probability . . . . . . . . . . . . . . . . . . . . . . On the second- and higher-order shape derivative of the probability . . . . . . . . . . . . . .

5

7 9 13

3 Shape dierentiability for multideterminant wave functions

21

4 Shape dierentiability for one-determinant wave functions

31

3.1 3.2 3.3 4.1 4.2 4.3

On the expression of the probability as a sum of Slater determinants . . . . . . . . . . . . . On the rst-order shape derivative of the probability . . . . . . . . . . . . . . . . . . . . . . On the second-order shape derivative of the probability . . . . . . . . . . . . . . . . . . . . On the expression of the probability with a single Slater determinant . . . . . . . . . . . . . On the rst-order shape derivative of the probability . . . . . . . . . . . . . . . . . . . . . . On the second-order shape derivative of the probability . . . . . . . . . . . . . . . . . . . .

5 Annexes 5.1

5.2

On the rst- and second-order derivatives of the determinant . . . . . . . . . . . . . . 5.1.1 About the signature of a permutation . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 About the determinant of a matrix . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.3 About the comatrix operator and the rst-order dierential of the determinant 5.1.4 About the dierential of the comatrix operator . . . . . . . . . . . . . . . . . . 5.1.5 About the properties associated with the Hessian of the determinant . . . . . . 5.1.6 About the particular cases of invertible, complex-valued, and unitary matrices On the rst- and second-order shape derivatives of a volume integral . . . . . . . . . . 5.2.1 Some denitions and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 About the dierentiability properties related to the inverse operator . . . . . . 5.2.3 About the dierentiability properties related to the Jacobian determinant . . . 5.2.4 About the dierentiability properties related to the composition operator . . . 5.2.5 About the shape derivative of a volume integral at any order . . . . . . . . . . 5.2.6 Proof of Theorem 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.7 Proof of Theorem 5.30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.8 On the second-order dierential of the associated map and further regularity . 5.2.9 About Lipschitz boundaries, C 1,1 domains, and complex valued functionals . .

4

. . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . .

21 24 26 31 34 37

44

44 44 46 47 50 52 57 61 63 65 66 67 70 78 80 84 88

1.1 On the various motivations concerning maximal probability domains On the one hand, the traditional chemical intuition i.e. the way chemists understand how molecules interact together have been deeply inuenced by a localized vision of electrons around the cores (electron pairs, shells, classication of bonds, other resonating structures). More precisely, we can say the following. • The chemical bonding theory has rst been rationalized by Lewis in [30], where he introduced the fundamental concept of electron pair shared between two atoms and his original model of the cubical atom, later supplemented by Langmuir's octet rule [29]. • Then, Pauling described in [35] two major mechanisms that can lead to a two-electron chemical bond between two atoms: the covalent bond coming from a spin-exchange resonance energy, which is a purely quantum eect [25]; the ionic bond resulting from the electrostatic interaction between ions. • Much more recently, Shaik and its collaborators identied in [47, 48, 50] a third distinct type of resonating structure called the charge-shift bond, that drives the bonding stabilization mechanism through a covalent-ionic resonance energy.

Modern classications can add more atoms (multi-center bonds) or less electrons (electron decient bonds) without signicantly altering the image of Lewis' localized vision of electrons and Pauling's classication of bonds. Indeed, these concepts are extremely fruitful in chemistry and rmly rooted to the models because they can simply explain many dierent experimental manifestations. On the other hand, Quantum Mechanics allows the electrons to be delocalized over the whole space. Indeed, according to the principles of Quantum Mechanics [41, 52], a molecular system of n electrons is completely characterized by its wave function, which is a priori dened everywhere. To be more precise, under the clamped nuclei and the Coulomb Hamiltonian approximations, the solution of the electronic Schrödinger equation of a n-electron molecule allows to obtain, in principle, a full statistical description of the positions of electrons in the real three-dimensional space, via the squared norm of the wave function in the coordinate representations. Therefore, although the accurate multi-electron wave function contains all the information relevant to a molecular system, it remains a complex high-dimensional mathematical object that lacks of immediate and intuitive visual representation of chemical bonds in the real space. Consequently, there is a loss of chemical informations that Quantum Chemistry tries to recover in several manners. • Interpretative methods (valence bond theory, molecular orbitals) enable a connection between the results of accurate quantum mechanical calculations and the expectations of chemists by working in the Fock space. Hence, they deliver images that are valid in this space, but usually implicitly assume to depict what is happening in the real space. Moreover, they are dicult to generalize in the case where the correlation between electrons plays an important role. We can also mention the non-covalent interaction method formulated in [28], which allows to recover semi-quantitative informations about intermolecular interactions [13]. • As an alternative, topological approaches try to partition the physical three-dimensional space into chemically interesting domains. It has thus been for a long time a matter of research to nd simple and clear ways to separate the space into regions with a chemical meaning. Of course, the choice is not unique and there is some exibility in choosing the criterion dening the region of interest. Some methods are based on the density (the minima of radial density [40], the atomic basins of Bader [3], the basins of the Laplacian of the electron density [3], the basins of the electron localization function [4, 44, 49]) while others directly work with the wave function (the loges of Daudel [15]).

Hence, one way to reconnect the chemical viewpoint with the quantum mechanical one consists in removing the problematical high-dimensionality of the wave function by averaging it correctly over the positions of electrons. Working directly with the wave function instead of the density ensures no information is lost in the integration process. Moreover, the analysis of such electron number probability distributions have shown to provide a wealth of chemical bonding informations [12, 18, 19, 20, 21, 22, 23, 36, 37, 38, 39]. Furthermore, although the experience and the models can provide some non-overlapping regions with clear chemical signicance, the number of electrons within them is only dened as an averaged number, which is not as sharp as an observable, since the electron population local operator does not commute with the molecular Hamiltonian. In particular, the computed number of electrons within one of these domains can be a non-integer. It is more logical though, instead of using regions of space generated by quite dierent interpretative and topological methods, to search directly for the regions of space maximizing the probability to nd a given number ν of electrons. Such regions are called Maximal Probability Domains and stand for the domains solving (1) i.e. maximizing the probability to nd exactly ν electrons inside it. The model of Maximal Probability Domains was proposed by Savin in [43] to analyse the distribution of electrons in regions of space. The idea goes back to the early work of Daudel and its coworkers [15].

5

Savin shows in [43] that the solution of (1) can provide radii which describe very well a spatial separation into atomic shells, a feature which is not obtained for heavy atoms when using other widespread methods. In this eld of methods, the Maximum Probability Domains (MPDs) analysis is conceptually appealing but has not been extensively applied yet. It has been shown to provide vivid images of cores and valence regions of atoms [8, 43], lone and bonding pairs [33], and sharp domains in which can move the electrons in a simple molecule [24, 31], a liquid [1], a crystal [10, 11], or an inorganic compound [9]. Therefore, the Maximal Probability Domains may become a rigorous entry point to recover standard chemical concepts from wave functions, which gather all the quantum information of the system. Moreover, they deliver images in the real space and the model is kept as simple as possible in the sense that the denition and physical meaning of the interpretative quantities computed with the method can be readily understood by most experimental chemists. For example, the domains that locally maximize the probability to nd exactly two electrons can be directly related to the Lewis' concept of electron pair. Furthermore, it can generate explanatory models which rationalize a whole set of data and allow predictions to be made. Maximal Probability Domains are the global/local maximizers and the critical points of (1). The theoretical/numerical study of this shape optimization problem relies on the concept of shape derivative. The mathematical existence and regularity of maximizers for (1) are dicult and open problems, even for analytic and simple wave functions such as the case of a two-electron molecule. Indeed, the direct method of Calculus of Variations does not apply. Roughly speaking, there is a lack of continuity and compactness due to the poor control we have on the perimeter of a minimizing sequence. The boundary can oscillate severely, reveal some cut/cusps, or simply become unbounded. For example, the half-space maximizes the probability to nd exactly one electron for the fundamental state of the dihydrogen molecule [32]. From a numerical point of view, it is still an on-going eort to develop algorithms and programs that are able to eciently optimize the domains solving (1) for molecules and solids. For example, the probability pν cannot be naively computed in a reasonable time. Indeed, as shown in Section 3.1, the simple evaluation of a typical wave function Ψ requires the calculus of thousands of (n × n)-determinants, where |Ψ|2 is the integrand of n three-dimensional integrals with n quite large. In order to estimate pν , an interesting alternative consists in using Quantum Monte-Carlo methods, which have been carried out in [45, 46]. However, as shown in Section 2.2, the shape gradient needed in the optimization process of steepest descent is only dened on the boundary of the domain, which has zero measure from a probabilistic point of view. Hence, Quantum Monte-Carlo techniques are of little interest for implementing a classical shape optimization algorithm. However, the shape gradient given in (19) can be computed via Quantum MonteCarlo techniques since it is a 3(n − 1)-dimensional volume integral. All these observations also hold true for (32)(33) concerning the second-order shape derivative and the setting-up of Newton-like methods. In the specic case of a single-Slater-determinant wave function, the evaluation of the probability and the implementation of a shape optimization process have been described in the breakthrough article [8]. Indeed, it gives an ecient formula for computing the shape gradient and setting up an optimization algorithm via a level-set method. In Section 3.1, we extend the expression of the probability obtained in [8, Appendix] to the Slater-multi-determinant wave functions. Moreover, a dierent proof concerning the recursive formula (2) of the shape gradient is shown in Section 4.2, which does not assume the simplicity of the eigenvalues of the overlap matrix as it is case in [8, Section 2.2]. Finally, one of the main contributions of this paper is to derive in Section 4.3 an ecient recursive formula (3) for the second-order shape derivative.

6

2 Shape dierentiability in the case of general wave functions 2.1 On the expression of the probability for general wave functions Let n > 2 be an integer henceforth set. In this article, we consider a quantum system of n electrons whose state is assumed to be completely characterized by a given well-dened wave function [41, Section 1.1.1]: 
n −→ C Ψ: R3 × − 12 , 12 

x1 σ1



 ,...,

xn σn



 7−→

Ψ

x1 σ1



 ,...,

xn σn

(4)

 ,

where xi and σi respectively refers to the space and spin variables of the i-th electron, for any i ∈ J1, nK. Since we are dealing with fermions, we assume the antisymmetry of the wave function [52, Section 2.1.3], and we set L2 ((R3 × {− 21 , 21 })n , C) as the complex separable Hilbert space of all possible quantum states. However, we do not impose here a unitary L2 -norm condition on the wave function as it is often the case. In other words, we make the following hypothesis.

Assumption 2.1.

The map Ψ is a skew-symmetric form i.e. for any (i, j) ∈ J1, nK2 such that i 6= j , and for any (σi , σj ) ∈ {− 12 , 21 }2 and any (xi , xj ) ∈ R3 × R3 , we have:             xi xj xj xi ,... . (5) ,..., , . . . = −Ψ . . . , ,..., Ψ ..., σi σj σj σi Moreover, the map Ψ is measurable and square integrable i.e. for any (σ1 , . . . , σn ) ∈ {− 21 , 12 }n , the following map belongs to L2 ((R3 )n , C): Ψ(σ1 ,...,σn ) :

R3 × . . . × R3

−→

C 

(x1 , . . . , xn )

7−→

Ψ

x1 σ1



 ,...,

xn σn

 .

(6)

Finally, in addition to be well-dened and nite, we assume that the following normalizing constant is a positive quantity i.e. it is not equal to zero:     2 Z X x1 xn c0 := Ψ ,..., dx1 . . . dxn . (7) σ1 σn n (R3 )n (σ1 ,...,σn )∈{− 1 ,1 2 2} Hence, assuming that the wave function Ψ given in (4) satises Assumption 2.1, we can now use the traditional probabilistic interpretation of the wave function [41, Section 1.1.1] in order to dene the shape functional in which we will be interested throughout the article. Indeed, the probability to nd for any i ∈ J1, nK the electron i of spin σi in a domain Ωi is proportional to: Z 2 Ψ(σ1 ,...,σn ) , Ω1 ×...×Ωn

where Ψ(σ1 ,...,σn ) is dened by (6). Since Ψ(σ1 ,...,σn ) is measurable, the above quantity is well dened for any (Lebesgue) measurable subsets Ω1 , . . . , Ωn of R3 , and it is nite since Ψ(σ1 ,...,σn ) is square integrable. In particular, the probability to nd n electrons in a measurable subset Ω of R3 , regardless of their spins, is proportional to:     2 Z X x1 xn dx1 . . . dxn . Ψ ,..., σ1 σn Ωn 1 n (σ1 ,...,σn )∈{− 1 , 2 2} The constant of proportionality is determined by the fact that we expect to nd n electrons in the whole space R3 with probability one. Hence, let c0 > 0 be as in (7) and the probability pn (Ω) to nd n electrons in a measurable subset Ω of R3 is given by the following well-dened quantity:     2 Z X 1 x1 xn pn (Ω) := Ψ ,..., dx1 . . . dxn . (8) σ1 σn c0 Ωn 1 n (σ1 ,...,σn )∈{− 1 , } 2 2 Similarly, the probability p0 (Ω) to nd zero electron in a measurable subset Ω of R3 is dened as:
p0 (Ω) := pn R3 \Ω .

(9)

We now set ν ∈ J1, n − 1K and search for the probability to nd exactly ν electrons in Ω, where exactly means that the n − ν remaining ones are located in the complement R3 \Ω. The associated event can be

7

interpreted as the reunion of the events nding exactly electrons i1 , . . . , iν in Ω, taken among all the subsets {i1 , . . . , iν } of ν pairwise distinct elements of J1, nKQ . Hence, for any subset Iν ⊂ J1, nK of ν elements i.e. such that card Iν = ν , we introduce the set ΩIν := ni=1 Ai , where Ai = Ω if i ∈ Iν otherwise Ai = R3 \Ω. Following the same arguments than for pn and p0 , the probability pν (Ω) to nd exactly ν electrons in a measurable subset Ω of R3 is given by the following well-dened quantity:     2 Z X 1 x1 xn [ pν (Ω) := Ψ ,..., dx1 . . . dxn . (10) σ1 σn c0 Ω Iν n (σ1 ,...,σn )∈{− 1 ,1 2 2} Iν ⊂J1,nK card Iν =ν

Then, we observe that such a nite reunion is disjoint i.e. ΩIν ∩ ΩJν = ∅ if Iν 6= Jν so we have:       2 Z X X   1 x1 xn  pν (Ω) = Ψ ,..., dx1 . . . dxn  .  σ σ c0 1 n Ω n Iν 1 1 I ⊂J1,nK ν (σ1 ,...,σn )∈{− 2 , 2 } card I =ν ν

R R (σ ,...,σp (n) ) 2 Iν Since the wave function Ψ satises (5), we get ΩI |Ψ(σ1 ,...,σn ) |2 = Ων ×(R3 \Ω)n−ν |Ψ pIν (1) | , ν (J1, νK) = I . Rearranging the summation on the map satisfying p where pIν : J1, nK → J1, nK is a bijective ν Iν R R P P (σ1 ,...,σn ) 2 (˜ σ1 ,...,˜ σn ) 2 spin variables σ ˜i = σpIν (i) , we get | = | , σ1 ,...,σn ΩIν |Ψ σ ˜ 1 ,...,˜ σn Ων ×(R3 \Ω)n−ν |Ψ which does not depend on Iν any longer. It can thus be removed from the corresponding sum for which
n! we know that card{Iν ⊂ J1, nK, card Iν = ν} = ν!(n−ν)! := nν . We deduce that: 1 pν (Ω) = c0

! n ν



Z

X n

(σ1 ,...,σn )∈{− 1 ,1 2 2}

Ων ×(R3 \Ω)

n−ν

Ψ

x1 σ1



 ,...,

xn σn



2

dx1 . . . dxn .

(11)

Although (10) should be the original denition for pν in the sense that it is clear with (10) that pν ∈ [0, 1] as it is the case for (8) and (9), we will however use the more practical formula (11) for pν in the remaining part of the article. In other words, we are now in position to properly dene the shape functional in which we will be interested throughout the article.

Denition 2.2.

Assume that the wave function Ψ given in (4) satises Assumption 2.1. Let M be the set of all (Lebesgue) measurable subsets of R3 and c0 > 0 as in (7). Then, for any ν ∈ J0, nK, the following shape functional is a well-dened map: pν :

M Ω

−→ 7−→

[0, 1] pν (Ω) ,

where the probability pν (Ω) to nd exactly ν electrons in the domain Ω is well dened by (11) if ν ∈ J1, n−1K, by (8) if ν = n, and by (9) if ν = 0.

Remark 2.3.

Note that with the convention A0 × B = B × A0 = B for any sets A and B , we can deduce (8)(9) from (11) by setting ν = 0 or ν = n in (11). We will adopt this convention in the rest of the article, in order to simplify the proofs and not to have to treat specically the cases ν = 0 and ν = n. Finally, we give a rst result concerning the symmetry property of the probability. It states as follows.

Lemma 2.4.

Let M be the class of all measurable subsets of R3 . We assume that the wave function Ψ given in (4) satises Assumption 2.1. Then, the map pν : Ω ∈ M 7→ pν (Ω) ∈ [0, 1] of Denition 2.2 is well dened and we have:

pn R3 = p0 (∅) = 1 and ∀Ω ∈ M, pν (Ω) = pn−ν R3 \Ω . In particular, the whole space R3 (respectively the empty set ∅) is optimal for pn (respectively for p0 ). Moreover, if Ω∗ is optimal for pν , then R3 \Ω∗ is optimal for pn−ν i.e.
∃Ω∗ ∈ M, pν (Ω∗ ) = max pν (Ω) =⇒ pn−ν R3 \Ω∗ = max pn−ν (Ω) . Ω∈M

Ω∈M

In other words, the shape optimization problem (1) only needs to be studied for integers ν 6 has an obvious global maximizer if ν = 0 or if ν = n.

8

n+1 2

and it

Proof. Let M contain all the measurable subsets of R3 . We assume that the wave function Ψ given in (4) satises Assumption 2.1. Hence, from the foregoing, the map pν : Ω ∈ M 7→ pν (Ω) ∈ [0, 1] introduced in Denition 2.2 is well dened. First, we get pn (R3 ) = p0 (∅) = 1 if we consider (8)(9) with Ω = R3 and Ω = ∅. Then, let Ω ∈ M. From (11), we get: ! Z X 1 n pν (Ω) = |Ψ(σ1 ,...,σn ) |2 , n−ν c0 ν ν 3 Ω × R \Ω ( ) 1,1 n (σ1 ,...,σn )∈{− 2 2} where Ψ(σ1 ,...,σn ) and c0 > 0 are respectively dened by (6) and (7). Using the alternating property (5) R R of the wave function, we deduce that Ων ×(R3 \Ω)n−ν |Ψ(σ1 ,...,σn ) |2 = (R3 \Ω)n−ν ×Ων |Ψ(σν+1 ,...,σn ,σ1 ,...,σν ) |2 .

n n! = n−ν Observing that nν := ν!(n−ν)! and re-indexing the summation on the spin variables σ ˜i := σν+i for any i ∈ J1, n − νK and σ ˜i := σi−n+ν for any i ∈ Jn − ν + 1, nK, we obtain: ! Z X
1 n pν (Ω) = |Ψ(˜σ1 ,...,˜σn ) |2 = pn−ν R3 \Ω . n−ν c0 n − ν 3 ν ×Ω n (R \Ω) (˜ σ1 ,...,˜ σn )∈{− 1 ,1 2 2} Finally, if we assume that there exists Ω∗ ∈ M such that pν (Ω∗ ) = maxΩ∈M pν (Ω), then for any Ω ∈ M, we deduce from the above symmetry property:

pn−ν (Ω) = pν R3 \Ω 6 max pν (A) = pν (Ω∗ ) = pn−ν R3 \Ω∗ . A∈M

Hence, we get pn−ν (R \Ω ) = maxΩ∈M pn−ν (Ω), concluding the proof of Lemma 2.4. 3

∗

2.2 On the rst-order shape derivative of the probability We refer to Section 5.2 for further details concerning the denition of rst-order shape dierentiability and the notation associated to it, especially Denition 5.28, Theorem 5.29, and Section 5.2.1. Before stating our results concerning the rst-order shape derivative of pν : Ω 7→ pν (Ω) i.e. the rst-order dierential at the origin of the associated map pν,Ω : θ 7→ pν [(I + θ)(Ω)], we rst study the continuity properties of pν,Ω .

Proposition 2.5.

Let n > 2 be any integer, ν ∈ J0, nK, and M contain all the measurable subsets of R3 . We assume that the wave function Ψ given by (4) satises Assumption 2.1. Then, the shape functional pν : Ω ∈ M 7→ pν (Ω) ∈ [0, 1] of Denition 2.2 is well dened, and for any Ω ∈ M, the associated map pν,Ω : θ ∈ C 0,1 (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is well dened on the open unit ball of C 0,1 (R3 , R3 ) centred at the origin denoted by B0,1 := {θ ∈ C 0,1 (R3 , R3 ), kθkC 0,1 (R3 ,R3 ) < 1}. Moreover, for any Ω ∈ M, the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is continuous at any point of B0,1 ∩ W 1,∞ (R3 , R3 ). Proof. Let n > 2 be any integer, ν ∈ J0, nK, and M3 contain all the measurable subsets of R3 . We assume that the wave function Ψ given by (4) satises Assumption 2.1. From Denition 2.2, the shape functional pν : Ω ∈ M3 7→ pν (Ω) ∈ [0, 1] is a well-dened map. Then, let B30,1 = {θ ∈ C 0,1 (R3 , R3 ), kθkC 0,1 (R3 ,R3 ) < 1} be the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. Consider any θ ∈ B30,1 . By Proposition 5.31, the Lipschitz continuous map I3 + θ : x ∈ R3 7→ x + θ(x) ∈ R3 is bijective and its inverse (I3 + θ)−1 is also Lipschitz continuous. In particular, (I3 + θ)−1 is a measurable map and for any Ω ∈ M3 , we get (I3 + θ)(Ω) ∈ M3 . Hence, for any Ω ∈ M3 , the map pν,Ω : θ ∈ C 0,1 (R3 , R3 ) 7→ pν [(I3 + θ)(Ω)] ∈ [0, 1] is well dened on B30,1 . We now study its continuity. For this purpose, we introduce an higher dimensional version of pν in order to get back to the known case of a shape functional associated with a volume integral: p˜ν :

M3n

−→

R

e Ω

7−→

e := p˜ν (Ω)

1 c0

! n ν

Z

X (σ1 ,...,σn )∈{− 1 ,1 2 2}

n

|Ψ(σ1 ,...,σn ) |2 ,

(12)

e Ω

where M3n refers to the class of all measurable subsets of (R3 )n , and where Ψ(σ1 ,...,σn ) is dened by (6). e are measurable, the above integral is well dened. Moreover, it is a nite quantity Since Ψ(σ1 ,...,σn ) and Ω (σ1 ,...,σn ) since Ψ is square integrable. Hence, the shape function p˜ν is a well-dened map. We can thus apply ˜ Ω)] e ∈ M3n , the map p˜ e : θ˜ ∈ C 0,1 ((R3 )n , (R3 )n ) 7→ p˜ν [(I3n + θ)( e ∈R Proposition 5.39 to p˜ν and for any Ω ν,Ω

is therefore well dened on the open unit ball of C 0,1 ((R3 )n , (R3 )n ) centred at the origin denoted by 0,1 B3n ((R3 )n , (R3 )n ), kθkC 0,1 ((R3 )n ,(R3 )n ) < 1}. Moreover, for any Ω ∈ M3n , the following map 0,1 = {θ ∈ C 1,∞ is continuous at any point of B3n ((R3 )n , (R3 )n ): 0,1 ∩ W
p˜ν,Ω W 1,∞ (R3 )n , (R3 )n −→ R e : h  i (13) ˜ ˜ e . θ 7−→ p˜ν,Ω ˜ν I3n + θ˜ (Ω) e (θ) := p

9

Then, we want to relate p˜ν and pν . For this purpose, we consider the following map:
f : W 1,∞ (R3 , R3 ) −→ W 1,∞ (R3 )n , (R3 )n (θ : x 7→ θ(x)) 7−→ f (θ) := (x1 , . . . , xn ) 7→ (θ (x1 ) , . . . , θ (xn )) ,

(14)

which is well dened and clearly linear. It is also continuous since one can check by direct calculations:  √  kf (θ) kW 1,∞ ((R3 )n ,(R3 )n ) 6 nkθkW 1,∞ (R3 ,R3 ) (15) ∀θ ∈ W 1,∞ (R3 , R3 ),  kf (θ) kC 0,1 ((R3 )n ,(R3 )n ) = kθkC 0,1 (R3 ,R3 ) . Moreover, let Ω ∈ M3 . Since the set Ων × (R3 \Ω)n−ν belongs to M3n , we get: pν (Ω) = p˜ν [Ων × (R3 \Ω)n−ν ]

(16)

pν,Ω = p˜ν,Ων ×(R3 \Ω)n−ν ◦ f.

and

1,∞ Since f is continuous and p˜ν,Ων ×(R3 \Ω)n−ν is continuous on B3n ((R3 )n , (R3 )n ), we deduce from 0,1 ∩ W 3 1,∞ 3 3 (15)(16) that pν,Ω is continuous on B0,1 ∩ W (R , R ), concluding the proof of Proposition 2.5.

Theorem 2.6.

Let n > 2 be any integer, let ν ∈ J0, nK, and let Ω be any measurable subset of R3 . First, we assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the shape functional pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W 1,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 12 , 21 }n . Then, the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is well dened around the origin and it is dierentiable at the origin. Its dierential at the origin is given by the following continuous linear form:  ! Z X 1 n 1,∞ 3 3
 ∀θ ∈ W R , R , D0 pν,Ω (θ) = ν c0 ν Ω n 1 1 (σ1 ,...,σn )∈{ 2 , 2 } Z  E D  ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (x1 ) n−ν Ων−1 ×(R3 \Ω)  + |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (x1 ) dx2 . . . dxn  dx1

(17)

 Z

Z

+ (n − ν) R3 \Ω



D Ων ×(R3 \Ω)

n−ν−1

  E ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (x1 ) 

(σ1 ,...,σn ) 2

+ |Ψ

| (x1 , . . . , xn ) div θ (x1 ) dx2 . . . dxn  dx1 .

In other words, the shape functional pν : Ω 7→ pν (Ω) of Denition 2.2 is shape dierentiable at any measurable subset of R3 . If in addition, we now assume that Ω is an open bounded subset of R3 with a Lipschitz boundary as in Denition 5.46, then the shape derivative of pν at Ω takes the following form:  ! Z X 1 n 1,∞ 3 3
 ∀θ ∈ W R , R , D0 pν,Ω (θ) = c0 ν n ∂Ω 1 1 Z (σ1 ,...,σn )∈{ 2 , 2 } ν |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) dx2 . . . dxn (18) n−ν Ων−1 ×(R3 \Ω)  Z (σ1 ,...,σn ) 2 − (n − ν) |Ψ | (x1 , . . . , xn ) dx2 . . . dxn  hθ (x1 ) | nΩ (x1 )i dA (x1 ) , n−ν−1 Ων ×(R3 \Ω) where the integration on the boundary is done with respect to the two-dimensional Hausdor measure referred to as A(•), and where nΩ (x) denotes the unit vector normal to the boundary ∂Ω at the point x pointing outwards Ω. In other words, for any open bounded subset of R3 with Lipschitz boundary, the shape functional pν : Ω 7→ pν (Ω) of Denition 2.2 has a well-dened shape gradient, which is uniquely determined up to a set of zero A(• ∩ ∂Ω)-measure, and which is given by: ! Z X 1 n ∂pν (Ω) : x ∈ ∂Ω 7→ ν |Ψ(σ1 ,...,σn ) |2 (x, x2 , . . . , xn )dx2 . . . dxn n−ν ∂Ω c0 ν Ων−1 ×(R3 \Ω) n 1 1 (σ1 ,...,σn )∈{ 2 , 2 } Z − (n − ν) Ων ×(R3 \Ω)

n−ν−1

|Ψ(σ1 ,...,σn ) |2 (x, x2 , . . . , xn ) dx2 . . . dxn ,

(19) where the boundary values of Ψ(σ1 ,...,σn ) ∈ W 1,2 ((R3 )n , C) have to be understood inRthe sense of trace. Finally, we recall the conventions A0 × B = B × A0 = A, A−1 × B = B × A−1 = ∅ and ∅ f (x, y)dy = f (x), which have to be used to interpret (17)(18)(19) if ν ∈ {0, 1, n − 1, n}.

10

Proof. The method and the notation are the same than the ones used in the proof of Proposition 2.5. The idea consists in using an higher dimensional version (13) of pν in order to get back to the known case of the shape derivative associated with a volume integral. Let n > 2 be any integer, ν ∈ J0, nK, and M3 contain all the measurable subsets of R3 . First, we assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the functional pν : Ω ∈ M3 7→ pν (Ω) ∈ [0, 1] of Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W 1,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 12 , 12 }n . Note that higher regularity is required on the wave function in order to get shape dierentiability. Then, let Ω ∈ M3 and set B30,1 := {θ ∈ C 0,1 (R3 , R3 ), kθkC 0,1 (R3 ,R3 ) < 1} to be the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. We obtain from Proposition 2.5 that the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I3 + θ)(Ω)] is well dened and continuous on B30,1 ∩ W 1,∞ (R3 , R3 ). We now show it is dierentiable at the origin. In the proof of Proposition 2.5, we proved that the map e ∈ M3n 7→ p˜ν (Ω) e ∈ R given by (12) is well dened, where we recall that M3n refers to the class p˜ν : Ω e ∈ M3n , the of all measurable subsets of (R3 )n . We can thus apply Theorem 5.29 to p˜ν and for any Ω 1,∞ 3 n 3 n ˜ ˜ e map p˜ν,Ωe : θ ∈ W ((R ) , (R ) ) 7→ p˜ν [(I3n + θ)(Ω)] ∈ R is well dened around the origin and it is dierentiable at the origin. Its dierential at the origin is given by the following continuous linear form: ! Z   X
1 n ˜ ∀θ˜ ∈ W 1,∞ (R3 )n , (R3 )n , D0 p˜ν,Ω ( θ) = div |Ψ(σ1 ,...,σn ) |2 θ˜ . (20) e c0 ν e n Ω 1 1 (σ1 ,...,σn )∈{− 2 , 2 } Then, we want to relate the shape derivative of p˜ν and pν . Let Ω ∈ M3 . Hence, Ων × (R3 \Ω)n−ν ∈ M3n . Considering the continuous linear map f given in (14), we deduce from (15)(16) and Lemma 5.36, that the map pν,Ω = p˜ν,Ων ×(R3 \Ω)n−ν ◦ f is dierentiable at the origin and its (rst-order) dierential at the origin is given by the continuous linear form on W 1,∞ (R3 , R3 ):
D0 pν,Ω = D0 p˜ν,Ων ×(R3 \Ω)n−ν ◦ f = Df (0) p˜ν,Ων ×(R3 \Ω)n−ν ◦ D0 f = D0 p˜ν,Ων ×(R3 \Ω)n−ν ◦ f. (21) Next, we observe that for any (x1 , . . . , xn ) ∈ R3 × . . . × R3 , for any (σ1 , . . . , σn ) ∈ {− 21 , 12 }n , and for any θ ∈ W 1,∞ (R3 , R3 ), we have: n D     E X ∇xi |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (xi ) div |Ψ(σ1 ,...,σn ) |2 f (θ) (x1 , . . . , xn ) = i=1 (22)  +|Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (xi ) . Combining (20)(21)(22), we deduce that the rst-order shape derivative of pν takes the following form for any θ ∈ W 1,∞ (R3 , R3 ): ! Z n D   E X X 1 n D0 pν,Ω (θ) = ∇xi |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (xi ) c0 ν n Ων ×(R3 \Ω)n−ν i=1 (σ1 ,...,σn )∈{− 1 ,1 2 2}  +|Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (xi ) dx1 . . . dxn .

(23) Finally, we can use the alternating property (5) of the wave function Ψ in order to get for any i ∈ J1, nK, for any (x1 , . . . , xn ) ∈ R3 × . . . R3 , and for any (σ1 , . . . , σn ) ∈ {− 21 , 12 }n :     (24) ∇xi |Ψ(σ1 ,...,σi ,...,σn ) |2 (x1 , . . . , xi , . . . , xn ) = ∇x1 |Ψ(σi ,...,σ1 ,...,σn ) |2 (xi , . . . , x1 , . . . , xn ) . Inserting (24) in (23) and rearranging the summation on the spins variables in (23) by setting σ ˜ i = σ1 , σ ˜1 = σi , and σ˜j = σj for any j ∈ J1, nK\{1, i}, we obtain that (17) holds true. It remains to study the Lipschitz case. Hence, we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in Denition 5.46. First, we consider any θ ∈ W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) in (17). Using the fact that for any measurable subset A of (R3 )n−1 , and for any g ∈ W 1,1 ((R3 )n , C) we have in the sense of distributions thus for almost every x ∈ R3 : Z  Z ∇x1 (g)(x, x2 , . . . , xn )dx2 . . . dxn = ∇x1 g (•, x2 , . . . , xn ) dx2 . . . dxn (x) , (25) A

A

we can apply the Trace Theorem [17, Section 4.3] in (17). We deduce that (18) holds true by observing that the unit outer normal to the boundary ∂Ω = ∂(R3 \Ω) satises nR3 \Ω = −nΩ . More precisely, (18) holds true for any θ ∈ W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ). We obtain that (18) holds true for any θ ∈ W 1,∞ (R3 , R3 ) by using the density result of Lemma 5.41. This approximating argument is a standard procedure, which is detailed in Section 5.2.6 for treating the Lipschitz case in Theorem 5.29. We refer to (119) and below for further details. We emphasize the fact that the boundary values of Ψ(σ1 ,...,σn ) ∈ W 1,2 ((R3 )n , C) have to be understood in the sense of trace. In particular, they are uniquely determined up to a set of zero A(• ∩ ∂Ω)measure, which implies that the shape gradient of pν is unique and well dened by (19), concluding the proof of Theorem 2.6.

11

In fact, under the same assumptions than the ones of Theorem 2.6, we can obtain that the map pν,Ω associated with pν is continuously dierentiable around the origin. The result states as follows.

Corollary 2.7.

Let n > 2 be any integer, let ν ∈ J0, nK, and let Ω be any measurable subset of R3 . First, we assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the shape functional pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W 1,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 12 , 21 }n . Then, the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I +θ)(Ω)] ∈ [0, 1] is well dened and continuously dierentiable at any point of W 1,∞ (R3 , R3 ) ∩ B0,1 , where B0,1 := {θ ∈ C 0,1 (R3 , R3 ), kθkC 0,1 (R3 ,R3 ) < 1} refers to the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. Moreover, its (rst-order) dierential is well dened by the following continuous map:


D• pν,Ω : W 1,∞ R3 , R3 ∩ B0,1 −→ Lc W 1,∞ R3 , R3 , R   (26) θ0 7−→ Dθ0 pν,Ω : θ 7→ D0 pν,(I+θ0 )(Ω) θ ◦ (I + θ0 )−1 , where D0 pν,(I+θ0 )(Ω) is the shape derivative of pν at (I + θ0 )(Ω) dened by (17). If in addition, we assume that Ω is an open bounded subset of Rn with a Lipschitz boundary as in Denition 5.46, then the same result still holds true but we can now use the expression (18) to dene D0 pν,(I+θ0 )(Ω) in (26). Proof. The notation are the ones used in the proofs of Proposition 2.5 and Theorem 2.6. The idea consists in comparing (13) with pν in order to use the results of Section 5.2 for a volume integral. Let n > 2 be any integer, ν ∈ J0, nK, and M3 contain all the measurable subsets of R3 . First, we assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the functional pν : Ω ∈ M3 7→ pν (Ω) ∈ [0, 1] of Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W 1,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 21 , 12 }n . Then, let Ω ∈ M3 and B30,1 denote the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. From Theorem 2.6, the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I3 + θ)(Ω)] is well dened on B30,1 ∩ W 1,∞ (R3 , R3 ) and dierentiable at the origin. We now show that it is continuously dierentiable at any point of B30,1 ∩ W 1,∞ (R3 , R3 ). In the proof of Proposition 2.5, we have established e ∈ M3n 7→ p˜ν (Ω) e ∈ R given by (12) is well dened, where M3n refers to the class of that the map p˜ν : Ω 3 n e ∈ M3n , the map all measurable subsets of (R ) . We can thus apply Proposition 5.43 to p˜ν and for any Ω 1,∞ 3 n 3 n ˜ ˜ e p˜ν,Ω ((R ) , (R ) ) 7→ p˜ν [(I3n + θ)(Ω)] ∈ R is well dened and continuously dierentiable at e : θ ∈ W any point W 1,∞ ((R3 )n , (R3 )n ) ∩ B3n 0,1 . Moreover, its (rst-order) dierential is well dened by the following continuous map:


1,∞ −→ Lc W 1,∞ (R3 )n , (R3 )n , R (R3 )n , (R3 )n ∩ B3n D• p˜ν,Ω e : W 0,1   −1  (27) ˜ ˜ ˜ ˜ν,(I3n +θ˜0 )(Ω θ˜0 7−→ Dθ˜0 p˜ν,Ω , e : θ 7→ D0 p e ) θ ◦ I3n + θ0 e given by (20). Then, we want to where D0 p˜ν,(I3n +θ˜0 )(Ωe ) is the shape derivative of p˜ν at (I3n + θ˜0 )(Ω)

relate the dierential of p˜ν,Ωe and pν,Ω . Let Ω ∈ M3 . Hence, we have Ων × (R3 \Ω)n−ν ∈ M3n . Considering the continuous linear map f given in (14), we deduce from (15)(16) and Lemma 5.36, that the map pν,Ω = p˜ν,Ων ×(R3 \Ω)n−ν ◦ f is continuous dierentiable at any point θ0 ∈ W 1,∞ (R3 , R3 ) ∩ B30,1 and we have for any θ ∈ W 1,∞ (R3 , R3 ):
Dθ0 pν,Ω (θ) = Dθ0 p˜ν,Ων ×(R3 \Ω)n−ν ◦ f (θ) = Df (θ0 ) p˜ν,Ων ×(R3 \Ω)n−ν [Dθ0 f (θ)]   D0 p˜ν,[I3n +f (θ0 )](Ων ×(R3 \Ω)n−ν ) f (θ) ◦ (I3n + f (θ0 ))−1 ,

=

where we have used the fact that Dθ0 f = f and (27) to obtain the last equality. From Proposition 5.31, we have [I3n + f (θ0 )](Ων × (R3 \Ω)n−ν ) = [(I3 + θ0 )(Ω)]ν × [R3 \(I3 + θ0 )(Ω)]n−ν . We can also check that f (θ) ◦ [I3n + f (θ0 )]−1 = f [θ ◦ (I3 + θ0 )−1 ] so we deduce that: 
 Dθ0 pν,Ω (θ) = D0 p˜ν,[(I +θ )(Ω)]ν × R3 \(I +θ )(Ω) n−ν f θ ◦ (I3 + θ0 )−1 [ ] 3 0 3 0 =

Df (0) p˜ν,[(I

=

 D0 p˜ν,[(I

=

  D0 pν,(I3 +θ0 )(Ω) θ ◦ (I3 + θ0 )−1 .

[R3 \(I3 +θ0 )(Ω)]n−ν

ν 3 +θ0 )(Ω)] ×

  ◦ D0 f θ ◦ (I3 + θ0 )−1 

ν 3 +θ0 )(Ω)] ×

[R3 \(I3 +θ0 )(Ω)]

n−ν

◦f



θ ◦ (I3 + θ0 )−1



Hence, we have proved that the rst-order dierential of pν,Ω is well dened by the continuous map (26). Finally, if we assume that Ω is a open bounded subset of R3 with Lipschitz boundary as in Denition 5.46, then (I3 + θ0 )(Ω) is also a open bounded Lipschitz domain which satises ∂[(I3 + θ0 )(Ω)] = (I3 + θ0 )(∂Ω). Moreover, we have θ ◦ (I3 + θ0 )−1 ∈ W 1,∞ so we deduce that the expression (18) denes well D0 pν,(I+θ0 )(Ω) in (26), concluding the proof of Corollary 2.7.

12

2.3 On the second- and higher-order shape derivative of the probability We refer to Section 5.2 for details on the second-order shape dierentiability and the notation related to it, especially Denition 5.28, Theorem 5.30, and Section 5.2.1. We rst state some results about the secondorder shape derivative of pν : Ω 7→ pν (Ω) i.e. the second-order dierential at the origin of the associated map pν,Ω : θ 7→ pν [(I + θ)(Ω)]. Then, we obtain a formula for the shape derivative of pν at any order.

Theorem 2.8.

Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any measurable subset of R3 . First, we assume the wave function Ψ given by (4) satises Assumption 2.1 and the shape functional pν : Ω 7→ pν (Ω) of Denition 2.2 is well dened. Moreover, we require that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W 2,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 21 , 21 }n . Then, pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is a well-dened and dierentiable map around the origin, its (rst-order) dierential map D• pν,Ω being well dened around the origin by (26). Moreover, the map (26) is dierentiable at the origin. Hence, pν,Ω is twice dierentiable at the origin and its second-order dierential at the origin is given by the following ˜ ∈ W 1,∞ (R3 , R3 ) × W 1,∞ (R3 , R3 ) by: continuous symmetric bilinear form dened for any (θ, θ) ! Z X ˜ = 1 n D02 pν,Ω (θ, θ) ν c0 ν Ω n (σ1 ,...,σn )∈{ 1 ,1 2 2}  Z   E D  Hessx1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) θ (x1 ) | θ˜ (x1 ) n−ν Ων−1 ×(R3 \Ω) D   E + ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (x1 ) div θ˜ (x1 ) + θ˜ (x1 ) div θ (x1 )  i h  + |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (x1 ) div θ˜ (x1 ) − trace Dx1 θDx1 θ˜ dx2 . . . dxn  dx1  Z

Z

+ (n − ν) R3 \Ω



D Ων ×(R3 \Ω)

n−ν−1

  E Hessx1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) θ (x1 ) | θ˜ (x1 )

  E ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (x1 ) div θ˜ (x1 ) + θ˜ (x1 ) div θ (x1 )  i h  + |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (x1 ) div θ˜ (x1 ) − trace Dx1 θDx1 θ˜ dx2 . . . dxn  dx1 +

D

 Z

3 X

Z

+ ν (ν − 1)

 Ω×Ω

Ων−2 ×(R3 \Ω)

+

D

n−ν

  2 ∂(x |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) θk (x1 ) θ˜l (x2 ) 1 )k ,(x2 )l

k,l=1  E ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (x1 ) div θ˜ (x2 ) + θ˜ (x1 ) div θ (x2 )  + |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (x1 ) div θ˜ (x2 ) dx3 . . . dxn  dx1 dx2

 Z +2ν (n − ν) Ω×(R3 \Ω)

3 X

Z 

  2 ∂(x |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) θk (x1 ) θ˜l (x2 ) 1 )k ,(x2 )l

n−ν−1

Ων−1 ×(R3 \Ω)

+

D



k,l=1 (σ1 ,...,σn ) 2

∇x1 |Ψ

|



E (x1 , . . . , xn ) | θ (x1 ) div θ˜ (x2 ) + θ˜ (x1 ) div θ (x2 ) 

+ |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (x1 ) div θ˜ (x2 ) dx3 . . . dxn  dx1 dx2 Z + (n − ν) (n − ν − 1) 3 \Ω)×(R3 \Ω) (R Z 3   X 2  ∂(x |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) θk (x1 ) θ˜l (x2 ) 1 )k ,(x2 )l n−ν−2 Ων ×(R3 \Ω) k,l=1 E D   (σ1 ,...,σn ) 2 + ∇x1 |Ψ | (x1 , . . . , xn ) | θ (x1 ) div θ˜ (x2 ) + θ˜ (x1 ) div θ (x2 ) 

+ |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (x1 ) div θ˜ (x2 ) dx3 . . . dxn  dx1 dx2 .

(28) In other words, the shape functional pν : Ω 7→ pν (Ω) of Denition 2.2 is twice shape dierentiable at any measurable subset of R3 . If in addition, we now assume that Ω is an open bounded subset Ω of R3 with Lipschitz boundary as in Denition 5.46, then pν,Ω : θ ∈ W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 ) 7→ pν [(I +θ)(Ω)] ∈ [0, 1]

13

is a well-dened and dierentiable map around the origin, its (rst-order) dierential being given by:




D• pν,Ω : W 1,∞ R3 , R3 ∩ C 1 R3 , R3 −→ Lc W 1,∞ R3 , R3 ∩ C 1 R3 , R3 , R  (29) θ0 7−→ Dθ0 pν,Ω : θ 7→ D0 pν,(I+θ0 )(Ω) θ ◦ (I + θ0 )−1 , where D0 pν,(I+θ0 )(Ω) is the shape derivative of pν at (I + θ0 )(Ω) dened by (18). Moreover, the map (29), which is well dened around the origin, is also dierentiable at the origin. Hence, pν,Ω is twice dierentiable at the origin and its second-order dierential at the origin is given by the following continuous symmetric ˜ ∈ (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) × (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) by: bilinear form dened for any (θ, θ) ! Z X 1 n 2 2 ˜ ˜ D0 pν,Ω (θ, θ) = D0 pν,Ω (θ, θ) = c0 ν n ∂Ω (σ1 ,...,σn )∈{ 1 ,1 2 2}  Z   ED E D ν ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (x1 ) θ˜ (x1 ) | nΩ (x1 ) n−ν Ων−1 ×(R3 \Ω) h i E D + |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) div θ (x1 ) θ˜ (x1 ) − Dx1 θ θ˜ (x1 ) | nΩ (x1 ) dx2 . . . dxn Z

  ED E D ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | θ (x1 ) θ˜ (x1 ) | nΩ (x1 ) Ων ×(R3 \Ω)  h i E D (σ1 ,...,σn ) 2 + |Ψ | (x1 , . . . , xn ) div θ (x1 ) θ˜ (x1 ) − Dx1 θ θ˜ (x1 ) | nΩ (x1 ) dx2 . . . dxn  dA (x1 ) − (n − ν)

n−ν−1

Z

Z ν (ν − 1)

+

3 Ων−2 Z ×(R \Ω)

∂Ω×∂Ω

−2ν (n − ν) (

Ων−1 ×

n−ν

|Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) dx3 . . . dxn

n−ν−1 R3 \Ω

|Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) dx3 . . . dxn

)

!

Z + (n − ν) (n − ν − 1) Ων ×(R3 \Ω)

n−ν−2

|Ψ

(σ1 ,...,σn ) 2

| (x1 , . . . , xn ) dx3 . . . dxn

hθ (x1 ) | nΩ (x1 )i hθ˜ (x2 ) | nΩ (x2 )idA (x1 ) dA (x2 ) . (30) If in addition, we now assume that Ω is an open bounded subset of R3 with a boundary of class C 1,1 in the sense of Denition 5.47, then the second-order shape derivative of pν at Ω takes the following form for any ˜ ∈ (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) × (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )): (θ, θ)  ! Z Z h X n 1 2 ˜ = ν HΩ (x1 ) |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) D0 pν,Ω (θ, θ) n−ν c0 ν ν−1 3 ∂Ω Ω × R \Ω n ( ) (σ1 ,...,σn )∈{ 1 ,1 2 2} D   E i + ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | nΩ (x1 ) dx2 . . . dxn Z − (n − ν)

h n−ν−1

Ων ×(R3 \Ω)

+

D   E i ∇x1 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) | nΩ (x1 ) dx2 . . . dxn  θn (x1 ) θ˜n (x1 ) dA (x1 )

Z

Z ν (ν − 1)

+

HΩ (x1 ) |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) 

Ων−2 ×

∂Ω×∂Ω

n−ν R3 \Ω

(

|Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) dx3 . . . dxn

)

Z −2ν (n − ν)

|Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) dx3 . . . dxn + (n − ν) (n − ν − 1) ! (σ1 ,...,σn ) 2 |Ψ | (x1 , . . . , xn ) dx3 . . . dxn θn (x1 ) θ˜n (x2 ) dA (x1 ) dA (x2 )

Ων−1 ×(R3 \Ω)

Z n−ν−2

n−ν−1

Ων ×(R3 \Ω)

Z −

Z ν

∂Ω

(

Ων−1 ×

n−ν R3 \Ω

|Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) dx2 . . . dxn

)

!

Z

(σ1 ,...,σn ) 2

− (n − ν) |Ψ | (x1 , . . . , xn ) dx2 . . . dxn n−ν−1 ν 3 h  Ω ×(R \Ω ) D E D Ei IIΩ θ˜∂Ω , θ∂Ω + ∇∂Ω (θn ) | θ˜∂Ω + ∇∂Ω (θ˜n ) | θ∂Ω (x1 ) dA (x1 ) ,

(31)

14

where (•)n := h(•) | nΩ i is the normal component of a vector eld, where (•)∂Ω := (•) − (•)n nΩ is the tangential component of a vector eld, which means that in particular ∇∂Ω (•) := ∇(•) − h∇(•) | nΩ inΩ is the tangential component of the gradient operator, where IIΩ (•, •) := −hD∂Ω nΩ (•) | (•)i is the second fundamental form associated with the C 1,1 -surface ∂Ω, which is a symmetric bilinear form on the tangent space, with D∂Ω (•) := D(•) − D(•)nΩ [nΩ ]T denoting the tangential component of the dierential operator on vector elds, where HΩ := div∂Ω (nΩ ) is the scalar mean curvature associated with the C 1,1 -surface ∂Ω, with div∂Ω (•) := div(•) − hD(•)nΩ | nΩ i denoting the tangential component of the divergence operator. Finally, we emphasize the structure (31) obtained for the second-order shape derivative. • The last term of (31) can be interpreted as a continuous linear form as follows: observing from (18) ˜ , where Z that we can write D0 pν,Ω (θ) = l1 (θn ), then the third term in (31) has the form l1 [Z(θ, θ)] ˜ is a vector eld which is is equal to zero if θ and θ are normal to the boundary ∂Ω. In this case, the last term of (31) is thus equal to zero. Moreover, if Ω is a critical shape for pν i.e. if D0 pν,Ω ≡ 0, then, l1 ≡ 0 and this term is also equal to zero. • The middle term of (31) is a continuous symmetric bilinear form, which takes the form of a kernel R KΩ (x, y)θn (x)θ˜n (y)dA(x)dA(y), where we have set: ∂Ω×∂Ω 1 KΩ : (x, y) ∈ ∂Ω × ∂Ω 7−→ c0

! n ν

Z

X

ν (ν − 1) n

(σ1 ,...,σn )∈{ 1 ,1 2 2}

Ων−2 ×(R3 \Ω)

n−ν

|Ψ(σ1 ,...,σn ) |2 (x, y, x3 , . . . , xn ) dx3 . . . dxn Z −2ν (n − ν) Ων−1 ×(R3 \Ω)

n−ν−1

Z + (n − ν) (n − ν − 1) Ων ×(R3 \Ω)

n−ν−2

|Ψ(σ1 ,...,σn ) |2 (x, y, x3 , . . . , xn ) dx3 . . . dxn

(32)

|Ψ(σ1 ,...,σn ) |2 (x, y, x3 , . . . , xn ) dx3 . . . dxn .

• The rst term of (31) is a continuous symmetric bilinear form, which can be interpreted as the shape R 2 Hessian part of the second-order shape derivative. Indeed, it takes the form ∂Ω ∂∂Ωp2ν (Ω)θn θ˜n dA, where the integrand can be considered as the shape Hessian of pν : ! Z X ∂ 2 pν 1 n ν (Ω) : x ∈ ∂Ω − 7 → n−ν ∂Ω2 c0 ν Ων−1 ×(R3 \Ω) 1 n (σ1 ,...,σn )∈{ 1 }  E D 2 , 2 HΩ (x) |Ψ(σ1 ,...,σn ) |2 (x, x2 , . . . , xn ) + ∇x1 |Ψ(σ1 ,...,σn ) |2 (x, x2 , . . . , xn ) | nΩ (x) dx2 . . . dxn Z − (n − ν) HΩ (x) |Ψ(σ1 ,...,σn ) |2 (x, x2 , . . . , xn ) n−ν−1 Ων ×(R3 \Ω)   E D + ∇x1 |Ψ(σ1 ,...,σn ) |2 (x, x2 , . . . , xn ) | nΩ (x) dx2 . . . dxn .

(33)

We recall that in (30)(31)(32)(33), the boundary values of Ψ(σ1 ,...,σn ) ∈ W 2,2 ((R3 )n , C) and also the ones of ∇x1 Ψ(σ1 ,...,σn ) ∈ W 1,2 ((R3 )n , C) have to be understood in the sense of trace.R We also recall the conventions A0 ×B = B ×A0 = A, A−1 ×B = B ×A−1 = A−2 ×B = B ×A−2 = ∅ and ∅ f (x, y)dy = f (x), which are used to interpret (26)(28)(29)(30)(31)(32)(33) if n ∈ {2, 3} and ν ∈ {0, 1, 2, n−1, n−2, n}. Proof. We use the notation of Proposition 2.5 and we proceed as in the proof of Theorem 2.6. Again, we compare (13) with pν in order to use the results of Section 5.2 available for a volume integral. Let n > 2 be any integer, ν ∈ J0, nK, and M3 contain all the measurable subsets of R3 . First, we assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the functional pν : Ω ∈ M3 7→ pν (Ω) ∈ [0, 1] of Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W 2,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 21 , 21 }n . Note that higher regularity is required on Ψ to get the second-order shape dierentiability. Then, let Ω ∈ M3 and B30,1 denote the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. From Corollary 2.7, the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I3 + θ)(Ω)] ∈ [0, 1] is well dened and dierentiable on B30,1 ∩ W 1,∞ (R3 , R3 ). We now show that it is twice dierentiable at the origin i.e. (26) is dierentiable at the origin. In the proof of Proposition 2.5, we have established that e ∈ M3n 7→ p˜ν (Ω) e ∈ R is well dened by (12), where M3n contains the measurable subsets the map p˜ν : Ω 3 n ˜ Ω)] e ∈R of (R ) . Applying Theorem 5.30 to p˜ν , the map p˜ν,Ωe : θ˜ ∈ W 1,∞ ((R3 )n , (R3 )n ) 7→ p˜ν [(I3n + θ)( e ∈ M3n . Moreover, it is twice dierentiable is well dened and dierentiable around the origin for any Ω at the origin and its second-order dierential at the origin is given by the following continuous symmetric

15

bilinear form dened for any (θ˜1 , θ˜2 ) ∈ W 1,∞ ((R3 )n , (R3 )n ) × W 1,∞ ((R3 )n , (R3 )n ) by: ! Z D    E X 1 n ˜ ˜ D02 p˜ν,Ω ( θ , θ ) := Hess |Ψ(σ1 ,...,σn ) |2 θ˜1 | θ˜2 1 2 e c0 ν e n Ω (σ1 ,...,σn )∈{− 1 ,1 2 2} Z D   E    + ∇ |Ψ(σ1 ,...,σn ) |2 | θ˜1 div θ˜2 + θ˜2 div θ˜1 e Ω Z h      i + |Ψ(σ1 ,...,σn ) |2 div θ˜1 div θ˜2 − trace D• θ˜1 D• θ˜2 . e Ω

(34) Then, we want to relate the second-order shape derivative of p˜ν with the one of pν as in (21). Let Ω ∈ M3 . We thus have Ων × (R3 \Ω)n−ν ∈ M3n . Considering the continuous linear map f given in (14), we deduce from (15)(16) and Proposition 5.37, that the map pν,Ω = p˜ν,Ων ×(R3 \Ω)n−ν ◦ f is twice dierentiable at the origin and its second-order dierential at the origin is given by the following continuous symmetric bilinear form dened for any (θ1 , θ2 ) ∈ W 1,∞ (R3 , R3 ) × W 1,∞ (R3 , R3 ) by:
D02 pν,Ω (θ1 , θ2 ) = D02 p˜ν,Ων ×(R3 \Ω)n−ν ◦ f (θ1 , θ2 ) =

  Df2 (0) p˜ν,Ων ×(R3 \Ω)n−ν [D0 f (θ1 ), D0 f (θ2 )] + Df (0) p˜ν,Ων ×(R3 \Ω)n−ν D02 f (θ1 , θ2 )

=

D02 p˜ν,Ων ×(R3 \Ω)n−ν [f (θ1 ), f (θ2 )] .

(35) Proceeding as in (22), we observe that for any (x1 , . . . , xn ) ∈ R3 ×. . .×R3 , for any (σ1 , . . . , σn ) ∈ {− 21 , 12 }n , and for any (θ1 , θ2 ) ∈ W 1,∞ (R3 , R3 ) × W 1,∞ (R3 , R3 ), we have: D

  E Hess |Ψ(σ1 ,...,σn ) |2 [f (θ1 )] | f (θ2 ) (x1 , . . . , xn ) =

n 3 X X

  2 |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) ∂(x i )k (xj )l

i,j=1 k,l=1

[θ1 (xi )]k [θ2 (xj )]l =

n D X

  E Hessxi |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) [θ1 (xi )] | θ2 (xi )

i=1 n X X

+

3 X

  2 ∂(x |Ψ(σ1 ,...,σn ) |2 (x1 , . . . , xn ) [θ1 (xi )]k [θ2 (xj )]l , i )k (xj )l

i=1 j∈J1,nK k,l=1 j6=i

where we have distinguished the cases i = j and i 6= j to obtain the last equality. Using the alternating property (5) of the wave function Ψ, we can proceed as in (24) in order to get: D   E Hess |Ψ(σ1 ,...,σn ) |2 [f (θ1 )] | f (θ2 ) (x1 , . . . xn ) n D X

=

  E Hessx1 |Ψ(σi ,...,σ1 ,...,σn ) |2 (xi , . . . , x1 , . . . , xn ) [θ1 (xi )] | θ2 (xi )

i=1

+

n X X

3 X

  2 (σi ,σj ,...,σ1 ,...,σ2 ,...σn ) 2 ∂(x |Ψ | (xi , xj , . . . , x1 , . . . , x2 , . . . xn ) ) (x ) 1 k 2 l

i=1 j∈J1,nK k,l=1 j6=i

[θ1 (xi )]k [θ2 (xj )]l .

Then, we can rearrange the summation on the spin variables by setting σ ˜1 = σ ˆ1 = σi , σ ˜2 = σj , σ ˆ2 = σ2 , σ ˜i = σ ˆ i = σ1 , σ ˜j = σ2 , and σ ˜m = σ ˆm = σm for any m ∈ J1, nK\{1, 2, i, j}. We deduce that: D   E X Hess |Ψ(σ1 ,...,σn ) |2 [f (θ1 )] | f (θ2 ) (x1 , . . . xn ) 1,1 n (σ1 ,...,σn )∈{− 2 2} n D   E X Hessx1 |Ψ(ˆσ1 ,...,ˆσn ) |2 (xi , . . . , x1 , . . . xn ) [θ1 (xi )] | θ2 (xi )

X

=

n

1,1 (ˆ σ1 ,...,ˆ σn )∈{− 2 2}

+

n X X

X (˜ σ1 ,...,˜ σn )∈{− 1 ,1 2 2}

i=1

n

3 X

  2 ∂(x |Ψ(˜σ1 ,...,˜σn ) |2 (xi , xj , . . . , x1 , . . . , x2 , . . . xn ) 1 )k (x2 )l

i=1 j∈J1,nK k,l=1 j6=i

[θ1 (xi )]k [θ2 (xj )]l . e := Ω × (R \Ω) Integrating this last relation on Ω , we can distinguish the cases (xi , xj ) ∈ Ω × Ω, (xi , xj ) ∈ (R3 \Ω) × (R3 \Ω), (xi , xj ) ∈ Ω × (R3 \Ω), and (xi , xj ) ∈ (R3 \Ω) × Ω, where the two last cases ν

3

n−ν

16

lead to the same expression by re-labelling the rst and second spin variables as before. Proceeding similarly for the other integrands of (34) applied to θ˜1 = f (θ1 ) and θ˜2 = f (θ2 ), we get from (34) that (28) holds true. It remains to study the Lipschitz case. Hence, we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary and we consider the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 ) 7→ pν [(I +θ)(Ω)] ∈ [0, 1]. Since the set W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 ) is also equipped with the W 1,∞ -norm, we deduce from Corollary 2.7 that pν,Ω is well dened and continuously dierentiable at any point of W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) ∩ B30,1 , its rst-order dierential being well dened by (29). Moreover, we get from the foregoing that (29) is dierentiable at the origin i.e. pν,Ω is twice dierentiable at the origin and its second-order dierential at the origin is well dened by (28) for any (θ1 , θ2 ) ∈ (W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 ))×(W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 )). We use the additional regularity of Ω, θ1 , and θ2 to improve the expression (28) of the second-order shape derivative of pν . First, we assume that (θ1 , θ2 ) ∈ W 2,∞ (R3 , R3 ) × W 2,∞ (R3 , R3 ). Arguing as in (25), we can apply the Trace Theorem [17, Section 4.3] in relation (28) in order to obtain that (30) holds true. More precisely, note that we have to apply the Trace Theorem for W 1,∞ -elds, which have already been established in the proof of Theorem 5.29 (see (119) and below) using the density result of Lemma 5.41. Hence, (30) holds true for any (θ1 , θ2 ) ∈ W 2,∞ (R3 , R3 ) × W 2,∞ (R3 , R3 ) by observing that the unit outer normal to the boundary ∂Ω = ∂(R3 \Ω) satises nR3 \Ω = −nΩ . As done in the proof of Theorem 5.30 (see (125) and below), we emphasize the fact that W 2,∞ -regularity is needed to obtain (30) from (28). Then, we use Lemma 5.42 to extend by a density argument relation (30) for any (θ1 , θ2 ) ∈ (W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 ))×(W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 )). This is a standard procedure, which has been detailed in the proof of Theorem 5.30 (see (125) and below) to study the Lipschitz case. Hence, we have proved that (30) holds true. Finally, if we assume that Ω is an open bounded subset of R3 with a C 1,1 -boundary, then we can proceed exactly as in the end of the proof of Theorem 5.30 to obtain (31) from (30) by distinguishing the normal/tangential parts of the operators/vector-elds. We emphasize the fact that in (30)(31), the boundary values of Ψ(σ1 ,...,σn ) ∈ W 2,2 ((R3 )n , C) and ∇x1 Ψ(σ1 ,...,σn ) have to be understood in the sense of trace. In particular, the kernel (32) and the shape Hessian (33) are uniquely determined up to a set of zero A(• ∩ ∂Ω)-measure, concluding the proof of Theorem 2.8. In fact, under the same assumptions than the ones of Theorem 2.8, we can obtain that the map pν,Ω associated with pν is twice continuously dierentiable around the origin. The result states as follows.

Corollary 2.9.

Let n > 2 be any integer, let ν ∈ J0, nK, and let Ω be any measurable subset of R3 . First, we assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the shape functional pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W 2,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 12 , 21 }n . Then, the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is well dened and twice continuously dierentiable at any point of W 1,∞ (R3 , R3 ) ∩ B0,1 , where B0,1 = {θ ∈ C 0,1 (R3 , R3 ), kθkC 0,1 (R3 ,R3 ) < 1} refers to the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. Moreover, its second-order dierential is well dened by the following continuous map:



D•2 pν,Ω : W 1,∞ R3 , R3 ∩ B0,1 −→ Bc W 1,∞ R3 , R3 × W 1,∞ R3 , R3 , R θ0 7−→ Dθ20 pν,Ω : (θ1 , θ2 ) 7→  (36)  D02 pν,(I+θ0 )(Ω) θ1 ◦ (I + θ0 )−1 , θ2 ◦ (I + θ0 )−1 , where D02 pν,(I+θ0 )(Ω) is the second-order shape derivative of pν at (I +θ0 )(Ω) dened by (28). If in addition, we assume that Ω is an open bounded subset of Rn with a Lipschitz boundary as in Denition 5.46, then the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is also twice continuously dierentiable at any point of W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) ∩ B0,1 and its second-order dierential is well dened by the following continuous map: 



2  D•2 pν,Ω : W 1,∞ R3 , R3 ∩ C 1 (R3 , R3 ) ∩ B0,1 −→ Bc W 1,∞ R3 , R3 ∩ C 1 R3 , R3 ,R θ0 7−→ Dθ20 pν,Ω : (θ1 , θ2 )7→  D02 pν,(I+θ0 )(Ω) θ1 ◦ (I + θ0 )−1 , θ2 ◦ (I + θ0 )−1 , (37) where D02 pν,(I+θ0 )(Ω) is now dened by (30). Finally, if we assume that Ω is an open bounded subset of R3 with a boundary of class C 1,1 as in Denition 5.47, then the last result still holds true but we can now use the expression (31) to dene D02 pν,(I+θ0 )(Ω) in (37). Proof. The notation are the ones used in the proof of Proposition 2.5. We proceed as in the proof of Corollary 2.9. Again, the idea consists in comparing (13) with pν in order to use the results of Section 5.2 available for a volume integral. Let n > 2 be any integer, ν ∈ J0, nK, and M3 contain all the measurable subsets of R3 . We assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the functional pν : Ω ∈ M3 7→ pν (Ω) ∈ [0, 1] of Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W 2,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 21 , 12 }n . Then, let Ω ∈ M3 and B30,1 denote the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. From Theorem 2.8, the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I3 + θ)(Ω)] is well dened on B30,1 ∩ W 1,∞ (R3 , R3 ) and twice dierentiable

17

at the origin. We now show that it is twice continuously dierentiable at any point of B30,1 ∩ W 1,∞ (R3 , R3 ). e ∈ M3n 7→ p˜ν (Ω) e ∈ R given by In the proof of Proposition 2.5, we have established that the map p˜ν : Ω (12) is well dened, where M3n refers to the class of all measurable subsets of (R3 )n . We can thus apply ˜ Ω)] e ∈ M3n , the map p˜ e : θ˜ ∈ W 1,∞ ((R3 )n , (R3 )n ) 7→ p˜ν [(I3n +θ)( e ∈R Proposition 5.44 to p˜ν and for any Ω ν,Ω is well dened and twice continuously dierentiable at any point of W 1,∞ ((R3 )n , (R3 )n ) ∩ B3n . Moreover, 0,1 its second-order dierential is well dened by the following continuous map:



1,∞ D•2 p˜ν,Ω (R3 )n , (R3 )n ∩ B3n −→ Bc W 1,∞ (R3 )n , (R3 )n × W 1,∞ (R3 )n , (R3 )n , R e : W 0,1 ˜ ˜ θ˜0 7−→ Dθ2˜0 p˜ν,Ω e : (θ1 , θ2 ) 7→   −1   −1 ˜ ˜ , , θ˜2 ◦ I3n + θ˜0 D02 p˜ν,(I3n +θ˜0 )(Ω e ) θ1 ◦ I3n + θ0 (38) e given by (34). Then, where D02 p˜ν,(I3n +θ˜0 )(Ωe ) is the second-order shape derivative of p˜ν at (I3n + θ˜0 )(Ω) we want to relate the second-order dierential of p˜ν,Ωe with the one of pν,Ω . Let Ω ∈ M3 . Hence, we have Ων × (R3 \Ω)n−ν ∈ M3n . Considering the continuous linear map f given in (14), we deduce from (15)(16) and Proposition 5.37, that the map pν,Ω = p˜ν,Ων ×(R3 \Ω)n−ν ◦ f is twice continuously dierentiable at any point θ0 ∈ W 1,∞ (R3 , R3 ) ∩ B30,1 and we have for any (θ1 , θ2 ) ∈ W 1,∞ (R3 , R3 ) × W 1,∞ (R3 , R3 ):
Dθ20 pν,Ω (θ1 , θ2 ) = Dθ20 p˜ν,Ων ×(R3 \Ω)n−ν ◦ f (θ1 , θ2 ) = Df2 (θ0 ) p˜ν,Ων ×(R3 \Ω)n−ν [Dθ0 f (θ1 ) , Dθ0 f (θ2 )]   + Df (θ0 ) p˜ν,Ων ×(R3 \Ω)n−ν Dθ20 f (θ1 , θ2 ) = Df2 (θ0 ) p˜ν,Ων ×(R3 \Ω)n−ν [f (θ1 ) , f (θ2 )]   = D02 p˜ν,[I3n +f (θ0 )](Ων ×(R3 \Ω)n−ν ) f (θ1 ) ◦ (I3n + f (θ0 ))−1 , f (θ2 ) ◦ (I3n + f (θ0 ))−1 ,

where we have used the fact that Dθ0 f = f and (38) to obtain the last equality. From Proposition 5.31, we have [I3n + f (θ0 )](Ων × (R3 \Ω)n−ν ) = [(I3 + θ0 )(Ω)]ν × [R3 \(I3 + θ0 )(Ω)]n−ν . We can also check that f (θi ) ◦ [I3n + f (θ0 )]−1 = f [θ ◦ (I3 + θ0 )−1 ] for i ∈ {1, 2} so we deduce that: 

 Dθ20 pν,Ω (θ1 , θ2 ) = D02 p˜ν,[(I +θ )(Ω)]ν × R3 \(I +θ )(Ω) n−ν f θ1 ◦ (I3 + θ0 )−1 , f θ2 ◦ (I3 + θ0 )−1 [ ] 3 0 3 0 =

  D0 f θ1 ◦ (I3 + θ0 )−1 ,  
D0 f θ2 ◦ (I3 + θ0 )−1   +Df (0) p˜ν,[(I +θ )(Ω)]ν × R3 \(I +θ )(Ω) n−ν ◦ D02 f θ1 ◦ (I3 + θ0 )−1 , θ2 ◦ (I3 + θ0 )−1 [ ] 3 0 3 0

=

D02

=

  D02 pν,(I3 +θ0 )(Ω) θ1 ◦ (I3 + θ0 )−1 , θ2 ◦ (I3 + θ0 )−1 .

Df2 (0) p˜ν,[(I

 p˜ν,[(I

ν 3 3 +θ0 )(Ω)] ×[R \(I3 +θ0 )(Ω)]

n−ν

 [R3 \(I3 +θ0 )(Ω)]n−ν

ν 3 +θ0 )(Ω)] ×

◦f



θ1 ◦ (I3 + θ0 )−1 , θ2 ◦ (I3 + θ0 )−1



We thus have proved that the second-order dierential of pν,Ω is well dened by the continuous map (36). Finally, if we now assume that Ω is a open bounded subset of R3 with a Lipschitz boundary, then (I3 +θ0 )(Ω) is also an open bounded Lipschitz domain which satises ∂[(I3 + θ0 )(Ω)] = (I3 + θ0 )(∂Ω) for any θ0 ∈ B30,1 . Moreover, we have θ ◦ (I3 + θ0 )−1 ∈ W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) for any (θ, θ0 ) ∈ W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) such that θ0 ∈ B30,1 . We deduce that the expression (28) denes well D0 pν,(I+θ0 )(Ω) in (37). To conclude, the proof of Corollary 2.9, if Ω is an open bounded subset of R3 with a C 1,1 -boundary, then we can now use the expression (31) to dene D0 pν,(I+θ0 )(Ω) in (37). To conclude this section, we can generalize the previous arguments to obtain the C k -regularity of pν,Ω . In particular, we get a formula for the shape derivative of pν at any order.

Theorem 2.10.

Let k0 > 1 and n > 2 be two integers, ν ∈ J0, nK, and Ω be any measurable subset of R3 . We denote by B0,1 := {θ ∈ C 0,1 (R3 , R3 ), kθkC 0,1 (R3 ,R3 ) < 1} as the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. First, we assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the shape functional pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W k0 ,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 12 , 21 }n . Then, the associated map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is k0 times dierentiable at the origin and for any k ∈ J1, k0 K, its dierential of order k at the origin is given by the following continuous symmetric

18

k-linear form dened for any (θ1 , . . . , θk ) ∈ W 1,∞ (R3 , R3 ) × . . . × W 1,∞ (R3 , R3 ) by: ! n 3 k X X X X X 1 n k D0 pν,Ω (θ1 , . . . , θk ) := c0 ν i ,...,i =1 m ,...,m =1 l=0 Il ⊆J1,kK (σ1 ,...,σn )∈{− 1 , 1 }n 1 1 k k 2 2 card Il =l     k−l (σ1 ,...,σn ) 2 Z ∂ |Ψ |  Y 
 
Q (x1 , . . . , xn )  θj x i j m   n−ν j ∂ xij m Ων ×(R3 \Ω) j∈J1,kK j j∈J1,kK j ∈I / l

j ∈I / l

   

 X

s p−1 Il ◦ p ◦ p Il

p:Il 7→Il p bijective


Y

h i Iij ip(j) Dxij θj

j∈Il

mj mp(j)

  dx1 . . . dxn , 

(39) where s : Sl → {−1, 1} is the signature map introduced in Denition 5.4, where pIl : J1, lK → Il is the unique strictly increasing map of Lemma 5.20, and where Ikl = 1 if and only if k = l. In other words, the functional pν of Denition 2.2 is k0 times shape dierentiable at any measurable subset of Rn , and its shape derivative of order k is given by (39) for any k ∈ J1, k0 K. Moreover, the associated map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is k0 times continuously dierentiable at any point of W 1,∞ (R3 , R3 ) ∩ B0,1 and for any k ∈ J1, k0 K, its k-th order dierential is well dened by the following continuous map:
D•k pν,Ω : W 1,∞ (R3 , R3 ) ∩ B0,1 −→ Lkc W 1,∞ (R3 , R3 )k , R   θ0 7−→ (θ1 , ..., θk ) 7→ D0k pν,(I+θ0 )(Ω) θ1 ◦ (I + θ0 )−1 , ..., θk ◦ (I + θ0 )−1 , (40) where D0k pν,(I+θ0 )(Ω) is the k-th order shape derivative of pν at (I + θ0 )(Ω) given by (39), and where Lkc refers to the class of continuous k-linear maps.

Proof. First, from Theorems 2.62.8 and Corollaries 2.72.9, the result is true for k0 = 1 and k0 = 2. We use the notation of Proposition 2.5 and proceed as in the proof of Theorems 2.62.8. Again, we relate (13) with pν in order to use the results of Section 5.2 available for a volume integral. Let n > 2 and k0 > 3 be two integers, ν ∈ J0, nK, and M3 contain all the measurable subsets of R3 . We assume that the wave function Ψ given by (4) satises Assumption 2.1. In particular, the functional pν : Ω ∈ M3 7→ pν (Ω) ∈ [0, 1] of Denition 2.2 is well dened. Moreover, we assume that the map Ψ(σ1 ,...,σn ) dened by (6) belongs to W k0 ,2 ((R3 )n , C) for any (σ1 , . . . , σn ) ∈ {− 21 , 12 }n . Note that W k0 ,2 -regularity is required on Ψ to get the k0 -th order shape dierentiability. Then, let Ω ∈ M3 and B30,1 denote the open unit ball of C 0,1 (R3 , R3 ) centred at the origin. From Corollary 2.9, the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I3 +θ)(Ω)] ∈ [0, 1] is well dened and twice dierentiable on B30,1 ∩ W 1,∞ (R3 , R3 ). We now show that it is k0 times dierentiable at e ∈ M3n 7→ p˜ν (Ω) e ∈ R is the origin. In the proof of Proposition 2.5, we have established that the map p˜ν : Ω well dened by (12), where M3n contains the measurable subsets of (R3 )n . Applying Theorem 5.40 to p˜ν , ˜ Ω)] e ∈ R is well dened and k0 times dierentiable at the map p˜ν,Ωe : θ˜ ∈ W 1,∞ ((R3 )n , (R3 )n ) 7→ p˜ν [(I3n + θ)( the origin and for any k ∈ J1, k0 K, its k-th order dierential at the origin is given by the following continuous symmetric k-linear form dened for any (θ˜1 , . . . , θ˜k ) ∈ W 1,∞ ((R3 )n , (R3 )n ) × . . . × W 1,∞ ((R3 )n , (R3 )n ) by: ! n 3 k X X X X X 1 n k ˜ ˜ D0 p˜ν,Ω e (θ1 , . . . , θk ) := c0 ν n i ,...,i =1 m ,...,m =1 l=0 Il ⊆J1,kK 1 1 k k (σ1 ,...,σn )∈{− 1 ,1 2 2} card Il =l     Z ∂ k−l |Ψ(σ1 ,...,σn ) |2 h i Y   
Q (x1 , . . . , xn )  θ˜j (x1 , . . . , xn )  3(ij −1)+mj  ∂ xi e Ω

   

j

j∈J1,kK j ∈I / l 

X p:Il →Il p bijective

mj

j∈J1,kK j ∈I / l

   ∂ θ˜j Y
  −1  s p Il ◦ p ◦ p Il     ∂ xip(j) j∈Il

mp(j)

  (x1 , . . . , xn ) 3(ij −1)+mj

   dx1 . . . dxn . 

(41) Then, we want to relate the k-th order shape derivative of p˜ν with the one of pν . Let Ω ∈ M3 . We thus have Ων × (R3 \Ω)n−ν ∈ M3n . Considering the continuous linear map f given in (14), for which any of its higher-than-one-order dierential is zero, we deduce from (15)(16) and Lemma 5.36 that the map pν,Ω = p˜ν,Ων ×(R3 \Ω)n−ν ◦ f is k0 times dierentiable at the origin and for any k ∈ J1, k0 K, its k-th order dierential at the origin is given by the following continuous symmetric k-linear form dened for any

19

(θ1 , . . . , θk ) ∈ W 1,∞ (R3 , R3 ) × . . . × W 1,∞ (R3 , R3 ) by: D0k pν,Ω (θ1 , . . . , θk )

=


D0k p˜ν,Ων ×(R3 \Ω)n−ν ◦ f (θ1 , . . . , θk )

=

Dfk(0) p˜ν,Ων ×(R3 \Ω)n−ν [D0 f (θ1 ), . . . , D0 f (θk )]

=

D0k p˜ν,Ων ×(R3 \Ω)n−ν [f (θ1 ), . . . , f (θk )] .

Observing that [f (θj )(x1 , . . . , xn )]3(ij −1)+mj = [θj (xij )]mj , we deduce that [∂(xi )mp(j) f (θj )]3(ij −1)+mj p(j) is equal to zero if ip(j) 6= ij otherwise it is equal to [∂mp(j) θj (xij )]mj . Introducing the notation Iip(j) ij , which is equal to one if and only if ip(j) = ij , we deduce from this foregoing observations inserted in (41) e ∈ M3n , that the that (39) holds true. Considering the second part of Theorem 5.40, we get for any Ω 1,∞ 3 n 3 n ˜ ˜ e map p˜ν,Ωe : θ ∈ W ((R ) , (R ) ) 7→ p˜ν [(I3n + θ)(Ω)] ∈ R is well dened and k0 times continuously dierentiable at any point of W 1,∞ ((R3 )n , (R3 )n ) ∩ B3n 0,1 . Moreover, for any k ∈ J1, k0 K, its k -th order dierential is well dened by the following continuous map: 

k  1,∞ D•k p˜ν,Ω (R3 )n , (R3 )n ∩ B3n Lkc W 1,∞ (R3 )n , (R3 )n , R e : W 0,1 −→ ˜ ˜ θ˜0 7−→ Dθ2˜0 p˜ν,Ω e : (θ1 , . . . , θk ) 7→ h i 2 ˜ ˜ −1 , ..., θ˜k ◦ (I3n + θ˜0 )−1 , D0 p˜ν,(I3n +θ˜0 )(Ω e ) θ1 ◦ (I3n + θ0 ) (42) e given by (41). Then, where D0k p˜ν,(I3n +θ˜0 )(Ωe ) is the k-th order shape derivative of p˜ν at (I3n + θ˜0 )(Ω) we want to relate the k-th order dierential of p˜ν,Ωe with the one of pν,Ω . Let Ω ∈ M3 . Hence, we have Ων × (R3 \Ω)n−ν ∈ M3n . Considering the continuous linear map f given in (14), for which any of its higher-than-one-order dierential is zero, we deduce from (15)(16) and Lemma 5.36 that the map pν,Ω = p˜ν,Ων ×(R3 \Ω)n−ν ◦ f is k0 times continuously dierentiable at any point θ0 ∈ W 1,∞ (R3 , R3 ) ∩ B30,1 and we have for any k ∈ J1, k0 K and any (θ1 , . . . , θk ) ∈ W 1,∞ (R3 , R3 ) × . . . × W 1,∞ (R3 , R3 ):
Dθk0 pν,Ω (θ1 , . . . θk ) = Dθk0 p˜ν,Ων ×(R3 \Ω)n−ν ◦ f (θ1 , . . . , θk ) = Dfk(θ0 ) p˜ν,Ων ×(R3 \Ω)n−ν [Dθ0 f (θ1 ) , . . . , Dθ0 f (θk )] = Df2 (θ0 ) p˜ν,Ων ×(R3 \Ω)n−ν [f (θ1 ) , . . . , f (θk )]  = D0k p˜ν,[I3n +f (θ0 )](Ων ×(R3 \Ω)n−ν ) f (θ1 ) ◦ (I3n + f (θ0 ))−1 ,

 . . . , f (θ2 ) ◦ (I3n + f (θ0 ))−1 ,

where we have used the fact that Dθ0 f = f and (42) to obtain the last equality. From Proposition 5.31, we have [I3n + f (θ0 )](Ων × (R3 \Ω)n−ν ) = [(I3 + θ0 )(Ω)]ν × [R3 \(I3 + θ0 )(Ω)]n−ν . We can also check that f (θi ) ◦ [I3n + f (θ0 )]−1 = f [θ ◦ (I3 + θ0 )−1 ] for any i ∈ J1, kK so we deduce that: 
Dθk0 pν,Ω (θ1 , . . . , θk ) = D0k p˜ν,[(I +θ )(Ω)]ν × R3 \(I +θ )(Ω) n−ν f θ1 ◦ (I3 + θ0 )−1 , [ ] 3 0 3 0
 . . . , f θk ◦ (I3 + θ0 )−1 =

Dfk(0) p˜ν,[(I

=

 D0k p˜ν,[(I

[R3 \(I3 +θ0 )(Ω)]n−ν

ν 3 +θ0 )(Ω)] ×

  D0 f θ1 ◦ (I3 + θ0 )−1 ,  
. . . , D0 f θk ◦ (I3 + θ0 )−1 

3 +θ0

)(Ω)]ν ×

[

R3 \(I

]

3 +θ0 )(Ω)

n−ν

◦f



θ1 ◦ (I3 + θ0 )−1 , . . . , θk ◦ (I3 + θ0 )−1

=



  D02 pν,(I3 +θ0 )(Ω) θ1 ◦ (I3 + θ0 )−1 , . . . , θk ◦ (I3 + θ0 )−1 .

We thus have proved that the k-th order dierential of pν,Ω is well dened by the continuous map (40) for any k ∈ J1, k0 K, concluding the proof of Theorem 2.10.

20

3 Shape dierentiability for multideterminant wave functions 3.1 On the expression of the probability as a sum of Slater determinants Let m > 1 be a xed integer. Henceforth, we assume that the wave function Ψ given in (4) is a sum of m Slater determinants of (spin-)orbitals. This means there exists m non-zero constants (c1 , . . . , cm ) ∈ (C∗ )m ±1/2 ±1/2 ±1/2 and also 2nm well-dened maps called orbitals φ±1/2 , . . . , φn1 , . . . , φ1m , . . . , φnm : R3 → C that are 11 measurable, square integrable, and such that for any i ∈ J1, nK and any (xi , σi ) ∈ R3 × {− 21 , 12 }, we have:  Ψ

x1 σ1





xn σn

,...,



m m X X cr cr √ det [Φσr 1 (x1 ) , ..., Φσr n (xn )] = √ = n! n! r=1 r=1

where we have introduced the vectorial maps:  ∀r ∈ J1, mK, ∀σ ∈ − 12 , 12 , Φσr :

R3

−→

φσ1r1 (x1 ) · · · φσ1rn (xn ) .. .. , (43) . . σ1 σn φnr (x1 ) · · · φnr (xn )

Cn

 φσ1r (x)   .. Φσr (x) :=  . . σ φnr (x) 

x

7−→

Remark 3.1. Observe that relation (5) is satised for the specic form (43) of the wave function Ψ. Indeed, the determinant is a skew-symmetric n-linear form (see e.g. Lemma 5.5 and Corollary 5.7). Moreover, each orbital belongs to L2 (R3 , C) so each determinant in (43) is measurable and square integrable. Hence, the wave function Ψ given by (43) is measurable and square integrable (see e.g. relation (46) for details). We now introduce some more notation to express the probability pν of Denition 2.2 in a simpler way. 2 Given any measurable subset Ω of R3 and any (r, s) ∈ J1, mK2 , we consider the (n×n)-matrix S rs (Ω) ∈ Cn : Z X φσir (x) φσjs (x)dx, ∀(i, j) ∈ J1, nK2 , [S rs (Ω)]ij := (44) Ω 1 , σ∈{− 1 2 2} which is a well-dened and nite quantity because of the L2 -assumption made on the orbitals. Moreover, for any xed subset Iν ⊆ J1, nK of ν pairwise distinct indices i.e. such that card Iν = ν , we set the 2 (n × n)-matrix SIrsν (Ω) ∈ Cn as: ( rs [S if j ∈ Iν 2 rs  rs (Ω)]3 ij
 (45) ∀(i, j) ∈ J1, nK , [SIν (Ω)]ij := R \Ω ij otherwise. S We are now in position to state a rst result, which is the multi-determinant version of [8, Appendix].

Proposition 3.2.

±1/2 Let m > 1, n > 2, ν ∈ J0, nK, (c1 , . . . , cm ) ∈ (C∗ )m , and φ±1/2 , . . . , φnm ∈ L2 (R3 , C). 11 We consider the wave function Ψ given by (43) and we assume that the constant c0 dened by (7) is positive. Then, the wave function Ψ satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Moreover, for any measurable subset Ω of R3 , we have:

pν (Ω) =

m X cr cs rs pν (Ω) c0 r,s=1

where

X

prs ν (Ω) :=

det SIrsν (Ω) ,

Iν ⊆J1,nK cardIν =ν

and where the matrix SIrsν (Ω) is dened by (45).

Corollary 3.3.

Using the assumptions and notation of Proposition 3.2, we have for any (r, s) ∈ J1, mK2 : ∀t ∈ C,

n 
 X ν det tS rs (Ω) + S rs R3 \Ω = prs ν (Ω) t , ν=0

where the matrix S (Ω) is dened by (44). In particular, the following relation holds true: rs

∀t ∈ C,

m n X X 
 cr cs det tS rs (Ω) + S rs R3 \Ω = pν (Ω) tν . c 0 ν=0 r,s=1

21

±1/2 Proof of Proposition 3.2. Let m > 1, n > 2, ν ∈ J0, nK, (c1 , ..., cm ) ∈ (C∗ )m and φ±1/2 ...φnm ∈ L2 (R3 , C). 11 We consider the wave function Ψ given by (43) and we rst check that the normalizing constant c0 given by (7) is a well-dened and nite quantity. Let (x1 , ..., xn ) ∈ R3 × ... × R3 and (σ1 , ..., σn ) ∈ {− 21 , 12 }n . Combining (43) with Denition 5.8, we have successively:     2         x1 xn x1 xn x1 xn Ψ ,..., = Ψ ,..., Ψ ,..., σ1 σn σ1 σn σ1 σn

=

m X cr √ det [Φσr 1 (x1 ) , ..., Φσr n (xn )] n! r=1 m X cr cs = n! r,s=1

X

s(p)

n Y

!

m X cs √ det [Φσs 1 (x1 ) , ..., Φσs n (xn )] n! s=1

! i φσp(i)r

(xi )

i=1

p∈Sn

X q∈Sn

s(q)

n Y

! σj φq(j)s

(xj )

,

j=1

and using the fact that s(q) = s(q) ∈ {−1, 1}, we obtain: !     2 m n X Y cr cs X x1 xn σi σi Ψ ,..., = s(p)s(q) φp(i)r (xi ) φq(i)s (xi ) . σ1 σn n! r,s=1

!

(46)

i=1

p,q∈Sn

Since we assumed φσir ∈ L2 (R3 , C) for any (i, r, σ) ∈ J1, nK × J1, mK × {− 21 , 12 }, we get φσir φσjs ∈ L1 (R3 , C) for any (i, j, r, s, σ) ∈ J1, nK2 × J1, mK2 × {− 12 , 12 } so the product of n such terms belongs in L1 ((R3 )n , C). In particular, we get from (46) that the map Ψ(σ1 ,...,σn ) given by (6) is measurable and square integrable for any (σ1 , . . . , σn ) ∈ {− 21 , 12 }n . Hence, the normalizing constant c0 given by (7) is a well-dened and nite quantity. In addition, we now assume c0 > 0. Therefore, considering Remark 3.1, Assumption 2.1 is satised for the wave function Ψ given by (43) and the shape functional pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Let Ω be any measurable subset of R3 . Combining (11) and (46), we can thus express the probability introduced in Denition 2.2 as:     2 Z X 1 n! x1 xn pν (Ω) := Ψ ,..., dx1 . . . dxn σ1 σn c0 ν!(n − ν)! Ων ×(R3 \Ω)n−ν 1 n (σ1 ,...,σn )∈{− 1 , 2 2} =

1 c0

X (σ1 ,...,σn )∈{− 1 ,1 2 2}

n

n! ν!(n − ν)!

"

Z Ων ×(R3 \Ω)

n−ν

m X cr cs X s(p)s(q) n! p,q∈S r,s=1 n !# n Y σi σi φp(i)r (xi ) φq(i)s (xi ) dx1 . . . dxn i=1

=

m X cr cs 1 X s(p)s(q) c0 r,s=1 ν!(n − ν)! p,q∈S n

X n

1,1 (σ1 ,...,σn )∈{− 2 2}

n Y

Z Ων ×(R3 \Ω)

n−ν

! i i φσp(i)r (xi ) φσq(i)s (xi ) dx1 . . . dxn .

i=1

Then, we can apply the Fubini's Theorem [17, Section 1.4] on the measurable set Ων × (R3 \Ω)n−ν and we have successively: ! m ν Z X X Y cr cs 1 X σi σi s(p)s(q) φp(i)r (xi ) φq(i)s (xi )dxi pν (Ω) = c0 r,s=1 ν!(n − ν)! p,q∈S Ω 1 n i=1 n , (σ1 ,...,σn )∈{− 1 2 2} ! Z n Y σj σj φp(j)r (xj ) φq(j)s (xj )dxj j=ν+1

=

R3 \Ω

   Z m ν X X Y X 1 cr cs    i i s(p)s(q)   φσp(i)r (xi )dxi  (xi ) φσq(i)s c0 r,s=1 ν!(n − ν)! p,q∈S Ω 1,1 i=1 n σi ∈{− 2 2}    Z n Y X σj σj    φp(j)r (xj ) φq(j)s (xj )dxj  .   R3 \Ω 1 1 j=ν+1 σj ∈{− 2 , 2 }

Considering the notation introduced in (44), the previous equality can be rewritten as: ! ! m ν n X Y Y
1 X cr cs rs rs 3 pν (Ω) = s(p)s(q) S (Ω)p(i)q(i) S R \Ω p(j)q(j) . c0 r,s=1 ν!(n − ν)! p,q∈S i=1 j=ν+1 n

22

The map q : J1, nK → J1, nK is bijective thus we can make a change of indices k = q(i) in the previous relation, from which we deduce:     m X Y  Y 
1 X cr cs pν (Ω) = s(p)s(q)  S rs (Ω)p◦q−1 (k)k   S rs R3 \Ω p◦q−1 (l)l   . c0 ν!(n − ν)! r,s=1

p,q∈Sn

k∈q(J1,νK)

l∈J1,nK l∈q(J1,νK) /

Then, we make a partition of the set Sn by subsets of maps xing ν elements. Using (45), we get: ! m Y rs X X X 1 X cr cs pν (Ω) = s(p)s(q) S (Ω)(p◦q−1 )(k)k c0 r,s=1 ν!(n − ν)! p∈S q∈Sn k∈Iν n Iν ⊆J1,nK cardIν =ν q(J1,νK)=Iν 



 Y 
 S rs R3 \Ω (p◦q−1 )(l)l    l∈J1,nK l∈I / ν

=

m X 1 X cr cs c0 r,s=1 ν!(n − ν)! p∈S

n

X

X

Iν ⊆J1,nK cardIν =ν

q∈Sn q(J1,νK)=Iν

s(p)s(q)

n Y

! SIrsν

(Ω)(p◦q−1 )(k)k

.

k=1

From Corollary 5.3, we have s(q)s(q −1 ) = s(q◦q −1 ) = s(Id) = 1. Since s(q) ∈ {−1, 1}, we get s(q) = s(q −1 ). We deduce that s(p)s(q) = s(p)s(q −1 ) = s(p ◦ q −1 ). Furthermore, for any xed p ∈ Sn , note that the map q ∈ Sn 7→ p ◦ q −1 ∈ Sn is bijective. We can thus make a change of indices q˜ := p ◦ q −1 and we obtain: ! n m Y X X X 1 X cr cs SIrsν (Ω)q˜(k)k . s(˜ q) pν (Ω) = c0 r,s=1 ν!(n − ν)! p∈S q˜∈S k=1 n

Iν ⊆J1,nK cardIν =ν

n

p(J1,νK)=˜ q (Iν )

Finally, we again partition the set Sn into subsets of maps xing ν elements, which gives:   pν (Ω) =

1 c0

m X r,s=1

cr cs ν!(n − ν)!

X Jν ⊆J1,nK cardJν =ν

X

  

p∈Sn p(J1,νK)=Jν

|

 X  1 

{z

=ν!(n−ν)!

X

Iν ⊆J1,nK cardIν =ν

s(˜ q)

q˜∈Sn q˜(Iν )=Jν

n Y k=1

 !

 SIrsν (Ω)q˜(k)k  .

}

Indeed, in the last sum of the previous relation, we have substituted p(J1, νK) by Jν . Hence, we can remove the dependence in p from the product and the last two sums. The cardinal of {p ∈ Sn , p(J1, νK) = Jν } is equal to the number of permutations of the set Jν times the number of permutations of the set J1, nK\Jν i.e. ν!(n − ν)! possibilities. Recognizing again a partition of Sn into subsets of maps xing ν elements and using Denition 5.8, we deduce successively that:   ! m n X X X X Y   cr cs  pν (Ω) = s(˜ q) SIrsν (Ω)q˜(k)k    c 0 r,s=1 q˜∈S k=1 Iν ⊆J1,nK cardIν =ν

=

m X cr cs c0 r,s=1

Jν ⊆J1,nK n cardJν =ν q˜(Iν )=Jν

" X

X

Iν ⊆J1,nK cardIν =ν

q˜∈Sn

s(˜ q)

 =

n Y

!# SIrsν

(Ω)q˜(k)k

k=1



m X X cr cs    c0 r,s=1

Iν ⊆J1,nK cardIν =ν

 det SIrsν (Ω) ,

which concludes the proof of Proposition 3.2. ±1/2 Proof of Corollary 3.3. Let m > 1, n > 2, ν ∈ J0, nK, (c1 , ..., cm ) ∈ (C∗ )m and φ±1/2 ...φnm ∈ L2 (R3 , C). 11 3 2 Consider any measurable subset Ω of R and (r, s) ∈ J1, mK . These assumptions ensures that the (n × n)2 matrices S rs (Ω) and S rs (R3 \Ω) given by (44) are well-dened elements of Cn . Using Denition 5.8, we

23

have successively for any t ∈ C: 
 det tS rs (Ω) + S rs R3 \Ω

=

X

s(p)

n h Y

p∈Sn

i=0

X

n X

tS rs (Ω)p(i)i + S rs R3 \Ω

i


p(i)i



 =

s(p)

ν=0

p∈Sn

! X

Y

Iν ⊆J1,nK cardIν =ν

i∈Iν

tS rs (Ω)p(i)i

  Y
 S rs R3 \Ω p(j)j    j∈J1,nK j ∈I / ν



 =

n X ν=0

! tν

X

X

Y

s(p)

Iν ⊆J1,nK p∈Sn cardIν=ν

S rs (Ω)p(i)i

i∈Iν

  Y
 S rs R3 \Ω p(j)j  .  j∈J1,nK j ∈I / ν

Then, we use the notation introduced in (45) and Denition 5.8 to obtain for any t ∈ C:   ! n n X X X Y rs  ν 
  det tS rs (Ω) + S rs R3 \Ω = s(p) SIν (Ω)p(i)i   t . ν=0

i=1

p∈Sn

Iν ⊆J1,nK cardIν=ν



 =

n X  X  

ν=0

Iν ⊆J1,nK cardIν=ν

n X  ν ν prs det SIrsν (Ω) ν (Ω) t , t = ν=0

the last equality coming from the notation used in Proposition 3.2. Finally, observing that the expression of pν established in Proposition 3.2 holds true, we get: ! m n n m X X X X 
 cr cs cr cs rs ∀t ∈ C, det tS rs (Ω) + S rs R3 \Ω = pν (Ω) tν = pν (Ω) tν , c c 0 0 r,s=1 ν=0 r,s=1 ν=0 which concludes the proof of Corollary 3.3.

3.2 On the rst-order shape derivative of the probability We refer to Section 5.2 for further details concerning the denition of rst-order shape dierentiability and the notation associated to it, especially Denition 5.28 and Theorem 5.29. First, we can apply the concept of rst-order shape derivative to the coecients of the matrices S rs (Ω) given by (44). Hence, we ±1/2 ±1/2 ±1/2 , . . . , φn1 , . . . , φ1m , . . . , φnm ∈ W 1,2 (R3 , C). set m > 1 and we consider 2nm well-dened orbitals φ±1/2 11 Compared to Section 3.1, note that higher regularity is required here for the orbitals in order to get the rst-order shape dierentiability. Let (r, s, i, j) ∈ J1, mK2 × J1, nK2 and M be the set of all measurable subsets of R3 . Then, the following functional is well-dened: rs Sij :

M

−→

C

Ω

7−→

rs Sij (Ω) := S rs (Ω)ij =

X 1,1 σ∈{− 2 2}

Z

φσir (x) φσjs (x)dx. Ω

P ±1/2 ±1/2 σ σ 1,1 Moreover, since we have φir , φjs ∈ W 1,2 (R3 , C), we deduce that (R3 , C). 1 , 1 } φir φjs ∈ W σ∈{− 2 2 rs Therefore, let Ω ∈ M and we apply Theorem 5.29 to the map Sij , which is thus shape dierentiable at Ω. More precisely, the following map is well dened around the origin and it is dierentiable at the origin:
rs Sij,Ω : W 1,∞ R3 , R3 −→ C Z X rs θ 7−→ Sij,Ω (θ) := S rs [(I + θ) (Ω)]ij = φσir (x) φσjs (x)dx. (I+θ)(Ω) 1 1 σ∈{− 2 , 2 }

In addition, its rst-order (shape) derivative at the origin is given by the following continuous linear form: Z X
rs ∀θ ∈ W 1,∞ (R3 , R3 ), D0 Sij,Ω (θ) = div φσir φσjs θ . (47) Ω 1,1 σ∈{− 2 2} If we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in Denition 5.46, then we get: Z X rs ∀θ ∈ W 1,∞ (R3 , R3 ), D0 Sij,Ω (θ) = φσir (x) φσjs (x) hθ (x) | nΩ (x)i dA (x) . (48) ∂Ω 1,1 σ∈{− 2 } 2

24

We also mention that the boundary values of the functions φσir φσjs in (48) have to be understood in the sense of trace. In particular, they are uniquely determined up to a set of zero A(• ∩ ∂Ω)-measure. In other rs words, the following map is well dened and it is the shape gradient of Sij at Ω: rs ∂Sij (Ω) : ∂Ω

∂Ω

−→

x

7−→

C X

φσir (x) φσjs (x).

σ∈{− 1 ,1 2 2}

Then, we want to express the shape derivative of pν so more notation are introduced for this purpose. Let n > 2 and ν ∈ J0, nK. For any subset Iν ⊆ J1, nK of ν pairwise disjoint elements i.e. for which card Iν = ν , we consider the following map: ε Iν :

J1, nK

i

−→

{−1, 1}

7−→

εIν (i) :=



1 −1

(49)

if i ∈ Iν otherwise.

We are now in position to state a rst result concerning the shape dierentiability of multi-determinant wave functions. We refer to Section 5.1.3 for some denitions of the comatrix operator denoted by Com.

Proposition 3.4.

±1/2 Let m > 1, n > 2, ν ∈ J0, nK, (c1 , . . . , cm ) ∈ (C∗ )m and φ±1/2 , . . . φnm ∈ W 1,2 (R3 , C). 11 We consider the wave function Ψ given by (43) and we assume that the constant c0 given by (7) is positive. Then, the wave function Ψ satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) of Denition 2.2 is well dened. Moreover, for any measurable subset Ω of R3 , the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] is well dened around the origin, dierentiable at the origin, and its (rst-order) derivative at the origin is given by the following continuous linear form dened for any θ ∈ W 1,∞ (R3 , R3 ) by:

D0 pν,Ω (θ) =

m n X cr cs X rs rs Aij (Ω) D0 Sij,Ω (θ) , c 0 r,s=1 i,j=1

(50)

rs (θ) and Ars where the quantities D0 Sij,Ω ij (Ω) are respectively dened by (47) and: X ∀(i, j, r, s) ∈ J1, nK2 × J1, mK2 , Ars εIν (j) [Com SIrsν (Ω)]ij , ij (Ω) :=

(51)

Iν ⊆J1,nK card Iν =ν

and where εIν and SIrsν (Ω) are respectively dened by the map (49) and (45). In other words, the functional pν : Ω 7→ pν (Ω) of Denition 2.2 is shape dierentiable for any measurable subset Ω of R3 and its rst-order shape derivative at Ω is given for any θ ∈ W 1,∞ (R3 , R3 ) by:   Z m n X X cr cs X  X  σ σ
 D0 pν,Ω (θ) = εIν (j) [Com SIrsν (Ω)]ij    div φir φjs θ (x) dx. (52) c 0 Ω 1,1 r,s=1 i,j=1 Iν ⊆J1,nK σ∈{− 2 2} card I =ν ν

If in addition, we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in rs Denition 5.46, then the same formula (50) still holds true but with D0 Sij,Ω (θ) now given by (48) (and rs 1,∞ 3 3 Aij (Ω) still by (51)), which gives for any θ ∈ W (R , R ):   Z m n X X X X   σ cr cs σ  D0 pν,Ω (θ) = εIν (j) [Com SIrsν (Ω)]ij    φir (x) φjs (x)θn (x) dA (x) . c0 i,j=1 ∂Ω 1 , 1 r,s=1 I ⊆J1,nK ν σ∈{− 2 2} card I =ν ν

where the integration is done with respect to the two-dimensional Hausdor measure referred to as A(•), where θn := hθ | nΩ i is the normal component of θ, with nΩ (x) denoting the unit vector normal to the boundary ∂Ω at the point x pointing outwards Ω. In this case, the functional pν : Ω 7→ pν (Ω) has a well-dened shape gradient, which is unique up to a set of zero A(• ∩ ∂Ω)-measure, and given by: m n rs X ∂Sij ∂pν cr cs X rs (Ω) = Aij (Ω) (Ω) , ∂Ω c0 i,j=1 ∂Ω r,s=1

or equivalently  ∂pν (Ω) : x ∈ ∂Ω 7→ ∂Ω

X

m X

σ∈{− 1 , 1 r,s=1 2 2}

cr cs c0

n X i,j=1

  

 X Iν ⊆J1,nK card Iν =ν

 σ σ εIν (j) [Com SIrsν (Ω)]ij   φir (x) φjs (x),

where the boundary values of φσir φσjs have to be understood in the sense of trace.

25

(53)

±1/2 Proof. Let m > 1, n > 2, ν ∈ J0, nK, (c1 , . . . , cm ) ∈ (C∗ )m and φ±1/2 , . . . φnm ∈ W 1,2 (R3 , C). We consider 11 the wave function Ψ given by (43) and we assume that the constant c0 given by (7) is positive. First, from Proposition 3.2, the wave function Ψ satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Then, let Ω be any measurable subset of R3 . On the one hand, we get from Theorem 2.6 that the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] is well dened around the origin and dierentiable at the origin. On the other hand, one can notice the matrices SIrsν (Ω) given by (45) are also shape dierentiable at Ω. Indeed, the shape dierentiability of SIrsν (Ω) is equivalent to the one of each of its coecients. Similar arguments than the one used in the beginning of Section 3.2 for the matrices S rs (Ω) gives for any Iν ⊆ J1, nK of ν pairwise distinct elements:
  rs (θ) , (54) ∀(r, s) ∈ J1, mK2 , ∀(i, j) ∈ J1, nK2 , ∀θ ∈ W 1,∞ R3 , R3 , D0 SIrsν ,Ω (θ) ij = εIν (j) D0 Sij,Ω

where εIν is dened by (49). From Proposition 5.11, the determinant map is dierentiable at any matrix so Lemma 5.36 ensures that the map Ω 7→ det ◦ SIrsν (Ω) is shape dierentiable at Ω. In other words, from the foregoing, we can correctly shape dierentiate at Ω the relation established in Proposition 3.2. Using the linearity of the dierential, Lemma 5.36 and (54), we deduce that:
∀θ ∈ W 1,∞ R3 , R3 , D0 pν,Ω (θ)

=

=

=

=

m X cr cs c0 r,s=1

m X cr cs c0 r,s=1

m X cr cs c0 r,s=1

X

  D0 det ◦ SIrsν ,Ω (θ)

Iν ⊆J1,nK cardIν =ν

X

  DSIrs ,Ω (0) det D0 SIrsν ,Ω (θ) ν

Iν ⊆J1,nK cardIν =ν

X

n X 

Com SIrsν ,Ω (0)

  ij

D0 SIrsν ,Ω (θ)

 ij

Iν ⊆J1,nK i,j=1 cardIν =ν

m n X cr cs X c0 i,j=1 r,s=1

X

rs (θ) . [Com SIrsν (Ω)]ij εIν (j) D0 Sij,Ω

Iν ⊆J1,nK cardIν =ν

Hence, we have proved that (50) holds true. Inserting the explicit expression (47) in (50) and using the linearity of the integral, we deduce that (52) also holds true. Finally, if we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in Denition 5.46, then we can insert (48) in (50) to deduce with the linearity of the integral the expected expression (53) for the shape gradient of pν , concluding the proof of Proposition 3.4.

3.3 On the second-order shape derivative of the probability We refer to Section 5.2 for further details concerning the denition of second-order shape dierentiability and the notation associated to it, especially Denition 5.28 and Theorem 5.30. The method we follow is similar to the one used in Section 3.2. First, we apply the concept of second-order shape derivative to the coecients of the matrices S rs (Ω) given by (44). Hence, we set m > 1 and we consider 2nm well-dened ±1/2 ±1/2 ±1/2 orbitals φ±1/2 , . . . , φn1 , . . . , φ1m , . . . , φnm ∈ W 2,2 (R3 , C). Again, in order to get the second-order 11 shape dierentiability, we emphasize the fact that the regularity required here for the orbitals is higher than the one assumed in Sections 3.1-3.2. Let (r, s, i, j) ∈ J1, mK2 × J1, nK2 and M be the set of all measurable subsets of R3 . Then, as in Section 3.2, the following functional is well dened: rs Sij :

M

−→

C

Ω

7−→

rs Sij (Ω) := S rs (Ω)ij =

Z

X σ∈{

−1 ,1 2 2

}

φσir (x) φσjs (x)dx. Ω

P ±1/2 ±1/2 σ σ 2,1 Moreover, since we have φir , φjs ∈ W 2,2 (R3 , C), we deduce that (R3 , C). 1 , 1 } φir φjs ∈ W σ∈{− 2 2 rs Therefore, let Ω ∈ M and we apply Theorem 5.30 to the map Sij , which is thus twice shape dierentiable at Ω. More precisely, the following map is well dened and dierentiable around the origin:
rs : W 1,∞ R3 , R3 −→ C Sij,Ω Z X rs θ 7−→ Sij,Ω (θ) := S rs [(I + θ) (Ω)]ij = φσir (x) φσjs (x)dx. (I+θ)(Ω) 1 1 σ∈{− 2 , 2 }

Its rst-order dierential is given by the following map, which is well dened around the origin:


rs D• Sij,Ω : W 1,∞ R3 , R3 −→ Lc W 1,∞ R3 , R3 , C   rs rs θ0 7−→ Dθ0 Sij,Ω : θ 7→ D0 Sij,(I+θ θ ◦ (I + θ0 )−1 , 0 )(Ω)

26

(55)

rs rs where D0 Sij,(I+θ is the shape derivative of Sij at (I + θ0 )(Ω) dened by (47). By Proposition 5.31, 0 )(Ω) note that (I + θ0 ) is a bijective map for θ0 small enough and its inverse (I + θ0 )−1 is a well-dened Lipschitz continuous map. In particular, (I + θ0 )(Ω) is measurable for θ0 small enough. Moreover, the rs map (55) is dierentiable at the origin i.e. Sij,Ω is twice dierentiable at the origin and its second-order (shape) derivative at the
origin is given by
the following continuous symmetric bilinear form dened for ˜ ∈ W 1,∞ R3 , R3 × W 1,∞ R3 , R3 by: any (θ, θ)  Z D E Z D E X

2 rs ˜ + θdiv ˜ (θ) ˜ =  Hess φσir φσjs (θ) | θ˜ + ∇ φσir φσjs | θdiv(θ) D0 Sij,Ω (θ, θ) Ω Ω 1,1 σ∈{− 2 2}  (56) Z i h  ˜ − trace D• θD• θ˜  . φσir φσjs div (θ) div(θ) + Ω

If we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in Denition 5.46, then similar results hold true in a slightly more regular space. Hence, the following map is well dened and dierentiable around the origin:

rs Sij,Ω : W 1,∞ R3 , R3 ∩ C 1 R3 , R3 −→ C rs θ 7−→ Sij,Ω (θ) := S rs [(I + θ) (Ω)]ij . Its rst-order dierential is given by the following map, which is well dened around the origin:




rs −→ Lc W 1,∞ R3 , R3 ∩ C 1 R3 , R3 , C : W 1,∞ R3 , R3 ∩ C 1 R3 , R3 D• Sij,Ω  rs rs θ0 7−→ Dθ0 Sij,Ω : θ 7→ D0 Sij,(I+θ θ ◦ (I + θ0 )−1 , 0 )(Ω)

(57)

rs rs is the shape derivative of Sij at (I + θ0 )(Ω) now dened by (48). Moreover, the where D0 Sij,(I+θ 0 )(Ω) rs map (57) is dierentiable at the origin i.e. Sij is twice dierentiable at the origin and its second-order (shape) derivative at the origin is given
by the following continuous symmetric bilinear form dened for


˜ ∈ (W 1,∞ R3 , R3 ∩ C 1 R3 , R3 ) × (W 1,∞ R3 , R3 ∩ C 1 R3 , R3 ) by: any (θ, θ) Z  E D E X
D

rs ˜ = ˜ | nΩ dA D02 Sij,Ω (θ, θ) ∇ φσir φσjs | θ θ˜ | nΩ + φσir φσjs div(θ)θ˜ − D• θ[θ] ∂Ω 1,1 σ∈{− 2 2} (58) Z D E E D X σ σ
σ σ ˜ ˜ ˜ = ∇ φir φjs | θ hθ | nΩ i + φir φjs div(θ)θ − D• θ[θ] | nΩ dA. ∂Ω 1 σ∈{− 1 , } 2 2

Finally, if we now assume that Ω is an open bounded subset of R3 with a boundary of class C 1,1 as in Denition 5.47, then the at the origin
takes the following form for any

shape derivative (58)
second-order ˜ ∈ (W 1,∞ R3 , R3 ∩ C 1 R3 , R3 ) × (W 1,∞ R3 , R3 ∩ C 1 R3 , R3 ): (θ, θ)  Z 


 rs ˜ = HΩ φσir φσjs + ∇ φσir φσjs | nΩ θn θ˜n dA (θ, θ) D02 Sij,Ω ∂Ω   Z (59) h i D E D E − φσir φσjs IIΩ θ˜∂Ω , θ∂Ω + ∇∂Ω (θn ) | θ˜∂Ω + ∇∂Ω (θ˜n ) | θ∂Ω  dA, ∂Ω rs where we refer to Theorems 1.25.30 for the notation. In other words, in this case, the functional Sij has a well-dened shape Hessian, which is uniquely determined up to a set of zero A(• ∩ ∂Ω)-measure, and given by: rs ∂ 2 Sij (Ω) : ∂Ω −→ C 2 (60) ∂Ω


 x 7−→ HΩ (x) φσir (x) φσjs (x) + ∇ φσir φσjs (x) | nΩ (x) .

We also mention that the boundary values of the functions φσir φσjs and ∇(φσir φσjs ) in (58)(60) have to be understood in the sense of trace. We are now in position to state the result concerning the second-order shape dierentiability for multideterminant wave functions. This proposition is of little interest from the numerical viewpoint since the
implementation of the formulas we derive scales as m2 n4 nν . However, they will be fundamental in the particular case m = 1, and more precisely to obtain (2)(3). We refer to Sections 5.1.45.1.5 for some denitions and properties of the Hessian associated with the determinant map. This operator is denoted by Hess det and introduced in Denition 5.15.

27

Proposition 3.5.

±1/2 Let m > 1, n > 2, ν ∈ J0, nK, (c1 , . . . , cm ) ∈ (C∗ )m and φ±1/2 , . . . φnm ∈ W 2,2 (R3 , C). 11 We consider the wave function Ψ given by (43) and we assume that the constant c0 given by (7) is positive. Then, the wave function Ψ satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) of Denition 2.2 is well dened. Moreover, for any measurable subset Ω of R3 , the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] is well dened and dierentiable around the origin, twice dierentiable at the origin, and its second-order dierential at the origin is given by the following continuous symmetric bilinear form dened for any ˜ ∈ W 1,∞ (R3 , R3 ) × W 1,∞ (R3 , R3 ) by: (θ, θ)  

˜ = D02 pν,Ω (θ, θ)

m n X cr cs X  2 rs ˜ Ars  ij (Ω) D0 Sij,Ω (θ, θ) + c 0 r,s=1 i,j=1

X (k,l)∈J1,nK2 k6=i,l6=j

 rs rs rs ˜ , Bij,kl (Ω) D0 Sij,Ω (θ) D0 Skl,Ω (θ) 

(61) rs rs rs ˜ , Ars where the quantities D0 Sij,Ω (θ), D02 Sij,Ω (θ, θ) ij (Ω) and Bij,kl (Ω) are respectively dened by (47), (56), (51), and: X rs (62) ∀(i, j, k, l, r, s) ∈ J1, nK4 × J1, mK2 , Bij,kl (Ω) := εIν (j) εIν (l) [Hess det (SIrsν (Ω))]ij,kl , Iν ⊆J1,nK card Iν =ν

with εIν , SIrsν (Ω), and Hess det respectively dened by (49), (45), and Denition 5.15. In other words, the functional pν : Ω 7→ pν (Ω) of Denition 2.2 is twice shape dierentiable for any measurable subset Ω of R3 ˜ ∈ W 1,∞ (R3 , R3 ) × W 1,∞ (R3 , R3 ) by: and its second-order shape derivative at Ω is given for any (θ, θ)    Z m n E X X cr cs X  X  D σ σ rs ˜ = ˜   (Ω)] Hess(φ ε (j) [Com S φ )(θ) | θ D02 pν,Ω (θ, θ) I ir I ν js ν ij  c0 i,j=1  Ω Iν ⊆J1,nK σ∈{− 1 , 1 r,s=1 2 2} card Iν =ν i  E h  D
˜ + θdiv(θ) ˜ ˜ − trace D• θD• θ˜ + φσir φσjs div(θ)div(θ) + ∇ φσir φσjs | θdiv(θ) 

 Z Z

X

+ Ω

Ω

m X

σ∈{− 1 , 1 r,s=1 2 2}

cr cs c0

n X

X

i,j=1 (k,l)∈J1,nK2 k6=i,l6=j

  

X Iν ⊆J1,nK card Iν =ν

 εIν (j) εIν (l) [Hess det (SIrsν (Ω))]ij,kl  


˜ div φσir φσjs θ (x)div(φσkr φσls θ)(y)dxdy. (63) If in addition, we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in Denition 5.46, then the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) 7→ pν [(I + θ)(Ω)] is well dened and dierentiable around the origin, twice dierentiable at the origin, and its second-order derivative at the ˜ ∈ (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) × W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ), origin is also given by (50) for any (θ, θ) rs rs 2 rs (Ω), Ars but with D0 Sij,Ω and D0 Sij,Ω (θ) now given respectively by (58) and (48) (and Bij,kl ij (Ω) still respectively by (62) and (51)). In other words, the shape functional pν : Ω 7→ pν (Ω) is twice shape dierentiable at any open bounded subset Ω of R3 with Lipschitz boundary and its second-order shape derivative at Ω is given by the following continuous symmetric bilinear form dened for any pair of maps ˜ ∈ (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) × (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) by: (θ, θ)    Z m n X X X X  

 cr cs rs ˜ =   D02 pν,Ω (θ, θ) ε (j) [Com S (Ω)] ∇(φσir φσjs ) | θ θ˜n Iν Iν ij   c 0 ∂Ω 1 , 1 r,s=1 i,j=1 Iν ⊆J1,nK σ∈{− 2 2} card Iν =ν E D σ σ ˜ ˜ +φir φjs div(θ)θ − D• θ(θ) | nΩ dA  Z

Z

+ ∂Ω

∂Ω

X 1,1 σ∈{− 2 2}

m n X cr cs X c0 i,j=1 r,s=1

X

  

 X

 εIν (j) εIν (l) [Hess det (SIrsν (Ω))]ij,kl  

Iν ⊆J1,nK (k,l)∈J1,nK2 card Iν =ν k6=i,l6=j φσir (x) φσjs (x)φσkr

(y) φσls (y)θn (x) θ˜n (y) dA (x) dA (y) . (64) If in addition, we now assume that Ω is an open bounded subset of R3 with a boundary of class C 1,1 as in Denition 5.47, then the second-order shape derivative of pν at Ω takes the following form for any

28

˜ ∈ (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) × (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )): (θ, θ)   Z m n X X cr cs X  X  ˜ =  D02 pν,Ω (θ, θ) εIν (j) [Com SIrsν (Ω)]ij    c0 i,j=1 ∂Ω 1 , 1 r,s=1 Iν ⊆J1,nK σ∈{− 2 2} card I =ν ν  


 HΩ φσir φσjs + ∇ φσir φσjs | nΩ θn θ˜n dA 

 Z

Z

X

+ ∂Ω

∂Ω

m X

1 , 1 r,s=1 σ∈{− 2 2}

cr cs c0

n X

X

  

X

 εIν (j) εIν (l) [Hess det (SIrsν (Ω))]ij,kl  

i,j=1 (k,l)∈J1,nK2 Iν ⊆J1,nK card Iν =ν k6=i,l6=j σ φir (x) φσjs (x)φσkr

(y) φσls (y)θn (x) θ˜n (y) dA (x) dA (y) .

 Z

X

− ∂Ω



m X

1 , 1 r,s=1 σ∈{− 2 2}

cr cs c0

n X i,j=1

(65)

 X

  

Iν ⊆J1,nK card Iν =ν

 σ σ εIν (j) [Com SIrsν (Ω)]ij   φir φjs

  D E D E IIΩ θ˜∂Ω (x) , θ∂Ω (x) + ∇∂Ω [θn ] (x) | θ˜∂Ω (x) + ∇∂Ω [θ˜n ] (x) | θ∂Ω (x) dA (x) ,

where (•)∂Ω := (•) − (•)n nΩ is the tangential component of a vector eld, which means that in particular ∇∂Ω (•) := ∇(•) − h∇(•) | nΩ inΩ is the tangential component of the gradient operator, where IIΩ (•, •) := −hD∂Ω nΩ (•) | (•)i is the second fundamental form associated with the C 1,1 -surface ∂Ω, which is a symmetric bilinear form on the tangent space, with D∂Ω (•) = D(•) − D(•)nΩ [nΩ ]T denoting the tangential component of the dierential operator on vector elds, where HΩ := div∂Ω (nΩ ) is the scalar mean curvature associated with the C 1,1 -surface ∂Ω, with div∂Ω (•) := div(•) − hD(•)nΩ | nΩ i denoting the tangential component of the divergence operator. Finally, we emphasize the structure (65) obtained for the second-order shape derivative. • The last term of (65) can be interpreted as a continuous linear form l1 as follows: observing from ˜ , where (74) that we can write D0 pν,Ω (θ) = l1 (θn ), then the third term in (3) has the form l1 [Z(θ, θ)] ˜ Z is a vector eld which is is equal to zero if θ and θ are normal to the boundary ∂Ω. In this case, the last term of (3) is thus equal to zero. Moreover, if Ω is a critical shape for pν i.e. if D0 pν,Ω ≡ 0, then, l1 ≡ 0 and this term is also equal to zero. • The middle term of (65) is a continuous symmetric bilinear form, which takes the form of a kernel R KΩ (x, y)θn (x)θ˜n (y)dA(x)dA(y), where we have set: ∂Ω×∂Ω KΩ : (x, y) ∈ ∂Ω × ∂Ω 7−→

X 1,1 σ∈{− 2 2}

m n X cr cs X X c0 i,j=1 r,s=1

X

k∈J1,nK l∈J1,nK k6=i l6=j

   

 X

Iν ⊆J1,nK card Iν =ν

 εIν (j) εIν (l) [Hess det (SIrsν (Ω))]ij,kl   φσir (x) φσjs (x)φσkr (y) φσls (y). (66)

• The rst term of (65) is a continuous symmetric bilinear form, which can be interpreted as the shape R 2 Hessian part of the second-order shape derivative. Indeed, it takes the form ∂Ω ∂∂Ωp2ν (Ω)θn θ˜n dA, where the integrand can be considered as the shape Hessian of pν :   ∂ 2 pν (Ω) : x ∈ ∂Ω 7−→ ∂Ω2

X σ∈{− 1 ,1 2 2}

m n X cr cs X   c0 i,j=1  r,s=1

X

 εIν (j) [Com SIrsν (Ω)]ij  

Iν ⊆J1,nK card Iν =ν 




 HΩ φσir φσjs + ∇ φσir φσjs | nΩ .

(67)

To conclude, we recall that in (64)(65)(66)(67), the boundary values of ∇(φσi φσj ) ∈ W 1,1 (R3 , C) have to be understood in the sense of trace.

29

φσi φσj

∈ W

2,1

(R , C) and 3

±1/2 Proof. Let m > 1, n > 2, ν ∈ J0, nK, (c1 , . . . , cm ) ∈ (C∗ )m and φ±1/2 , . . . φnm ∈ W 2,2 (R3 , C). We consider 11 the wave function Ψ given by (43) and we assume that the constant c0 given by (7) is positive. First, from Proposition 3.2, the wave function Ψ satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Then, let Ω be any measurable subset of R3 . On the one hand, we get from Proposition 2.8 that the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] is well dened and dierentiable around the origin. It is also twice dierentiable at the origin. On the other hand, one can notice the matrices SIrsν (Ω) given by (45) are also twice shape dierentiable at Ω. Indeed, the second-order shape dierentiability of SIrsν (Ω) is equivalent to the one of each of its coecients. Similar arguments than the one used in the beginning of Section 3.3 for the matrices S rs (Ω) gives for any Iν ⊆ J1, nK of ν pairwise distinct elements: h i
rs ˜ ∈ W 1,∞ R3 , R3 2 , ˜ ∀(r, s) ∈ J1, mK2 , ∀(i, j) ∈ J1, nK2 , ∀(θ, θ) (θ) , D02 SIrsν ,Ω (θ, θ) = εIν (j) D02 Sij,Ω ij

(68) where εIν is dened by (49). From Corollary 5.17, the determinant map is twice dierentiable at any matrix so Proposition 5.37 ensures that the map Ω 7→ det ◦ SIrsν (Ω) is shape dierentiable at Ω. In other words, from the foregoing, we can correctly twice shape dierentiate at Ω the relation established in Proposition 3.2. Using the
linearity of the dierential, Proposition 5.37 and (54)(68), we deduce that for
˜ ∈ W 1,∞ R3 , R3 × W 1,∞ R3 , R3 we have successively: any ∀(θ, θ) ˜ D02 pν,Ω (θ, θ)

=

=

m X cr cs c0 r,s=1

m X cr cs c0 r,s=1

X

  ˜ D02 det ◦ SIrsν ,Ω (θ, θ)

Iν ⊆J1,nK cardIν =ν

h i ˜ DSIrs ,Ω (0) det D02 SIrsν ,Ω (θ, θ)

X

ν

Iν ⊆J1,nK cardIν =ν

h i ˜ + DS2 Irs ,Ω (0) det D0 SIrsν ,Ω (θ), D0 SIrsν ,Ω (θ) ν

=

m X cr cs c0 r,s=1

X

n X 

Com SIrsν ,Ω (0)

 h ij

Iν ⊆J1,nK i,j=1 cardIν =ν

X

+

i ˜ D02 SIrsν ,Ω (θ, θ)

ij

i     h ˜ Hess det SIrsν ,Ω (0) ij,kl D0 SIrsν ,Ω (θ) ij D0 SIrsν ,Ω (θ)

kl

(k,l)∈J1,nK2 k6=i,l6=j

=

m X cr cs c0 r,s=1

X

n X 

Com SIrsν ,Ω (0)

Iν ⊆J1,nK i,j=1 cardIν =ν X

+

 ij

rs ˜ εIν (j) D02 Sij,Ω (θ, θ)

  rs rs ˜ Hess det SIrsν ,Ω (0) ij,kl εIν (j) εIν (l) D0 Sij,Ω (θ)D0 Skl,Ω (θ)

(k,l)∈J1,nK2 k6=i,l6=j

Hence, we have proved that (61) holds true. Inserting the explicit expression (47) and (56) in (61) and using the linearity of the integral, we deduce that (63) also holds true. If we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in Denition 5.46, then the same arguments holds true with the space W 1,∞ (R3 , R3 ) replaced by W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 ). We can thus twice shape dierentiate at Ω the relation established in Proposition 3.2 in this new space. Inserting (48) and (58) in (61), we deduce from the linearity of the integral the relation (64). Finally, if we now assume that Ω is an open bounded subset of R3 with a boundary of class C 1,1 as in Denition 5.47, then we can insert (48) and (59) in (61) to deduce with the linearity of the integral the expected expression (65) for the second-order shape derivative of pν , concluding the proof of Proposition 3.4.

30

4 Shape dierentiability for one-determinant wave functions 4.1 On the expression of the probability with a single Slater determinant Here, we assume that m = 1 and cm = 1 in (43) i.e. the wave function Ψ in (4) is a single Slater determinant. ±1/2 Simplifying a bit the notation of Section 3, there exists 2n orbitals φ±1/2 , . . . , φn : R3 → C, that are 1 measurable and square integrable, and such that for any i ∈ J1, nK and any (xi , σi ) ∈ R3 × {− 12 , 21 }:  Ψ



x1 σ1

 ,...,

xn σn



φσ1 1 (x1 ) · · · φσ1 n (xn ) 1 1 .. .. = √ det [Φσ1 (x1 ) , . . . , Φσn (xn )] = √ , . n! n! σ1 . σn φn (x1 ) · · · φn (xn )

where we have introduced the vectorial maps:  ∀σ ∈ − 21 , 12 , Φσ : R3

−→

(69)

Cn

 φσ1 (x)   .. Φσ (x) :=  . . σ φn (x) 

x

7−→

In the single determinant case, for any measurable subset Ω of R3 , we can consider the so-called overlap 2 (n × n)-matrix S(Ω) ∈ Cn dened as: Z X φσi (x) φσj (x)dx, (70) ∀(i, j) ∈ J1, nK2 , S (Ω)ij := Ω 1,1 σ∈{− 2 } 2 which is a well-dened and nite quantity because of the L2 -assumptions made on the orbitals. Similarly, 2 dropping the upper indices, we can dene SIν (Ω) ∈ Cn as in (45) for any xed subset Iν ⊆ J1, nK of ν pairwise distinct indices i.e. such that card Iν = ν : ( S (Ω)ij
if j ∈ Iν 2 (71) ∀(i, j) ∈ J1, nK , SIν (Ω)ij := S R3 \Ω ij otherwise. We now state the single determinant version of Proposition 3.2 and Corollary 3.3 that were already proved in [8, Appendix]. Note that the authors of [8] assumes that c0 = 1, which is not the case here.

Proposition 4.1 (2004 [8, Appendix]).

Let n > 2 be any integer and ν ∈ J0, nK. We consider 2n well-dened maps φ1±1/2 , . . . , φn±1/2 ∈ L2 (R3 , C) and the wave function Ψ given by (69). We assume that the constant c0 given by (7) is positive. Then, the wave function Ψ satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) of Denition 2.2 is well dened. Moreover, for any measurable subset Ω of R3 , we have: pν (Ω) =

1 c0

X

det SIν (Ω)

and

∀t ∈ C,

Iν ⊆J1,nK cardIν =ν

n 
 X 1 det tS (Ω) + S R3 \Ω = pν (Ω) tν , c0 ν=0

where the matrix S(Ω) and SIν (Ω) are respectively dened by (70) and (71). Moreover, the authors of [8, Section 2.1] obtained a more convenient formula for the computation of pν , which is recursive and has a lower algorithmic complexities in O(n2 ) operations. However, it only works in the case of single-determinant wave functions because the arguments are strongly based on the fact that the overlap matrix S(Ω) is Hermitian and on the following standard result from linear algebra.

Lemma 4.2.

2

Let m be a positive integer. We consider an Hermitian (m × m)-matrix A ∈ Cm i.e. satisfying A = [A]T , where [•]T and (•) respectively refers to the transpose and conjugate operators. Then, 2 A can be diagonalized by a unitary (m × m)-matrix U ∈ Cm , i.e. such that U [U ]T = [U ]T U = Im , and T the resulting diagonal matrix [U ] AU has only real diagonal coecients. Proof. We proceed by induction on m. First, it is true for m = 1. Then, let m > 2 and assume that 2 the result holds for m − 1. Let A ∈ Cm be an Hermitian (m × m)-matrix. Consider an eigenvalue λ and a corresponding normalized eigenvector x. Then, Cn = Cx ⊕ x⊥ is an orthogonal decomposition which is A-invariant. Therefore, we can apply the induction hypothesis on the well-dened restriction map e ∈ C(m−1)2 , which u ∈ x⊥ 7→ Au ∈ x⊥ , which is represented by an hermitian (m − 1) × (m − 1)-matrix A is diagonalizable by the induction hypothesis. Finally, we have:
λ − λ = λ − λ |x|2 = hx | λxi − hλx | xi = hx | Axi − hAx | xi = 0, from which we deduce that all the eigenvalues of A are real. We conclude that any Hermitian matrix is diagonalizable by a unitary matrix and it has only real eigenvalues.

31

2

Hence, we can apply Lemma 4.2 to the overlap (n × n)-matrix S(Ω) ∈ Cn , which is clearly Hermitian 2 according to (70). We deduce that S(Ω) can be diagonalized by a unitary (n × n)-matrix U (Ω) ∈ Cn , T and the resulting diagonal(n × n)-matrix D(Ω) := [U (Ω)] S(Ω)U (Ω) has a real k-th diagonal coecient λk (Ω) for any k ∈ J1, nK. In other words, for any k ∈ J1, nK, the k-th column of U (Ω) is the normalized eigenvector associated to λk (Ω) i.e. we have: (72)

∀k ∈ J1, nK, S(Ω)U•k (Ω) = λk (Ω)U•k (Ω) and |U•k (Ω)| = 1.

Moreover, the family of n vectors (U•1 (Ω), . . . , U•n (Ω)) forms an orthonormal basis of C . n

Finally, the recursive formulas of [8, Section 2.1] are also strongly based on the fact that S(R3 ) is equal to the identity (n × n)-matrix I i.e. the (spin-integrated) orbitals forms an orthonormal basis of L2 (R3 , C). In other words, we now make the following hypothesis.

Assumption 4.3.

We assume that the orbitals φσi ∈ L2 (R3 , C) for any (i, σ) ∈ J1, nK × {− 21 , 12 } and that they satisfy the following relations:  Z X 1 if i = j ∀(i, j) ∈ J1, nK, φσi (x) φσj (x)dx = 0 otherwise. 3 R 1,1 σ∈{− 2 2} In particular, the overlap matrix (70) is well dened and the relation S(R3 ) = I holds true. We are now in position to state the result of [8, Section 2.1], which is a consequence of Proposition 4.1 under Assumption 4.3.

Corollary 4.4 (2004 [8, Section 2.1]). Let n > 2 be an integer and ν ∈ J0, nK. Consider 2n well-dened ±1/2 : R3 → C satisfying Assumption 4.3. Then, the wave function Ψ given by (69) , . . . , φn maps φ±1/2 1 satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) of Denition 2.2 is well dened. Moreover, we have c0 = 1, where c0 is the normalizing given by (7), and for any measurable subset Ω of R3 , we have:   ! X Y   Y [1 − λj (Ω)] pν (Ω) = λi (Ω)  ,  i∈Iν

Iν ⊆J1,nK cardIν =ν

j∈J1,nK j ∈I / ν

2

where λ1 (Ω), . . . , λn (Ω) are the real eigenvalues of the Hermitian (n × n)-matrix S(Ω) ∈ Cn given by (70). ±1/2 : R3 7→ C satisfying , . . . , φn Proof. Let n > 2 and ν ∈ J0, nK. We consider 2n well-dened maps φ±1/2 1 Assumption 4.3 and the wave function Ψ dened by (69). First, we check that the normalizing constant c0 given by (7) is a well-dened and nite quantity. Let (x1 , ..., xn ) ∈ R3 ×...×R3 and (σ1 , ..., σn ) ∈ {− 21 , 12 }n . Combining (69) with Denition 5.8, we have successively:

 Ψ

x1 σ1



 ,...,

xn σn



2

 = Ψ 

=

x1 σ1



 ,...,

xn σn



 Ψ

1 √ det [Φσ1 (x1 ) , ..., Φσn (xn )] n!

=

1 n!

X p∈Sn

s(p)

n Y



x1 σ1



i=1

(xi )

xn σn



1 √ det [Φσ1 (x1 ) , ..., Φσn (xn )] n!

! i φσp(i)

 ,...,

X q∈Sn

s(q)

n Y



! σj φq(j)

(xj ) ,

j=1

and using the fact that s(q) = s(q) ∈ {−1, 1}, we obtain:  Ψ

x1 σ1



 ,...,

xn σn



2

! n Y 1 X σi σi = s(p)s(q) φp(i) (xi ) φq(i) (xi ) . n! p,q∈S i=1

(73)

n

Since we assumed φσi ∈ L2 (R3 , C), we get φσi φσj ∈ L1 (R3 , C) for any (i, j, σ) ∈ J1, nK2 × {− 12 , 21 } so the product of n such terms belongs in L1 ((R3 )n , C). In particular, we get from (73) that the map Ψ(σ1 ,...,σn ) given by (6) is measurable and square integrable for any (σ1 , . . . , σn ) ∈ {− 12 , 12 }n . Hence, the normalizing constant c0 given by (7) is a well-dened and nite quantity. Let us now prove c0 = 1. From (7) and (73),

32

we have successively: c0

:=

(σ1 ,...σn )∈{− 1 ,1 2 2}

Z

(σ1 ,...σn )∈{− 1 ,1 2 2}

=

(R3 )n

n

X

=



Z

X

n

1 X s (p) s (q) n! p,q∈S n

Ψ



x1 σ1

 ,...,

xn σn



2

dx1 . . . dxn

n Y 1 X i i s (p) s (q) (xi ) φσq(i) (xi )dx1 . . . dxn φσp(i) (R3 )n n! p,q∈Sn i=1 n Y

Z

X 1,1 (σ1 ,...σn )∈{− 2 2}

n R3

( )

n

i i φσp(i) (xi ) φσq(i) (xi )dx1 . . . dxn .

i=1

Then, we can apply the Fubini's Theorem [17, Section 1.4] and we get:  n Z X Y 1 X i i s (p) s (q) (x)dx c0 = φσp(i) (x) φσq(i) n! p,q∈S R3 n i=1 n (σ1 ,...σn )∈{− 1 ,1 2 2}

=

1 n!

 n Y  s (p) s (q) 

X

i=1

p,q∈Sn

 Z

X

,1 σ∈{− 1 2 2}

R3

 φσp(i) (x) φσq(i) (x)dx .

Using the denition (70) of the overlap matrix in the case Ω = R3 , the previous relation can be rewritten in the following form: n X Y
S R3 p(i)q(i) . c0 = s (p) s (q) i=1

p,q∈Sn

Since we have assumed that S(R ) = I , the product in the above relation is dierent from zero if and only if p(i) = q(i) for any i ∈ J1, nK i.e. i p = q , from which we deduce that: 3

c0 =

n Y
card Sn 1 X s(p)2 S R3 p(i)p(i) = = 1. n! p∈S | {z } i=1 | n! {z } n =1

=1

Finally, we can apply Proposition 4.1 in the case c0 = 1. In particular, the wave function Ψ satises Assumption 2.1 and the shape functional pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. Moreover, for any measurable subset Ω of R3 and for any t ∈ C, we have: n X

pν (Ω) tν =

ν=0


 
 1 det tS (Ω) + S R3 \Ω = det tS (Ω) + S R3 − S (Ω) = det [tS (Ω) + I − S (Ω)] . c0 2

We now use the fact the overlap (n × n)-matrix S(Ω) ∈ Cn is well dened and hermitian according to (70). 2 Applying Lemma 4.2, we deduce that S(Ω) can be diagonalized by a unitary (n × n)-matrix U (Ω) ∈ Cn , T and the resulting diagonal(n × n)-matrix D(Ω) := [U (Ω)] S(Ω)U (Ω) has a real k-th diagonal coecient λk (Ω) for any k ∈ J1, nK. Therefore, we have using Corollary 5.22: n X

pν (Ω) tν

=

det [tS (Ω) + I − S (Ω)]

=

  det U (Ω) [tD (Ω) + I − D (Ω)] [U (Ω)]T

=

  det U (Ω) [U (Ω)]T det [tD (Ω) + I − D (Ω)] = det [tD (Ω) + I − D (Ω)]

ν=0

=

n Y

[tλi (Ω) + 1 − λi (Ω)] = 

=

ν=0

  

X

Y

[tλi (Ω)]

ν=0 Iν ⊆J1,nK i∈Iν card Iν =ν 

i=1

n X

n X

X

Y

Iν ⊆J1,nK i∈Iν card Iν =ν

λi (Ω)

Y j∈J1,nK j ∈I / ν

Y

[1 − λj (Ω)]

j∈J1,nK j ∈I / ν

 ν [1 − λj (Ω)] t .

By identifying the coecients of the two polynomials, we obtain the expected formula for pν , which concludes the proof of Corollary 4.4.

33

4.2 On the rst-order shape derivative of the probability Let us now state the equivalent of Proposition 3.4 in the case of a single-Slater-determinant wave function. In this particular situation, as for pν , a more convenient formula for the computation
of the shape derivative is available. It is recursive [8, Sections 2.1 (5)(6)] and scales as n5 instead of n3 nν operations. Such a formula has already been given in [8, Section 2.2 (16)], but the authors had to assume that the eigenvalues of the overlap matrix S(Ω) are shape dierentiable, which is more or less equivalent to require the simplicity of the eigenvalues associated with the Hermitian matrix S(Ω). In the present proof, we proceed in a completely dierent manner and we do not need this technical hypothesis, which can be restrictive in the applications. However, we recall that it only works here because the overlap matrix S(Ω) given in (70) is Hermitian thus from Lemma 4.2 it has n real eigenvalues, and also because we are going to assume S(R3 ) = I i.e. the set of considered (spin-integrated) orbitals forms an orthonormal basis of L2 (R3 , C). Therefore, the matrices S(Ω) and S(R3 \Ω) = I − S(Ω) commute and can be simultaneously diagonalized. Of course, these two properties does not hold any longer in the multi-determinant case. That is why the implementation of an ecient shape optimization algorithm is currently available only at the Hartree-Fock level i.e. for single-Slater-determinant wave functions. Some eorts have been made to overpass these diculties and an algorithm with a working program should be soon available for Slater-multi-determinant wave functions. The results will soon be published in a forthcoming article. Finally, Proposition 1.1 is a particular case of the following result.

Proposition 4.5 (2004 [8, Section 2.2]).

Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any measurable ±1/2 ∈ W 1,2 (R3 , C) satisfying Assumption 4.3. , . . . , φn subset of R3 . Consider 2n well-dened maps φ±1/2 1 Then, the wave function Ψ given by (69) satises Assumption 2.1 and the map pν : Ω 7−→ pν (Ω) of Denition 2.2 is well dened. We also have c0 = 1, where c0 is the normalizing constant dened by (7), and the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I + θ)(Ω)] ∈ [0, 1] is well dened around the origin. Moreover, it is dierentiable at the origin and its (rst-order) dierential at the origin is given by the following continuous linear form dened for any θ ∈ W 1,∞ (R3 , R3 ) by:    n X n  X Y Y  X   D0 pν,Ω (θ) = εIν (l) λk (Ω) [1 − λk (Ω)]   Ω  Iν ⊆J1,nK  k∈Iν k∈J1,nK σ∈{− 1 , 1 } i,j=1 l=1 Z

X

2 2

k6=l

card Iν =ν

k∈I / ν k6=l


Uil (Ω)Ujl (Ω) div φσi φσj θ (x) dx,

where εIν (l) is given by (49), and where (λk (Ω))16k6n are the real eigenvalues of the Hermitian matrix S(Ω) given by (70), and where U (Ω) is a unitary matrix such that [U (Ω)]T S(Ω)U (Ω) is a diagonal matrix whose k-th diagonal coecient is λk (Ω) for any k ∈ J1, nK. In other words, the functional pν : Ω 7→ pν (Ω) of Denition 2.2 is shape dierentiable for any measurable subset Ω of R3 . If in addition, we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in Denition 5.46, then the shape derivative of pν at Ω takes the following form for any θ ∈ W 1,∞ (R3 , R3 ):    n X n  X Y Y  X    Uil (Ω)Ujl (Ω) ε (l) [1 − λ (Ω)] λ (Ω) I k k ν   ∂Ω   k∈Iν Iν ⊆J1,nK k∈J1,nK σ∈{− 1 , 1 } i,j=1 l=1

Z D0 pν,Ω (θ) =

X

2 2

k6=l

card Iν =ν

k∈I / ν k6=l

φσi (x) φσj (x) hθ (x) | nΩ (x)i dA (x) , (74) where the integration is done with respect to the two-dimensional Hausdor measure referred to as A(•), and where nΩ (x) denotes the unit vector normal to the boundary ∂Ω at the point x pointing outwards Ω In other words, for any open bounded subset of R3 with Lipschitz boundary, the functional pν : Ω 7→ pν (Ω) of Denition 2.2 has a well-dened shape gradient, which is unique up to a set of zero A(• ∩ ∂Ω)-measure, and given by:   ∂pν (Ω) : x ∈ ∂Ω 7−→ ∂Ω

 n X n  X Y Y  X   εIν (l) λk (Ω) [1 − λk (Ω)]    Iν ⊆J1,nK  k∈Iν k∈J1,nK σ∈{− 1 , 1 i,j=1 l=1 2 2} k6=l card I =ν X

ν

k∈I / ν k6=l

Uil (Ω)Ujl (Ω) φσi (x) φσj (x),

where the boundary value of φσi φσj in (74)(75) have to be understood in the sense of trace.

34

(75)

Proof. Let n > 2 be an integer, ν ∈ J0, nK, and Ω be any measurable subset of R3 . We consider 2n well-dened maps φ1±1/2 , . . . , φn±1/2 ∈ W 1,2 (R3 , C) satisfying Assumption 4.3. Then, from Corollary 4.4, the wave function Ψ given by (69) satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. We also have c0 = 1, where c0 is the normalizing constant dened by (7). Moreover, we can apply Proposition 3.4 to the case m = 1. Dropping the upper indices r and s, we notice that the only dierence between the statement of Proposition 3.4 with m = 1 and Proposition 4.5 is the expression of Aij (Ω) := A11 ij (Ω) given in (51). Hence, if we able to prove the result stated in the next proposition, then Proposition 4.5 holds true, concluding the proof.

Proposition 4.6.

Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any measurable subset of R3 . We consider 2n well-dened maps φ1±1/2 , . . . , φ±1/2 ∈ L2 (R3 , C) satisfying Assumption 4.3. Then, we have for n any (i, j) ∈ J1, nK2 : X Aij (Ω) := εIν (j) [Com SIν (Ω)]ij Iν ⊆J1,nK card Iν =ν

 =



 n  Y Y X  X    Uil (Ω)Ujl (Ω) , (l) λ (Ω) ε [1 − λ (Ω)] Iν k k    k∈Iν l=1  Iν ⊆J1,nK k∈J1,nK k6=l

card Iν =ν

k∈I / ν k6=l

where εIν (l) and SIν (Ω) are respectively given by (49) and (71), and where (λk (Ω))16k6n are the real eigenvalues of the Hermitian matrix S(Ω) given by (70), and where U (Ω) is a unitary matrix such that [U (Ω)]T S(Ω)U (Ω) is a diagonal matrix whose k-th diagonal coecient is λk (Ω) for any k ∈ J1, nK.

Proof. Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any measurable subset of R3 . We consider 2n welldened maps φ1±1/2 , . . . , φn±1/2 ∈ L2 (R3 , C) satisfying Assumption 4.3. In particular, the overlap matrix S(Ω) given in (70) is well dened and it is Hermitian. By Lemma 4.2, there exists a unitary matrix 2 U (Ω) ∈ Cn such that [U (Ω)]T S(Ω)U (Ω) is a diagonal matrix whose k-th diagonal coecient λk (Ω) is real for any k ∈ J1, nK. Let Iν ⊆ J1, nK a set of ν pairwise distinct indices i.e. such that card Iν = ν . We 2 2 introduce the (n × n)-matrices SIν (Ω) ∈ Cn and UIν (Ω) ∈ Cn respectively dened by (71) and: ∀(i, j) ∈ J1, nK2 ,

[UIν (Ω)]ij := εIν (i)εIν (j) [U (Ω)]ij ,

(76)

where εIν is given by (49). One can check by direct calculation that UIν (Ω) is also a unitary matrix and that det UIν (Ω) = det U (Ω). Then, we compute the product of matrices SIν (Ω)UIν (Ω). Let (i, j) ∈ J1, nK2 . We have: [SIν (Ω)UIν (Ω)]ij

=

n X

[SIν (Ω)]ik [UIν (Ω)]kj

k=1



= ε Iν

 X X   (j)  εIν (k) [S(Ω)]ik [U (Ω)]kj + εIν (k)([S(R3 )]ik −[S(Ω)]ik )[U (Ω)]kj  | {z } {z } | {z } | k∈I ν

k∈I / ν

=1

=−1

= ε Iν

=Iik





  n  X X   Iik [U (Ω)]ij  , (j)  [S(Ω)]ik [U (Ω)]kj −   k∈I / ν  k=1 {z } | =[S(Ω)U (Ω)]ij

from which we deduce that: [SIν (Ω)UIν (Ω)]ij =

  εIν (j) λj (Ω)[U (Ω)]ij 

εIν (j) [λj (Ω) − 1] [U (Ω)]ij

if i ∈ Iν

(77) otherwise.

Therefore, we can compute the comatrix of SIν (Ω)UIν (Ω) with the relation (77) and Denition 5.10 in

35

mind. Let (i, j) ∈ J1, nK2 . It comes successively: Y X s(p) [SIν (Ω)UIν (Ω)]p(k)k [Com (SIν (Ω)UIν (Ω))]ij = p∈Sn p(j)=i

=

k∈J1,nK k6=j

X

Y 

s(p)

p∈Sn p(j)=i

εIν (k) λk (Ω)U (Ω)p(k)k



k∈J1,nK k6=j p(k)∈Iν

Y 

εIν (k) (λk (Ω) − 1)U (Ω)p(k)k



k∈J1,nK k6=j p(k)∈I / ν

=

X

Y

s(p)

p∈Sn p(j)=i

Y

εIν (k)

k∈J1,nK k6=j

{z

|

}

Y

λk (Ω)

k∈J1,nK k6=j p(k)∈Iν

Y

[λk (Ω) − 1]

k∈J1,nK k6=j p(k)∈I / ν

U (Ω)p(k)k .

k∈J1,nK k6=j

=εIν (j)(−1)n−ν

Next, we partition the set of permutation p ∈ Sn into subsets xing ν elements. We deduce that: X X Y Y (−1)n−ν s(p) [Com (SIν (Ω)UIν (Ω))]ij = εIν (j) λk (Ω) [λk (Ω) − 1] | {z } p∈S k∈J Q J ⊆J1,nK k∈J1,nK n

ν

card Jν =ν

n k=1

p(j)=i = p(Jν )=Iν

ν

εJν (k)

k6=j

k6=j k∈J / ν

Y

U (Ω)p(k)k

k∈J1,nK k6=j

=

εIν (j)

X

εJν (j)

Jν ⊆J1,nK card Jν =ν

Y k∈Jν k6=j

εJν (k) λk (Ω) | {z } =1

Y

εJν (k) [λk (Ω) − 1] | {z }

k∈J1,nK =−1 k6=j k∈J /X ν

Y

s(p)

p∈Sn p(j)=i p(Jν )=Iν

U (Ω)p(k)k ,

k∈J1,nK k6=j

and we obtain: [Com (SIν (Ω)UIν (Ω))]ij = εIν (j)

X

Y

εJν (j)

Jν ⊆J1,nK card Jν =ν

λk (Ω)

k∈Jν k6=j

Y

[1 − λk (Ω)]

k∈J1,nK k6=j k∈J / ν X

Y

s(p)

p∈Sn p(j)=i p(Jν )=Iν

U (Ω)p(k)k .

(78)

k∈J1,nK k6=j

We emphasize the fact that εIν and εJν are distinct notation. Note also that the only dependence in Iν in relation (78) comes from εIν and the last sum on p ∈ Sn . Finally, We use Corollaries 5.225.13 in order to get: i h  X X Aij (Ω) := εIν (j) [Com SIν (Ω)]ij = εIν (j) Com SIν (Ω)UIν (Ω)[UIν (Ω)]T Iν ⊆J1,nK card Iν =ν

=

X

εIν (j)

Iν ⊆J1,nK card Iν =ν

=

X Iν ⊆J1,nK card Iν =ν

ij

Iν ⊆J1,nK card Iν =ν n X

 h iT  [Com (SIν (Ω)UIν (Ω))]il Com UIν (Ω) lj

l=1

εIν (j)

n X

i h  . [Com (SIν (Ω)UIν (Ω))]il Com UIν (Ω) jl

l=1

From Corollary 5.27, we get [Com(UIν (Ω))]jl = det UIν (Ω)[UIν (Ω)]jl = det U (Ω)εIν (j)εIν (l)Ujl (Ω), which is inserted in the above relation to obtain: Aij (Ω) = det U (Ω)

n X l=1

Ujl (Ω)

X Iν ⊆J1,nK card Iν =ν

εIν (j)2 εIν (l) [Com (SIν (Ω)UIν (Ω))]il . | {z }

36

=1

(79)

Combining (78) and (79), we deduce successively that: Aij (Ω)

=

det U (Ω)

n X

X

Ujl (Ω)

Iν ⊆J1,nK card Iν =ν

l=1

X

εIν (l)2 | {z }

Jν ⊆J1,nK card Jν =ν

=1

Y

εJν (l)

λk (Ω)

k∈Jν k6=l

Y

[1 − λk (Ω)]

k∈J1,nK k6=l k∈J / ν X

Y

s(p)

p∈Sn p(l)=i p(Jν )=Iν

=

det U (Ω)

n X l=1

X

Ujl (Ω)

Y

εJν (j)

Jν ⊆J1,nK card Jν =ν

Y

λk (Ω)

k∈Jν k6=l

[1 − λk (Ω)]

k∈J1,nK k6=l k∈J / ν X

X

Iν ⊆J1,nK card Iν =ν

=

det U (Ω)

n X l=1

Ujl (Ω)]

X

Y

εJν (j)

Jν ⊆J1,nK card Jν =ν

λk (Ω)

k∈Jν k6=l

Y

U (Ω)p(k)k

k∈J1,nK k6=j

[1 − λk (Ω)]

|

s(p)

Y

U (Ω)p(k)k

k∈J1,nK k6=j

{z

}

=[Com(U )]il =det  U (Ω)Uil (Ω)



=1

Y

k∈J1,nK k6=l k∈J / ν

p∈Sn p(l)=i

 n X  Uil (Ω)Ujl (Ω)  |det U (Ω)|2  | {z } l=1 

s(p)

p∈Sn p(l)=i p(Jν )=Iν

X

=

U (Ω)p(k)k

k∈J1,nK k6=j

X Jν ⊆J1,nK card Jν =ν

εJν (l)

Y k∈Jν k6=l

λk (Ω)

  [1 − λk (Ω)] ,  k∈J1,nK Y

k6=l k∈J / ν

where in the two last equalities we used again Corollary 5.27, concluding the proof of Proposition 4.6

4.3 On the second-order shape derivative of the probability We refer to Section 5.2 for further details concerning the denition of second-order shape dierentiability and the notation associated to it, especially Denition 5.28 and Theorems 5.30. The method we follow is similar to the one used in Section 4.2. We now state the equivalent of Proposition 3.5 in the case of a single-Slater-determinant wave function. In this particular situation, as for pν and its rst-order shape derivative, a more convenient formula for the computation of
the shape derivative is obtained. It is recursive [8, Sections 2.1 (5)(6)] and scales as n8 instead of n5 nν operations. Such a formula is new and extends the results of [8, Section 2.2 (16)]. However, we recall that it only works here because the overlap matrix S(Ω) given in (70) is Hermitian so it has n real eigenvalues by Lemma 4.2, and also because we assume S(R3 ) = I i.e. the set of (spin-integrated) orbitals forms an orthonormal basis of L2 (R3 , C). Therefore, the matrices S(Ω) and S(R3 \Ω) = I − S(Ω) commute and can be simultaneously diagonalized. These two properties does not hold in the multi-determinant case. We refer to Sections 5.1.45.1.5 for some denitions and properties of the Hessian associated with the determinant map. This operator is denoted by Hess det and introduced in Denition 5.15. Note that Theorem 1.2 a particular of the following result.

Theorem 4.7.

Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any measurable subset of R3 . We consider 2n well-dened maps φ1±1/2 , . . . , φn±1/2 ∈ W 2,2 (R3 , C) satisfying Assumption 4.3. Then, the wave function Ψ given by (69) satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) of Denition 2.2 is well dened. We also have c0 = 1, where c0 is the normalizing constant (7), and pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) 7→ pν [(I +θ)(Ω)] ∈ [0, 1] is a well-dened and dierentiable map around the origin, which is also twice dierentiable at the origin. Its second-order derivative at the origin is given by the following continuous symmetric bilinear form dened

37

˜ ∈ W 1,∞ (R3 , R3 ) × W 1,∞ (R3 , R3 ) by: for any (θ, θ)  ˜ = D02 pν,Ω (θ, θ)

Z



 n X n  X Y Y  X    Uil (Ω)Ujl (Ω) (l) λ (Ω) ε [1 − λ (Ω)] Iν k k   Ω   1 1 i,j=1 l=1 k∈I k∈J1,nK I ⊆J1,nK ν ν σ∈{− 2 , 2 } k6=l card Iν =ν k∈I / ν k6 = l D E
Hess φσi φσj (x) [θ (x)] | θ˜ (x) D E
+ ∇ φσi φσj (x) | θ (x) div θ˜ (x) + θ˜ (x) div θ (x) h  i  dx +φσi (x) φσj (x) div θ (x) div θ˜ (x) − trace Dx θDx θ˜ X



 n Y Y X X X   X    (q) λ (Ω) (p) ε [1 − λ (Ω)] ε m I m I ν ν   Ω   1 1 m∈Iν m∈J1,nK Iν ⊆{1,...n} σ∈{− , } i,j=1 k∈J1,nK p=1 q∈J1,nK

Z Z + Ω



n X

X

2 2

q6=p

l∈J1,nK k6=i l6=j

m6=p,q

card Iν =ν

m∈I / ν m6=p,q

  Uip (Ω)Ukq (Ω) − Upk (Ω)Uiq (Ω) Ujp (Ω) Ulq (Ω) − Upl (Ω) Ujq (Ω)
˜ div φσi φσj θ (x)div(φσk φσl θ)(y)dxdy, (80) where εIν is given by (49), and where (λk (Ω))16k6n are the real eigenvalues of the Hermitian matrix S(Ω) given by (70), and where U (Ω) is a unitary matrix such that [U (Ω)]T S(Ω)U (Ω) is a diagonal matrix whose k-th diagonal coecient is λk (Ω) for any k ∈ J1, nK. In other words, the functional pν : Ω 7→ pν (Ω) of Denition 2.2 is twice shape dierentiable at any measurable subset Ω. If in addition, we now assume that Ω is an open bounded subset of R3 with Lipschitz boundary as in Denition 5.46, then the map pν,Ω : θ ∈ W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 ) 7→ pν [(I + θ)(Ω)] is well dened and dierentiable around the origin, twice dierentiable at the origin, and its second-order derivative at the origin is the following continuous ˜ ∈ (W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 ))×(W 1,∞ (R3 , R3 )∩C 1 (R3 , R3 )) symmetric bilinear form dened for any (θ, θ) by:   

˜ = D02 pν,Ω (θ, θ)

 n X n  X Y Y  X    Uil (Ω)Ujl (Ω) ε (l) λ (Ω) [1 − λ (Ω)] Iν k k   ∂Ω  Iν ⊆J1,nK  k∈Iν k∈J1,nK σ∈{− 1 , 1 } i,j=1 l=1

Z

X

2 2

k6=l

card Iν =ν

k∈I / ν k6=l



E
D ∇ φσi φσj (x) | θ (x) θ˜ (x) | nΩ (x) D E σ σ ˜ ˜ +φi (x) φj (x) div θ (x) θ (x) − Dx θ[θ (x)] | nΩ (x) dA





 Z Z n n X X X X X  Y Y   X  + εIν (p) εIν (q) λm (Ω) [1 − λm (Ω)]   ∂Ω ∂Ω Iν ⊆{1,...n}  m∈Iν m∈J1,nK σ∈{− 1 , 1 } i,j=1 k∈J1,nK p=1 q∈J1,nK 2 2

q6=p

l∈J1,nK k6=i l6=j



card Iν =ν

m6=p,q

m∈I / ν m6=p,q

  Uip (Ω)Ukq (Ω) − Upk (Ω)Uiq (Ω) Ujp (Ω) Ulq (Ω) − Upl (Ω) Ujq (Ω)

φσi (x) φσj (x)φσk (y) φσl (y) hθ (x) | nΩ (x)i hθ˜ (y) | nΩ (y)idA (x) dA (y) . (81) If in addition, we now assume that Ω is an open bounded subset of R3 with a boundary of class C 1,1 as in Denition 5.47, then the second-order shape derivative of pν at Ω takes the following form for any

38

˜ ∈ (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )) × (W 1,∞ (R3 , R3 ) ∩ C 1 (R3 , R3 )): (θ, θ)  Z

˜ = D02 pν,Ω (θ, θ)



 n X n  X Y Y  X    Uil (Ω)Ujl (Ω) (l) λ (Ω) ε [1 − λ (Ω)] Iν k k   ∂Ω   1 1 i,j=1 l=1 k∈I k∈J1,nK I ⊆J1,nK ν ν σ∈{− , } X

2 2

k6=l

card Iν =ν



k∈I / ν k6=l

H∂Ω (x)φσi (x)φσj (x) + ∇ φσi φj (x) | nΩ (x)


σ





θn (x) θ˜n (x) dA (x)

 Z

Z

+ ∂Ω



 n X  Y Y X X  X    ε (p) ε (q) λ (Ω) [1 − λ (Ω)] Iν Iν m m   ∂Ω   1 1 p=1 i,j=1 m∈I q∈J1,nK I ⊆{1,...n} m∈J1,nK k∈J1,nK ν ν σ∈{− , } X

2 2

n X

q6=p

l∈J1,nK k6=i l6=j

m6=p,q

card Iν =ν

m∈I / ν m6=p,q

   Uip (Ω)Ukq (Ω) − Upk (Ω)Uiq (Ω) Ujp (Ω) Ulq (Ω) − Upl (Ω) Ujq (Ω) φσi (x)φσj (x)φσk (y)φσl (y)θn (x) θ˜n (y) dA (x) dA (y) 

 Z −

 n X n  Y Y X   X  Uil (Ω)Ujl (Ω) φσi (x)φσ (x)  (l) ε λ (Ω) [1 − λ (Ω)] I k k ν j   ∂Ω   k∈Iν Iν ⊆J1,nK k∈J1,nK σ∈{− 1 , 1 } i,j=1 l=1 X

2 2

k6=l

card Iν =ν

k∈I / ν

k6=l    D E D E ˜ IIΩ θ∂Ω (x) , θ∂Ω (x) + ∇∂Ω [θn ] (x) | θ˜∂Ω (x) + ∇∂Ω [θ˜n ] (x) | θ∂Ω (x) dA (x) ,

(82) where (•)n := h(•) | nΩ i is the normal component of a vector eld, where (•)∂Ω := (•) − (•)n nΩ is the tangential component of a vector eld, which means that in particular ∇∂Ω (•) := ∇(•) − h∇(•) | nΩ inΩ refers to the tangential component of the gradient operator, where IIΩ (•, •) := −hD∂Ω nΩ (•) | (•)i is the second fundamental form associated with the C 1,1 -surface ∂Ω, which is a symmetric bilinear form on the tangent space, with D∂Ω (•) = D(•) − D(•)nΩ [nΩ ]T denoting the tangential component of the dierential operator on vector elds, and where HΩ := div∂Ω (nΩ ) is the scalar mean curvature associated with the C 1,1 surface ∂Ω, with div∂Ω (•) := div(•) − hD(•)nΩ | nΩ i denoting the tangential component of the divergence operator. Finally, we emphasize the structure (82) obtained for the second-order shape derivative. • The last term of (82) can be interpreted as a continuous linear form l1 as follows: observing from ˜ , where (74) that we can write D0 pν,Ω (θ) = l1 (θn ), then the third term in (3) has the form l1 [Z(θ, θ)] Z is a vector eld which is is equal to zero if θ and θ˜ are normal to the boundary ∂Ω. In this case, the last term of (3) is thus equal to zero. Moreover, if Ω is a critical shape for pν i.e. if D0 pν,Ω ≡ 0, then, l1 ≡ 0 and this term is also equal to zero. • The middle term of (82) is a continuous symmetric bilinear form, which takes the form of a kernel R KΩ (x, y)θn (x)θ˜n (y)dA(x)dA(y), where we have set: ∂Ω×∂Ω KΩ : (x, y) ∈ ∂Ω × ∂Ω 7−→

X

n X

X

n X X X

1 , 1 i,j=1 k∈J1,nK l∈J1,nK p=1 q∈J1,nK σ∈{− 2 2} k6=i l6=j q6=p 



 X  Y Y     ε (p) ε [1 − λ (Ω)] (q) λ (Ω) Iν Iν m m   Iν ⊆{1,...n}  m∈Iν m∈J1,nK card Iν =ν



m6=p,q

m∈I / ν

  m6=p,q Uip (Ω)Ukq (Ω) − Upk (Ω)Uiq (Ω) Ujp (Ω) Ulq (Ω) − Upl (Ω) Ujq (Ω) φσi (x)φσj (x)φσk (y)φσl (y)

(83) • The rst term of (82) is a continuous symmetric bilinear form, which can be interpreted as the shape R 2 Hessian part of the second-order shape derivative. Indeed, it takes the form ∂Ω ∂∂Ωp2ν (Ω)θn θ˜n dA,

39

where the integrand can be considered as the shape Hessian of pν :  n X n  X  ∂ 2 pν  (Ω) : x ∈ ∂Ω − 7 →  2 ∂Ω  i,j=1 l=1

X

εIν (l)

Iν ⊆J1,nK card Iν =ν

Y k∈Iν k6=l



λk (Ω)



  [1 − λk (Ω)]  Uil (Ω)Ujl (Ω)  k∈J1,nK Y

k∈I / ν k6=l




 H∂Ω (x)φσi (x)φσj (x) + ∇ φσi φσj (x) | nΩ (x) .

(84)

To conclude, we recall that in (81)(82)(83)(84), the boundary values of ∇(φσi φσj ) ∈ W 1,1 (R3 , C) have to be understood in the sense of trace.

φσi φσj

∈ W

2,1

(R , C) and 3

Proof. Let n > 2 be an integer, ν ∈ J0, nK, and Ω be any measurable subset of R3 . We consider 2n welldened maps φ1±1/2 , . . . , φn±1/2 ∈ W 2,2 (R3 , C) satisfying Assumption 4.3. From Corollary 4.4, the wave function Ψ given by (69) satises Assumption 2.1 and the map pν : Ω 7→ pν (Ω) introduced in Denition 2.2 is well dened. We also have c0 = 1, where c0 is the normalizing constant dened by (7). Moreover, we can apply Proposition 3.5 to the case m = 1. Dropping the upper indices r and s, we notice that the only dierence between the statement of Proposition 3.5 with m = 1 and Theorem 4.7 is the expressions of 11 Aij (Ω) := A11 ij (Ω) and Bij,kl (Ω) := Bij,kl (Ω) respectively given in (51)(62). Hence, with Proposition 4.6, if we able to prove the result stated in the next proposition, then Proposition 3.5 holds true, concluding the proof.

Proposition 4.8.

Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any measurable subset of R3 . We ∈ L2 (R3 , C) satisfying Assumption 4.3. Then, we have for consider 2n well-dened maps φ1±1/2 , . . . , φ±1/2 n 4 any (i, j, k, l) ∈ J1, nK : X Bij,kl (Ω) := εIν (j) εIν (l) [Hess det (SIrsν (Ω))]ij,kl Iν ⊆J1,nK card Iν =ν

 =



 n X X  Y Y  X    ε (p) ε (q) λ (Ω) [1 − λ (Ω)] I I m m ν ν    p=1 q∈J1,nK Iν ⊆{1,...n} m∈Iν m∈J1,nK q6=p

m6=p,q

card Iν =ν



m∈I / ν

 m6=p,q  Uip (Ω)Ukq (Ω) − Upk (Ω)Uiq (Ω) Ujp (Ω) Ulq (Ω) − Upl (Ω) Ujq (Ω) ,

where εIν , Hess det and SIν (Ω) are respectively given by (49), Denition 5.15 and (71), and where (λk (Ω))16k6n are the real eigenvalues of the Hermitian matrix S(Ω) given by (70), and where U (Ω) is a unitary matrix such that [U (Ω)]T S(Ω)U (Ω) is a diagonal matrix whose k-th diagonal coecient is λk (Ω) for any k ∈ J1, nK.

Proof. The method of the proof in very similar to the one used in the proof of Proposition 4.6. We refer to Sections 5.1.45.1.55.1.6 for the properties of the Hessian associated with the determinant map, denoted by Hess det. Let n > 2 be any integer, ν ∈ J0, nK, and Ω be any measurable subset of R3 . We consider ±1/2 ±1/2 2n well-dened maps φ1 , . . . , φn ∈ L2 (R3 , C) satisfying Assumption 4.3. In particular, the overlap matrix S(Ω) given in (70) is well dened and it is Hermitian. By Lemma 4.2, there exists a unitary matrix 2 U (Ω) ∈ Cn such that [U (Ω)]T S(Ω)U (Ω) is a diagonal matrix whose k-th diagonal coecient λk (Ω) is real for any k ∈ J1, nK. Let Iν ⊆ J1, nK a set of ν pairwise distinct indices i.e. such that card Iν = ν . We 2 2 introduce the (n × n)-matrices SIν (Ω) ∈ Cn and UIν (Ω) ∈ Cn respectively dened by (71) and (76). One can check by direct calculation that UIν (Ω) is also a unitary matrix and that det UIν (Ω) = det U (Ω). Then, we can compute the product of matrices SIν (Ω)UIν (Ω) and obtain (77) as in the proof of Proposition 4.6. We now compute Hess detSIν (Ω)UIν (Ω) with the relation (77) and Denition 5.15 in mind. Let n > 3

40

and (i, j) ∈ J1, nK2 . We consider (k, l) ∈ J1, nK2 such that k 6= i and l 6= j . It comes successively: X Y s(p) [SIν (Ω)UIν (Ω)]p(m)m [Hess det (SIν (Ω)UIν (Ω))]ij,kl = p∈Sn p(j)=i p(l)=k

=

X

m∈J1,nK m6=j,l

Y

s(p)

p∈Sn p(j)=i p(l)=k



εIν (m) λm (Ω)U (Ω)p(m)m



m∈J1,nK m6=j,l p(m)∈Iν

Y



εIν (m) (λm (Ω) − 1)U (Ω)p(m)m



m∈J1,nK m6=j,l p(m)∈I / ν

=

X

Y

s(p)

p∈Sn p(j)=i p(l)=k

Y

εIν (m)

m∈J1,nK m6=j,l

|

{z

λm (Ω)

m∈J1,nK m6=j,l p(m)∈Iν

}

=εIν (j)εIν (l)(−1)n−ν

Y

[λm (Ω) − 1]

m∈J1,nK m6=j,l p(m)∈I / ν

Y

U (Ω)p(m)m .

m∈J1,nK m6=j

Next, we partition the set of permutation p ∈ Sn into subsets xing ν elements. We deduce that: X X [Hess det (SIν (Ω)UIν (Ω))]ij,kl = εIν (j)εIν (l) (−1)n−ν s(p) | {z } p∈S Q J ⊆J1,nK n

ν

card Jν =ν

Y

λm (Ω)

m∈Jν m6=j,l

=

n m=1

Y

m∈J1,nK m6=j,l m∈J / ν

U (Ω)p(m)m

m∈J1,nK m6=j,l

Y m∈Jν m6=j,l

εJν (m) [λm (Ω) − 1] | {z } =−1

Y

[λm (Ω) − 1]

εJν (j)εJν (l)

Jν ⊆J1,nK card Jν =ν

Y

εJν (m)

m∈J1,nK m6=j,l m∈J / ν

X

εIν (j)εIν (l)

p(j)=i = p(l)=k p(Jν )=Iν

εJν (m) λm (Ω) | {z } =1

X

s(p)

p∈Sn p(j)=i p(l)=k p(Jν )=Iν

Y

U (Ω)p(m)m ,

m∈J1,nK m6=j,l

and we obtain: [Hess det (SIν (Ω)UIν (Ω))]ij,kl = εIν (j)εIν (l)

X Jν ⊆J1,nK card Jν =ν

εJν (j)εJν (l)

Y

λm (Ω)

m∈Jν m6=j,l

X p∈Sn p(j)=i p(l)=k p(Jν )=Iν

Y

[1 − λm (Ω)]

m∈J1,nK m6=j,l m∈J / ν

s(p)

Y

U (Ω)p(m)m .

m∈J1,nK m6=j

(85) We emphasize the fact that εIν and εJν are distinct notation. Note also that the only dependence in Iν in relation (85) comes from εIν and the last sum on p ∈ Sn . Finally, We use Corollaries 5.225.19 in order

41

to get: Bij,kl (Ω)

:=

X

εIν (j)εIν (l) [Hess det SIν (Ω)]ij,kl

Iν ⊆J1,nK card Iν =ν

=

h  i εIν (j)εIν (l) Hess det SIν (Ω)UIν (Ω)[UIν (Ω)]T

X Iν ⊆J1,nK card Iν =ν

=

X

εIν (j)εIν (l)

n X X

ij,kl

[Hess det (SIν (Ω)UIν (Ω))]ip,kq

p=1 q∈J1,nK q6=p

Iν ⊆J1,nK card Iν =ν

 Hess det

h

iT  UIν (Ω) pj,ql

=

X

εIν (j)εIν (l)εIν (l)

n X X

[Hess det (SIν (Ω)UIν (Ω))]ip,kq

p=1 q∈J1,nK q6=p

Iν ⊆J1,nK card Iν =ν

h  i Hess det UIν (Ω)

From Corollary 5.27, we get: h  i Hess det UIν (Ω)

jp,lq

. jp,lq

= det U (Ω)εIν (j)εIν (l)εIν (p)εIν (q)[Ujp (Ω)Ulq (Ω) − Ulp (Ω)Ujq (Ω)],

which is inserted in the penultimate relation to obtain: Bij,kl (Ω) = det U (Ω)

n X X

[Ujp (Ω)Ulq (Ω) − Ulp (Ω)Ujq (Ω)]

p=1 q∈J1,nK q6=p

X Iν ⊆J1,nK card Iν =ν

[εIν (j)εIν (l)]2 εIν (p)εIν (q) [Hess det (SIν (Ω)UIν (Ω))]ip,kq . | {z } =1

(86)

Combining (85) and (86), we deduce successively that: Bij,kl (Ω)

=

det U (Ω)

n X X p=1 q∈J1,nK q6=p

X

Iν ⊆J1,nK card Iν =ν

εJν (p)εJν (q)

det U (Ω)

Y

n X X

Y

λm (Ω)

m∈Jν m6=p,q

Jν ⊆J1,nK card Jν =ν

=

det U (Ω)

Y

[1 − λm (Ω)]

m∈Jν m6=p,q

λm (Ω)

X

Y

U (Ω)r(m)m

m∈J1,nK m6=p,q

εJν (p)εJν (q)

m∈J1,nK m6=p,q m∈J / ν

X

X

Iν ⊆J1,nK card Iν =ν

r∈Sn r(p)=i r(q)=k r(Jν )=Iν

X

[Ujp (Ω)Ulq (Ω) − Ulp (Ω)Ujq (Ω)]

p=1 q∈J1,nK q6=p

Y

s(r)

Jν ⊆J1,nK card Jν =ν

λm (Ω)

n X X

=1

r∈Sn r(p)=i r(q)=k r(Jν )=Iν

m∈J1,nK m6=p,q m∈J / ν

[Ujp (Ω)Ulq (Ω) − Ulp (Ω)Ujq (Ω)]

m∈Jν m6=p,q

=

[εIν (p)εIν (q)]2 {z } |

X

[1 − λm (Ω)]

p=1 q∈J1,nK q6=p

Y

X

[Ujp (Ω)Ulq (Ω) − Ulp (Ω)Ujq (Ω)]

Y m∈J1,nK m6=p,q m∈J / ν

s(r)

X

s(r)

r∈Sn r(p)=i r(q)=k

|

42

Y

U (Ω)r(m)m

m∈J1,nK m6=p,q

{z

=[Hess det U (Ω)]ip,kq

m∈J1,nK m6=p,q

εJν (p)εJν (q)

Jν ⊆J1,nK card Jν =ν

[1 − λm (Ω)]

Y

}

U (Ω)r(m)m

Finally, we use again Corollary 5.27 to have: [Hess det U (Ω)]ip,kq = det U (Ω)[Ujp (Ω)Ulq (Ω) − Ulp (Ω)Ujq (Ω)],

which is inserted in the penultimate relation to obtain the expected formula:  Bij,kl (Ω)

=



 n Y Y X X    X   (q) λ (Ω) [1 − λ (Ω)] (p)ε ε m m J J ν ν     p=1 q∈J1,nK m∈Jν m∈J1,nK Jν ⊆J1,nK q6=p

m6=p,q

card Jν =ν

m6=p,q m∈J / ν

|det U (Ω)|2 [Ujp (Ω)Ulq (Ω) − Ulp (Ω)Ujq (Ω)] [Ujp (Ω)Ulq (Ω) − Ulp (Ω)Ujq (Ω)], {z } | =1

concluding the proof of Proposition 4.8.

43

5 Annexes In this section, we present and derive all the technical material that is needed in the rest of the article. Roughly speaking, if a molecular system is chemically characterized by a wave function dened as a sum of Slater determinants, then the probability to nd a xed number of electrons in a given domain is a sum of determinants of matrices whose coecients are dened as volume integrals on this domain. Since we are here interested in establishing formulas for the rst- and second-order shape derivatives of this probability, we need to examine the dierentiability of the composition between the determinant map and matrices with shape-depending coecients. For this purpose, we thus have to study the dierentials of the determinant and the shape derivatives of a volume integral. Hence, this section is divided into two parts. On the one hand, we consider the rst- and second-order derivatives of the determinant, which involve the comatrix operator, its dierential, and their properties. On the other hand, we study the rst-, second-, and higher-order shape derivatives of a volume integral with precise regularity results depending on the integrand and the boundary of the admissible domain.

5.1 On the rst- and second-order derivatives of the determinant First, we introduce the signature of a permutation in order to properly dene the determinant of a matrix as a map. Then, we consider the comatrix operator, interpreted as the gradient of the determinant map, and we study the dierent properties of the Hessian associated with this determinant map. Finally, we give simpler expressions of all these operators in the case of invertible matrices and we show how to extend the statements of Section 5.1 from real- to complex-valued matrices. We conclude by analysing the particular case of the unitary matrices and their sub-determinant properties.

5.1.1 About the signature of a permutation Let n > 2 be an integer henceforth set. We set J1, nK := {1, . . . , n} and the set of all permutations is referred to as Sn i.e. the set of bijective maps from J1, nK into J1, nK. Endowed with the composition operator (p, q) 7→ p ◦ q , the set Sn is a non-commutative nite group of cardinal n! := n(n − 1) . . . 1. Let k ∈ J2, nK. A k-cycle is a permutation p ∈ Sn such that there exists k dierent indices {i1 , . . . , ik } satisfying p(ik ) = i1 , p(ij−1 ) = ij for any j ∈ J2, kK, and p(j) = j for any j ∈ / {i1 , . . . , ik }. Such a k-cycle is denoted by (i1 , . . . , ik ). With this notation, we have (i1 , . . . , ik ) = (ij , . . . , ik , i1 , . . . , ij−1 ) for any j ∈ J2, kK. Note also that pk = Id with k > 2 the smallest integer satisfying this equation. Two cycles (i1 , . . . , ik ) and (j1 , . . . , jl ) are said to be disjoint if {i1 , . . . , ik } ∩ {j1 , . . . , jl } = ∅. A 2-cycle is called a transposition.

Lemma 5.1.

First, two disjoint cycles commute. Then, the identity map can be considered as the unique 1-cycle. Moreover, any permutation which is dierent from the identity can be decomposed into a product of pairwise-disjoint cycles and this decomposition is unique up to the order of the cycles in the decomposition. Finally, any permutation can be decomposed into a (non-unique) product of transpositions. Proof. First, let p = (i1 , . . . , ik ) and q = (j1 , . . . , jl ) be two disjoint cycles of Sn . We consider any m ∈ J1, nK and we distinguish the following three dierent cases: (i) if m ∈ {i1 , . . . , ik }, then p(m) ∈ {i1 , . . . , ik }, and since p and q are disjoint, we have m ∈ / {j1 , . . . , jl } and p(m) ∈ / {j1 , . . . , jl }, from which we deduce p ◦ q(m) := p[q(m)] = p(m) = q[p(m)] = q ◦ p(m); (ii) if m ∈ {j1 , . . . , jl }, then it is similar to case (i) by exchanging the role of p and q ; (iii) if m ∈ / {i1 , . . . , ik , j1 , . . . , jl }, then p(m) = m = q(m) thus p ◦ q(m) = m = q ◦ p(m). Hence, two cycles commute if they are disjoint. Then, let p ∈ Sn . Before proving the decomposition for p, we need to establish the three following properties: (1) for any i ∈ J1, nK such that p(i) 6= i, there exists ki > 2 such that αi := (i, . . . , pki −1 (i)) is a ki -cycle whose support Ci := {i, . . . , pki −1 (i)} contains i; (2) conversely, any cycle whose support contains i is equal to αi ; (3) for any (i, j) ∈ J1, nK2 such that p(i) 6= i and p(j) 6= j , either Ci ∩ Cj = ∅ or Ci = Cj .

Let i ∈ J1, nK be such that p(i) 6= i. We introduce the subset Ci := {pk (i), k ∈ N} of the nite set J1, nK so there exists two non-negative integers k > l such that pk (i) = pl (i) i.e. pk−l (i) = i. We can thus consider the smallest positive integer ki such that pki (i) = i and it comes Ci = {i, . . . , pki −1 (i)}. Since p(i) 6= i, note that ki > 2 hence the map αi := (i, . . . , pki −1 (i)) is a ki -cycle. Property (1) is thus proved. Conversely, if i is contained in the support of a cycle (j1 , . . . , jl ), then either i = j1 which directly gives (j1 , . . . , jl ) = αi , or there exists l0 ∈ J2, lK such that i = jl0 = pl0 −1 (j1 ) and we also get: (j1 , . . . , jl ) = (jl0 , . . . , jl , j1 , . . . , jl0 −1 ) = (i, . . . , pl−1 (i)) = αi .

44

Hence, Property (2) is established. Now consider another integer j ∈ J1, nK such that p(j) 6= j . If there exists k ∈ Ci ∩ Cj , then k = pr (i) = ps (j) for some r ∈ J1, ki K and s ∈ J1, kj K. We distinguish three cases: (a) if r = s, then i = j thus of course Ci = Cj ; (b) if r > s, then pr−s (i) = j so j ∈ Ci and Ci = Cj using Property (2); (c) if r < s, then it is similar to case (b) by exchanging the role of i and j . Therefore, the properties (1)-(2)-(3) hold true, from which we can get the expected decomposition for p. Indeed, if p is the identity map, then it can be considered as the unique 1-cycle. If p 6= Id, then there exists i1 ∈ J1, nK such that p(i1 ) 6= i1 . Considering Property (1), either p = αi1 or p 6= αi1 . If p = αi1 , then such a decomposition is unique because Property (2) ensures that any other decomposition p = βj1 ◦ . . . ◦ βjl has to satisfy Ci1 = Cj1 = . . . = Cjl i.e. l = 1 by Property (3), from which we deduce p = αi1 = βj1 . Otherwise p 6= αi1 and there exists i2 ∈ J1, nK\Ci1 such that p(i2 ) 6= i2 . In particular, Ci2 ∩ Ci1 = ∅ using Property (3). Again, repeating the arguments, either p = αi2 ◦ αi1 or p 6= αi2 ◦ αi1 , etc. Since the set J1, nK is nite and the Cij are not empty and pairwise disjoint, the process has to end in at most b n2 c iterations, which gives the expected decomposition for p into a product of pairwise-disjoint cycles, and it is unique up to reordering. Finally, if (i1 , . . . , ik ) is a k-cycle, then one can check that: (i1 , . . . , ik ) = (i1 , ik ) ◦ . . . ◦ (i1 , i2 ) = (i1 , i2 ) ◦ . . . ◦ (ik−1 , ik ).

Hence, any cycle can be decomposed into a non-unique product of transpositions, and since any permutation can be decomposed into a (unique) product of pairwise-disjoint cycles, we deduce that any permutation can be decomposed into a non-unique product of transpositions, concluding the proof of Lemma 5.1.

Proposition 5.2.

Let p ∈ Sn \{Id}. If we write p as a product of transpositions in two dierent ways: p = (a1 , a ˜1 ) ◦ . . . ◦ (ak , a ˜k ) = (b1 , ˜b1 ) ◦ . . . ◦ (bl , ˜bl ),

then k and l have the same parity i.e. either k and l are both even, or both odd. Proof. Let p ∈ Sn \{Id}. First, note that from Lemma 5.1, p can be decomposed into a (non-unique) product of transpositions. Let us consider two of them p = (a1 , a ˜1 ) ◦ . . . ◦ (ak , a ˜k ) = (b1 , ˜b1 ) ◦ . . . ◦ (bl , ˜bl ). We get: Id = p−1 ◦ p = (ak , a ˜k ) ◦ . . . ◦ (a1 , a ˜1 ) ◦ (b1 , ˜b1 ) ◦ . . . ◦ (bl , ˜bl ). Hence, it is enough to prove that the identity map can only be written as a product of an even number of transpositions, in which case k + l is even i.e. k and l have the same parity. Therefore, we consider any decomposition of the identity map into a product of transpositions: Id = (c1 , c˜1 ) ◦ . . . ◦ (cm , c˜m ),

(87)

where m > 1 and ci 6= c˜i for any i ∈ J1, mK. Let us prove by induction that m is even. We cannot have m = 1 since a transposition is dierent from the identity map, but we could have m = 2. Let m > 3 and assume by induction that any decomposition of the identity map into a product of fewer than m transpositions can only involve an even number of them. Note also that one of the (ci , c˜i ), i ∈ J2, mK, has to move c1 , otherwise the overall product on the right-hand side of (87) cannot be equal to the identity map. We deduce that there exists i ∈ J2, mK such that ci = c1 , after interchanging the role of ci and c˜i if necessary. Moreover, from Lemma 5.1, we have for any dierent indices c, c˜, d, and d˜: ˜ ◦ (c, c˜) = (c, c˜) ◦ (d, d) ˜ (d, d)

and we have also

(d, c˜) ◦ (c, c˜) = (c, d) ◦ (˜ c, d),

which shows that any product of two transpositions in which the second factor moves c and the rst factor does not move c can be rewritten as a product of two transpositions in which the rst factor moves c and the second factor does not move c. Consequently, without changing the number of transpositions in the right-hand side of (87), we can push the position of the second most left transposition that moves c1 to the position right after (c1 , c˜1 ) i.e. we can assume c2 = c1 . On the one hand, if c˜1 = c˜2 , then the product (c1 , c˜1 ) ◦ (c2 , c˜2 ) is the identity map and we can remove it. This reduces (87) to a product of m − 2 transpositions, for which we know it is an even number by the induction hypothesis. Hence, m is also even. On the other hand, if c˜1 6= c˜2 , then (c1 , c˜1 ) ◦ (c1 , c˜2 ) = (c1 , c˜2 ) ◦ (˜ c1 , c˜2 ) so (87) can be rewritten as: Id = (c1 , c˜2 ) ◦ (˜ c1 , c˜2 ) ◦ (c3 , c˜3 ) ◦ . . . ◦ (cm , c˜m ),

where we have only modied the rst two factors in the right-hand side of the above decomposition. It still involves m transpositions but they are fewer of them in the product that moves c1 , since we used to have (c1 , c˜1 ) and (c1 , c˜2 ) in (87), and we now have (c1 , c˜2 ) and (˜ c1 , c˜2 ). Some transpositions other than (c1 , c˜2 ) in the new product must move c1 , so by the same arguments as before, either we are able to reduce the number of transpositions by two and we are done by induction, or we are able to rewrite the product with the same total number of transpositions but drop by one the number of them that move c1 . Finally, this rewriting process has to fall into the case where the rst two transpositions cancel out, since we cannot end up with the identity map written as a product of transpositions where only the rst one moves c1 . Hence, we have proved by induction that m is even, which concludes the proof of Proposition 5.2.

45

Corollary 5.3.

There exists a unique non-trivial morphism of group from Sn into {−1, 1}. This map is called the signature.

Proof. Let s(Id) := 1 and for any p ∈ Sn such that p 6= Id, we set s(p) := (−1)k , where k is the number of transpositions whose product is a decomposition for p. Such a decomposition exists by Lemma 5.1 and moreover, Proposition 5.2 ensures that the number s(p) does not depend on the chosen decomposition i.e. the map s : Sn → {−1, 1} is well dened. Furthermore, one can check that s is a morphism of group i.e s(p ◦ q) = s(p)s(q) for any (p, q) ∈ Sn × Sn . Let us now consider another morphism of group s˜ : Sn → {−1, 1}. First, we have s˜(Id) = s˜(Id ◦ Id) = s˜(Id)2 = 1 = s(Id). Then, we have two possibilities. On the one hand, if s˜(p) = 1 for any transposition p ∈ Sn , then from Lemma 5.1 we deduce s˜(p) = 1 for any p ∈ Sn i.e. s˜ ≡ 1 is trivial. On the other hand, if there exists (i, j) ∈ J1, nK2 with i 6= j such that the transposition (i, j) satises s˜[(i, j)] 6= 1 i.e. s˜[(i, j)] = −1, then either n = 2 and we have obviously s˜ = s, or n > 3 and we have:  2  s˜[(i, k)] = s˜ [(j, k) ◦ (i, j) ◦ (j, k)] = s˜[(j, k)] s˜[(i, j)] = −1 ∀(k, l) ∈ (J1, nK\{i, j}) × (J1, nK\{i, j}),  s˜[(l, j)] = s˜ [(i, l) ◦ (i, j) ◦ (i, l)] = s˜[(i, l)]2 s˜[(i, j)] = −1. We deduce that for any (k, l) ∈ (J1, nK\{i}) × (J1, nK\{i, j}) such that k 6= l, we have also: s˜[(k, l)] = s˜ [(l, j) ◦ (i, l) ◦ (i, j) ◦ (i, k) ◦ (i, l)] = s˜[(l, j)]˜ s[(i, l)]2 s˜[(i, j)]˜ s[(i, k)] = −1.

We get that for any transposition p ∈ Sn , s˜(p) = −1. Combined with Lemma 5.1 and the fact that s˜ is a morphism of group from Sn into {−1, 1}, we obtain: s˜(p) = (−1)k = s(p),

for any p ∈ Sn \{Id} that can be decomposed into a product of k transpositions. To conclude the proof of Corollary 5.3, for any morphism of group s˜ : Sn → {−1, 1}, either s˜ ≡ 1 i.e. s˜ is trivial or s˜ = s i.e. the non-trivial morphism s is unique.

Denition 5.4. Let n > 2 be an integer. The signature map, whose existence and uniqueness is guaranteed by Corollary 5.3, is denoted by s : Sn → {−1, 1}. It is well dened by s(Id) := 1 and: s(p) := (−1)k ,

∀p ∈ Sn \{Id},

where k is the number of transpositions whose product is a decomposition for p.

5.1.2 About the determinant of a matrix The space Rn is endowed with its usual Euclidean structure and (e1 , . . . , en ) refers to its canonical basis. A n-linear form on Rn is a well-dened map L : (u1 , . . . , un ) ∈ Rn × . . . × Rn 7→ L(u1 , . . . , un ) ∈ R which is linear with respect to any of its variables. A n-linear form is said to be alternating if L(u1 , . . . , un ) = 0 for any (u1 , . . . , un ) ∈ Rn × . . . × Rn such that there exists (i, j) ∈ J1, nK2 satisfying i 6= j and ui = uj .

Lemma 5.5.

Let p ∈ Sn and L be an alternating n-linear form on Rn . Then, we have:
∀ (u1 , . . . , un ) ∈ Rn × . . . × Rn , L up(1) , . . . , up(n) = s(p)L (u1 , . . . , un ) ,

where s : Sn → {−1, 1} is the signature map introduced in Denition 5.4. Proof. Let L be an alternating n-linear form on Rn and (u1 , . . . , un ) ∈ Rn × . . . × Rn . We consider two indices (i, j) ∈ J1, nK2 such that i < j . First, we have from the alternating property: L(. . . , ui + uj , . . . , ui + uj , . . .) = 0. | {z } | {z } i−th position

j−th position

Then, combining the linearity of the i-th and j -th variables with the fact that L(. . . , ui , . . . , ui , . . .) = 0 and L(. . . , uj , . . . , uj , . . .) = 0, we get L(. . . , uj , . . . , ui , . . .) = −L(u1 , . . . , un ). Finally, Lemma 5.5 clearly holds true for p = Id, and for any p ∈ Sn \{Id}, Lemma 5.1 ensures we can decompose p into a product of k transpositions. From the foregoing, we deduce recursively that:
∀ (u1 , . . . , un ) ∈ Rn × . . . × Rn , L up(1) , . . . , up(n) = (−1)k L (u1 , . . . , un ) = s(p)L (u1 , . . . , un ) , which concludes the proof of Lemma 5.5.

Proposition 5.6.

In the canonical basis of Rn , the representation of a non-zero alternating n-linear form is proportional to any other alternating n-linear form.

46

Proof. Let L be a non-zero alternating n-linear form on Rn and (u1 , . . . , un ) ∈ Rn × . . . × Rn . Considering the canonical basis (e1 , . . . , en ) of Rn and the n-linearity of L, we have successively: ! ! n n n n X X Y X 

L (ei1 , ..., ein ) . hun | ein i ein = uj | eij L (u1 , ..., un ) = L hu1 | ei1 i ei1 , . . . , i1 ,...,in =1

in =1

i1 =1

j=1

Note that if there exists (j, k) ∈ J1, nK2 such that j 6= k and ij = ik , then the alternating property of L gives that L(ei1 , . . . , ein ) = 0. Hence, the above sum can be reduced to the indices such that the map j ∈ J1, nK 7→ ij ∈ J1, nK is injective. Since such an injective map is bijective, we obtain: ! " !# n n X Y X Y




 L (u1 , ..., un ) = L ep(1) , ..., ep(n) = s(p) L (e1 , ..., en ) , uj | ep(j) uj | ep(j) p∈Sn

j=1

j=1

p∈Sn

where the last equality comes from Lemma 5.5. The constant L(e1 , . . . , en ) cannot be zero otherwise L e , we have: would be identically zero. We deduce that for any other alternating n-linear form L ∀ (u1 , . . . , un ) ∈ Rn × . . . × Rn ,

e e (u1 , . . . , un ) = L (e1 , . . . , en ) L (u1 , . . . , un ) , L L (e1 , . . . , en )

e are proportional, concluding the proof of Proposition 5.6. which means that L and L

Corollary 5.7.

There exists a unique alternating n-linear form on Rn which is equal to one when its arguments are the canonical basis of Rn . This map is called the determinant. Proof. Considering the canonical basis (e1 , . . . , en ) of Rn , we introduce the following map: L:

Rn × . . . × Rn

−→ 7−→

(u1 , . . . , un )

R L (u1 , . . . , un ) :=

X

s(p)

n Y

ui | ep(i)

! 

.

i=1

p∈Sn

One can check that L is a well-dened alternating n-linear form on Rn satisfying L(e1 , . . . , en ) = 1. Then, e is another alternating n-linear form on Rn such that L(e e 1 , . . . , en ) = 1, then from Proposition 5.6, we if L e e = CL. But in this case, L = L e since get that L and L are proportional i.e. there exists C > 0 such that L e 1 , . . . , en ) = 1. To conclude the proof of Corollary 5.7, there exists a unique C = C L(e1 , . . . , en ) = L(e alternating n-linear form on Rn which is equal to one when its arguments are the canonical basis of Rn .

Denition 5.8.

Let n > 2 be an integer. We consider the canonical basis (e1 , . . . , en ) of Rn . Then, the 2 identication map (u1 , . . . , un ) ∈ Rn × . . . × Rn 7−→ U = (Uij )16i,j6n := (huj | ei i)16i,j6n ∈ Rn is well dened and bijective. In particular, the determinant map, whose existence and uniqueness is guaranteed 2 by Corollary 5.7, can be considered as an operator acting on the space Rn of (n × n)-matrices with real coecients and it is well dened by the following map: det :

2

Rn

U

−→ 7−→

R det (U ) :=

X p∈Sn

s(p)

n Y

! Up(i)i

,

i=1

where s : Sn → {−1, 1} is the signature map introduced in Denition 5.4.

5.1.3 About the comatrix operator and the rst-order dierential of the determinant 2

The space Rn of (n×n)-matrices with real coecients is equipped with an addition (A+B)ijP:= Aij +Bij , an external multiplication (λA)ij := λAij , and its usual internal multiplication (AB)ij := nk=1 Aik Bkj . P 2 We introduce the trace operator denoted by trace : A ∈ Rn 7→ ni=1 Aii ∈ R. The identity matrix is 2 2 denoted by I . We also dened the transpose operator [•]T : A ∈ Rn 7→ [A]T := (Aji )16i,j6n ∈ Rn . Hence, n2 R is a non-commutative unitary algebra, which is endowed with an additional Euclidean structure: v uX n   X p u n n2 n2 T ∀(A, B) ∈ R × R , hA | Bi := trace [A] B = Aij Bij and kAk := hA | Ai = t |Aij |2 . i,j=1

i,j=1

We refer to Section 5.2.1 for notation and denitions associated with the dierentiability in Banach spaces. First, the following standard properties can be checked by direct simple calculations from the denitions: 2 2 trace(A) = trace([A]T ), trace(AB) = trace(BA), and [AB]T = [B]T [A]T for any (A, B) ∈ Rn × Rn . Then, we prove that the transpose operator is invariant under the action of the determinant. Finally, we dene the comatrix operator in order to express the dierential of the determinant in a simpler way.

47

Lemma 5.9.

2

We have det(A) = det([A]T ) for any A ∈ Rn . 2

Proof. Let A ∈ Rn . First, we get from Denition 5.8: n n   X Y X Y det [A]T = s(p) [A]Tp(i)i = s(p) Aip(i) . i=1

p∈Sn

i=1

p∈Sn

Then, we note that 1 = s(Id) = s(p ◦ p ) = s(p)s(p ), with s valued in {−1, 1}, hence s(p) = s(p−1 ). Making a change of indices j = p(i) in the above product, we obtain: n n   X X Y
Y det [A]T = s p−1 Ap−1 (j)j = s (q) Aq(j)j = det (A) , −1

−1

j=1

p∈Sn

j=1

q∈Sn

where we have rearranged the summation on p by the one on q := p−1 since the map p ∈ Sn 7→ p−1 ∈ Sn is bijective, concluding the proof of Lemma 5.9.

Denition 5.10. dened by:

2

2

Let n > 2 be an integer. The comatrix operator Com : A ∈ Rn → 7 Com(A) ∈ Rn is   2

∀A ∈ Rn , ∀(i, j) ∈ J1, nK2 ,

X

[Com (A)]ij :=

p∈Sn p(j)=i

 Y  s(p)  Ap(k)k   , k∈J1,nK k6=j

where s : Sn → {−1, 1} is the signature map introduced in Denition 5.4.

Proposition 5.11.

2

Let A ∈ Rn . Then, we have: 

 2

∀H ∈ Rn ,

det (A + H) = det (A) +

n X i,j=1



 Y   X   s(p)  Ap(k)k    Hij + kHk R(H),  p∈Sn p(j)=i

k∈J1,nK k6=j

where |R(H)| → 0 as kHk → 0. In other words, the determinant map is dierentiable at any point A ∈ Rn 2 and its dierential at A is given by the following continuous linear form on Rn : n   X 2 [Com (A)]ij Hij . ∀H ∈ Rn , DA det (H) = trace [Com A]T H =

2

i,j=1

In particular, the comatrix operator of Denition 5.10 can be interpreted as the gradient of the determinant 2 map det : Rn → R introduced in Denition 5.8. 2

2

Proof. Let (A, H) ∈ Rn × Rn . Then, we have successively: det (A + H)

=

X

n Y

s(p)

p∈Sn

j=1

X

n Y

X


Ap(j)j + Hp(j)j =

s(p)

p∈Sn

X

=

p∈Sn

s(p)

Ap(k)k +

X

j=1 p∈Sn

k=1



 = det (A) +

j=1 i=1

 X

X

s(p) 

Y





 Y  Hp(j)j   Ap(k)k    k∈J1,nK k∈I / l



 X  Y    s(p)  Ap(k)k     Hij p∈Sn p(j)=i

k∈J1,nK k6=j

n X

[Com (A)]ij Hij +

i,j=1



j∈Il

 X

X

s(p) 

l=2 Il ⊆J1,nK p∈Sn card Il =l

= det (A) +

k∈J1,nK k∈I / l

k∈J1,nK k6=j



+

n X

  Y Hp(j)j   Ap(k)k   



l=2 Il ⊆J1,nK p∈Sn card Il =l

n X n X

j∈Il





 Y  s(p)  Ap(k)k    Hp(j)j

n X

+

Y

l=0 Il ⊆J1,nK card Il =l

 n X





n X

n X

j∈Il

X

X

s(p) 





 Y  Hp(j)j   Ap(k)k    k∈J1,nK k∈I / l





l=2 Il ⊆J1,nK p∈Sn card Il =l

48

 Y

Y

j∈Il





 Y  Ap(k)k  Hp(j)j    . k∈J1,nK k∈I / l

We deduce that:   det (A + H) − det (A) − trace [Com(A)]T H

n X

Y Y |Hp(j)j | |Ap(k)k | |s(p)| | {z } j∈I l=2 Il ⊆J1,nK p∈Sn k∈J1,nK l =1 | {z } k∈I card Il =l / l {z } | 6kHkl

6

X

X

6kAkn−l

n X n(n − 1) . . . (n − l + 1) n! kHkl kAkn−l 6 l! l=2

n(n − 1) n! (kAk + kHk)n−2 kHk2 . 2

6 2

Since H ∈ Rn 7→ trace([Com(A)]T H) ∈ R is a continuous linear form, the uniqueness of the dierential 2 ensures that det : Rn → R is dierentiable at A and that its gradient is given by the comatrix operator, concluding the proof of Proposition 5.11.

Proposition 5.12.

2

Let A ∈ Rn . Then, we have the following expression for the comatrix operator:   ∀(i, j) ∈ J1, nK2 , [Com (A)]ij = (−1)i+j det A(i,j) ,

2

with A(i,j) ∈ R(n−1) the (n − 1) × (n − 1)-matrix obtained by deleting the i-th row and j -th column of A. 2

2

Proof. Let A ∈ Rn and (i, j) ∈ J1, nK2 . First, we dene the (n × n)-matrix B ∈ Rn by Bpq := Apq and Bpj := Ipi for any p ∈ J1, nK and any q ∈ J1, nK\{j}. In other words, we introduce the matrix:   A11 ... A1(j−1) 0 A1(j+1) ... A1n  . .. .. .. ..   .. . . . .      A(i−1)1 . . . A(i−1)(j−1) 0 A(i−1)(j+1) . . . A(i−1)n    ... Ai(j−1) 1 Ai(j+1) ... Ain  . B :=  Ai1   A(i+1)1 . . . A(i+1)(j−1) 0 A(i+1)(j+1) . . . A(i+1)n   .. .. .. ..   ..   . . . . .  An1 ... An(j−1) 0 An(j+1) ... Ann Then, note that from Denition 5.10, we get with this notation for any (i, j) ∈ J1, nK2 :     X

[Com (A)]ij =

p∈Sn p(j)=i

 Y  s(p)  Ap(k)k    = k∈J1,nK k6=j

X p∈Sn

 Y  s(p)  Bp(k)k    I|p(j)i {z } k∈J1,nK

= det (B) .

=Bp(j)j

k6=j

Finally, we use the alternating property of the determinant to move the j -th column of B to the the rst position according to the cycle (1, . . . , j) = (1, 2) ◦ . . . ◦ (j − 1, j) if j > 1, followed by a shift of the i-th line to the rst one thanks to the cycle (1, . . . , i) = (1, 2) ◦ . . . ◦ (i − 1, i) if i > 1. Note that the alternating property of the determinant can also by used on the rows, as a consequence of Lemma 5.9. We deduce that [Com(A)]ij = (−1)(i−1)+(j−1) det(A(i,j) ), which concludes the proof of Proposition 5.12.

Corollary 5.13.

2

The following property holds true: Com([A]T ) = [Com(A)]T for any A ∈ Rn . 2

Proof. Let A ∈ Rn . Using Proposition 5.12 and its notation, we have successively for any (i, j) ∈ J1, nK2 :  h h  i (i,j)  iT    Com [A]T := (−1)i+j det [A]T = (−1)i+j det A(j,i) = (−1)j+i det A(j,i) , ij

where we have used Lemma 5.9 to obtain the last equality. We deduce that [Com([A]T )]ij = [Com(A)]ji for any (i, j) ∈ J1, nK2 , concluding the proof of Corollary 5.13.

Lemma 5.14.

2

For any A ∈ Rn , we have: [Com(A)]T A = A[Com(A)]T = det(A)I . In particular, if we 1 assume that A is an invertible matrix, then its inverse takes the following form: A−1 = det(A) [Com(A)]T . 2

Proof. Let A ∈ Rn . First, we prove [Com(A)]T A = det(A)I . We consider (i, j) ∈ J1, nK2 such that i 6= j . We have successively:   n  n n    X X X X Y    [Com (A)]T A [Com (A)]T Akj = [Com (A)]ki Akj = s (p) Ap(l)l  =   Akj . ij ik k=1

k=1

k=1

49

p∈Sn p(i)=k

l∈J1,nK l6=i

2

We now introduce a new (n × n)-matrix B ∈ Rn whose columns are the ones of A except the i-th one which is replaced by the j -th column of A: ∀p ∈ J1, nK, ∀q ∈ J1, nK\{i},

Bpq := Apq

and

Bpi := Apj .

Note that since i 6= j , the i-th and j -th columns of B are identical so in particular we have det(B) = 0 using the alternating property of the determinant. We deduce that: 

[Com (A)]T A

 ij

=

n X X k=1 p∈Sn p(i)=k

Y

s (p) Akj |{z}

Ap(l)l | {z }

n X X

=

s (p)

k=1 p∈Sn p(i)=k

l∈J1,nK =Bp(i)i l6=i =Bp(l)l

n Y

Bp(l)l .

l=1

In the above summation, we can recognize the fact that the set Sn is partitioned into subsets xing the i-th element. We thus have: n   X Y [Com (A)]T A = s (p) Bp(l)l = det (B) = 0. ij

p∈Sn

l=1

Therefore, we have proved that [Com(A)]T A is a diagonal matrix. Let us check that each of its diagonal coecient is equal to det(A). Let i ∈ J1, nK. The same computation as before leads to: n   X X [Com (A)]T A = s (p) Aki |{z} ii k=1 p∈Sn p(i)=k

Y

X

Ap(l)l =

s(p)

p∈Sn

=Ap(i)i l∈J1,nK l6=i

n Y

Ap(l)l = det (A) .

l=1

2

Hence, we have [Com(A)]T A = det(A)I for any A ∈ Rn . Then, applying this last identity to [A]T and combining it with Lemma 5.9 and Corollary 5.13, we deduce that: h T h   iT  iT   det (A) I = det [A]T I = Com [A]T [A]T = ACom [A]T = A [Com (A)]T . 1 1 [Com(A)]T A = A( det(A) [Com(A)]T ). Finally, if A is now invertible, then det(A) 6= 0 and we get I = det(A) T −1 1 The uniqueness of the inverse ensures A = det(A) [Com(A)] , concluding the proof of Lemma 5.14.

5.1.4 About the dierential of the comatrix operator Denition 5.15.

2

4

Let n > 2 be an integer. The operator Hess det : A ∈ Rn 7→ Hess det(A) ∈ Rn , whose notation will be justied later in Corollary 5.17, is dened by: Y  X s (p) Ap(m)m if i 6= k and j = 6 l     p∈Sn m∈J1,nK  2 p(j)=i m6=j ∀A ∈ Rn , ∀(i, j, k, l) ∈ J1, nK4 , [Hess det (A)]ij,kl := m6=l p(l)=k      0 otherwise, where s : Sn → {−1, 1} is the signature map introduced in Denition 5.4.

Proposition 5.16.

2

2

Let A ∈ Rn . Then, we have for any (i, j) ∈ J1, nK2 and any H ∈ Rn :  

[Com (A + H)]i,j = [Com (A)]i,j +

 Y X   X    Hkl + kHk [R (H)] , s (p) A p(m)m ij     p∈Sn k∈J1,nK l∈J1,nK m∈J1,nK X k6=i

p(j)=i p(l)=k

l6=j

m6=j m6=l

2

2

where kR(H)k → 0 as kHk → 0. In other words, the comatrix operator Com : Rn → Rn introduced in 2 Denition 5.10 is dierentiable at any point A ∈ Rn and its dierential at A is given by the following n2 n2 continuous linear map from R into R : 2

∀H ∈ Rn , ∀(i, j) ∈ J1, nK2 ,

[DA Com (H)]ij

:=

n X

[Hess det (A)]ij,kl Hkl

k,l=1

=

X

X

k∈J1,nK l∈J1,nK k6=i l6=j

where Hess det is the operator introduced in Denition 5.15.

50

[Hess det (A)]ij,kl Hkl ,

2

2

Proof. Let (A, H) ∈ Rn × Rn and (i, j) ∈ J1, nK2 . First, from Denition 5.10, we have successively: Y X
s (p) [Com (A + H)]ij := Ap(m)m + Hp(m)m p∈Sn p(j)=i

=

X

m∈J1,nK m6=j

s (p)

!

n−1 X

Y

X

r=0 Ir ⊂J1,nK j ∈I / r cardIr =r

p∈Sn p(j)=i

Y

Hp(m) ˜ m ˜

m∈I ˜ r

Ap(m)m

m∈J1,nK m6=j m∈I / r

 =

X

Y

s (p)

p∈Sn p(j)=i

Ap(m)m

m∈J1,nK m6=j

+



 Y  X X    + s (p)  Ap(m)m   Hp(l)l m∈J1,nK  l∈J1,nK p∈Sn l6=j

p(j)=i

X

X

m6=j m6=l

!

n−1 X

s (p)

r=2 Ir ⊂J1,nK p∈Sn cardIr =r p(j)=i j ∈I / r

Y

Y

Hp(m) ˜ m ˜

m∈I ˜ r

Ap(m)m .

m∈J1,nK m∈I / r m6=j

Then, in the second term of the right-member associated with the last equality, the summation on p is partitioned by subsets xing p(l). We thus have:   [Com (A + H)]ij

=

[Com (A)]ij +

  X  X Y    s (p) A p(m)m  Hkl   l∈J1,nK k∈J1,nK  p∈Sn m∈J1,nK X l6=j

p(j)=i p(l)=k

k6=i

+

m6=j m6=l

!

n−1 X

X

X

s (p)

r=2 Ir ⊂J1,nK p∈Sn cardIr =r p(j)=i j ∈I / r

Y

Y

Hp(m) ˜ m ˜

m∈I ˜ r

Ap(m)m .

m∈J1,nK m∈I / r m6=j

Note that in the particular case n = 2, the above summation on r is empty so the rst-order Taylor expansion of the comatrix operator is exact in this situation. Hence, Proposition 5.16 holds true for n = 2. We now assume n > 3. Using the notation introduced in Denition 5.15, we deduce that: [Com (A + H)]ij − [Com (A)]ij −

n X

[Hess det (A)]ij,kl Hkl

k,l=1

6

n−1 X

! X

X

r=2 Ir ⊂J1,nK p∈Sn cardIr =r p(j)=i j ∈I / r

Y

|s (p) | | {z }

m∈I ˜ r

=1

|

Y

Hp(m) ˜ m ˜ {z

6kHkr

|

6

n−1 X r=2

6

Ap(m)m

m∈J1,nK } m∈I / r m6=j

{z

6kAkn−1−r

}

(n − 1) . . . (n − r) (n − 1)! kHkr kAkn−1−r r!

(n − 1)(n − 2) (n − 1)! kHk2 (kAk + kHk)n−3 . 2

Finally, we obtain: Com (A + H) − Com (A) −

n X

[Hess det (A)]••,kl Hkl

k,l=1

6

(n − 1)(n − 2) n!kHk2 (kAk + kHk)n−3 . 2

P 2 2 Since the map H ∈ Rn 7→ nk,l=1 [Hess det (A)]••,kl Hkl ∈ Rn is linear and continuous, the uniqueness of the dierential ensures that the comatrix operator is dierentiable at A, and that this map is the dierential of the comatrix operator at A, concluding the proof of Proposition 5.16.

51

Corollary 5.17.

2

Let A ∈ Rn . Then, the following expansion holds true:

2

2

∀(H, K) ∈ Rn × Rn ,

n X

DA+K det (H) = DA det (H) +

[Hess det (A)]ij,kl Hij Kkl + kKkR (H, K) ,

i,j,k,l=1

where supH∈Rn2 \{0}

|R(H,K)| kHk

→ 0 as kKk → 0. In other words, the determinant map is twice dierentiable

n2

at any point A ∈ R and its second-order dierential at A is given by the following continuous symmetric 2 2 bilinear form on Rn × Rn : n   X 2 2 2 ∀ (H, K) ∈ Rn × Rn , DA det (H, K) := trace [DA Com(K)]T H = [Hess det (A)]ij,kl Hij Kkl i,j,k,l=1

=

n X n X X

X

[Hess det (A)]ij,kl Hij Kkl .

i=1 j=1 k∈J1,nK l∈J1,nK k6=i l6=j 2

4

In particular, the operator Hess det : Rn → Rn of Denition 5.15 can be considered as the Hessian 2 associated with the determinant map det : Rn → R introduced in Denition 5.8. 2

2

2

Proof. Let (A, H, K) ∈ Rn × Rn × Rn . Considering the expressions obtained in Proposition 5.11 and Proposition 5.16, we have successively:   DA+K det (H) − DA det (H) − trace [DA Com (K)]T H   = trace [Com (A + K) − Com (A) − DA Com (K)]T H . Using the Cauchy-Schwarz inequality, we deduce that:   DA+K det (H) − DA det (H) − trace [DA Com (K)]T H sup kHk n2 H∈R H6=0

6 kCom (A + K) − Com (A) − DA Com (K) k.

From Proposition 5.16, the comatrix operator is dierentiable at A. Hence, for any ε > 0, there exists δ > 0 2 such that for any K ∈ Rn such that kKk < δ , we have kCom (A + K) − Com (A) − DA Com (K) k 6 εkKk. 2 We deduce that for any K ∈ Rn such that kKk < δ , we have:   DA+K det (H) − DA det (H) − trace [DA Com (K)]T H 6 εkKk, sup kHk n2 H∈R H6=0

2

2

2

so the map D• det : Rn → Lc (Rn , R) is dierentiable at A i.e. det : Rn → R is twice dierentiable at A. To conclude the proof of Corollary 5.17, its second-order dierential is given by the continuous bilinear form 2 2 2 det : (H, K) ∈ Rn × Rn 7→ trace([DA Com(K)]T H) ∈ R, which is symmetric since by Denition 5.15, DA one can check the symmetry property [Hess det(A)]ij,kl = [Hess det(A)]kl,ij for any (i, j, k, l) ∈ J1, nK4 .

5.1.5 About the properties associated with the Hessian of the determinant First, we introduce the following well-dened map referred to as the sign function: sgn :

R

−→

x

7−→

{−1, 0, 1}   1 −1 sgn(x) :=  0

if x > 0 if x < 0 otherwise.

(88)

As we did in Proposition 5.12 with the comatrix operator, up to a sign, we can express the Hessian of the determinant map as the determinant of second-order minors. Then, we want to study the behaviour of our operators with respect to the multiplication of matrices. For this purpose, we need to prove an extended version of the Binet-Cauchy Formula.

Proposition 5.18.

2

Let n > 3 be an integer and A ∈ Rn . Then, the Hessian of the determinant map introduced in Denition 5.15 is also given by the following quantity:   ∀(i, j, k, l) ∈ J1, nK4 , Hess det (A)ij,kl = (−1)i+j+k+l sgn [(k − i)(l − j)] det A{i,k},{j,l} , 2

where sgn is the sign function dened in (88), and where A{i,k},{j,l} ∈ R(n−2) is the (n−2)×(n−2)-matrix obtained from A by deleting its i-th row, its k-th row, its j -th column, and its l-th column. Moreover, the above relation still holds true in the case n = 2 by setting det(A{i,k},{j,l} ) := 1 for any (i, j, k, l) ∈ J1, 2K4 .

52

2

Proof. Let n > 3 be an integer and A ∈ Rn . Consider any (i, j) ∈ J1, nK2 and (k, l) ∈ J1, nK2 such that 2 i < k and j < l. First, we introduce the (n × n)-matrix B ∈ Rn dened by: ∀p ∈ J1, nK, ∀q ∈ J1, nK\{j, l},

Bpq := Apq

and

From Denition 5.15 and Denition 5.8, we thus have: X Y [Hess det (A)]ij,kl := s (p) Ap(m)m = p∈Sn p(j)=i p(l)=k

=

X

s (p)

Bpj := Ipi

X p∈Sn

m∈J1,nK m6=j m6=l n Y

and

Bpl := Ipk .

s (p) Ip(j)i Ip(l)k | {z } | {z }

Y

Ap(m)m | {z }

m∈J1,nK =Bp(j)j =Bp(l)l m6=j =Bp(m)m m6=l

Bp(m)m = det (B) .

m=1

p∈Sn

Hence, the evaluation of the Hessian of the determinant is reduced to  A11 ... A1(j−1) 0 A1(j+1) ... A1(l−1)  .. .. .. .. ..  . . . . .   A  (i−1)1 . . . A(i−1)(j−1) 0 A(i−1)(j+1) . . . A(i−1)(l−1)  A ... Ai(j−1) 1 Ai(j+1) ... Ai(l−1) i1   A  (i+1)1 . . . A(i+1)(j−1) 0 A(i+1)(j+1) . . . A(i+1)(l−1)  .. .. .. .. .. det   . . . . .   A(k−1)1 . . . A(k−1)(j−1) 0 A(k−1)(j+1) . . . A(k−1)(l−1)   Ak1 ... Ak(j−1) 0 Ak(j+1) ... Ak(l−1)   A(k+1)1 . . . A(k+1)(j−1) 0 A(k+1)(j+1) . . . A(k+1)(l−1)   .. .. .. .. ..  . . . . . An1 ... An(j−1) 0 An(j+1) ... An(l−1)

the computation of det(B) i.e. of:  0 A1(l+1) ... A1n  .. .. ..  . . .  0 A(i−1)(l+1) . . . A(i−1)n    0 Ai(l+1) ... Ain  0 A(i+1)(l+1) . . . A(i+1)n    .. .. .. .  . . .  0 A(k−1)(l+1) . . . A(k−1)n   1 Ak(l+1) ... Akn   0 A(k+1)(l+1) . . . A(k+1)n    .. .. ..  . . . 0 An(l+1) ... Ann

Then, we use the alternating property of the determinant to move the j -th column of B to the the rst position according to the cycle (1, . . . , j) = (1, 2) ◦ . . . ◦ (j − 1, j) if j > 1, followed by a shift of the l-th column of B to the second position with respect to the cycle (2, . . . , l) = (2, 3) ◦ . . . ◦ (l − 1, l) if l > 2. Note that the alternating property of the determinant map can also be used on rows, as a consequence of Lemma 5.9. Hence, we can continue our re-ordering procedure by shifting the i-th line to the rst position thanks to the cycle (1, . . . , i) = (1, 2) ◦ . . . ◦ (i − 1, i) if i > 1, followed by a shift of the k-th line of B to the second position with respect to the cycle (2, . . . , k) = (2, 3) ◦ . . . ◦ (k − 1, k) if k > 2. We deduce that in the case where i < k and j < l, we have:   [Hess det (A)]ij,kl = (−1)(j−1)+(l−2)+(i−1)+(k−2) det A{i,k},{j,l} =

  (−1)i+j+k+l sgn [(k − i)(l − j)] det A{i,k},{j,l} .

In the case where i > k and j > l, we can apply the previous case by exchanging the role of (i, j) and (k, l). We thus have: X Y [Hess det (A)]ij,kl := s (p) Ap(m)m = [Hess det (A)]kl,ij p∈Sn p(j)=i p(l)=k

m∈J1,nK m6=j m6=l

=

  (−1)k+l+i+j sgn [(i − k)(j − l)] det A{k,i},{l,j}

=

  (−1)i+j+k+l sgn [(k − i)(l − j)] det A{i,k},{j,l} .

Note that the above equality also holds true in the case where i = k or l = j since the sign function (88) and the Hessian of the determinant are both equal to zero in this case. Therefore, we now have to prove that the above equality holds true in the case where i < k and j > l. The same arguments hold but before using the alternating property of the determinant, we need to proceed rst to an additional transposition exchanging the j -th column and the l-th one, in order to be back in the previous case i < k and j < l. This transposition add a negative sign to the overall procedure. Hence, if i < k and j > l, we get:   [Hess det (A)]ij,kl = −(−1)(j−1)+(l−2)+(i−1)+(k−2) det A{i,k},{j,l} =

  (−1)i+j+k+l sgn [(k − i)(l − j)] det A{i,k},{j,l} .

53

Finally, it remains to study the case i > k and j < l, which is deduced from the case i < k and j > l in the same way that the case i > k and j > l was obtained from the case i < k and j < l, using the symmetry of Hess det(A). Hence, we have proved that [Hess det(A)]ij,kl = (−1)i+j+k+l sgn[(k − i)(l − j)]det(A{i,k},{j,l} ) for any (i, j, k, l) ∈ J1, nK4 . In the case where n = 2, if we set det(A{i,k},{j,l} ) := 1 by convention, then one can check by direct calculation that the above relation still holds true. In particular, we have the exact second-order Taylor expansion:   1 2 det (A + H) − det(A) − trace [Com (A)]T H = DA det(H, H) = det (H) 2 =

1 2

n X

[Hess det (A)]ij,kl Hij Hkl ,

i,j,k,l=1

concluding the proof of Proposition 5.18.

Corollary 5.19.

2

Let A ∈ Rn . Then, we have for any (i, j, k, l) ∈ J1, nK4 : h  i [Hess det (A)]ij,kl = [Hess det (A)]kl,ij and Hess det [A]T

ij,kl

= [Hess det (A)]ji,lk ,

where Hess det is the operator introduced in Denition 5.15. 2

Proof. Let A ∈ Rn and (i, j, k, l) ∈ J1, nK4 . First, we use the notation and the result of Proposition 5.18 in order to have:   [Hess det (A)]ij,kl = (−1)i+j+k+l sgn [(k − i) (l − j)] det A{i,k},{j,l} =

  (−1)k+l+i+j sgn [(i − k) (j − l)] det A{k,i},{l,j} = [Hess det (A)]kl,ij .

Similarly, applying Lemma 5.9 and Proposition 5.18, we obtain:  h  i {i,k},{j,l}  Hess det [A]T = (−1)i+j+k+l sgn [(k − i) (l − j)] det [A]T ij,kl h iT  = (−1)i+j+k+l sgn [(k − i) (l − j)] det A{j,l},{i,k}   = (−1)j+i+l+k sgn [(l − j) (k − i)] det A{j,l},{i,k} = [Hess det (A)]ji,lk , which concludes the proof of Corollary 5.19.

Lemma 5.20. Let n > 2 be an integer and l ∈ J1, nK. For any subset I ⊆ J1, nK of l pairwise distinct elements i.e. such that card I = l, there exists a unique bijective map from J1, lK into I that is strictly increasing. This map is thus uniquely associated with I and it is denoted by pI : J1, lK → I .

Proof. Let n > 2 be an integer. First, we prove by induction that p(k) > k for any k ∈ J1, nK and any well-dened map p : J1, nK → N∗ which is strictly increasing. We have obviously p(1) > 1. Assuming p(k) > k, we get from the fact that p is strictly increasing p(k + 1) > p(k) > k and since p is integer valued, we get p(k + 1) > k + 1. Hence, p(k) > k for any k ∈ J1, nK. Then, let l ∈ J1, nK and consider a subset I ⊆ J1, nK of l pairwise distinct elements i.e. such that card I = l. We dene recursively the map pI : J1, lK → I by setting p(1) = min I and p(k) = min I\{p(1), . . . , p(k − 1)} for any k ∈ J2, lK. This is possible since card I = l and any non-empty subset of N has a minimum. One can check that p is an injective map between two sets of l pairwise distinct elements thus it is bijective, and it is strictly increasing by construction. Finally, consider any other strictly increasing bijective map q : J1, lK → I . −1 Then, the map p−1 I ◦ q : J1, lK → J1, lK is also strictly increasing, from which we deduce that pI ◦ q(k) > k i.e. q(k) > pI (k) for any k ∈ J1, lK. By exchanging the role of pI and q , the map q −1 ◦ pI is also strictly increasing so pI (k) > q(k) for any k ∈ J1, lK. Hence, we get pI = q ensuring the uniqueness of the map pI , and concluding the proof of Lemma 5.20.

Proposition 5.21.

Let n1 > 2, n2 > 2, and n3 > 2 be three integers. We consider l ∈ J1, min{n1 , n2 , n3 }K and also two subsets I ⊆ J1, n1 K and J ⊆ J1, n3 K of l pairwise distinct elements i.e. such that we have card I = card J = l. For any (n1 × n2 )-matrix A ∈ Rn1 n2 and for any (n2 × n3 )-matrix B ∈ Rn2 n3 , the following formula holds true:   X det [AB]I,J = det (AI,K ) det (AK,J ) , K⊆J1,n2 K card K=l 2

where the notation AI,K ∈ Rl refers to the (l × l)-matrix obtained from A by deleting the rows that are not labelled in I and the columns that are not labelled in J i.e. (AI,K )rs := ApI (r)pK (s) for any (r, s) ∈ J1, lK2 , with the notation pI : J1, lK → I referring to the map introduced in Lemma 5.20.

54

Proof. Let n1 > 2, n2 > 2, and n3 > 2 be three integers. We consider l ∈ J1, min{n1 , n2 , n3 }K and also two subsets I ⊆ J1, n1 K and J ⊆ J1, n3 K of l pairwise distinct elements i.e. such that card I = card J = l. Let A ∈ Rn1 n2 be a (n1 × n2 )-matrix and B ∈ Rn2 n3 be a (n2 × n3 )-matrix. Using the notation introduced in the statement of Proposition 5.21, we have successively:         [AB]I,J [AB]I,J 11  1l          .. .. det [AB]I,J := det  , . . . ,    .   .        [AB]I,J [AB]I,J l1

ll



=

=





 (AB)pI (1)pJ (1) (AB)pI (1)pJ (l)     .. ..  ,...,  det      . . (AB)pI (l)pJ (1) (AB)pI (l)pJ (l)  X n2 ApI (1)k1 Bk1 pJ (1)   k1 =1   .. det  .  n2  X  ApI (l)k1 Bk1 pJ (1)

 X n2 ApI (1)kl Bkl pJ (l)     kl =1     .. ,..., .   n2   X   ApI (l)kl Bkl pJ (l) 

k1 =1

=

k1 =1

=

n2 X

n2 X

l Y

kl =1

m=1

...

X

Bkm pJ (m)

p:J1,lK→J1,n2 K p is injective

kl =1

ApI (l)kl



   ApI (1)k1 ApI (1)kl     .. .. det  ,...,  . . ApI (l)k1 ApI (l)kl !

l Y

X kl ∈J1,n2 K kl ∈{k / 1 ,...,kl−1 }


Bkm pJ (m) det ApI (•)k1 , . . . , ApI (•)kl .

m=1

!
Bp(m)pJ (m) det ApI (•)p(1) , . . . , ApI (•)p(l) .

l Y

X

=

its alternating property to obtain:    ApI (1)kl n2 X    .. Bkl pJ (l)   ,..., .

!

...

k1 =1 k2 ∈J1,n2 K k2 6=k1

     .   

kl =1

We now use the multi-linearity of the determinant and   ApI (1)k1 n2   X   .. det [AB]I,J = det  Bk1 pJ (1)  . k1 =1 ApI (l)k1 n2 X



m=1

Then, we observe that instead of summing on every injective map p : m ∈ J1, lK → km ∈ J1, n2 K, we can partition them by xing their imaging being a subset K ∈ J1, n2 K of l pairwise distinct elements i.e. such that card K = l. Hence, we get: ! l   X X Y
det [AB]I,J = Bp(m)pJ (m) det ApI (•)p(1) , . . . , ApI (•)p(l)

=

=

=

K⊆J1,n2 K card K=l

p:J1,lK→K p is bijective

X

X

K⊆J1,n2 K card K=l

p:J1,lK→K p is bijective

X

X

K⊆J1,n2 K card K=l

p:J1,lK→K p is bijective

X

X

K⊆J1,n2 K card K=l

p:J1,lK→K p is bijective

m=1

l Y

!

s p−1 K ◦ p det ApI (•)pK (1) , . . . , ApI (•)pK (l)

Bp(m)pJ (m)

m=1

l Y

!

s p−1 K ◦ p det [AI,K ]•1 , . . . , [AI,K ]•n

Bp(m)pJ (m)

m=1

l Y

! Bp(m)pJ (m)


s p−1 K ◦ p det (AI,K ) .

m=1

−1 Since the map (p : J1, lK → K) 7−→ (p−1 K ◦ p) ∈ Sl is bijective, we can make a change of indices q = pK ◦ p

55

in the above summation on p. We deduce that:   det [AB]I,J

X

:=

det (AI,K )

X

s (q)

l Y

K⊆J1,n2 K card K=l

q∈Sl

m=1

X

X

l Y

=

det (AI,K )

X

=

s (q)

q∈Sl

K⊆J1,n2 K card K=l

! BpK [q(m)]pJ (m)

! [BK,J ]q(m)m

m=1

det (AI,K ) det (BK,J ) ,

K⊆J1,n2 K card K=l

concluding the proof of Proposition 5.21.

Corollary 5.22.

2

2

Let (A, B) ∈ Rn × Rn . Then, we have det(AB) = det(A)det(B) = det(BA) and Com(AB) = Com(A)Com(B). Moreover, for any (i, j, k, l) ∈ J1, nK4 , we also have: [Hess det(AB)]ij,kl

=

n−1 X

n X

[Hess det(A)]ip,kq [Hess det(B)]pj,ql

p=1 q=p+1

=

n 1X X [Hess det(A)]ip,kq [Hess det(B)]pj,ql 2 p=1 q∈J1,nK q6=p

=

n 1 X [Hess det (A)]ip,kq [Hess det (B)]pj,ql , 2 p,q=1

where the determinant map, its associated Hessian, and the comatrix operator are respectively introduced in Denitions 5.8, 5.15, and 5.10. 2

2

Proof. Let (A, B) ∈ Rn × Rn . First, we can apply Proposition 5.21 with l = n1 = n2 = n3 = n and I = J = J1, nK. Using the notation of Proposition 5.21, we thus have:   X

det (AB) = det [AB]J1,nK,J1,nK = det AJ1,nK,K det BK,J1,nK K⊆J1,nK card K=n



det AJ1,nK,J1,nK det BJ1,nK,J1,nK = det (A) det (B) .

=

Hence, we get det(AB) = det(A)det(B) = det(B)det(A) = det(BA). Then, let (i, j) ∈ J1, nK2 . We can apply Proposition 5.21 with l + 1 = n1 = n2 = n3 = n, I = J1, nK\{i}, and J = J1, nK\{j}, which is combined with Proposition 5.12 to get:     [Com (AB)]ij = (−1)i+j det [AB](i,j) = (−1)i+j det [AB]J1,nK\{i},J1,nK\{j} =

(−1)i+j

X K⊆J1,nK card K=n−1

=

n X



det AJ1,nK\{i},K det BK,J1,nK\{j}



(−1)i+j+2k det AJ1,nK\{i},J1,nK\{k} det BJ1,nK\{k},J1,nK\{j}

k=1

=

n X

n     X (−1)i+k det A(i,k) (−1)k+j det A(k,j) = [Com (A)]ik [Com (B)]kj

k=1

=

k=1

[Com (A) Com (B)]ij .

Therefore, we have proved Com(AB) = Com(A)Com(B). Finally, we assume n > 3. Let (i, j) ∈ J1, nK2 and (k, l) ∈ J1, nK2 such that k 6= i and l 6= j . We can again apply Proposition 5.21 with n1 = n2 = n3 = n, card I = card J = n − 2, I = J1, nK\{i, k}, and J = J1, nK\{j, l}, which is combined with Proposition 5.18

56

and the observation sgn(xy) = sgn(x)sgn(y) in order to get successively:   [Hess det (AB)]ij,kl = (−1)i+j+k+l sgn [(k − i) (l − j)] det [AB]{i,k},{k,l} =

  (−1)i+j+k+l sgn [(k − i) (l − j)] det [AB]J1,nK\{i,k},J1,nK\{j,l}

=

(−1)i+j+k+l sgn [(k − i) (l − j)]

X K⊆J1,nK card K=n−2

det AJ1,nK\{i,k},K




det BK,J1,nK\{j,l} =

=

n−1 X

n X

n−1 X

n X




  (−1)i+j+k+l+2p+2q sgn (k − i) (q − p)2 (l − j) p=1 q=p+1

det AJ1,nK\{i,k},J1,nK\{p,q} det BJ1,nK\{p,q},J1,nK\{j,l}

  (−1)i+p+k+q sgn [(k − i) (q − p)] det A{i,k},{p,q}

p=1 q=p+1

  (−1)p+j+q+l sgn [(q − p) (l − j)] det B {p,q},{j,l}

=

n−1 X

n X

[Hess det (A)]ip,kq [Hess det (B)]pj,ql .

p=1 q=p+1

Note that with the conventions of Denition 5.15, the last equality also holds true if i = k or if j = l. Moreover, one can check by direct calculations from Denition 5.15 that in the case n = 2, the above equality is also valid. Hence, we have established that: ∀(i, j, k, l) ∈ J1, nK4 ,

[Hess det (AB)]ij,kl =

n−1 X

n X

[Hess det (A)]ip,kq [Hess det (B)]pj,ql ,

p=1 q=p+1

concluding the proof of Corollary 5.22.

5.1.6 About the particular cases of invertible, complex-valued, and unitary matrices First, we give simpler expressions for the derivatives of the determinant and the comatrix operator in the 1 case of invertible matrices. This is based on the fact that Com(A) = det(A) [(A)−1 ]T provided det(A) 6= 0, which has been already proved in Lemma 5.14. Then, we show how to extend the results of Section 5.1 from real- to complex-valued matrices. Finally, we conclude Section 5.1 by considering the particular case of unitary matrices and their sub-determinant properties.

Lemma 5.23.

2

Let A ∈ Rn be any invertible (n × n)-matrix. Then, the (n × n)-matrix A + H is also 2 invertible for any H ∈ Rn such that kHk < (kA−1 k)−1 and the following expansion holds true: (A + H)−1 = A−1 − A−1 HA−1 + kHkR(H), 2

2

where kR(H)k → 0 as kHk → 0. In other words, the invertible operator (•)−1 : A ∈ Rn → A−1 ∈ Rn is 2 well dened around any invertible matrix A ∈ Rn . Moreover, it is dierentiable at A and its dierential 2 2 at A is given by the continuous linear map DA (•)−1 : H ∈ Rn 7→ −A−1 HA−1 ∈ Rn . Proof. First, we start by considering the dierentiability of the invertible operator at the identity map. 2 Let H ∈ Rn satisfy kHk < 1. Considering its associated continuous linear map h : x ∈ Rn 7→ Hx ∈ Rn , one notice that |||h||| 6 kHk < 1. We now introduce the partial sum P with kthek Cauchy-Schwarz inequality k sn = n (−1) h for any n ∈ N , where h = h ◦ . . . ◦ h is the k-times successive k=1 P composition of h. Since |||h||| < 1 and ||| • ||| is a Banach norm on Lc (Rn , Rn ), we get that the series n∈N (−1)n hn is normally convergent in a complete space thus it is converging as a Cauchy sequence to a continuous linear map referred to as s : Rn → Rn . Then, one can check that: ∀n ∈ N,

sn ◦ (Id + h) = (Id + h) ◦ sn = Id − (−1)n+1 hn+1 .

Using the continuity of the composition operator and the fact that sn+1 − sn = (−1)n+1 hn+1 tends to zero as n → +∞, we can correctly let n → +∞ in the above expression in order to obtain the equalities

57

s ◦ (Id + h) = (Id + h) ◦ s = Id i.e. Id + h is invertible and its inverse is precisely given by s. Getting back √ √ to the space of matrices, one can check that kHk 6 n|||h||| 6 nkHk and we thus have: k (I + H)−1 − I + Hk 6

√

n |||(Id + h)−1 − Id + h||| 6

2

√ √ +∞ √ X n |||h|||2 n kHk2 6 n |||h|||n = 1 − |||h||| 1 − kHk n=2

2

Since the map H ∈ Rn 7→ −H ∈ Rn is linear and continuous, the uniqueness of the dierential ensures that it is the dierential at I of the invertible operator (•)−1 : A 7→ A−1 , which is thus well dened around 2 2 the identity map and dierentiable at I . Then, let A ∈ Rn be an invertible matrix. Then, for any H ∈ Rn −1 −1 −1 −1 −1 such that kHk < (kA k) , we have |||a ◦ h||| 6 |||a ||| |||h||| 6 kA k kHk < 1, so we deduce from the foregoing that Id + a−1 ◦ h : x ∈ Rn 7→ (I + A−1 H)x ∈ Rn is invertible. Since A is also invertible, we get that A(I + A−1 H) = A + H is invertible and we have successively:   k(A + H)−1 − A−1 + A−1 HA−1 k = k (I + A−1 H)−1 − I + A−1 H A−1 k 6

h i
−1 √ n ||| Id + a−1 ◦ h − Id + a−1 ◦ h ◦ a−1 |||

6


−1 √ − Id + a−1 ◦ h||| |||a−1 ||| n ||| Id + a−1 ◦ h √

6

√ |||a−1 |||3 |||h|||2 n |||a−1 ◦ h|||2 |||a−1 ||| 6 n −1 1 − |||a ◦ h||| 1 − |||a−1 ||| |||h|||

√ 6

n kA−1 k3 kHk2 . 1 − kA−1 k kHk

2

2

Since H ∈ Rn 7→ −A−1 HA−1 ∈ Rn is a continuous and linear map, the uniqueness of the dierential ensures that it is the dierential at A of the invertible operator (•)−1 : A 7→ A−1 , which is thus well dened around any invertible matrix A and dierentiable at A, concluding the proof of Lemma 5.23.

Proposition 5.24.

2

Let A ∈ Rn be an invertible (n × n)-matrix. Then, we have Com(A) = det(A)[A−1 ]T and the dierential of the determinant and the comatrix, respectively introduced in Proposition 5.11 and Proposition 5.16, can take the following form: 
−1   DA det (H) = det (A) trace A H 2 ∀H ∈ Rn ,  T  T  T 
   DA Com (H) = det (A) trace A−1 H A−1 − A−1 [H]T A−1 Moreover, the second-order dierential of the determinant given in Corollary 5.17 has the following form: 


 2 2 2 ∀ (H, K) ∈ Rn × Rn , DA det (H, K) = det (A) trace A−1 K trace A−1 H − trace A−1 KA−1 H . 2

1 Proof. Let A ∈ Rn be an invertible matrix. First, we use Lemma 5.14 to get A−1 = det(A) [Com(A)]T i.e. Com(A) = det(A)[A−1 ]T , which is combined with Proposition 5.11 to deduce that the determinant is dierentiable at A with the following dierential:  
2 ∀H ∈ Rn , DA det (H) := trace [Com (A)]T H = det (A) trace A−1 H .

Then, Propositions 5.16 and Proposition 5.11 are combined with Lemma 5.23 in order to ensure that we can dierentiate at A the equality Com(A) = det(A)[A−1 ]T . Using successively the dierential of a product, Lemma 5.36, and Lemma 5.23, we obtain: h i  T 2 ∀H ∈ Rn , DA Com (H) = DA det (H) A−1 + det (A) DA (•)T ◦ (•)−1 (H) =


 T   det (A) trace A−1 H A−1 + DA−1 (•)T DA (•)−1 (H) .

Since the transpose operator is a continuous linear map, its dierential at any point is the map itself so we have: 
 T  T  2 ∀H ∈ Rn , DA Com (H) = det (A) trace A−1 H A−1 + DA (•)−1 (H) =


 T  T  T  det (A) trace A−1 H A−1 − A−1 [H]T A−1 .

58

Finally, from Corollary 5.17, the determinant is twice dierentiable at A. We can thus insert the above relation in the expression of the second-order dierential of the determinant given in Corollary 5.17 to get:   2 2 ∀(H, K) ∈ Rn × Rn , D2A det (H, K) := trace [DA Com (K)]T H 


 = det (A) trace A−1 K trace A−1 H − trace A−1 KA−1 H ,

concluding the proof of Proposition 5.24.

Remark 5.25. All the results of Section 5.1 still hold true with complex-valued matrices instead of realvalued matrices. The proofs are rigorously identical. The only modication occurs in the rst paragraph of 2 Section 5.1.3, where the (hermitian) product associated with the space Cn of (n × n)-matrices with complex entries has now to be dened as: 2

2

∀(A, B) ∈ Cn × Cn ,

hA | Bi := trace



A

T

B



n X

=

Aij Bij ,

i,j=1 2

where z ∈ C → z ∈ C is the standard conjugate operator, and where A ∈ Cn is the (n × n)-matrix dened 2 as Aij := Aij for any A ∈ Cn and any (i, j) ∈ J1, nK2 . Finally, one can check the following properties by direct calculations: A + B = A + B , λA = λ A, AB = A B , [ A ]T = [A]T , trace(A) = trace(A), 2 2 det(A) = det(A), Com(A) = Com(A), and Hess det(A) = Hess det(A) for any (A, B, λ) ∈ Cn × Cn × C.

Proposition 5.26.

2

Let k ∈ J1, n−1K and U ∈ Cn be a unitary (n×n)-matrix i.e. a matrix which satises U [U ] = [U ] U = I . We consider the following block decomposition:   A B U := , C D T

T

2

2

where A ∈ Ck is a (k × k)-matrix, where D ∈ C(n−k) is a (n − k) × (n − k)-matrix, where B ∈ Ck(n−k) is a k × (n − k)-matrix, and where C ∈ C(n−k)k is (n − k) × k-matrix. Then, we have: det(D) = det(U )det(A). 2

Proof. Let U ∈ Cn be a unitary (n × n)-matrix with the same the block decomposition than the one given in Proposition 5.26. With the same notation, we thus have:   T     T A B [A] [C]T A[A]T + B[B]T A[C]T + B[D]T U U = = T T T T T T C D [B] [D] C[A] + D[B] C[C] + D[D] Using U [U ]T = I , we deduce A[A]T +B[B]T = Ik , C[C]T +D[D]T = In−k , and C[A]T +D[B]T = 0(n−k)×k . With these equalities in mind, we thus get: !      T [A]T 0k×(n−k) 0k×(n−k) A[A]T + B[B]T B A B [A] U  T = = C D [B]T In−k C[A]T + D[B]T D B In−k  =

Ik 0(n−k)×k

 B . D

Finally, consider the determinant of each member in the last equality, use Corollary 5.22 and Lemma 5.9, and we obtain: " !# !   [A]T 0k×(n−k) [A]T 0k×(n−k) Ik B det (D) = det = det U  T = det (U ) det  T 0(n−k)×k D B In−k B In−k =


det (U ) det [A]T = det (U ) det(A),

concluding the proof of Proposition 5.26.

Corollary 5.27.

2

Let U ∈ Cn be a unitary (n × n)-matrix i.e. satisfying U [U ]T = [U ]T U = I . Then, we have |det(U )| = 1, Com(U ) = det(U )U , and also:
∀(i, j, k, l) ∈ J1, nK4 , [Hess det (U )]ij,kl = det (U ) Uij Ukl − Ukj Uil .

59

2

Proof. Let n > 3 be an integer and U ∈ Cn be a unitary (n × n)-matrix. First, we use Corollary 5.22 and Lemma 5.9 to get:     1 = det(I) = det U [U ]T = det(U )det [U ]T = det(U )det(U ) = |det(U )|2 . Then, since U [U ]T = [U ]T U = I , the uniqueness of the inverse ensures that U is invertible with U −1 = [U ]T . Considering Proposition 5.24, we deduce Com(U ) = det(U )[U −1 ]T = det(U )U . Finally, let (i, j) ∈ J1, nK2 . We consider any (k, l) ∈ J1, nK2 such that k > i and l > j . We introduce the following (n × n)-matrix with complex entries:   Uij Uil Ui1 . . . Ui(j−1) Ui(j+1) . . . Ui(l−1) Ui(l+1) . . . Uin  Ukj Ukl Uk1 . . . Uk(j−1) Uk(j+1) . . . Uk(l−1) Uk(l+1) . . . Ukn      .. ..   . .     U U (i−1)l   (i−1)j   U   (i+1)j U(i+1)l , V :=  . .   . .   . .   {i,k},{j,l}   U(k−1)j U(k−1)l U     U(k+1)j U(k+1)l     .. ..   . . Unj

Unl

where U {i,k},{j,l} is the (n − 2) × (n − 2) obtained from U by deleting the i-th row, the k-th row, the j -th column, and the l-th column. Note that V is obtained from U by the successive four procedures: (i) A shift of the j -th column to the rst one thanks cycle (1, ..., j) = (1, 2) ◦ ... ◦ (j − 1, j) if j > 1; (ii) A shift of the l-th column to the second one thanks cycle (2, ..., l) = (2, 3) ◦ ... ◦ (l − 1, l) if l > 2; (iii) A shift of the i-th row to the rst one thanks to the cycle (1, ..., i) = (1, 2) ◦ ... ◦ (i − 1, i) if i > 1; (iv) A shift of the k-th row to the second one thanks to the cycle (2, ..., k) = (2, 3) ◦ ... ◦ (k − 1, k) if k > 2. Denoting by Ppq the (n × n)-matrix (of permutation) exchanging the p-th column and the q -th one, we thus have: V = P23 . . . P(k−1)k P12 . . . P(i−1)i U P(j−1)j . . . P12 P(l−1)l . . . P23 . Since det(Ppq ) = −1, we get after simplications according to Corollary 5.22 det(V ) = (−1)i+j+k+l det(U ). Moreover, since permutation matrices are unitary, and since the space of unitary matrices is stable under the multiplication of matrices, we deduce that V is also unitary. We can thus apply Proposition 5.26 to V with k = 2, in order to obtain:    
Uij Uil = (−1)i+j+k+l det(U ) Uij Ukl − Ukj Uil . det U {i,k},{j,l} = det (V ) det Ukj Ukl In the case where i > k and j > l, the arguments are the same but before doing procedure (i) we have to exchange the column i and k and then j and l in order to be back in our previous case. Hence, this does not aect the sign of the result. Similarly, the case i < k and j > l and the case i > k and j < l are studied by adding a transposition matrix to get back to our rst case before doing procedure (i). However, this add a negative sign to the result. Gathering all these cases in a synthetic form, we obtain:  
det U {i,k},{j,l} = (−1)i+j+k+l sgn [(k − i)(l − j)] det(U ) Uij Ukl − Ukj Uil . Finally, we combine the last estimation with Proposition 5.18 in order to get:   [Hess det (U )]ij,kl = (−1)i+j+k+l sgn [(k − i)(l − j)] det U {i,k},{j,l} =


det(U ) Uij Ukl − Ukj Uil .

Note that the last estimation also holds true in the case where i = k or j = l since both the left and right members are equal to zero in this case. To conclude the proof of Proposition 5.27, one can check that the last equality still holds true in the case n = 2.

60

5.2 On the rst- and second-order shape derivatives of a volume integral Let n > 2 be a positive integer henceforth set. In this section, Ω denotes a subset of Rn , and θ : Rn → Rn refers to a well-dened vector eld. First, we recall some terminology about shape dierentiability.

Denition 5.28.

Let us assume that the following shape functional is a well-dened map for a certain class of admissible shapes: F : Ω 7−→ F (Ω) .

By abuse of terminology, we say that F is shape dierentiable at Ω if the following associated functional is well dened around the origin and (Frechet) dierentiable at the origin: FΩ : θ 7−→ FΩ (θ) := F [(I + θ) (Ω)] .

If this is the case, then the dierential D0 FΩ of the map FΩ at the origin is called the (rst-order) shape derivative of F at Ω. If in addition, FΩ is dierentiable around the origin and twice dierentiable at the origin, then we say that F is twice dierentiable at Ω and similarly, the second-order dierential D02 FΩ of the map FΩ at the origin is called the second-order shape derivative of F at Ω. Moreover, by analogy with the nite dimensional case, let us assume that there exists a unique well-dened map fΩ : ∂Ω → R such that: Z D0 FΩ (θ) = fΩ (x) θn (x) dA (x) , ∂Ω

where the integration is done with respect to the (n − 1)-dimensional Hausdor measure referred to as A(•), and where θn (x) := hθ(x) | nΩ (x)i is the normal component of θ at the point x with nΩ (x) denoting the unit vector normal to the boundary ∂Ω at the point x pointing outwards Ω. Then, the function fΩ , eventually depending on Ω, is denoted by abuse of notation ∂F (Ω) and called the shape gradient of F at Ω. Similarly, ∂Ω let us now assume that in addition to the existence of a shape gradient, there exists a unique well-dened map f˜Ω : ∂Ω → R such that: Z Z  h i D E ∂F ˜ = (Ω) IIΩ θ˜∂Ω , θ∂Ω + ∇∂Ω (θn ) | θ˜∂Ω D02 FΩ (θ, θ) f˜Ω θn θ˜n dA − ∂Ω ∂Ω ∂Ω E D + ∇∂Ω (θ˜n ) | θ∂Ω dA, where (•)∂Ω := (•) − (•)n nΩ is the tangential component of a vector eld, which means that in particular ∇∂Ω (•) := ∇(•) − h∇(•) | nΩ inΩ refers to the tangential component of the gradient operator, and where IIΩ (•, •) := −hD∂Ω nΩ (•) | (•)i is the second fundamental form associated with the (hyper)-surface ∂Ω, which is a symmetric bilinear form on the tangent space, with D∂Ω (•) = D(•) − D(•)nΩ [nΩ ]T denoting the tangential component of the dierential operator on vector elds. Then, the function f˜Ω , eventually 2 depending on Ω, is denoted by ∂∂ΩF2 (Ω) and called the shape Hessian of F at Ω. Note that in the above expression of D02 FΩ , the contribution of tangential components vanishes if θ and ˜ θ are orthogonal to the boundary or if ∂F (Ω) = 0 on ∂Ω, which is typically the case if Ω is a critical point ∂Ω for F i.e. if D0 FΩ = 0. Such a structure is expected from shape derivatives (see e.g. [27, Section 5.9.4] and [34]). Moreover, in Denition 5.28, note that we did not clearly specify on which spaces are dened F and FΩ because we want to play between the required regularity for Ω and θ in order to get similar shape dierentiability results in slightly dierent spaces, depending on the problem under consideration. Then, let f ∈ L1 (Rn , R). We now consider the specic shape functional F : Ω 7→ F (Ω), dened for any (Lebesgue) measurable subset Ω of Rn by: Z F (Ω) := f (x) dx, (89) Ω

where the integration is done with respect to the n-dimensional Lebesgue measure. In this setting, the map F : Ω 7→ F (Ω) is well dened. In this section, our goal is to study the shape dierentiability of (89). More precisely, we are going to establish the following standard results from the shape sensitivity analysis (see e.g. [16, Chapter 9], [27, Chapter 5], and [51, Section 2.16]).

Theorem 5.29.

Let n > 2 be any integer, f ∈ W 1,1 (Rn , R), and Ω be any measurable subset of Rn . Then, the functional FΩ : θ ∈ W 1,∞ (Rn , Rn ) 7→ F [(I + θ)(Ω)] ∈ R is well dened around the origin and the following expansion holds true for any θ ∈ W 1,∞ (Rn , Rn ) small enough: Z Z Z f = f + div (f θ) + kθkW 1,∞ (Rn ,Rn ) R (θ) , (I+θ)(Ω)

Ω

Ω

where |R (θ) | → 0 as kθkW 1,∞ (Rn ,Rn ) → 0. Hence, the functional FΩ is dierentiable at the origin and its dierential at the origin is given by the following continuous linear form on W 1,∞ (Rn , Rn ): Z Z Z ∀θ ∈ W 1,∞ (Rn , Rn ), D0 FΩ (θ) := div (f θ) = h∇f (x) | θ(x)idx + f (x)trace(Dx θ)dx. (90) Ω

Ω

61

Ω

In other words, the functional F introduced in (89) is shape dierentiable at any measurable subset of Rn . If in addition, we now assume that Ω is an open bounded subset of Rn with a Lipschitz boundary in the sense of Denition 5.46, then the shape derivative of F at Ω now takes the following form: Z ∀θ ∈ W 1,∞ (Rn , Rn ), D0 FΩ (θ) = f (x) hθ (x) | nΩ (x)i dA (x) . (91) ∂Ω

In other words, for any open bounded subset of R with Lipschitz boundary, the map F dened by (89) has a well-dened shape gradient, which is uniquely determined up to a set of zero A(• ∩ ∂Ω)-measure, and which is given by: ∂F (Ω) : x ∈ ∂Ω 7−→ f (x) ∈ R, ∂Ω where the boundary values of f ∈ W 1,1 (Rn , R) have to be understood in the sense of trace. n

Theorem 5.30.

Let n > 2 be any integer, f ∈ W 2,1 (Rn , R), and Ω be any measurable subset of Rn . Then, the functional FΩ : θ ∈ W 1,∞ (Rn , Rn ) 7→ F [(I + θ)(Ω)] ∈ R is well dened and dierentiable around the origin, its (rst-order) dierential being well dened around the origin by the following map:
D• FΩ : W 1,∞ (Rn , Rn ) −→ Lc W 1,∞ (Rn , Rn ) , R   θ0 7−→ Dθ0 FΩ : θ 7→ D0 F(I+θ0 )(Ω) θ ◦ (I + θ0 )−1 , where D0 F(I+θ0 )(Ω) is the shape derivative of F at (I + θ0 )(Ω) dened by (90). Moreover, the following expansion holds true for any θ0 ∈ W 1,∞ (Rn , Rn ) small enough: Z Z ∀θ ∈ W 1,∞ (Rn , Rn ) , Dθ0 FΩ (θ) = D0 FΩ (θ) + hHess f (θ0 ) | θi + h∇f | θ0 div (θ) + θ div (θ0 )i Z Ω Ω + f [div (θ) div (θ0 ) − trace (D• θ0 D• θ)] + kθ0 kW 1,∞ (Rn ,Rn ) R (θ0 , θ) , Ω

where supθ∈W 1,∞ (Rn ,Rn )\{0}

|R(θ0 ,θ)| kθkW 1,∞ (Rn ,Rn )

→ 0 as kθ0 kW 1,∞ (Rn ,Rn ) → 0. Hence, the functional FΩ is

twice dierentiable at the origin and its second-order dierential at the origin is given by the following continuous symmetric bilinear form dened for any (θ1 , θ2 ) ∈ W 1,∞ (Rn , Rn ) × W 1,∞ (Rn , Rn ) by: Z Z Z D02 FΩ (θ1 , θ2 ) := hHess f (θ1 ) | θ2 i + h∇f | θ1 i div (θ2 ) + h∇f | θ2 i div (θ1 ) Ω Ω Z Ω Z (92) f div (θ1 ) div (θ2 ) − f trace (D• θ1 D• θ2 ) . + Ω

Ω

In other words, the function F dened by (89) is twice shape dierentiable at any measurable subset of Rn . If in addition, we now assume that Ω is an open bounded subset of Rn with Lipschitz boundary in the sense of Denition 5.46, then FΩ : θ ∈ W 1,∞ (Rn , Rn ) ∩ C 1 (Rn , Rn ) 7→ F [(I + θ)(Ω)] ∈ R is also well dened and dierentiable around the origin, its (rst-order) dierential being well dened around the origin by the following map:
D• FΩ : W 1,∞ (Rn , Rn ) ∩ C 1 (Rn , Rn ) −→ Lc W 1,∞ (Rn , Rn ) ∩ C 1 (Rn , Rn ) , R  (93) θ0 7−→ Dθ0 FΩ : θ 7→ D0 F(I+θ0 )(Ω) θ ◦ (I + θ0 )−1 , where D0 F(I+θ0 )(Ω) is the shape derivative of F at (I + θ0 )(Ω) now given by (91). Moreover, the dierential map given by (93) is dierentiable at the origin. Hence, FΩ is twice dierentiable at the origin and its second-order dierential at the origin is now given by the following continuous symmetric bilinear form dened for any (θ1 , θ2 ) ∈ (W 1,∞ (Rn , Rn ) ∩ C 1 (Rn , Rn )) × (W 1,∞ (Rn , Rn ) ∩ C 1 (Rn , Rn )) by:  Z  D02 FΩ (θ1 , θ2 ) = h∇f (x) | θ1 (x)i + f (x) trace (Dx θ1 ) hθ2 (x) | nΩ (x)i dA (x) ∂Ω Z − hDx θ1 [θ2 (x)] | nΩ (x)i dA (x) ∂Ω



Z = ∂Ω

(94)

 h∇f (x) | θ2 (x)i + f (x) trace (Dx θ2 ) hθ1 (x) | nΩ (x)i dA (x) Z − hDx θ2 [θ1 (x)] | nΩ (x)i dA (x) . ∂Ω

If in addition, we now assume that Ω is an open bounded subset of Rn with a boundary of class C 1,1 in the sense of Denition 5.47, then the second-order shape derivative of F at Ω now takes the following form for ˜ ∈ (W 1,∞ (Rn , Rn ) ∩ C 1 (Rn , Rn )) × (W 1,∞ (Rn , Rn ) ∩ C 1 (Rn , Rn )) : any (θ, θ)  Z  Z h   D E ˜ = D02 FΩ (θ, θ) HΩ f + h∇f | nΩ i θn θ˜n dA − f IIΩ θ˜∂Ω , θ∂Ω + ∇∂Ω (θn ) | θ˜∂Ω (95) ∂Ω ∂Ω D Ei + ∇∂Ω (θ˜n ) | θ∂Ω dA,

62

where HΩ := div∂Ω (nΩ ) is denoting the scalar mean curvature associated with the C 1,1 -hypersurface ∂Ω, with div∂Ω (•) := div(•) − hD(•)nΩ | nΩ i referring to the tangential component of the divergence operator. In other words, for any open bounded subset of Rn with a C 1,1 -boundary, the functional F dened by (89) has a well-dened shape Hessian, which is uniquely determined up to a set of zero A(• ∩ ∂Ω)-measure, and which is given by: ∂2F (Ω) : x ∈ ∂Ω 7−→ HΩ (x) f (x) + h∇f (x) | nΩ (x)i ∈ R, ∂Ω2 where the boundary values of f ∈ W 2,1 (Rn , R) and ∇f have to be understood in the sense of trace. In Theorems 5.29-5.30, we only give the results concerning the shape dierentiability of F . However, in this section, we are going to prove more rened statements by studying the regularity of the map FΩ associated with F . The results and their precise references in the text are sum up in Table 3 below. Ω ⊆ Rn

Measurable Measurable Measurable Measurable Lipschitz Measurable Lipschitz C 1,1 -domain Measurable

Z

θ : Rn → Rn

Regularity of

L L1 L1 W 1,1 W 1,1 W 2,1 W 2,1 W 2,1

C 0,1 W 1,∞ W 1,∞ W 1,∞ W 1,∞ W 1,∞ ∩ C 1 W 1,∞ ∩ C 1

F : Ω 7→ Ω f is well dened by (89). FΩ is well dened on B0,1 . FΩ is C 0 on B0,1 ∩ W 1,∞ . FΩ is C 1 on B0,1 ∩ W 1,∞ . F has a well-dened shape gradient. FΩ is C 2 on B0,1 ∩ W 1,∞ . FΩ is C 2 on B0,1 ∩ W 1,∞ ∩ C 1 . F has a well-dened shape Hessian.

W k,1

W 1,∞

FΩ is C k on B0,1 ∩ W 1,∞ .

f : Rn → R 1

FΩ : θ 7→

f (I+θ)(Ω)

Proof

R

Proposition 5.39 Proposition 5.39 Proposition 5.43 Theorem 5.29 Proposition 5.44 Proposition 5.44 Theorem 5.30 Theorem 5.40

Table 3: Summary of the dierent regularity results concerning the functional FΩ : θ 7→

R (I+θ)(Ω)

f.

Section 5.2 is organized as follows. First, we recall some terminology about dierentiability in Banach spaces and we introduce the Sobolev norms in which we are interested. Then, we give some dierentiability results related to the inverse, the Jacobian determinant, the comatrix, and the composition operators. Finally, we study the continuity, the rst- and second-order dierentiability properties of the map FΩ associated with the map F given in (89). We conclude by examining the C k -regularity, recalling some denitions and properties of Lipschitz and C 1,1 -domains, and extending the results of Section 5.2 to the case where the map f in (89) is valued in another nite dimensional space than R, such as the complex 2 2 space C or the matrix spaces Rn and Cn .

5.2.1 Some denitions and notation Let n > 2 be an integer henceforth set. The space Rn is equipped with its usual Euclidean structure: v u n n X p uX n n |xk |2 . ∀ (x, y) ∈ R × R , hx | yi := xk yk and |x| := hx | xi = t k=1

k=1

More generally, any set E considered here is a real vector space provided with a norm denoted by k • kE . The space Lc (E, F ) of continuous linear maps between two such spaces is endowed with its operator norm: ∀u ∈ Lc (E, F ) ,

|||u||| := sup x∈E x6=0E

ku (x) kF . kxkE

This norm is complete as soon as k • kF is, and if in addition E = F , then Lc (E, E) is a unitary Banach algebra whose invertible elements contain the open unit ball centred at the identity map [42, Chapter 18]. We also recall that if E is a nite-dimensional real vector space, then all the norms dened on E are equivalent and complete [5, I Ÿ2 Section 3]. Hence, equipped with one of these norms, E is automatically a Banach space and moreover, any linear map u : E → F is automatically continuous, where F refers to any real normed vector space (not necessarily nite dimensional). Any well-dened map g : E → F is said to be dierentiable at a point x ∈ E if there exists a continuous linear map Lx ∈ Lc (E, F ) such that: ∀h ∈ E,

g (x + h) = g (x) + Lx (h) + khkE R (h) ,

where kR(h)kF → 0 as khkE → 0. If this is the case, the operator Lx is unique, denoted by Dx g , and called the dierential of g at the point x. We also use the notation Def (g) to refer to the set of points for which g is dierentiable. Moreover, if in addition, the map D• f : y ∈ E → Dy f ∈ Lc (E, F ) is well dened around the point x and dierentiable at x, then we say that f is twice dierentiable at x, and

63

the dierential of D• f at x is identied with a continuous bilinear map, denoted by Dx2 f and called the second-order dierential of f at x, via the following bijective linear isometry: Lc (E, Lc (E, F )) [x 7→ (ux : y 7→ ux (y))]

−→ 7−→

Bc (E × E, F ) [(x, y) 7→ ux (y)] ,

where Bc (E × E, F ) is the set of continuous bilinear maps from E × E into F equipped with the norm: ∀u ∈ Bc (E × E, F ) ,

|||u||| :=

sup (x,y)∈E×E x6=0E y6=0E

ku(x, y)kF . kxkE kykE

Note that a continuous linear map is dierentiable at any point and its dierential at any point is given by the map itself. In particular, its dierential map is constant thus any continuous linear map is twice dierentiable at any point and its second-order dierential at any point is identically equal to zero. Then, for any real p > 1, we denote by Lp the space of measurable maps from Rn into Rn whose p-th power is integrable, and by L∞ the space of measurable maps from Rn into Rn that are essentially bounded. They are respectively endowed with their usual norm: Z 1 p ∀ (f , g) ∈ Lp × L∞ , kf kp := |f (x) |p dx and kgk∞ := ess sup |g (x) |, x∈Rn

Rn

where the integration is done with respect to the usual n-dimensional Lebesgue measure denoted by Ln (•). We recall that each Lp and L∞ are Banach spaces [42, Ÿ3.11 Theorem] i.e. real vector spaces provided with complete norms. Moreover, for any measurable map f : Rn → R which is locally integrable, we say that f is weakly dierentiable if there exists a measurable map g : Rn → Rn which is locally integrable, and such that: Z Z ∀ϕ ∈ Cc∞ , hg (x) | ϕ (x)i dx = − f (x) div ϕ (x) dx, Rn

Rn

where Cc∞ refers to the set of smooth maps from Rn into Rn with compact support. If this is the case, the function g is unique, denoted by ∇f , and called the weak gradient of f . For any real p > 1, we can now introduce the Sobolev space W 1,p as the set of functions f ∈ Lp (Rn , R) that are weakly dierentiable and whose weak gradients ∇f are functions of Lp . Moreover, any W 1,p is a Banach space [6, Chapter IX] endowed with the norm: 1 Z p p 1,p |f (x) | dx + k∇f kp . ∀f ∈ W , kf k1,p := Rn

We also introduce the space C

of Lipschitz continuous maps from Rn into Rn equipped with the norm:

0,1

∀θ ∈ C 0,1 ,

kθk0,1 :=

sup (x,˜ x)∈Rn ×Rn ˜ x6=x

|θ (x) − θ (˜ x) | . ˜| |x − x

We emphasize the fact that C 0,1 is not a Banach space i.e. k • k0,1 is a real norm which is not complete. We recall that the space C 0,1 can be identied with the subspace of continuous maps from Rn into Rn whose weak partial derivatives are functions of L∞ [17, Section 4.2.3]. Moreover, any Lipschitz continuous map is dierentiable almost everywhere [17, Section 3.1.2] and we have: ∀θ ∈ C 0,1 ,

kθk0,1 = ess sup |||Dx θ|||. x∈Rn

We also consider the space W provided with the norm: We recall that W

1,∞

1,∞

∞

=L

∩C

0,1

of Lipschitz continuous bounded maps from Rn into Rn ,

∀θ ∈ W 1,∞ , kθk1,∞ := kθk∞ + kθk0,1 . is a Banach space [6, IX.1]. The identity map from Rn into Rn is referred to as I .

Finally, let k > 2 be an integer. We dene recursively the Sobolev spaces W k,p as the set of all maps f : Rn → R that are in W 1,p and such that each component of its weak gradient is a function of W k−1,p . It can be endowed with the norm: n X ∀f ∈ W k,p , kf kk,p := kf k1,p + k∂i f kk−1,p . i=1

Similarly, the space W is dened recursively as the set of all maps θ : Rn → Rn that are in W 1,∞ such that their partial derivatives are functions of W k−1,∞ . The space W k,∞ is equipped with the norm: k,∞

∀θ ∈ W k,∞ ,

kθkk,∞ := kθk1,∞ +

n X

k∂i θkk−1,∞ .

i=1

To conclude, it can be checked that W k,p and W k,∞ are Banach spaces for any integer k > 2 [6, IX.1].

64

5.2.2 About the dierentiability properties related to the inverse operator Proposition 5.31.

Let θ ∈ C 0,1 be such that kθk0,1 < 1. Then, I + θ is a (1 + kθk0,1 )-Lipschitz continuous 1 map which is invertible, and its inverse (I + θ)−1 is a 1−kθk -Lipschitz continuous map satisfying: 0,1 k (I + θ)−1 − Ik0,1 6

kθk0,1 . 1 − kθk0,1

(96)

In particular, the map θ ∈ C 0,1 7→ (I + θ)−1 ∈ C 0,1 is well dened on the open unit ball of C 0,1 centred at the origin, and it is continuous at the origin. If in addition, we assume that θ is bounded i.e. θ ∈ W 1,∞ , then we have (I + θ)−1 − I ∈ W 1,∞ and the following estimations hold true: kθk1,∞ , 1 − kθk0,1  2 kθk1,∞ 6 . 2

k (I + θ)−1 − Ik1,∞ 6 k (I + θ)−1 − I + θk∞

(97) (98)

In particular, the map θ ∈ W 1,∞ 7→ (I + θ)−1 − I ∈ W 1,∞ is well dened on the open unit ball of W 1,∞ centred at the origin, and it is continuous at the origin. Moreover, the map θ ∈ W 1,∞ 7→ (I +θ)−1 −I ∈ L∞ is dierentiable at the origin, its dierential being the opposite of the inclusion map from W 1,∞ into L∞ . Proof. Let θ ∈ C 0,1 be such that kθk0,1 < 1. First, from the triangle inequality, we get the (1 + kθk0,1 )Lipschitz continuity of I + θ. Then, for any z ∈ Rn , the map x ∈ Rn 7→ z − θ(x) ∈ Rn is a contraction thus the Banach Fixed-Point Theorem [6, Theorem 5.7] asserts there exists a unique point xz ∈ Rn such that z − θ(xz ) = xz i.e. I + θ is a bijective map. Moreover, since (I + θ)−1 = I − θ ◦ (I + θ)−1 , we obtain 1 -Lipschitz k(I + θ)−1 k0,1 6 1 + kθk0,1 k(I + θ)−1 k0,1 , from which we deduce that (I + θ)−1 is a 1−kθk 0,1 −1 −1 continuous map. Similarly, we have k(I + θ) − Ik0,1 = k − θ ◦ (I + θ) k0,1 6 kθk0,1 k(I + θ)−1 k0,1 so relation (96) holds true. Finally, if we now assume that θ is bounded, then we also get:  kθk1,∞ − kθk0,1 kθk∞ kθk1,∞ kθk0,1   = 6 k (I + θ)−1 − Ik1,∞ 6 kθk∞ +   1 − kθk0,1 1 − kθk0,1 1 − kθk0,1  2    k (I + θ)−1 − I + θk = kθ − θ ◦ (I + θ)−1 k 6 kθk kI − (I + θ)−1 k 6 kθk kθk 6 kθk1,∞ . ∞ ∞ 0,1 ∞ 0,1 ∞ 4

To conclude the proof of Proposition 5.31, (I − θ)−1 − I ∈ W 1,∞ and relations (96)-(97)-(98) hold true.

Lemma 5.32.

2

Let θ ∈ C 0,1 . Then, the dierential map D• θ : x ∈ Rn 7→ Dx θ ∈ Rn is well dened almost √ 2 everywhere, measurable, and it is essentially bounded by nkθk0,1 in the matrix space Rn . In particular, 2 the map θ ∈ C 0,1 7→ D• θ ∈ L∞ (Rn , Rn ) is well dened, linear, and continuous. Moreover, for almost n every point + θ) = I + Dx θ and if we assume that kθk0,1 < 1, then we also have  x ∈ R−1, we have Dx (I D(I+θ)(x) (I + θ) = (I + Dx θ)−1 for almost every point x ∈ Rn . Proof. Let θ ∈ C 0,1 and Def (θ) contain the points of Rn for which θ is dierentiable. First, Rademacher's Theorem [17, Section 3.1.2] ensures that θ is dierentiable almost everywhere i.e. Ln (Rn \Def (θ)) = 0. Then, the dierential Dx θ of θ at any point x ∈ Def (θ) is a well-dened linear map, thus identied with its (n × n)-matrix representation in the canonic basis of Rn denoted by (e1 , . . . , en ). Hence, the map 2 D• θ : x ∈ Def (θ) 7→ Dx θ ∈ Rn is measurable if and only if any [D• θ]ij : x ∈ Def (θ) 7→ ∂j θi (x) ∈ R is measurable, which is the case since they respectively are the pointwise limits of the continuous maps k (θij )k∈N : x ∈ Rn 7→ k[θi (x + k1 ej ) − θi (x)] ∈ R. Moreover, for any point x ∈ Def (θ), we have: v uX u n √ √ √ kDx θkRn2 := t |∂j θi (x) |2 6 n |Dx θ (ej0 ) | 6 n ess sup |||Dx θ||| := n kθk0,1 , (99) x∈Rn

i,j=1

2

where j0 ∈ J1, nK is such that |∂j0 θi (x)| = max16j6n |∂j θi (x)|. Hence, θ ∈ C 0,1 7→ D• θ ∈ L∞ (Rn , Rn ) is a well-dened map, which is also linear (thus continuous by (99)). Indeed, for any (θ1 , θ2 , λ) ∈ C 0,1 ×C 0,1 ×R, we have Dx θ1 + λDx θ2 = Dx (θ1 + λθ2 ) for any x ∈ Def (θ1 ) ∩ Def (θ2 ). Using Rademacher's Theorem, the last equality holds true almost everywhere i.e. D• (θ1 + λθ2 ) = D• θ1 + λD• θ2 . In particular, one can check that Def (θ) = Def (I + θ) and I + Dx θ = Dx (I + θ) for any x ∈ Def (θ). Using again Rademacher's Theorem, we deduce that the last equality holds true almost everywhere. Finally, we assume that kθk0,1 < 1 so Proposition 5.31 ensures that the Lipschitz continuous map (I + θ) has a Lipschitz continuous inverse. Hence, at any point x ∈ A := Def (I+θ)∩(I+θ)−1 hDef [(I+θ)−1 ]i, we can dierentiate (I+θ)−1 ◦(I+θ) = I and Lemma 5.36 gives D(I+θ)(x) [(I + θ)−1 ](I + Dx θ) = I . Since |||Dx θ||| 6 kθk0,1 < 1, the matrix I + Dx θ has an inverse [42, Ÿ18.3], which is multiplied to the last equality to get D(I+θ)(x) [(I + θ)−1 ] = (I + Dx θ)−1 . Combining [17, Section 2.4.1 Theorem 1] and [17, Section 2.2 Theorem 2] with Rademacher's Theorem, we deduce that Ln (Rn \A) = 0 i.e. D(I+θ)(x) [(I + θ)−1 ] = (I + Dx θ)−1 for almost every point x ∈ Rn , concluding the proof of Lemma 5.32.

65

Proposition 5.33.

2

Let θ ∈ C 0,1 be such that kθk0,1 < 1. Then, (I +D• θ)−1 : x ∈ Rn 7→ (I +Dx θ)−1 ∈ Rn is well dened almost everywhere and it is a measurable map satisfying for almost every point x ∈ Rn : √ n −1 k (I + Dx θ) kRn2 6 . (100) 1 − kθk0,1 2

In particular, the map θ ∈ C 0,1 7→ (I + D• θ)−1 ∈ L∞ (Rn , Rn ) is well dened on the open unit ball of C 0,1 centred at the origin. Moreover, it is dierentiable at the origin and its dierential at the origin is given 2 by the continuous linear map θ ∈ C 0,1 7→ −D• θ ∈ L∞ (Rn , Rn ). 2

Proof. Let θ ∈ C 0,1 be such that kθk0,1 < 1. First, from Lemma 5.32, the map D• θ : x ∈ Rn 7→ Dx θ ∈ Rn is well dened for almost every point x ∈ Rn and since |||Dx θ||| 6 kθk0,1 < 1, the matrix I + Dx θ is invertible [42, Ÿ18.3]. Hence, the map (I + D• θ)−1 is well dened almost everywhere. Moreover, it is measurable as the composition between D• θ, which is measurable by Lemma 5.32, and the map 2 2 A ∈ {B ∈ Rn , |||B||| < 1} 7→ (I + A)−1 ∈ Rn which is continuous [42, Ÿ18.4]. Then, we use successively −1 Lemma 5.32, relation (99) applied to (I + θ) , and Proposition 5.31 to get for almost every point x ∈ Rn : √   √ n k (I + Dx θ)−1 kRn2 6 kD• (I + θ)−1 kL∞ (Rn ,Rn2 ) 6 nk (I + θ)−1 k0,1 6 . 1 − kθk0,1 2

Hence, the map θ ∈ C 0,1 7→ (I + D• θ)−1 ∈ L∞ (Rn , Rn ) is well dened on the open unit ball of C 0,1 centred at the origin. It remains to prove that it is dierentiable at the origin. From Lemma 5.32, the 2 map f : θ ∈ C 0,1 7→ D• θ ∈ L∞ (Rn , Rn ) is well dened, linear and continuous. In particular, f is dierentiable at any point and its dierential at any point is the map f itself. In addition, the map 2 2 g : A ∈ {B ∈ Rn , |||B||| < 1} 7→ (I + A)−1 ∈ Rn is dierentiable at the origin [42, Ÿ18.4], its dierential 2 2 at the origin being the continuous linear map D0 g : A ∈ Rn 7→ −A ∈ Rn . Applying Lemma 5.36, we 2 deduce that the map g ◦ f : θ ∈ C 0,1 7→ (I + D• θ)−1 ∈ L∞ (Rn , Rn ) is dierentiable at the origin, and its dierential at the origin is given by the following continuous linear map: ∀θ ∈ C 0,1 ,

D0 (g ◦ f )(θ) = Df (0) g[D0 f (θ)] = D0 g[f (θ)] = −f (θ) = −D• θ,

concluding the proof of Proposition 5.33.

5.2.3 About the dierentiability properties related to the Jacobian determinant Proposition 5.34.

Let θ ∈ C 0,1 be such that kθk0,1 < 1. Then, the Jacobian determinant of I + θ i.e. the function x ∈ Rn 7→ det[Dx (I + θ)] ∈ R is well dened almost everywhere. In addition, it is a measurable map which satises for almost every point x ∈ Rn : |det [Dx (I + θ)] | 6

n! , 1 − kθk0,1

(101)

In particular, the map θ ∈ C 0,1 7→ det[D• (I + θ)] ∈ L∞ (Rn , R) is well dened on the open unit ball of C 0,1 centred at the origin. Moreover, it is dierentiable at the origin, and its dierential at the origin is given by the divergence operator i.e. the continuous linear map θ ∈ C 0,1 7→ div(θ) := trace(D• θ) ∈ L∞ (Rn , R). 2

Proof. Let θ ∈ C 0,1 be such that kθk0,1 < 1. From Lemma 5.32, the map D• θ : x ∈ Rn 7→ Dx θ ∈ Rn is well dened almost everywhere and measurable. Since the determinant map of Denition 5.8 is continuous by Proposition 5.11, and since D• (I + θ) = I + D• θ by Lemma 5.32, the Jacobian determinant of I + θ is well dened almost everywhere and measurable. Denoting by Sn the set of permutations of n elements i.e. the set of bijective maps from J1, nK into J1, nK, and introducing the signature map s : Sn → {−1, 1} of Denition 5.4, we can express the Jacobian determinant of I + θ according to Denition 5.8. It follows for almost every point x ∈ Rn : det [Dx (I + θ)]

= det (I + Dx θ)

=

X

s (p)

p∈Sn

n Y 

Ip(i)i + ∂i θp(i) (x)



(102)

i=1

Expanding the above product and using the fact that Ip(j)j = 1 if and only if j = p(j), we deduce that: det [Dx (I + θ)] =

n X k=0

X

X

Ik ⊆J1,nK cardIk =k

p∈Sn ∀j ∈I / k ,p(j)=j

s (p)

Y

∂i θp(i) (x) .

(103)

i∈Ik

Then, we observe that ∂i θp(i) (x) is the p(i)-th component of the vector Dx θ(ei ) where ei is the unit vector whose components are zero except the i-th one which is equal to one. Therefore, we have: v uX u n |∂i θp(i) (x) | = | [Dx θ (ei )]p(i) | 6 t [Dx θ (ei )]2j = |Dx θ (ei ) | 6 |||Dx θ||| |ei | 6 kθk0,1 . (104) j=1

66

Combining (103) and (104), we obtain: n X

|det [Dx (I + θ)] | 6

k=0

X

X

Ik ⊆J1,nK cardIk =k

p∈Sn ∀j ∈I / k ,p(j)=j

Y

|s (p) | |∂i θp(i) (x) | | {z } i∈I | {z } k

=1

6

n X

k−1 Y

k=0

i=0

! (n − i) kθkk0,1 .

6kθk0,1

Q The last inequality comes from the fact that i∈Ik kθk0,1 = kθkk0,1 does not depend on p and Ik . It can thus be removed from the corresponding sums, for which we know that card{p ∈ Sn | ∀j ∈ / Ik , p(j) = j} = k!
and card{Ik ⊆ J1, nK, card Ik = k} = nk = n(n−1)...(n−k+1) . Hence, we get: k! |det [Dx (I + θ)] | 6

n X

n(n − 1) . . . (n − k + 1)kθkk0,1 6 n!

k=0

n X

kθkk0,1 6 n!

k=0

+∞ X

kθkk0,1 =

k=0

n! . 1 − kθk0,1

P Note that the last inequality holds true because we assume kθk0,1 < 1 thus the geometric series k∈N kθkk0,1 0,1 converges. Hence, estimation (101) holds true. Finally, the map J : θ ∈ C 7→ det[D• (I +θ)] ∈ L∞ (Rn , R) is the composition of the determinant map det of Denition 5.8 with the ane map I + f , where f is its 2 linear part dened as f : θ ∈ C 0,1 7→ D• θ ∈ L∞ (Rn , Rn ). Since the determinant is dierentiable by Proposition 5.11 and since f is linear and continuous by Lemma 5.32, we deduce from Lemma 5.36 that J is dierentiable at the origin and its dierential at the origin is given by: ∀θ ∈ C 0,1 , D0 J(θ) = D0 [det ◦ (I + f )] (θ) = DI det[D0 (I + f )(θ)] = trace [f (θ)] = trace(D• θ) = div (θ) .

To conclude the proof, the divergence operator is the dierential of J = det ◦ (I + f ) at the origin.

Corollary 5.35.

Let θ ∈ C 0,1 be such that kθk0,1 < 1. Then, the map x ∈ Rn 7→ det([Dx (I + θ)]−1 ) ∈ R is well dened almost everywhere, measurable, and it satises for almost every point x ∈ Rn :
|det [Dx (I + θ)]−1 | 6

n! . (1 − kθk0,1 )n

(105)

In particular, the map θ ∈ C 0,1 7→ det([D• (I + θ)]−1 ) ∈ L∞ (Rn , R) is well dened on the open unit ball of C 0,1 centred at the origin. Proof. Let θ ∈ C 0,1 be such that kθk0,1 < 1. Considering Proposition 5.33 and Lemma 5.32, the map (I + D• θ)−1 = [D• (I + θ)]−1 is well dened almost everywhere and measurable . Since the determinant is a continuous map, we deduce that the map x ∈ Rn 7→ det([Dx (I + θ)]−1 ) ∈ R is well dened almost everywhere and measurable. Applying relation (102) to [D• (I + θ)]−1 , we get for almost every x ∈ Rn :
|det [Dx (I + θ)]−1 |

=

  |det (I + Dx θ)−1 | 6

X p∈Sn

6

X  p∈Sn

1 1 − |||Dx θ|||

n

 6

n Y   | (I + Dx θ)−1 p(i)i | |s (p) | | {z } i=1 | {z } =1

1 1 − kθk0,1

6|||(I+Dx θ)−1 |||

n

! X

1

p∈Sn

=

n! . (1 − kθk0,1 )n

Hence, relation (105) holds true and the map θ ∈ C 0,1 7→ det([D• (I + θ)]−1 ) ∈ L∞ (Rn , R) is well dened on the open unit ball of C 0,1 centred at the origin, concluding the proof of Corollary 5.35.

5.2.4 About the dierentiability properties related to the composition operator Lemma 5.36. Let E , F , and G be some real normed vector spaces. We consider two maps f : E → F and g : F → G, which are well dened respectively on an open neighbourhood U of a xed point a ∈ E , and on an open neighbourhood V of f (a) ∈ F . Then, the map g◦f : x ∈ E 7→ g[f (x)] ∈ G is well dened on f −1 (V )∩U . Moreover, if we assume that f and g are respectively dierentiable at a and f (a), then g ◦ f is dierentiable at a and its dierential at a is the continuous linear map Da (g ◦ f ) : h ∈ E 7→ Df (a) g[Da f (h)] ∈ G. Proof. We refer to [2, Ÿ2.6] for a proof of this standard result of dierential calculus.

Proposition 5.37. Let E , F , and G be some real normed vector spaces. We consider two maps f : E → F and g : F → G, which are well dened and dierentiable respectively on an open neighbourhood U of a xed point a ∈ E , and on an open neighbourhood V of f (a) ∈ F . Then, the map g ◦ f : x ∈ E 7→ g[f (x)] ∈ G is well dened and dierentiable on the open neighbourhood f −1 (V ) ∩ U of a. Moreover, if we assume that f and g are respectively twice dierentiable at a and f (a), then g ◦ f is twice dierentiable at a with the following second-order dierential at a: ∀(h, k) ∈ E × E,

Da2 (g ◦ f )(h, k) = Df2 (a) g[Da f (h), Da f (k)] + Df (a) g[Da2 f (h, k)].

67

e := f −1 (V ) ∩ U . First, note that a ∈ U e , which is thus not empty, and let us prove that U e Proof. We set U e , there exits ε > 0 such that Bε (f (x)) ⊆ V and Bε (x) ⊆ U . Since f is dierentiable is open. For any x ∈ U thus continuous at x, there exists r ∈]0, ε] such that f (Br (x)) ⊆ Bε (f (x)) and it comes successively: e. Br (x) ⊆ Bε (x) ∩ f −1 [f (Br (x))] ⊆ U ∩ f −1 [Bε (f (x))] ⊆ U ∩ f −1 (V ) := U e is an open neighbourhood of a so we can apply Lemma 5.36 to any point x ∈ U e . We get that Hence, U e . In other words, the following maps are well dened: g ◦ f is well dened and dierentiable on U D• f :

−→ 7−→

U x

Lc (E, F ) Dx f

D• g :

V y

−→ 7−→

Lc (F, G) Dy g

D• (g ◦ f ) :

e U z

−→ 7−→

Lc (E, G) Df (z) g ◦ Dz f.

e and Br (f (a)) ⊆ V . Moreover, there Then, using the hypothesis, there exists r > 0, such that Br (a) ⊆ U 2 2 exists Da f ∈ Bc (E × E, F ) and Df (a) g ∈ Bc (F × F, G) such that the well-dened maps: RDf : Br (0E ) h RDg : Br (0F ) k

→ 7→ → 7→

Lc (E, F ) Da+h f − Da f − Da2 f (•, h) Lc (F, G) Df (a)+k g − Df (a) g − Df2 (a) g(•, k)

Rf : Br (0E ) h

→ 7→

F f (a + h) − f (a) − Da f (h)

are negligible at the origin compared to the corresponding identity map (i.e. respectively on E and F ). We now show that g ◦ f is twice dierentiable at the point a. Let ε > 0. We set:   ε . h i  := min ε, (ε + |||Da f |||) 1 + |||Df2 (a) g||| (ε + |||Da2 f |||) + |||Df2 (a) g|||.|||Da f ||| + |||Df (a) g||| From the denition of Rf , RDf and RDg , there exists δ ∈]0, r] such that for any h ∈ Bδ (0E ) and k ∈ Bδ (0F ), we have kRf (h)kF 6 khkE , |||RDf (h)||| 6 khkE and |||RDg (k)||| 6 kkkF . We introduce the map: RD(g◦f ) :

Bη (0E ) h

−→ 7−→

Lc (E, G) , Da+h (g ◦ f ) − Da (g ◦ f ) − Df2 (a) g[Da f (•), Da f (h)] − Df (a) g[Da2 f (•, h)]

where η = min(, δ, |||Daδf |||+ ). Let us prove that RD(g◦f ) is negligible at the origin compared to the identity map on E . For this purpose, we set k = Da f (h) + Rf (h) and observe that: ∀h ∈ Bη (0E ),

kkkF 6 (|||Da f ||| + ) khkE < δ,

from which we deduce that the map RD(g◦f ) is well dened on Bη (0E ) since Da+h (g◦f ) = Df (a)+k g◦Da+h f and also the estimation: ∀h ∈ Bη (0E ),

|||RDg (k)||| 6 kkkF 6  (|||Da f ||| + ) khkE .

Finally, we get for any h ∈ Bη (0E ):   RD(g◦f ) (h) = Df (a) g ◦ RDf (h) + Df2 (a) g [Da f (•), Rf (h)] + Df2 (a) g Da2 f (•, h) + RDf (h), k + RDg (k),

so we have:

 |||RD(g◦f ) (h)||| 6 khkE |||Df (a) g||| + |||Df2 (a) g|||.|||Da f ||| + kkkF 1 + |||Df2 (a) g|||  + |||Da2 f ||| .

Combining the above inequality with the estimation on k, we obtain |||RD(g◦f ) (h)||| 6 εkhkE as required. ˜ ∈ E × E → D2 g[Da f (h), Da f (h)] ˜ + Df (a) g[Da2 f (h, h)] ˜ ∈ G is a To conclude the proof, Da2 (g ◦ f ) : (h, h) f (a) continuous (symmetric) bilinear map thus g ◦ f is twice dierentiable at a.

Proposition 5.38. Let us consider any f ∈ L1 (Rn , R). Then, the map θ ∈ C 0,1 7→ f ◦ (I + θ) ∈ L1 (Rn , R) is well dened on the open unit ball of C 0,1 centred at the origin. Moreover, the following composition map θ ∈ W 1,∞ 7→ f ◦ (I + θ) ∈ L1 (Rn , R) is continuous at the origin. If in addition, we assume that f ∈ W 1,1 , then the map θ ∈ W 1,∞ 7→ f ◦ (I + θ) ∈ L1 (Rn , R) is dierentiable at the origin and its dierential is given by the continuous linear map θ ∈ W 1,∞ 7→ h∇f | θi ∈ L1 (Rn , R). Proof. Let f ∈ L1 (Rn , R). First, we check that the map θ ∈ C 0,1 7→ f ◦ (I + θ) ∈ L1 (Rn , R) is well dened around the origin. Consider any θ ∈ C 0,1 such that kθk0,1 < 1. The function f ◦ (I + θ) is measurable as the composition of the Lipschitz continuous map (I + θ) with the measurable map f . Proposition 5.31 ensures that the map (I + θ) is bijective and its inverse (I + θ)−1 is also Lipschitz continuous. We deduce that for almost every point x ∈ Rn , we have:

1 = det (I) = det [Dx (I + θ)]−1 ◦ Dx (I + θ) = det [Dx (I + θ)]−1 det [Dx (I + θ)] .

68

Consequently, using this last observation, relation (105), and the change of variables formula valid for any Lipschitz continuous map [17, Section 3.3.3], we get: Z Z   |f [x + θ (x)] |dx = |f [x + θ (x)] det (I + Dx θ)−1 det [Dx (I + θ)] |dx Rn

Rn

n! (1 − kθk0,1 )n

6

Z n

|R

|f [x + θ (x)] det (Dx (I + θ)) |dx. {z } Z

(106)

|f (y) |dy < +∞

= Rn

Hence, we obtain f ◦ (I + θ) ∈ L1 (Rn , R) for any f ∈ L1 (Rn , R) and any θ ∈ C 0,1 such that kθk0,1 < 1. Then, we prove that the map θ ∈ W 1,∞ 7→ f ◦(I +θ) ∈ L1 (Rn , R) is continuous at the origin. Let θ ∈ W 1,∞ be such that kθk1,∞ 6 21 . We proceed by a density argument: there exists a sequence of smooth maps (fi )i∈N : Rn → R with compact support converging to f in L1 (Rn , R) [6, IV.23]. On the one hand, we have kθk0,1 6 21 < 1 so the foregoing holds true and we can apply the arguments of (106) to the map f − fi in order to get for any i ∈ N: Z n!kf − fi kL1 (Rn ,R) 6 n!2n kf −fi kL1 (Rn ,R) . kf ◦(I + θ)−fi ◦(I + θ) kL1 (Rn ,R) = |(f −fi ) [x + θ(x)] |dx 6 (1 − kθk0,1 )n Rn On the other hand, since the map fi is smooth with compact support, we have for any i ∈ N: Z |fi [x + θ (x)] − fi (x) |dx 6 kfi kC 0,1 (Rn ,R) kθk∞ Ln (supp fi ) . kfi ◦ (I + θ) − fi kL1 (Rn ,R) = Rn

Combining the triangle inequality with these two observations, we deduce that for any i ∈ N: kf ◦ (I + θ) − f kL1 (Rn ,R) 6 (1 + 2n n!)kf − fi kL1 (Rn ,R) + kfi kC 0,1 (Rn ,R) kθk1,∞ Ln (supp fi ) .

(107)

Let ε > 0. There exists I ∈ N such that kfI − f kL1 (Rn ,R) 6 We set:   1 ε δ := min , . 2 2kfI kC 0,1 (Rn ,R) Ln (supp fI ) ε . 2(1+2n n!)

For any θ ∈ W 1,∞ such that kθk1,∞ < δ , we get from (107) that kf ◦ (I + θ) − f kL1 (Rn ,R) 6 ε i.e. the map θ ∈ W 1,∞ 7→ f ◦ (I + θ) ∈ L1 (Rn , R) is continuous at the origin. We now assume that f ∈ W 1,1 and we prove that θ ∈ W 1,∞ 7→ f ◦ (I + θ) ∈ L1 (Rn , R) is dierentiable at the origin. First, note that the linear map θ ∈ W 1,∞ 7→ h∇f | θi ∈ L1 (Rn , R) is well dened and continuous since we get from the Cauchy-Schwarz inequality: Z Z Z |∇f (x) |dx 6 kθk1,∞ k∇f k1 < +∞. (108) | h∇f (x) | θ (x)i |dx 6 |∇f (x) | |θ (x) |dx 6 kθk∞ Rn

Rn

Rn

We want to show it is the dierential of θ ∈ W 1,∞ 7→ f ◦(I +θ) ∈ L1 (Rn , R) at the origin. For this purpose, we introduce the following map: Rf :

W 1,∞ θ

L1 (Rn , R) Rf (θ) := f ◦ (I + θ) − f − h∇f | θi.

−→ 7−→

From the foregoing observations (106) and (108), the map Rf is well dened on the open unit ball of W 1,∞ centred at the origin. Therefore, let θ ∈ W 1,∞ be such that kθk1,∞ < 1. Now, let us assume for a moment that f is a smooth map with compact support i.e. f ∈ Cc∞ (Rn , R). We consider x ∈ Rn and introduce the function ϕ : t ∈ [0, 1] 7→ f [x + tθ(x)]. Since ϕ is the composition of the ane map gx : t ∈ [0, 1] 7→ x + tθ(x) ∈ Rn with f , using Lemma 5.36, it is dierentiable on [0, 1] and we have:

 ∀t ∈ [0, 1], ϕ0 (t) = Dt (f ◦ gx ) = Dgx (t) f ◦ Dt gx = ∇f [gx (t)] | gx0 (t) = h∇f [x + tθ (x)] | θ (x)i . R1 From the Fundamental Theorem of Calculus [42, Ÿ7.16], we get 0 [ϕ0 (t) − ϕ0 (0)]dt = ϕ(1) − ϕ(0) − ϕ0 (0) thus for any x ∈ Rn , we obtain: Z 1 h∇f [x + tθ (x)] − ∇f (x) | θ (x)i dt = f [x + θ (x)] − f (x) − h∇f (x) | θ (x)i . 0

Then, it comes successively: Z |Rf (θ)|

Z

1

Z

h∇f [x + tθ (x)] − ∇f (x) | θ (x)i dt dx

=

Rn

Rn

1

Z

Z 6

0

Rn

 | h∇f [x + tθ (x)] − ∇f (x) | θ (x)i |dt dx

0

6

1

Z

Z |θ (x) | Rn

 |∇f [x + tθ (x)] − ∇f (x) |dt dx.

0

69

Hence, using the Fubini-Tonelli Theorem [42, Ÿ8.8 Theorem], we have established that:  Z 1 Z ∞ n |∇f [x + tθ (x)] − ∇f (x) |dt dx. ∀f ∈ Cc (R , R), kRf (θ)kL1 (Rn ,R) 6 kθk1,∞ 0

Rn

(109)

We now assume that f ∈ W 1,1 and we show that (109) still holds true by a density argument. Indeed, there exists a sequence (fi )i∈N of smooth maps with compact support converging to f in the W 1,1 -norm [7, Theorem 9.2]. Let i ∈ N. We have from the triangle inequality and relation (109) applied to fi :  Z 1 Z |∇fi [x + tθ (x)] − ∇fi (x) |dt dx . kRf (θ)kL1 (Rn ,R) 6 kRf (θ) − Rfi (θ)kL1 (Rn ,R) + kθk1,∞ Rn |0 {z } e f (θ) :=R i

On the one hand, we combine relation (106) applied to the maps fi − f and θ, with observation (108) applied to ∇(fi − f ) in order to obtain:   n! kRf (θ) − Rfi (θ)kL1 (Rn ,R) 6 1 + kfi − f kL1 (Rn ,R) + kθk∞ k∇fi − ∇f k1 . (1 − kθk0,1 )n On the other hand, we combine the triangle inequality with relation (106) applied ∇(fi − f ) and tθ to get:     Z 1 n! n! ef (θ) 6 R ef (θ) + 1 + ef (θ) + 1 + R dt k∇f − ∇f k 6 R k∇fi − ∇f k1 . i 1 n n i (1 − kθk0,1 ) 0 (1 − ktθk0,1 ) Therefore, from these two last inequalities, we deduce that:   ef (θ) + 1 + 2kθk1,∞ + n! 1 + kθk1,∞ n kfi − f k1,1 . kRf (θ)kL1 (Rn ,R) 6 kθk1,∞ R (1 − kθk1,∞ ) By letting i → +∞, we get that (109) hold true for any f ∈ W 1,1 . Finally, it only remains to prove that ef (θ)| → 0 as kθk1,∞ → 0 so as to conclude on the dierentiability of θ ∈ W 1,∞ 7→ f ◦ (I + θ) ∈ L1 (Rn , R) |R at the origin. Again, we use a density argument. Let f ∈ W 1,1 and θ ∈ W 1,∞ be such that kθk1,∞ 6 21 . There exists a sequence (fi )i∈N of smooth maps with compact support converging to f in W 1,1 . Let i ∈ N. As before, we get from the triangle inequality and relation (106) applied to the maps ∇(fi − f ) and tθ:   n! n ef (θ)| 6 R ef (θ) + 1 + e |R n kfi − f k1,1 6 Rfi (θ) + (1 + 2 n!)kfi − f k1,1 . i (1 − kθk1,∞ ) Moreover, since fi is smooth with compact support, we have:  Z 1 Z ef (θ) := R |∇f [x + tθ (x)] − ∇f (x) |dt dx 6 Ln (supp fi ) kθk1,∞ k∇fi k0,1 . i i i 0

Rn

We set δ = min{ 21 , 2Ln (supp fεI )k∇fI k0,1 }. ef (θ)| 6 ε as required. From the two last estimations, for any θ ∈ W such that kθk1,∞ < δ , we obtain |R ef (θ)| → 0 as kθk1,∞ → 0 so the map θ ∈ W 1,∞ 7→ f ◦ (I + θ) ∈ L1 (Rn , R) is dierentiable To conclude, |R at the origin, and its dierential is the continuous linear map θ ∈ W 1,∞ 7→ h∇f | θi ∈ L1 (Rn , R). Let ε > 0. There exists I ∈ N such that kfI − f k1,1 6

ε . 2(1+2n n!)

1,∞

5.2.5 About the shape derivative of a volume integral at any order General results about the structure of shape derivatives (see e.g. [34] and the references therein) often assume a priori the shape dierentiability of the functional. In this section, we consider the particular case of a volume integral and we prove by induction its shape dierentiability at any order. More precisely, under the W k,1 -regularity of the integrand and the measurability of the domain, we get the C k -regularity around the origin of the map FΩ : θ ∈ W 1,∞ 7→ F [(I + θ)(Ω)] associated with (89), and also an explicit formula for the shape derivative of a volume integral at any order. First, we state a continuity result.

Proposition 5.39.

Let n R> 2 be any integer, f ∈ L1 (Rn , R), and Ω be any measurable subset of Rn . Then, the map θ ∈ C 0,1 7→ (I+θ)(Ω) f ∈ R is well dened on the open unit ball of C 0,1 centred at the origin R denoted by B0,1 . Moreover, θ ∈ W 1,∞ 7→ (I+θ)(Ω) f ∈ R is continuous at any point of B0,1 ∩ W 1,∞ . Proof. Let n > 2 be any integer, f ∈ L1 (Rn , R), and Ω be any measurable subset of Rn . We consider any θ ∈ C 0,1 such that kθk0,1 < 1. From Proposition 5.31, the map I +Rθ has a Lipschitz continuous inverse. We deduce that (I + θ)(Ω) is measurable so the map θ ∈ C 0,1 7→ (I+θ)(Ω) f ∈ R is well dened on the R open unit ball of C 0,1 centred at the origin denoted by B0,1 . We now prove θ ∈ W 1,∞ 7→ (I+θ)(Ω) f ∈ R is

70

continuous at the origin. Let θ ∈ W 1,∞ be such that kθk1,∞ < 1. We use the change of variables formula valid for any Lipschitz continuous map [17, Section 3.3.3] and the (reverse) triangle inequality to get: Z Z Z Z f− f [f ◦ (I + θ) − f ] |det [D• (I + θ)] | + f (|det [D• (I + θ)] | − 1) 6 (I+θ)(Ω)

Ω

Ω

6

Ω

kdet [D• (I + θ)] kL∞ (Rn ,R) kf ◦ (I + θ) − f kL1 (Rn ,R) + kdet [D• (I + θ)] − 1kL∞ (Rn ,R) kf kL1 (Rn ,R) .

Combining (101) and the continuity at the origin of the Jacobian determinant of (I + θ) ensured by Proposition 5.34 with the one of θ ∈ W 1,∞ 7→ f ◦ (I + θ) ∈ L1 (Rn , R) ensured by Proposition 5.38, we can R let kθk1,∞ → 0 in the above inequality. We thus have proved that the map θ ∈ W 1,∞ 7→ (I+θ)(Ω) f ∈ R is continuous at the origin. Finally, let θ ∈ W 1,∞ ∩ B0,1 , where W 1,∞ ∩ B0,1 endowed with the W 1,∞ -norm. From Proposition 5.31, the Lipschitz continuous map I + θ is bijective and its inverse (I + θ)−1 is also a Lipschitz continuous map. In particular, we deduce that (I + θ)(Ω) is measurable. Moreover, we observe that for any h ∈ W 1,∞ such that khk1,∞ < 1 − kθk0,1 , we have kθ + hk0,1 < 1 so we can write: Z Z Z Z f− f= f− f, (110) (I+θ+h)(Ω)

(I+θ)(Ω)

(I+hθ )(Ωθ )

Ωθ

where we have set Ωθ := (I + θ)(Ω) and hθ := h ◦ (I + θ)−1 . Since we have khθ k∞ 6 khk∞ and also khk khθ k0,1 6 khk0,1 k(I + θ)−1 k0,1 6 1−kθk0,1 by Proposition 5.31, we deduce that: 0,1 khθ k1,∞ 6 khk∞ +

khk1,∞ − khk∞ kθk0,1 khk1,∞ khk0,1 = 6 , 1 − kθk0,1 1 − kθk0,1 1 − kθk0,1

(111)

and in particular R we deduce khθ k1,∞ → 0 as khk1,∞ → 0. Considering the continuity at the origin of h ∈ W 1,∞ 7→ (I+h)(Ω ) f , by letting khθ k1,∞ → 0 in (110), we get the continuity at θ. We have thus θ R obtained the continuity of θ ∈ W 1,∞ 7→ (I+θ)(Ω) f at any point of W 1,∞ ∩ B0,1 , concluding the proof.

Theorem 5.40.

Let n > 2 and k0 > 1 be two integers. We consider any f ∈ W k0 ,1 (Rn , R) and any measurable subset Ω of Rn . We denote by RB0,1 the open unit ball of C 0,1 (Rn , Rn ) centred at the origin. Then, the map FΩ : θ ∈ W 1,∞ (Rn , Rn ) 7→ (I+θ)(Ω) f ∈ R is k0 times dierentiable at the origin and for any k ∈ J1, k0 K, its dierential of order k at the origin is given by the following continuous symmetric k-linear form dened for any (θ1 , . . . , θk ) ∈ W 1,∞ (Rn , Rn ) × . . . × W 1,∞ (Rn , Rn ) by:   Z n k k−l X X X Y   ∂ f Q D0k FΩ (θ1 , . . . , θk ) := [θj (x)]ij  (x)    ∂x i Ω j i ,...,i =1 l=0 1

k

Il ⊆J1,kK card Il =l

   

j∈J1,kK j ∈I / l

j∈J1,kK j ∈I / l

 X

p:Il →Il p bijective





(112)


Y  s p−1 [Dx θj ]ij ip(j)  Il ◦ p ◦ pIl  dx, j∈Il

where s : Sl → {−1, 1} is the signature map introduced in Denition 5.4, and where pIl : J1, lK → Il is the unique strictly increasing map of Lemma 5.20. In other words, the functional (89) is k0 times shape dierentiable at any measurable subset of Rn , and its shape derivative of order R k is given by (112) for any k ∈ J1, k0 K. Moreover, the associated map FΩ : θ ∈ W 1,∞ (Rn , Rn ) 7→ (I+θ)(Ω) f ∈ R is k0 times continuously dierentiable at any point of W 1,∞ (Rn , Rn ) ∩ B0,1 and for any k ∈ J1, k0 K, its k-th order dierential is well dened by the following continuous map:
D•k FΩ : W 1,∞ (Rn , Rn ) ∩ B0,1 −→ Lkc W 1,∞ (Rn , Rn )k , R   θ0 7−→ (θ1 , ..., θk ) 7→ D0k F(I+θ0 )(Ω) θ1 ◦ (I + θ0 )−1 , ..., θk ◦ (I + θ0 )−1 , (113) where D0k F(I+θ0 )(Ω) is the k-th order shape derivative of F at (I +θ0 )(Ω) given by (112), and where Lkc refers 1 ,...,θk )| to the class of continuous k-linear maps, with the norm |||•||| := sup(θ1 ,...,θk )∈(W 1,∞ \{0})k kθ1|(•)(θ . k1,∞ ...kθk k1,∞ Proof. WeP will prove Q this result by induction on the integer k ∈ N. First, recalling the usual conventions ∂ 0 f = f , i∈∅ = 0, i∈∅ = 1, and Dθ00 FΩ = FΩ (θ0 ) = F(I+θ0 )(Ω) (0), we deduce from Proposition 5.39 that Theorem 5.40 is true for k = 0. Let us assume that it is also true for some integer k > 0. Let n > 2 be an integer, f ∈ W k+1,1 , and Ω be any measurable subset of Rn . The induction hypothesis ensures that the map FΩ : θ ∈ W 1,∞ 7→ F [(I + θ)(Ω)] ∈ R is k times continuously dierentiable at any point of W 1,∞ ∩ B0,1 , where B0,1 refers to the open unit ball of C 0,1 centred at the origin. We now show that the additional regularity assumption we made on f allows the function (113) to be dierentiable at the origin.

71

Let (θ0 , θ1 , . . . , θk ) ∈ (W 1,∞ )k+1 be such that kθ0 k1,∞ < 1. First, we express the k-th order dierential in a simpler form, using the change of variables formula for Lipschitz continuous maps [17, Section 3.3.3]:   Z n k X X X  Y    Dθk0 FΩ (θ1 , . . . , θk ) := ∂ k−lQ ij f  θj ◦ (I + θ0 )−1 i    i1 ,...,ik =1

l=0

(I+θ0 )(Ω)

Il ⊆J1,kK card Il =l

j

j∈J1,kK j ∈I / l

j∈J1,kK j ∈I / l

 X

s

p−1 Il




◦ p ◦ p Il 

p:Il →Il p bijective

 Y

  D• θj ◦ (I + θ0 )−1 i

j ip(j)



 =

n X

k X

X

i1 ,...,ik =1 l=0 Il ⊆J1,kK cardIl =l

Z

(∂ k−lQ

Ω

ij f ) j∈J1,kK j ∈I / l

  Y [θj ]ij  ◦ (I + θ0 )   |det[D• (I + θ0 )]|  j∈J1,kK j ∈I / l

 X



j∈Il

 Y

 s p−1 Il ◦ p ◦ p Il


p:Il →Il p bijective

D(I+θ0 )(•)



 θj ◦ (I + θ0 )−1 i

j ip(j)

.

j∈Il

For any θ ∈ W 1,∞ , we set A := Def (θ) ∩ Def (θ0 ) ∩ (I + θ0 )−1 [Def (θ ◦ (I + θ0 )−1 )] . From Lemmas 5.36 and 5.32, we get for almost every point x ∈ Rn (more precisely for any x ∈ A since Ln (Rn \A) = 0):   Dx θ (I + Dx θ0 )−1 = D(I+θ0 )(x) θ ◦ (I + θ0 )−1 . Furthermore, since det[D• (I + θ)] → 1 for the L∞ -norm as kθk1,∞ → 0, there exists δ ∈]0, 1[ such that for any kθ0 k1,∞ < δ , the Jacobian determinant of (I + θ0 ) is positive. Combining these two observations, we obtain from the foregoing for any θ0 ∈ W 1,∞ such that kθ0 k1,∞ < δ :   Z n k X X X Y   Dθk0 FΩ (θ1 , . . . , θk ) = (∂ k−lQ ij f ) ◦ (I + θ0 )  [θj ]ij    det [D• (I + θ0 )] i1 ,...,ik =1 l=0 Il ⊆J1,kK card Il =l

Ω

j∈J1,kK j ∈I / l

j∈J1,kK j ∈I / l

 X

s

p−1 Il




◦ p ◦ p Il 

p:Il →Il p bijective

 Y

D• θj (I + D• θ0 )

−1  ij ip(j)

.

j∈Il

(114) Then, we introduce the continuous (k + 1)-linear form (112), which is symmetric i.e. for any permutation p ∈ Sk+1 and (θ1 , . . . , θk+1 ) ∈ (W 1,∞ )k+1 , D0k+1 FΩ (θp(1) , . . . , θp(k+1) ) = D0k+1 FΩ (θ1 , . . . , θk+1 ) . Hence, it is a good candidate for the (k + 1)-th order dierential of FΩ . For this purpose, we express it dierently. We emphasize the fact that we have not (yet) proved that (112) is the (k + 1)-th order dierential of FΩ but we use its notation for convenience. We set θk+1 := θ0 to keep this in mind. We thus have:   Z k+1 n X X X  Y   Q D0k+1 FΩ (θ1 , . . . , θk , θ0 ) := ∂ k+1−l [θj ]ij  ij f   i1 ,...,ik ,ik+1 =1 l=0 Il ⊆J1,k+1K card Il =l

Ω

j∈J1,k+1K j ∈I / l

X p:Il →Il p bijective

j∈J1,k+1K j ∈I / l 

 s p−1 Il ◦ p ◦ p Il


Y



[D• θj ]ij ip(j)  .

j∈Il

We split the above expression into two disjoint situations, the last one being itself splitted into two subcases. In the rst situation, we assume k + 1 ∈ / Il . In this particular case, the sum on l can stop at k since we are assuming that Il has at most k elements. Moreover, we can explicit the indice ik+1 and the subset Il is included in J1, kK. In the second situation, we assume k + 1 ∈ Il . Similarly, the sum on l can start from one since we are assuming that Il is not empty. Then, two subcases follow. On the one hand, we assume p(k + 1) = k + 1. In this case, this is equivalent to search only for subsets Il−1 ⊆ J1, kK of l − 1 pairwise distinct elements, and also bijective maps q : Il−1 → Il−1 , then set Il := Il−1 ∪ {k + 1} and p := q on Il−1 with p(k + 1) := k + 1. On the other hand, we assume p(k + 1) 6= k + 1 so we can make a partition on the

72

bijections p : Il → Il by xing the element k + 1. We thus have:   Z n k n X X X X   D0k+1 FΩ (θ1 , . . . , θk , θ0 ) = [θ0 ]ik+1 ∂ik+1 ∂ k−lQ  Ω

i1 ,...,ik =1 l=0 Il ⊆J1,kK card Il =l

   Y     [θj ]ij  ij f    

j∈J1,kK j ∈I / l

ik+1 =1

j∈J1,kK j ∈I / l

 X p:Il →Il p bijective 

 n X

+

k+1 X

i1 ,...,ik =1 l=1

Z

X

k−(l−1) Q ij f j∈J1,kK j ∈I / l−1

∂ Ω

Il−1 ⊆J1,kK card Il−1 =l−1 Il :=Il−1 ∪{k+1}

p−1 Il

s

n X




◦ p ◦ p Il 

+

i1 ,...,ik =1 l=1

Z

X

k−(l−1) Q ij f j∈J1,kK j ∈I / l−1

∂ Ω

Il−1 ⊆J1,kK card Il−1 =l−1 Il :=Il−1 ∪{k+1}

 X j0 ∈Il−1

X

[D• θ0 ]ik+1 ik+1  

 Y

[D• θj ]ij iq(j) 

j∈Il−1



 Y   [θj ]ij    j∈J1,kK j ∈I / l−1



n X


 s p−1 Il ◦ p ◦ pIl

p:Il →Il p bijective p(j0 )=k+1 p(k+1)6=k+1



ik+1 =1

 k+1 X

[D• θj ]ij ip(j) 

j∈Il

j∈J1,kK j ∈I / l−1

q:Il−1 →Il−1 q bijective p:=q on Il−1 p(k+1):=k+1 n X

 Y

  Y  [θj ]ij    

X

 s p−1 Il ◦ p ◦ p Il


[D• θj0 ]ij

i 0 k+1

ik+1 =1





 Y  [D• θj ]ij ip(j)  [D• θ0 ]ik+1 ip(k+1)    . j∈Il−1 j6=j0

In the second integral, p(k + 1) = k + 1 so the number of transpositions needed to decompose p−1 Il ◦ p ◦ pIl is −1 −1 the same than for p−1 Il−1 ◦ q ◦ pIl−1 . From Denition 5.4, we deduce that s(pIl ◦ p ◦ pIl ) = s(pIl−1 ◦ q ◦ pIl−1 ). Moreover, in the last integral, we make a change of indices r :=  p◦t, wheret is only exchanging k +1 and j0 . −1 −1 ◦r◦p ) = s p−1 From Corollary 5.3, we can also notice that s(p−1 I l Il ◦ p ◦ pIl s(pIl ◦t◦pIl ) = −s(pIl ◦p◦pIl ). Il Indeed, t permutes two indices of Il thus p−1 Il ◦ t ◦ pIl ∈ Sl is a transposition, whose signature is −1. We are then back to a summation on r for which r(k + 1) = k + 1 i.e. in the situation of the second integral. We can thus do the same foregoing procedure. Consequently, we obtain:   * + Z n k X X X  Y  D0k+1 FΩ (θ1 , . . . , θk , θ0 ) = ∇(∂ k−lQ ij f ) | θ0  [θj ]ij    i1 ,...,ik =1 l=0 Il ⊆J1,kK card Il =l

Ω

j∈J1,kK j ∈I / l

X

n X

+

k+1 X

i1 ,...,ik =1 l=1

X

∂

Il−1 ⊆J1,kK card Il−1 =l−1 Il :=Il−1 ∪{k+1}

Ω

k−(l−1) Q ij f j∈J1,kK j ∈I / l−1

 s p−1 Il ◦ p ◦ pIl


p:Il →Il p bijective 

 Z

j∈J1,kK j ∈I / l

 Y

 Y   [θj ]ij    j∈J1,kK j ∈I / l−1

 X

s



p−1 Il−1

q:Il−1 →Il−1 q bijective p:=q on Il−1 p(k+1):=k+1 

+

n X

k+1 X

i1 ,...,ik =1 l=1

Z

X Il−1 ⊆J1,kK card Il−1 =l−1 Il :=Il−1 ∪{k+1}

Ω



◦ q ◦ pIl−1 div (θ0 ) 

 Y

[D• θj ]ij iq(j) 

j∈Il−1



 Y  k−(l−1) ∂ Q ij f  [θj ]ij    j∈J1,kK j ∈I / l−1

j∈J1,kK j ∈I / l−1

 X

[D• θj ]ij ip(j) 

j∈Il

X





s p−1 Il−1 ◦ q ◦ pIl−1 [−D• θj0 D• θ0 ]ij

i 0 q(j0 )

j0 ∈Il−1 q:Il−1 →Il−1 q bijective r:=q on Il−1 r(k+1)=k+1

73



 Y   [D• θj ]ij iq(j)   . j∈Il−1 j6=j0

Note that in the previous last product, we have replaced ir[t(j)] by iq(j) since they coincide on Il−1 \{j0 }. Finally, we can notice that in the two last integrals, we have expressed everything in terms of Il−1 and q and so we can drop the notation Il , p, and r. Re-indexing the summation on l in the two last integrals by m = l − 1, we get from all these observations:   * + Z n k X X X   Y  Q D0k+1 FΩ (θ1 , . . . , θk , θ0 ) = ∇(∂ k−m [θj ]ij  ij f ) | θ0   i1 ,...,ik =1 m=0 Im ⊆J1,kK card Im =m

Ω

j∈J1,kK j ∈I / m

j∈J1,kK j ∈I / m

! X

+

k X

Z

X

Ω

i1 ,...,ik =1 m=0 Im ⊆J1,kK card Im =m

◦ q ◦ pIl

Y




q:Im →Im q bijective 

 n X

s

p−1 Il

[D• θj ]ij iq(j)

j∈Im

 Y   [θj ]ij   

Q ∂ k−m

ij f j∈J1,kK j ∈I / m

j∈J1,kK j ∈I / m

! X

s

q:Im →Im q bijective  n X

+

k X

Z

X

Ω

i1 ,...,ik =1 m=0 Im ⊆J1,kK card Im =m

p−1 Im

Y




◦ q ◦ pIm div (θ0 )

[D• θj ]ij iq(j)

j∈Im



 Y   [θj ]ij   

Q ∂ k−m

ij f j∈J1,kK j ∈I / m

j∈J1,kK j ∈I / m

 X

X


s p−1 Im ◦ q ◦ pIm [−D• θj0 D• θ0 ]ij

i 0 q(j0 )

j0 ∈Im q:Im →Im q bijective



Y   [D• θj ]ij iq(j)   . j∈Im j6=j0

(115) We now introduce some more notation in order to handle the quantities (112)(115) we want to estimate. For this purpose, we set:  Q a0 := (∂ k−m  ij f ) ◦ (I + θ0 )    j∈J1,kK  h i   j ∈I / m  −1    a := D θ (I + D θ ) j • • 0 p (j)   Im   i ip   Im (j) q[pIm (j)]   k−m    b0 := ∂ Q ij f  j∈J1,kK , and ∀j ∈ J1, mK, bj := [D• θpIm (j) ]ipI (j) iq[pI (j)] j ∈I / m   m m         * +      cj := −[D• θpIm (j) D• θ0 ]ipI (j) iq[pI (j)]   k−m m m Q  c ∇(∂ f ) | θ 0 := 0  ij    j∈J1,kK j ∈I / m

and we also set am+1 := det[D• (I + θ0 )], bm+1 := 1, and cm+1 := div(θ0 ). Then, we introduce the following map: Rk (θ0 , . . . , θk ) := Dθk0 FΩ (θ1 , . . . , θk ) − D0k FΩ (θ1 , . . . , θk ) − D0k+1 FΩ (θ1 , . . . , θk , θ0 ) . Considering the expressions (112) and (114)(115) we have established, in each product/sum concerning j ∈ Im , we make a change of indices u := p−1 Im (j) so as to be able to order the product/sum from u = 1 to u = m. We obtain with our notation:   Z n k X X X  Y  X
 Rk (θ0 , . . . , θk ) = [θj ]ij  s p−1 Im ◦ q ◦ p m l   i1 ,...,ik =1 m=0 Im ⊆J1,kK card Im =m

Ω

q:Im →Im bijective

j∈J1,kK j ∈I / m

q   

 n X

=

k X

X

i1 ,...,ik =1 m=0 Im ⊆J1,kK card  Im =m

m+1 X

u0 −1

u0 =0

u=0

Y

Z Ω

bu

j∈J1,kK j ∈I / m

(au0 − bu0 − cu0 )

 au −

u=0

m+1 Y

bu −

u=0

m+1 X

cu0

u0 =0

Y u∈J0,m+1K u6=u0

 bu  .



 Y   [θj ]ij   

!

m+1 Y

m+1 Y

X

s p−1 Im ◦ q ◦ p m l




q:Im →Im q bijective

! au

+ cu0

u=u0 +1

74

m+1 X

l−1 Y

l=u0 +1

u=u0 +1

! au

(al − bl )

m+1 Y u=l+1

! bu  .

Therefore, we can now estimate each term in the last Qequality which is combined with the triangle inequality in order to obtain |Rk (θ0 , . . . , θk )| 6 R(n, f, k, θ0 ) kj=0 kθj k1,∞ with |R(n, f, k, θ0 )| → 0 as kθ0 k1,∞ → 0. Let us detail this procedure. First, we can apply Proposition 5.38 to the maps ∂•k−m f ∈ W 1,1 , m ∈ J0, kK, then use the Cauchy-Schwarz inequality with (99), and combine the relations (100) and (101) with the fact that kθ0 k1,∞ < δ < 1. We deduce that: ! !  m+1 Y m Y n (a0 − b0 − c0 ) au am+1 6 (n − 1)! kθj k0,1 kθ0 k1,∞ R(θ0 ), 1−δ 1 n u=1 j∈I L (R ,R)

m

where |R(θ0 )| → 0 as kθ0 k1,∞ → 0. Hence, we have estimated the rst term of the rst sum. We can proceed similarly for the other ones. Using the L1 -norm for the maps ∂•k−m f ∈ W 1,1 , m ∈ J0, kK, and the L∞ -norm for the remaining terms, we get from the Cauchy-Schwarz inequality with (99), relations (100)(101) with kθ0 k1,∞ < δ < 1, and Proposition 5.33: ! !  m u0 −1 m m Y X Y √ n au am+1 b0 bu (au0 − bu0 − cu0 ) 6 m n(n − 1)! kf kk,1 1−δ u=u0 +1 u0 =1 u=1 L1 (Rn ,R) ! Y kθj k0,1 kθ0 k0,1 R(θ0 ), j∈Im

where |R(θ0 )| → 0 as kθ0 k0,1 → 0. The same arguments and Proposition 5.34 also give: ! ! m Y Y b0 bu (am+1 − bm+1 − cm+1 ) 6 kf kk,1 kθj k0,1 kθ0 k0,1 R(θ0 ), u=1

j∈Im

L1 (Rn ,R)

where |R(θ0 )| → 0 as kθ0 k0,1 → 0. Next, we observe that m 6 k and δ only depends only on n (in fact one 1 ), where we recall that δ ∈]0, 1[ is such that the Jacobian determinant of (I + θ) can prove that δ = 1+n! is positive for any kθk1,∞ < δ . Gathering the three last estimations and these observations, we thus have obtained: ! ! m+1 m+1 0 −1 X uY Y Y bu (au0 − bu0 − cu0 ) au 6 C (n, k, kf kk,1 ) kθ0 k1,∞ R(θ0 ) kθj k1,∞ , u0 =0

u=0

u=u0 +1

j∈Im

L1 (Rn ,R)

(116) where C(n, k, kf kk,1 ) > 0 is a xed constant depending only on n, k, and kf kk,1 , and where |R(θ0 )| → 0 as kθ0 k1,∞ → 0. We continue our estimations. Considering (108) with θ0 and ∂•k−m f ∈ W 1,1 , m ∈ J0, kK, we use the Cauchy-Schwarz inequality with (99), relation (100) with kθ0 k1,∞ < δ < 1, and Proposition 5.33 to get: ! ! m−1  l−1 m m X Y Y √ n c0 au (al − bl ) bu bm+1 6m n kf kk+1,1 kθ0 k∞ R(θ0 ) 1−δ u=1 l=1 u=l+1 L1 (Rn ,R) ! Y kθj k0,1 , j∈Im

where |R(θ0 )| → 0 as kθk0,1 → 0. The same arguments combined with Proposition 5.34 give: ! !  m m Y Y n c0 au (am+1 − bm+1 ) 6m kf kk+1,1 kθj k0,1 kθ0 k∞ R(θ0 ), 1−δ 1 n u=1 j∈I L (R ,R)

m

where |R(θ0 )| → 0 as kθk0,1 → 0. Similarly, we get from Proposition 5.33: ! ! ! u0 −1 l−1 m m m X Y X Y Y b0 bu cu0 au (al − bl ) bu bm+1 u0 =1

u=1

l=u0 +1

u=u0 +1

u=l+1

√ 6n n



L1 (Rn ,R)

n 1−δ

m−2 !

m(m − 1)kf kk,1 kθ0 k0,1 R(θ0 )

Y

kθj k0,1

,

j∈Im

where |R(θ0 )| → 0 as kθk0,1 → 0, and also from Proposition 5.34: ! !  m−1 u0 −1 m m X Y Y n b0 bu cu0 au (am+1 − bm+1 ) 6n mkf kk,1 1−δ u =1 u=1 u=u +1 0

L1 (Rn ,R)

0

! kθ0 k0,1 R(θ0 )

Y j∈Im

75

kθj k0,1

,

where |R(θ0 )| → 0 as kθk0,1 → 0. Gathering the four last estimations and observing again that m 6 k and δ < 1 only depends on n, we obtain: ! ! ! m+1 l−1 m+1 m+1 0 −1 X Y Y X uY au (al − bl ) bu bu cu0 6 C(n, k, kf kk+1,1 )kθ0 k1,∞ u0 =0

u=0

l=u0 +1

u=u0 +1

u=l+1

L1 (Rn ,R)

! Y

R(θ0 )

kθj k1,∞

,

j∈Im

(117) where C(n, k, kf kk+1,1 ) > 0 is a xed constant depending only on n, k, and kf kk+1,1 , and where |R(θ0 )| → 0 as kθk1,∞ → 0. Finally, using (116)(117) in the last expression obtained for Rk (θ0 , θ1 , . . . , θk ), we deduce that: sup (θ1 ,...,θk )∈(W 1,∞ )k θ1 ,...,θk 6=0

k k X X |Rk (θ0 , θ1 , . . . , θk ) | 6 C (n, k, kf kk+1,1 ) kθ0 k1,∞ kθ1 k1,∞ . . . kθk k1,∞ i ,...,i =1 m=0 1

k

X

X

|R(θ0 )|.

Im ⊆J1,kK q:Im →Im cardIm =m q bijective

We emphasize the fact that even if the notation omitted it, the R(θ0 ) in (116) depends on n, k, and f , but also on i1 , . . . , ik , m, Im , and q . Since all the sums are nite, we can take the maximum of these R(θ0 ) for example, and we end up with |||Rk (θ0 , •, . . . , •)||| 6 kθ0 k1,∞ R(n, k, f, θ0 ), where |R(n, k, f, θ0 )| → 0 as kθ0 k1,∞ → 0 with R(n, k, f, θ) depending only on n, k, f and θR0 . Therefore, we have established that the map (113) is dierentiable at the origin i.e. FΩ : θ ∈ W 1,∞ → (I+θ)(Ω) f ∈ R is k + 1 times dierentiable at the origin for any measurable subset Ω of Rn . We now show that FΩ is k + 1 times dierentiable at any point of B0,1 ∩ W 1,∞ . Let θ0 ∈ W 1,∞ satisfy kθ0 k0,1 < 1. Proposition 5.31 ensures the Lipschitz continuous map I + θ0 is bijective with a Lipschitz continuous inverse (I + θ0 )−1 . In particular, we deduce that Ω0 := (I + θ0 )(Ω) is measurable. Consequently, from the foregoing, the functional FΩ0 : θ ∈ W 1,∞ → F [(I +θ)(Ω0 )] ∈ R is k +1 times dierentiable at the origin. Let ε > 0 and we set  := ε(1−kθ0 k0,1 )k+1 > 0. There exists δ ∈]0, 1] such that for any θ ∈ W 1,∞ such that kθk1,∞ < δ , we have: |||Dθk FΩ0 − D0k FΩ0 − D0k+1 FΩ0 (•, . . . , •, θ)||| 6 kθk1,∞ .

Proceeding as in (110), we observe that for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθ0 k0,1 ), we have kθ0 + hk0,1 < 1 so we can write for any (θ1 , . . . , θk ) ∈ W 1,∞ × . . . × W 1,∞ : Dθk0 +h FΩ (θ1 , . . . , θk ) − Dθk0 FΩ (θ1 , . . . , θk )  −D0k+1 FΩ0 θ1 ◦ (I + θ0 )−1 , . . . , θk ◦ (I + θ0 )−1 , h ◦ (I + θ0 )−1   = Dθk FΩ0 θ1 ◦ (I +θ0 )−1 , . . . , θk ◦ (I + θ0 )−1  −1 . , θk ◦ (I + θ0 )−1 − D0k FΩ0 θ1 ◦ (I  + θ0 ) , . . −1  k+1 − D0 FΩ0 θ1 ◦ (I + θ0 ) , . . . , θk ◦ (I + θ0 )−1 , θ ,

where we have set θ := h ◦ (I + θ0 )−1 . As in (111), we have kθk1,∞ 6

khk1,∞ 1−kθ0 k0,1

< δ so we deduce that:

|Dθk0 +h FΩ (θ1 , . . . , θk ) − Dθk0 FΩ (θ1 , . . . , θk )  −D0k+1 FΩ0 θ1 ◦ (I + θ0 )−1 , . . . , θk ◦ (I + θ0 )−1 , h ◦ (I + θ0 )−1 | 6 |||Dθk FΩ0 − D0k FΩ0 − D0k+1 FΩ0 (•, . . . , •, θ)|||

k Y

kθl ◦ (I + θ0 )−1 k1,∞

l=1

6 kh ◦ (I + θ0 )−1 k1,∞

k Y l=1

6

kθl k1,∞ 1 − kθ0 k0,1

k Y  khk1,∞ kθl k1,∞ . k+1 (1 − kθ0 k0,1 ) l=1 | {z } =ε

Consequently, we obtain for any h ∈ W such that khk1,∞ < δ(1 − kθk0,1 ):   |||Dθk0 +h FΩ − Dθk0 FΩ − D0k+1 FΩ0 (•) ◦ (I + θ0 )−1 , . . . , (•) ◦ (I + θ0 )−1 , h ◦ (I + θ0 )−1 ||| 6 εkhk1,∞ . 1,∞

Since (θ1 , . . . , θk , h) ∈ (W 1,∞ )k+1 7→ D0k+1 FΩ0 [θ1 ◦ (I + θ0 )−1 , . . . , θk ◦ (I + θ0 )−1 , h ◦ (I + Rθ0 )−1 ] ∈ R is a continuous symmetric (k + 1)-linear form, we have proved that the map FΩ : θ ∈ W 1,∞ → (I+θ)(Ω) f ∈ R is k + 1 times dierentiable at any point of W 1,∞ ∩ B0,1 and its dierential is well dened by (113) with

76

k + 1 instead of k. Then, we now show that the (k + 1)-th order dierential is continuous at the origin. 1 Let θ0 ∈ W 1,∞ be such that kθ0 k1,∞ < 1+n! . Since we have just proved that (113) holds true for k + 1, we can rigorously use the same arguments than we used in the beginning of the proof in order to get that (114) holds true for k + 1 instead of k. Moreover, considering the expressions (112) and (114) with k + 1 instead of k, in each product/sum concerning j ∈ Il , we make a change of indices u := p−1 Il (j) so as to be able to order the product/sum from u = 1 to u = l. Using again the previous notation we introduced below (115), we thus have for any (θ1 , . . . , θk+1 ) ∈ W 1,∞ × . . . × W 1,∞ : h

k+1 X

n X

i k+1 F − D F (θ1 , . . . , θk+1 ) = Dθk+1 Ω Ω 0 0

Z

X

Y Ω j∈J1,k+1K j ∈I / l

i1 ,...,ik+1 =1 l=0 Il ⊆J1,k+1K card Il =l

X

s

[θj ]ij

p−1 Il

◦ q ◦ p Il

" l+1
Y

q:Il →Il q bijective k+1 X

n X

=

Z

X

u0 =0

# bu

u=0

Y

[θj ]ij i1 ,...,ik+1 =1 l=0 Il ⊆J1,k+1K Ω j∈J1,k+1K card Il =l ∈I / l " l+1 u j−1 ! 0 X Y X
−1 bu (au0 s pIl ◦ q ◦ pIl

q:Il →Il q bijective

au −

u=0

l+1 Y

− bu0 )

u=0

l+1 Y

!# au

,

u=u0 +1

where (aj , bj )16j6l+1 are dened as before (see below (115) where m has been replaced by l), but where k is replaced by k + 1 in the denition of (a0 , b0 ). Therefore, we can now estimate each term in the last equality which is combined with the triangle inequality in order to get |||Dθk+1 FΩ − D0k+1 FΩ ||| → 0 as kθ0 k1,∞ → 0. 0 Let us detail this procedure. First, we can apply Proposition 5.38 to the maps ∂•k+1−l f ∈ L1 , l ∈ J0, k + 1K, use the Cauchy-Schwarz inequality with (99), and combine relations (100)(101) with kθ0 k1,∞ < δ < 1. We deduce that:   !  l+1 Y l Y n  6 (n − 1)! kθj k0,1  R(θ0 ), (a0 − b0 ) au al+1 1−δ 1 n j∈I u=1 L (R ,R)

l

where |R(θ0 )| → 0 as kθ0 k1,∞ → 0. Hence, we have estimated the rst term of the rst sum. We can proceed similarly for the other ones. Using the L1 -norm for the maps ∂•k+1−l f , l ∈ J0, k + 1K, and the L∞ norm for the remaining terms, we get from the Cauchy-Schwarz inequality with (99), relations (100)(101) with kθ0 k1,∞ < δ < 1, and Proposition (5.33):   ! !  l Y u0 −1 l l Y Y X √ n  au al+1 bu (au0 − bu0 ) b0 6 l n(n − 1)! kθj k0,1  1−δ u=u +1 u =1 u=1 j∈I L1 (Rn ,R)

0

0

l

kf kk+1,1 R(θ0 ),

where |R(θ0 )| → 0 as kθ0 k0,1 → 0. The same arguments combined with Proposition 5.34 also lead to:   ! l Y Y 6 kf kk+1,1  kθj k0,1  R(θ0 ), b0 bu (al+1 − bl+1 ) u=1

L1 (Rn ,R)

j∈Il

where |R(θ0 )| → 0 as kθ0 k0,1 → 0. Gathering the three last estimations and observing that l 6 k + 1 with δ only depending on n, we obtain:   ! ! u0 −1 l+1 l+1 X Y Y Y bu (au0 − bu0 ) au 6 C (n, k, kf kk+1,1 )  kθj k1,∞  R(θ0 ), u0 =0

u=0

u=u0 +1

j∈Il

L1 (Rn ,R)

where C(n, k, kf kk+1,1 ) > 0 is a xed constant depending only on n, k, and kf kk+1,1 , and where |R(θ0 )| → 0 as kθ0 k1,∞ → 0. Using this last inequality in the last expression obtained for Dθk+1 FΩ −D0k+1 FΩ , we deduce 0 that:   | Dθk+1 FΩ − D0k+1 FΩ (θ1 , . . . , θk+1 ) | 0 sup 6 C (n, k, kf kk+1,1 ) kθ1 k1,∞ . . . kθk+1 k1,∞ (θ1 ,...θk+1 )∈(W 1,∞ )k+1 θ1 ,...,θk+1 6=0 n X

k+1 X

X

X

i1 ,...,ik+1 =1 l=0 Il ⊆J1,k+1K q:Il →Il card Il =l q bijective

77

R(θ0 ).

We emphasize the fact that even if the notation omitted it, the R(θ0 ) in the previous estimations depends on n, k, and f , but also on i1 , . . . , ik , l, Il , and q . Since all the sums are nite, we can take the maximum of these R(θ0 ) for example, and we end up with |||Dθk+1 FΩ − D0k+1 FΩ ||| 6 R(n, k, f, θ0 ), where |R(n, k, f, θ0 )| → 0 0 as kθ0 k1,∞ → 0 with R(n, k, f, θ) depending only on n, k, f and θ0 . Therefore, we have established that the map D•k+1 FΩ : θ0 ∈ W 1,∞ 7→ Dθk+1 FΩ ∈ Lk+1 ((W 1,∞ )k+1 , R) is continuous at the origin i.e. c 0 R 1,∞ FΩ : θ ∈ W → (I+θ)(Ω) f ∈ R is k + 1 times continuously dierentiable at the origin for any measurable n subset Ω of R . Finally, it remains to show that D•k+1 FΩ is continuous at any point of B0,1 ∩ W 1,∞ . The arguments are the same than those used to obtain the (k + 1)-th order dierentiability at any point of W 1,∞ ∩ B0,1 from the one at the origin. Let θ0 ∈ W 1,∞ satisfy kθ0 k0,1 < 1. Proposition 5.31 ensures that the Lipschitz continuous map I + θ0 is bijective with a Lipschitz continuous inverse (I + θ0 )−1 . In particular, we deduce that Ω0 := (I +θ0 )(Ω) is measurable. Consequently, from the foregoing, the functional FΩ0 : θ ∈ W 1,∞ → F [(I + θ)(Ω0 )] ∈ R is k + 1 times continuously dierentiable at the origin. Let ε > 0 and we set  := ε(1 − kθ0 k0,1 )k+1 > 0. There exists δ ∈]0, 1] such that for any θ ∈ W 1,∞ such that kθk1,∞ < δ , we have: |||Dθk+1 FΩ0 − D0k+1 FΩ0 ||| 6 . Proceeding as in (110), we observe that for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθ0 k0,1 ), we have kθ0 + hk0,1 < 1 so we can write for any (θ1 , . . . , θk+1 ) ∈ W 1,∞ × . . . × W 1,∞ :   Dθk+1 FΩ (θ1 , . . . , θk+1 ) − Dθk+1 FΩ (θ1 , . . . , θk+1 ) = Dθk+1 FΩ0 θ1 ◦ (I + θ0 )−1 , . . . , θk+1 ◦ (I + θ0 )−1 0 +h 0   − D0k+1 FΩ0 θ1 ◦ (I + θ0 )−1 , . . . , θk+1 ◦ (I + θ0 )−1 , where we have set θ := h ◦ (I + θ0 )−1 . As in (111), we have kθk1,∞ 6

khk1,∞ 1−kθ0 k0,1

F − Dθk+1 FΩ ] (θ1 , . . . , θk+1 ) | 6 |||Dθk+1 FΩ0 − D0k+1 FΩ0 ||| |[Dθk+1 0 0 +h Ω

k+1 Y

< δ so we deduce that:

kθl ◦ (I + θ0 )−1 k1,∞

l=1 k+1

6

Y  kθl k1,∞ . (1 − kθ0 k0,1 )k+1 | {z } l=1 =ε

Consequently, we obtain |||Dθk+1 FΩ − Dθk+1 FΩ ||| 6 ε for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθk0,1 ). 0 +h 0 k+1 Hence, D• FΩ is continuous at any point W 1,∞ ∩ B0,1 . To conclude the proof of Theorem 5.40, we have proved that the theorem is true for k = 0, and that if it is true for an integer k > 0, then it is true for k + 1 provided that f ∈ W k+1,1 . Therefore, by induction, for any integer k0 > 1, if f ∈ W k0 ,1 , then we obtain recursively that for any k ∈ J1, k0 K, the shape functional (89) is k times dierentiable at any measurable subset of Rn , its k-th order shape derivative being well dened by (112), and moreover, FΩ : θ ∈ W 1,∞ 7→ F [(I + θ0 )(Ω)] ∈ R is k times continuously dierentiable at any point of W 1,∞ ∩ B0,1 , its k-th order dierential map begin well dened by (113).

5.2.6 Proof of Theorem 5.29 We are now in position to prove Theorem 5.29. We rst state a density lemma, which is a W 1,∞ -version of [17, Section 4.2 Theorem 2].

Lemma 5.41.

Let f ∈ W 1,∞ (Rn , R). Then, there exists a sequence (fi )i∈N in W 1,∞ (Rn , R) ∩ C 1 (Rn , R) converging to f strongly in L∞ (Rn , R), weakly-star in W 1,∞ (Rn , R) and uniformly on compact sets in Rn . Proof. First, we Rconsider the standard (unscaled) mollier η of [17, Section 4.2.1 Notation (ii)]). We have η ∈ C ∞ (Rn , R), Rn η = 1, and the support of η is the unit closed ball of Rn centred at the origin, denoted by B1 (0). Then, let ε > 0 and f ∈ W 1,∞ (Rn , R). We introduce the following map: Z ∀x ∈ Rn , fε (x) := η (y) f (x + εy) dy. Rn

Note that the map fε : R 7→ R is well dened since the integrand η(•)f (x+ε•) is continuous and integrable. Moreover, one can check that kfε kW 1,∞ (Rn ,R) 6 kf kW 1,∞ (Rn ,R) and supx∈Rn |(fε − f )(x)| 6 εkf kC 0,1 (Rn ,R) , from which we deduce fε ∈ W 1,∞ (Rn , R) and the convergence of fε to f as ε → 0 strongly in L∞ (Rn , R) and uniformly on any compact subset of Rn . Finally, x x ∈ Rn and a change of variables z = x + εy gives R )f as in [17, Section 4.2.1 Notation (iii)]. We can thus apply [17, Section 4.2.1 Theorem fε = Rn ε1n η( •−x ε 1] to get that ∇fε is a continuousR map that converges to ∇f almost everywhere. for any i ∈ J1, nK that ∂i fε = Rn ε1n η( •−x )∂i f . A direct calculation shows that: ε  
2 k∂i fε kW 1,∞ (Rn ,R) 6 1 + kηkC 0,1 (Rn ,R) Ln B1 (0) kf kC 0,1 (Rn ,R) , (118) ε n

78

In particular, the partial derivatives of fε exist and are continuous thus f ε is continuously dierentiable on Rn . To conclude, we have the estimation for almost every x ∈ Rn and any i ∈ J1, nK: ! Z
1 n ∞ n ∂i fε (x) − ∂i f (x) ∂i f (z) − ∂i f (x) dz . 6 L B1 (0) kηkL (R ,R) Ln (Bε (x)) Bε (x) Since ∂i f ∈ L∞ (Rn , R) ⊂ L1loc (Rn , R), we have that almost every point is a Lebesgue one [17, Section 1.7]. Using the above inequality, we thus get the almost everywhere convergence of the partial derivatives of fε to the one of f . Considering any g ∈ L1 (Rn , R), we can use again the Dominated Convergence Theorem as we did to obtain (??) in order to get the expected weak-star convergence of the partial derivatives of fε to the ones f in L∞ (Rn , R) as ε → 0, concluding the proof of Lemma 5.41.

Proof of Theorem 5.29. Let n > 2 be any integer, f ∈ W 1,1 , and Ω be any measurable subset of Rn . First, Proposition 5.39 ensures the map FΩ : θ ∈ W 1,∞ → F [(I + θ)(Ω) ∈ R is well dened around the origin 1 and it is continuous at the origin. Then, let θ ∈ W 1,∞ be such that kθk1,∞ < 1+n! . Corollary ?? ensures that det D• (I + θ) is positive almost everywhere, and since the change of variables formula is valid for any Lipschitz continuous map [17, Section 3.3.3], we get successively with the triangle inequality: Z Z Z Z Z f ◦ (I + θ) |det [D• (I + θ)] | − (f + h∇f | θi + f div θ) f− f− div (f θ) = | {z } Ω Ω (I+θ)(Ω)

Ω

Ω

> 0

Z 6 Ω

(f ◦ (I + θ) − f − h∇f | θi) det [D• (I + θ)] Z f (det [D• (I + θ)] − 1 − div θ) + Ω Z h∇f | θi (det [D• (I + θ)] − 1) + Ω

kdet [D• (I + θ)] kL∞ (Rn ,R) kf ◦ (I + θ) − f − h∇f | θi kL1 (Rn ,R)

6

+ kf k1,1 kdet [D• (I + θ)] − 1 − div θkL∞ (Rn ,R) + kθk1,∞ kf k1,1 kdet [D• (I + θ)] − 1kL∞ (Rn ,R) .

Then, we can combine the dierentiability at the origin of the maps θ ∈ W 1,∞ 7→ f ◦(I +θ) ∈ L1 (Rn , R) and θ ∈ C 0,1 7→ det[D• (I + θ)] ∈ L∞ (Rn , R), respectively ensured by Propositions 5.38 and 5.34, with relations (101)(??), i.e. the bound and continuity at the origin of the map θ ∈ C 0,1 7→ det[D• (I + θ)] ∈ L∞ (Rn , R) given by Proposition 5.34. We deduce that: Z Z Z f− f− div (f θ) 6 kθk1,∞ RFΩ (n, f, θ) , (I+θ)(Ω)

Ω

Ω

where |RFΩ (n, f, θ)| → 0 as kθk1,∞ → 0, with RFΩ (n, f, θ) depending only on n, f , and θ. Hence, the map FΩ : θ ∈ W 1,∞ → F [(I + θ)(Ω) ∈ R is dierentiable at the origin and its dierential at the origin is R given by the continuous linear form θ ∈ W 1,∞ 7→ Ω div(f θ) ∈ R. Finally, if we assume that Ω is an open bounded subset of Rn with Lipschitz boundary, then we can apply the Trace Theorem [17, Section 4.3] in order to obtain: Z Z ∀θ ∈ W 1,∞ ∩ C 1 , D0 FΩ (θ) := div(f θ) = f hθ | nΩ idA, (119) Ω

∂Ω

where the boundary values of function f in the last integral have to be understood in the sense of trace. In particular, they are uniquely determined up to a set of zero A(• ∩ ∂Ω)-measure, which implies that the shape gradient of F is unique and well dened by f . To conclude, we only need to extend the last equality of (119) for any θ ∈ W 1,∞ . This is done by an approximating argument using Lemma 5.41. Indeed, let θ = (θ1 , . . . , θn ) ∈ W 1,∞ . Applying Lemma 5.41 for any component θi ∈ W 1,∞ (Rn , R), there exists a sequence (θik )k∈N of maps in W 1,∞ (Rn , R) ∩ C 1 (Rn , R) converging to θi weakly-star in W 1,∞ (Rn , R) and uniformly on the compact set Ω. For any k ∈ N, we set θk = (θ1k , . . . θnk ) ∈ W 1,∞ ∩ C 1 so we have:  X Z n Z n Z X k k 1Ω f ∂i θi + 1Ω θi ∂i f = f θik (nΩ )i dA, (120) i=1

Rn

Rn

i=1

∂Ω

The weak-star convergence ensures we can let k → +∞ in the left member of (120). Concerning the right one, we get from the uniform convergence and continuity of the trace operator from W 1,1 (Ω) into L1 (∂Ω): Z X n Z n n   X X f θik − θi (nΩ )i dA 6 |f |dA kθik − θi kC 0 (Ω,R) 6 Ckf k1,1 kθik − θi kC 0 (Ω,R) , i=1

∂Ω

∂Ω

i=1

79

i=1

where C > 0 is a constant, more precisely the norm of this trace operator, depending only on Ω and n. Hence, we can let k → +∞ in (120) and we get (119) holds true for any θ ∈ W 1,∞ , concluding the proof of Theorem 5.29.

5.2.7 Proof of Theorem 5.30 As we did in Section 5.2.6, we rst state a density result very similar to Lemma 5.41, then we rene Theorem 5.29 by considering more dierentiability properties associated with the shape functional (89). Finally, we prove Theorem 5.30.

Lemma 5.42.

For any f ∈ W 1,∞ (Rn , R) ∩ C 1 (Rn , R), there exists a sequence (fk )k∈N of elements in W (R , R) such that fk and ∂i fk respectively converges to f and ∂i f uniformly on any compact subset of Rn as k → +∞ and for any i ∈ J1, nK. 2,∞

n

Proof. Let f ∈ W 1,∞ (Rn , R)∩C 1 (Rn , R). First, the beginning of the proof is rigorously identical to the one of Lemma 5.41 until (118). Using the same notation, we can explicitly build a function fε ∈ W 1,∞ (Rn , R) converging to f strongly in L∞ (Rn , R) and uniformly on any compact subset of Rn . Moreover, we get from (118) that ∇fε ∈ W 1,∞ i.e. fε ∈ W 2,∞ (Rn , R). Then, let i ∈ J1, nK. It remains to prove that ∂i fε converges to ∂i f uniformly on compact subsets of Rn . Let δ > 0 and K be any compact subset of Rn . Since ∂i f is e = {a + b, a ∈ K and b ∈ B1 (0)}, it is uniformly continuous on K e so continuous on the compact set K e e there exists  ∈]0, 1] such that for any (a, b) ∈ K × K satisfying |a − b| < , we have |∂i f (a) − ∂i f (b)| < δ . Consequently, from a change of variables y = z−x in (??), we obtain for any ε ∈]0, ] and any x ∈ K : ε Z η(y)|∂i f (x + εy) − ∂i f (x) |dy 6 δ. |∂i fε (x) − ∂i f (x)| 6 B1 (0)

Hence, (∂i fε )ε>0 converges to ∂i f uniformly on compact subsets of Rn , concluding the proof.

Proposition 5.43.

Let n > 2 be any integer, f ∈ W 1,1 (Rn , R), Ω be any measurable subset of Rn , and B0,1 refer to the open unit ball of C 0,1 centred at the origin. Then, the map FΩ : θ ∈ W 1,∞ 7→ F [(I + θ)(Ω)] ∈ R is well dened and continuously dierentiable at any point of W 1,∞ ∩ B0,1 . Moreover, its (rst-order) dierential is well dened by the following continuous map:
D• FΩ : B0,1 ∩ W 1,∞ −→ Lc W 1,∞ , R Z   (121) θ0 7−→ Dθ0 FΩ := θ 7→ div f θ ◦ (I + θ0 )−1 . (I+θ0 )(Ω)

If in addition, we now assume that Ω is an open bounded subset of Rn with a Lipschitz boundary as in Denition 5.46, then the rst-order dierential takes the following form:
D• FΩ : B0,1 ∩ W 1,∞ −→ Lc W 1,∞ , R Z

 (122) θ0 7−→ Dθ0 FΩ := θ 7→ f θ ◦ (I + θ0 )−1 | n(I+θ0 )(Ω) dA. (I+θ0 )(∂Ω)

Proof. Let n > 2 be any integer, f ∈ W 1,1 (Rn , R), and Ω be any measurable subset of Rn . We denote by B0,1 the open unit ball of C 0,1 centred at the origin. Let θ0 ∈ W 1,∞ be such that kθ0 k0,1 < 1. From Proposition 5.31, the Lipschitz continuous map I + θ0 is bijective with a Lipschitz continuous inverse. In particular, we deduce that Ω0 := (I + θ0 )(Ω) is measurable. Consequently, from Theorem 5.29, the functional FΩ0 : θ ∈ W 1,∞ 7→ F [(I + θ)(Ω0 )] ∈ R is dierentiable at the origin. Let ε > 0 and set  := ε(1 − kθ0 k0,1 ) > 0. There exists δ ∈]0, 1] such that for any θ ∈ W 1,∞ such that kθk1,∞ < δ , we have: |FΩ0 (θ) − FΩ0 (0) − D0 FΩ0 (θ)| 6 kθk1,∞ .

Proceeding as in (110), we observe that for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθ0 k0,1 ), we have kθ0 + hk0,1 < 1 so we can write: Z Z Z   f− f− div f h ◦ (I + θ0 )−1 = FΩ0 (θ) − FΩ0 (0) − D0 FΩ0 (θ), (I+θ0 +h)(Ω)

(I+θ0 )(Ω)

(I+θ0 )(Ω)

where we have set θ := h ◦ (I + θ0 )−1 . As in (111), we have kθk1,∞ 6 Z

Z f−

(I+θ0 +h)(Ω)

Z f−

(I+θ0 )(Ω)

khk1,∞ 1−kθ0 k0,1

< δ so we deduce that:

  khk1,∞ div f h ◦ (I + θ0 )−1 6 kh ◦ (I + θ0 )−1 k1,∞ 6  . 1 − kθ0 k0,1 (I+θ0 )(Ω) | {z } =εkhk1,∞

Hence, we have proved that the map θ ∈ W 1,∞ 7→ F [(I + θ)(Ω)] ∈ R is dierentiable at any point of B0,1 ∩ W 1,∞ and its rst-order dierential is well dened by (121). We now show that (121) is continuous

80

1 at the origin. Let θ ∈ W 1,∞ and θ0 ∈ W 1,∞ be such that kθ0 k1,∞ < 1+n! . First, we can express the dierential in a simpler form by using the change of variables formula valid for any Lipschitz continuous map [17, Section 3.3.3]: Z Z  


 f trace D• θ ◦ (I + θ0 )−1 Dθ0 FΩ (θ) := ∇f | θ ◦ (I + θ0 )−1 + (I+θ0 )(Ω)

(I+θ0 )(Ω)

Z = Ω

h∇f ◦ (I + θ0 ) | θi |det [D• (I + θ0 )] | Z  
+ f ◦ (I + θ0 ) trace D(I+θ0 )(•) θ ◦ (I + θ0 )−1 |det [D• (I + θ0 )] |. Ω

Proposition ?? and Lemma 5.36 give A := Def (θ0 ) ∩ Def (θ) = Def (θ0 ) ∩ (I + θ0 )−1 [Def (θ ◦ (I + θ0 )−1 )], and also for almost every point x ∈ Rn (and more precisely for any x ∈ A since Ln (Rn \A) = 0):   ∀θ ∈ W 1,∞ , Dx θ (I + Dx θ0 )−1 = D(I+θ0 )(x) θ ◦ (I + θ0 )−1 . (123) 1 Since kθ0 k1,∞ < 1+n! , we get from Corollary ?? that the Jacobian determinant of (I + θ0 ) is positive. Combining this observation with the fact that [Com(I + D• θ0 )]T = det[D• (I + θ0 )](I + D• θ0 )−1 coming from Lemmas 5.32 and 5.14, we obtain from the foregoing: Z Z   h∇f ◦ (I + θ0 ) | θi det [D• (I + θ0 )] + f ◦ (I + θ0 ) trace D• θ [Com (I + D• θ0 )]T . Dθ0 FΩ (θ) = Ω

Ω

(124) As a consequence, we have from the triangle inequality: Z h∇f [x + θ0 (x)] | θ (x)i (det [Dx (I + θ0 )] − 1) dx |Dθ0 FΩ (θ) − D0 FΩ (θ) | 6 Ω Z h∇f [x + θ0 (x)] − ∇f (x) | θ (x)i dx + Ω ! Z n X ∂j θi (x) [Com (I + Dx θ0 )]ij dx (f [x + θ0 (x)] − f (x)) + Ω

i,j=1

Z f (x)

+ Ω

n X

h i ∂j θi (x) Com (I + Dx θ0 )ij − Iij ,

i,j=1

and we obtain: |Dθ0 FΩ (θ) − D0 FΩ (θ) |

6

kdet [D• (I + θ0 )] − 1kL∞ (Rn ,R) kθk∞ +kθk∞

n X

n X

k∂i f ◦ (I + θ0 ) kL1 (Rn ,R)

i=1

k∂i f ◦ (I + θ0 ) − ∂i f kL1 (Rn ,R)

i=1

+kf ◦ (I + θ0 ) − f kL1 (Rn ,R) n2 kθk0,1 max ess sup | [Com (I + Dx θ0 )]ij | 16i,j6n

x∈Rn

+kf kL1 (Rn ,R) n2 kθk0,1 max ess sup | [Com (I + Dx θ0 )]ij − Iij |. 16i,j6n

x∈Rn

Using successively (??) with θ0 and (106) applied to ∂i f , then (??)(??) with the fact that kθ0 k1,∞ < 21 , we deduce that: sup θ∈W 1,∞ θ6=0

|Dθ0 FΩ (θ) − D0 FΩ (θ)| kθk1,∞

6

√

n2n+1 (n!)2 kf k1,1 kkθ0 k1,∞ +

n X

k∂i f ◦ (I + θ0 ) − ∂i f kL1 (Rn ,R)

i=1

+ 2n(n!)kf ◦ (I + θ0 ) − f kL1 (Rn ,R) .

Since f ∈ W 1,1 , we can apply Proposition 5.38 to f and ∂i f for any i ∈ J1, nK. We can thus correctly let kθ0 k1,∞ → 0 in the previous inequality, which gives |||Dθ0 FΩ − D0 FΩ ||| → 0. Hence, we have proved that the map given by (121) is continuous at the origin. Finally, it remains to prove the continuity of (121) at any point of W 1,∞ ∩ B0,1 . The arguments are similar to the ones used at the beginning of the proof to study the dierentiability at any point from the one at the origin. Let θ0 ∈ W 1,∞ be such that kθ0 k0,1 < 1. From Proposition 5.31, the Lipschitz continuous map I + θ0 is bijective and its inverse (I + θ0 )−1 is also Lipschitz continuous. In particular, we deduce that Ω0 := (I + θ0 )(Ω) is measurable. From the foregoing, the functional FΩ0 : θ ∈ W 1,∞ → F [(I + θ)(Ω0 )] ∈ R is continuously dierentiable at the origin. Let ε > 0 and set  = ε(1 − kθ0 k0,1 ) > 0. There exists δ ∈]0, 1] such that for any θ ∈ W 1,∞ such that kθk1,∞ < δ , we have |||Dθ FΩ0 − D0 FΩ0 ||| 6 . Proceeding as in (110), we observe that for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθ0 k0,1 ), we have kθ0 + hk0,1 < 1 so we can write for any θ˜ ∈ W 1,∞ : h i h i ˜ − Dθ FΩ (θ) ˜ = Dθ FΩ θ˜ ◦ (I + θ0 )−1 − D0 FΩ θ˜ ◦ (I + θ0 )−1 , Dθ0 +h FΩ (θ) 0 0 0

81

where we have set θ := h ◦ (I + θ0 )−1 . As in (111), we have kθk1,∞ 6

khk1,∞ 1−kθ0 k0,1

˜ 6 |||Dθ FΩ − D0 FΩ ||| kθ˜ ◦ (I + θ0 )−1 k1,∞ 6  ˜ − Dθ FΩ (θ)| |Dθ0 +h FΩ (θ) 0 0 0

< δ so we deduce that:

˜ 1,∞ kθk ˜ 1,∞ . = εkθk 1 − kθ0 k0,1

Consequently, we obtain |||Dθ0 +h FΩ −Dθ0 FΩ ||| 6 ε for any h ∈ W 1,∞ such that khk1,∞ < δ(1−kθ0 k0,1 ) and (121) is continuous at θ0 . We thus have proved that the map θ ∈ W 1,∞ 7→ F [(I +θ)(Ω)] ∈ R is continuously dierentiable at any point of W 1,∞ ∩ B0,1 and its dierential is well dened by (121). To conclude the proof of Proposition 5.43, if we now assume that Ω is an open bounded subset of Rn with Lipschitz boundary as in Denition 5.46, then we can apply the Trace Theorem [17, Section 4.3] as in the proof of Theorem 5.29 (see (119) and below). Indeed, we have established from a density argument given by Lemma 5.41, that the Trace Theorem [17, Section 4.3] is still valid for W 1,∞ -vector elds. More precisely, (119) holds true for any f ∈ W 1,1 and any θ ∈ W 1,∞ . From the hypothesis, we get f ∈ W 1,1 and θ ◦ (I + θ0 )−1 ∈ W 1,∞ , thus we can apply the Trace Theorem in (121) to get (122) since ∂[(I + θ0 )(Ω)] = (I + θ0 )(∂Ω).

Proof of Theorem 5.30. Let n > 2 be any integer, f ∈ W 2,1 , and Ω be any measurable subset of Rn . Proposition 5.43 ensures that the map FΩ : θ ∈ W 1,∞ 7→ F [(I + θ)(Ω)] ∈ R is well dened and continuously dierentiable at any point of W 1,∞ ∩ B0,1 , where B0,1 refers to the open unit ball of C 0,1 centred at the origin. We now show that the additional regularity assumed on f allows the map (121) to be dierentiable 1 at the origin. Let θ ∈ W 1,∞ and θ0 ∈ W 1,∞ be such that kθ0 k1,∞ < 1+n! . We set: Z Z hHess f (θ0 ) | θi − h∇f | θ0 i div (θ) RD• FΩ (θ0 , θ) := Dθ0 FΩ (θ) − D0 FΩ (θ) − Z Z ΩZ Ω − h∇f | θi div (θ0 ) − f div (θ) div (θ0 ) + f trace (D• θ0 D• θ) . Ω

Ω

Ω

Using the expression (124) obtained for the dierential of FΩ , we get from the triangle inequality: Z h∇f ◦ (I + θ0 ) | θi (det [D• (I + θ0 )] − 1 − div θ0 ) |RD• FΩ (θ0 , θ)| 6 Ω Z h∇f ◦ (I + θ0 ) − ∇f − Hessf (θ0 ) | θi + Ω Z h∇f ◦ (I + θ0 ) − ∇f | θi div (θ0 ) + Z h  Ω i f ◦ (I + θ0 ) trace D• θ [Com (I + D• θ0 )]T − I − div (θ0 ) I + D• θ0 + Ω Z [f ◦ (I + θ0 ) − f − h∇f | θ0 i] div (θ) + Z Ω [f ◦ (I + θ0 ) − f ] trace [D• θ (div (θ0 ) I − D• θ0 )] . + Ω

6

kθk∞ kdet [D• (I + θ0 )] − 1 − div θ0 kL∞ (Rn ,R) + kθk∞

n X

n X

k∂i f ◦ (I + θ0 ) kL1 (Rn ,R)

i=1

k∂i f ◦ (I + θ0 ) − ∂i f − h∇ (∂i f ) | θ0 i kL1 (Rn ,R)

i=1

+ nkθk∞ kθ0 k0,1

n X

k∂i f ◦ (I + θ0 ) − ∂i f kL1 (Rn ,R)

i=1

+ n2 kθk0,1 kf ◦ (I + θ0 )kL1 (Rn ,R) max ess sup | [Com (I + Dx θ0 )]ij − Iij 16i,j6n

x∈Rn

− div θ0 (x) Iij + [Dx θ0 ]ji | + nkθk0,1 kf ◦ (I + θ0 ) − f − h∇f | θ0 i kL1 (Rn ,R) + 2n2 kθk0,1 kθ0 k0,1 kf ◦ (I + θ0 ) − f kL1 (Rn ,R) .

Using successively (??), the fact that kθ0 k0,1 < sup θ∈W 1,∞ θ6=0

|RD• FΩ (θ0 , θ)| kθk1,∞

6

1 2

with (106), and relation (??), we deduce that:

n2n+1 (n!)2 kf k1,1 kθ0 k21,∞ + nkθ0 k1,∞

n X

k∂i ◦ (I + θ0 ) − ∂i f kL1 (Rn ,R)

i=1

+

n X

k∂i f ◦ (I + θ0 ) − ∂i f − h∇ (∂i f ) | θ0 i kL1 (Rn ,R)

i=1

+ nkf ◦ (I + θ0 ) − f − h∇f | θ0 i kL1 (Rn ,R) + 2n2 kθ0 k1,∞ kf ◦ (I + θ0 ) − f kL1 (Rn ,R) .

82

Since f ∈ W 2,1 , we can apply Proposition 5.38 to f and ∂i f for any i ∈ J1, nK. Therefore, we obtain that |||RD• FΩ (θ0 , •)||| 6 kθ0 k1,∞ R(n, f, θ0 ), where |R(n, f, θ0 )| → 0 as kθ0 k1,∞ → 0, with R(n, f, θ0 ) depending only on n, f , and θ0 . Since (92) is a continuous bilinear form on W 1,∞ × W 1,∞ , we thus have proved that FΩ : θ ∈ W 1,∞ → F [(I + θ)(Ω)] ∈ R is twice dierentiable at the origin and its second-order dierential at the origin is given by (92). Finally, we assume that Ω is an open bounded subset with Lipschitz boundary in the sense of Denition 5.46. First, we observe that for any (θ1 , θ2 ) ∈ W 2,∞ × W 2,∞ we can write: div [div (f θ1 ) θ2 − D• θ1 (θ2 )]

=

=

hHess f (θ1 ) | θ2 i + h∇f | θ1 div (θ2 ) + θ2 div (θ1 )i +f [div (θ1 ) div (θ2 ) − trace (D• θ1 D• θ2 )]

(125)

div [div (f θ2 ) θ1 − D• θ2 (θ1 )] .

We emphasize the fact that the W -assumption is needed to obtain the two last equalities, and more precisely the almost-everywhere existence of a second-order derivative for θ1 and θ2 . Integrating the two previous relations on Ω, we get from (92): Z Z D02 FΩ (θ1 , θ2 ) = div [div (f θ1 ) θ2 − D• θ1 (θ2 )] = div [div (f θ2 ) θ1 − f D• θ2 (θ1 )] . 2,∞

Ω

Ω

Then, we recall that in the proof of Theorem 5.29 (see (119) and below), we have established from a density argument given by Lemma 5.41, that the Trace Theorem [17, Section 4.3] is still valid for W 1,∞ -vector elds. More precisely, (119) holds true for any f ∈ W 1,1 and any θ ∈ W 1,∞ . From the hypothesis, we get f, div(f θ1 ), div(f θ2 ) ∈ W 1,1 and θ1 , θ2 , D• θ1 (θ2 ), D• θ2 (θ1 ) ∈ W 1,∞ , thus we can apply the Trace Theorem on the two last equalities in order to obtain: Z Z div (f θ1 ) hθ2 | nΩ i dA − f hD• θ1 (θ2 ) | nΩ i dA D02 FΩ (θ1 , θ2 ) = ∂Ω

∂Ω

Z

Z div (f θ2 ) hθ1 | nΩ i dA −

= ∂Ω

f hD• θ2 (θ1 ) | nΩ i dA. ∂Ω

Now considering Lemma 5.42, we can show by a density argument that the two previous relations hold true for any (θ1 , θ2 ) ∈ (W 1,∞ ∩C 1 )×(W 1,∞ ∩C 1 ). Since W 1,∞ ∩C 1 is a subset of W 1,∞ endowed with the k•k1,∞ norm, and since the map dened by (94) is a continuous bilinear form on W 1,∞ ∩ C 1 (but not on W 1,∞ because the trace of a L∞ -map is not well dened), we deduce that FΩ : θ ∈ W 1,∞ ∩C 1 7→ F [(I +θ)(Ω)] ∈ R is twice dierentiable at the origin and that its second-order dierential is well dened by the continuous symmetric bilinear form (94). We give a nal form for the second-order shape derivative. We decompose the operators in their tangential and normal part. The notation we use were introduced in Denition 5.28 ˜ ∈ (W 1,∞ ∩ C 1 ) × (W 1,∞ ∩ C 1 ): and Theorems 5.295.30. We obtain by direct calculation for any (θ, θ) D E D E ˜ | nΩ div(f θ)θ˜n − f D• θ(θ) = div∂Ω (f θ) θ˜n + hD (f θ) nΩ | nΩ i − f θ˜ | [D• θ]T nΩ =

D E div∂Ω (f θ) θ˜n + h∇f | nΩ i θn θ˜n + f θ˜n nΩ − θ˜ | [D• θ]T nΩ .

=

D   E div∂Ω (f θ) θ˜n + h∇f | nΩ i θn θ˜n − f D• θ θ˜∂Ω | nΩ .

Using the fact that D• θ(θ˜∂Ω ) = D∂Ω θ(θ˜∂Ω ) + D• θ(nΩ )hnΩ | θ˜∂Ω i = D∂Ω θ(θ˜∂Ω ) and also the notation ∂n (f ) = h∇f | nΩ i, we get: D E D E ˜ | nΩ = div∂Ω (f θ) θ˜n + ∂n (f ) θn θ˜n − f θ˜∂Ω | [D∂Ω θ]T nΩ . div(f θ)θ˜n − f D• θ(θ) We now assume that Ω is an open bounded subset with a boundary of class C 1,1 as in Denition 5.47. The normal vector is thus now a bounded Lipschitz continuous map on the surface. Hence, we can decompose θ = θ∂Ω + θn nΩ in the previous relation and we have successively: D E   D E ˜ | nΩ div(f θ)θ˜n − f D• θ(θ) = div∂Ω f θ˜n θ − f ∇∂Ω (θ˜n ) | θ + ∂n (f ) θn θ˜n D E − f θ˜∂Ω | [D∂Ω θ]T nΩ =

  E   D div∂Ω f θ˜n θ∂Ω + ∇∂Ω f θ˜n θn | nΩ + f div∂Ω (nΩ ) θn θ˜n D E + ∂n (f ) θn θ˜n − f ∇∂Ω (θ˜n ) | θ∂Ω D E − f θ˜∂Ω | ∇∂Ω (θn ) − [D∂Ω nΩ ]T (θ)

=

  div∂Ω f θ˜n θ∂Ω + [HΩ f + ∂n (f )] θn θ˜n D E D E − f ∇∂Ω (θ˜n ) | θ∂Ω − f ∇∂Ω (θn ) | θ˜∂Ω D E + f θ˜∂Ω | [D∂Ω nΩ ]T (θ∂Ω ) + θn [D∂Ω nΩ ]T (nΩ ) .

83

Since |nΩ |2 = 1, we get by dierentiating this relation [D∂Ω nΩ ]T nΩ = 0 and we deduce that: D E   ˜ | nΩ div(f θ)θ˜n − f D• θ(θ) = div∂Ω f θ˜n θ∂Ω + [HΩ f + h∇f | nΩ i] θn θ˜n E   D E D − f ∇∂Ω (θn ) | θ˜∂Ω − f ∇∂Ω (θ˜n ) | θ∂Ω − f IIΩ θ˜∂Ω , θ∂Ω . Integrating the previous relation on ∂Ω, we get from (94) that the left-member of the above equality is ˜ . Moreover, from Lemma 5.48, the rst term in the above right-member is equal to zero equal to D02 FΩ (θ, θ) ˜ ∈ (W 1,∞ ∩ C 1 )2 , and since one can check that f θ˜n θ∂Ω ∈ W 1,1 (∂Ω, Rn ) from the hypothesis f ∈ W 2,1 , (θ, θ) 1,1 also Ω is a C -domain. Hence, we have established that relation (95) holds true if Ω is a C 1,1 -domain. To conclude the proof of Theorem 5.30, the values of HΩ f + ∂n (f ) have to be understood in the sense of trace since f ∈ W 1,2 and Ω is a C 1,1 -domain. In particular, this map is uniquely determined up to a set of zero A(∂Ω ∩ •)-measure so from Denition 5.28, it denes well the shape Hessian of the function (89).

5.2.8 On the second-order dierential of the associated map and further regularity In this section, we prove that in fact, under the same hypothesis than the ones of Theorem 5.30, the map FΩ : θ ∈ W 1,∞ 7→ F [(I + θ)(Ω)] ∈ R is of class C 2 around the origin. Then, we generalize the result by induction to get the C k -regularity around the origin of this map FΩ associated with (89).

Proposition 5.44.

Let n > 2 be any integer, f ∈ W 2,1 , and Ω be any measurable subset of Rn . We consider the open unit ball B0,1 of C 0,1 centred at the origin. Then, the map FΩ : θ ∈ W 1,∞ 7→ F [(I + θ)(Ω)] ∈ R is twice continuously dierentiable at any point of B0,1 ∩ W 1,∞ . Moreover, its second-order dierential is well dened by the following continuous map:
D•2 FΩ : B0,1 ∩ W 1,∞ −→ Bc W 1,∞ × W 1,∞ , R   (126) θ0 7−→ Dθ20 FΩ : (θ1 , θ2 ) 7→ D02 F(I+θ0 )(Ω) θ1 ◦ (I + θ0 )−1 , θ2 ◦ (I + θ0 )−1 , where D02 F(I+θ0 )(Ω) is dened by (92) and corresponds to the second-order shape derivative of the map (89) at (I + θ0 )(Ω). If in addition, we now assume that Ω is an open bounded subset of Rn with a Lipschitz boundary as in Denition 5.46, then the map FΩ : θ ∈ W 1,∞ ∩C 1 7→ F [(I +θ)(Ω)] ∈ R is twice continuously dierentiable at any point of B0,1 ∩ W 1,∞ ∩ C 1 . Moreover, its second-order dierential is well dened by the following continuous map:
D•2 FΩ : B0,1 ∩ W 1,∞ ∩ C 1 −→ Bc (W 1,∞ ∩ C 1 ) × (W 1,∞ ∩ C 1 ), R  θ0 7−→ Dθ20 FΩ : (θ1 , θ2 ) 7→ D02 F(I+θ0 )(Ω) θ1 ◦ (I + θ0 )−1 , θ2 ◦ (I + θ0 )−1 , (127) where D02 F(I+θ0 )(Ω) is now dened by (94). Finally, if we assume that Ω is an open bounded subset of Rn with a C 1,1 -boundary as in Denition 5.47, then the previous result still holds true but we can now use the expression (95) to dene D02 F(I+θ0 )(Ω) in (127). Proof. Let n > 2 be any integer, f ∈ W 2,1 , and Ω be any measurable subset of Rn . We consider the open unit ball B0,1 of C 0,1 centred at the origin. The strategy to prove the second-order dierentiability at any point of B0,1 ∩ W 1,∞ from the one at the origin is the same than the arguments used in the proof of Propositions 5.39 and 5.43. Let θ0 ∈ W 1,∞ be such that kθ0 k0,1 < 1. From Proposition 5.31, the Lipschitz continuous map I + θ0 is bijective and its inverse (I + θ0 )−1 is also Lipschitz continuous. In particular, we deduce that Ω0 := (I + θ0 )(Ω) is measurable. Consequently, from Theorem 5.30, the functional FΩ0 : θ ∈ W 1,∞ 7→ F [(I + θ)(Ω0 )] ∈ R is twice dierentiable at the origin. Let ε > 0 and set  := ε(1 − kθ0 k0,1 )2 > 0. There exists δ ∈]0, 1] such that for any θ ∈ W 1,∞ such that kθk1,∞ < δ , we have: |||Dθ FΩ0 − D0 FΩ0 − D02 FΩ0 (•, θ)||| 6 kθk1,∞ .

Proceeding as in (110), we observe that for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθ0 k0,1 ), we have θ0 + h ∈ W 1,∞ and kθ0 + hk0,1 < 1 so we can write for any θ˜ ∈ W 1,∞ :     h i h i Dθ0 +h FΩ θ˜ − Dθ0 FΩ θ˜ − D02 FΩ0 θ˜ ◦ (I + θ0 )−1 , h ◦ (I + θ0 )−1 = Dθ FΩ0 θ˜ ◦ (I + θ0 )−1 h i − D0 FΩ0 θ˜ ◦ (I + θ0 )−1 h i − D02 FΩ0 θ˜ ◦ (I + θ0 )−1 , θ ,

84

where we have set θ := h ◦ (I + θ0 )−1 . As in (111), we have kθk1,∞ 6

khk1,∞ 1−kθ0 k0,1

< δ so we deduce that:

˜ − Dθ FΩ (θ) ˜ |Dθ0 +h FΩ (θ) 0 2 ˜ −D0 FΩ0 [θ ◦ (I + θ0 )−1 , h ◦ (I + θ0 )−1 ]| 6 |||Dθ FΩ0 − D0 FΩ0 − D02 FΩ0 (•, θ)||| kθ˜ ◦ (I + θ0 )−1 k1,∞ 6 kh ◦ (I + θ0 )−1 k1,∞ 6

˜ 1,∞ kθk 1 − kθ0 k0,1

 ˜ 1,∞ . khk1,∞ kθk (1 − kθ0 k0,1 )2 {z } | := ε

Consequently, we have obtained for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθk0,1 ): sup 1,∞ ˜ θ∈W θ6=0

˜ − D02 FΩ [θ˜ ◦ (I + θ0 )−1 , h ◦ (I + θ0 )−1 ]| ˜ − Dθ FΩ (θ) |Dθ0 +h FΩ (θ) 0 0 6 εkhk1,∞ . ˜ 1,∞ kθk

˜ ˜ h) ∈ W 1,∞ ×W 1,∞ 7→ D02 FΩ [θ◦(I + θ0 )−1 , h◦(I + θ0 )−1 ] ∈ R is a continuous symmetric Since the map (θ, 0 1,∞ bilinear form, we have proved that the map θ ∈ W 7→ F [(I + θ)(Ω)] ∈ R is twice dierentiable at any point of B0,1 ∩ W 1,∞ and its second-order dierential is well dened by (126). We now show that (126) 1 is continuous at the origin. Let (θ1 , θ2 ) ∈ W 1,∞ × W 1,∞ and θ0 ∈ W 1,∞ be such that kθ0 k1,∞ < 1+n! . First, we can express the second-order dierential (126) in a simpler form by using the change of variables formula valid for any Lipschitz continuous map [17, Section 3.3.3]: Z

   Dθ20 FΩ (θ1 , θ2 ) := Hess f θ1 ◦ (I + θ0 )−1 | θ2 ◦ (I + θ0 )−1 (I+θ Z0 )(Ω)

  
+ ∇f | θ1 ◦ (I + θ0 )−1 trace D• θ2 ◦ (I + θ0 )−1 (I+θZ0 )(Ω)

  
+ ∇f | θ2 ◦ (I + θ0 )−1 trace D• θ1 ◦ (I + θ0 )−1 (I+θ Z0 )(Ω)  
 
+ f trace D• θ1 ◦ (I + θ0 )−1 trace D• θ2 ◦ (I + θ0 )−1 (I+θ0 )(Ω) Z    
− f trace D• θ1 ◦ (I + θ0 )−1 D• θ2 ◦ (I + θ0 )−1 (I+θ0 )(Ω)

Z h(Hess f ) ◦ (I + θ0 ) [θ1 ] | θ2 i |det [D• (I + θ0 )] | Z  
+ h∇f ◦ (I + θ0 ) | θ1 i trace D(I+θ0 )(•) θ2 ◦ (I + θ0 )−1 |det [D• (I + θ0 )] | ΩZ  
+ h∇f ◦ (I + θ0 ) | θ2 i trace D(I+θ0 )(•) θ1 ◦ (I + θ0 )−1 |det [D• (I + θ0 )] | Z Ω  
 
+ f ◦ (I + θ0 )trace D(I+θ0 )(•) θ1 ◦ (I + θ0 )−1 trace D(I+θ0 )(•) θ2 ◦ (I + θ0 )−1 |det [D• (I + θ0 )] | Ω Z    
− f ◦ (I + θ0 ) trace D(I+θ0 )(•) θ1 ◦ (I + θ0 )−1 D(I+θ0 )(•) θ2 ◦ (I + θ0 )−1 |det [D• (I + θ0 )] |. =

Ω

Ω 1 Since kθ0 k1,∞ < 1+n! , we get from Corollary ?? that the Jacobian determinant of (I + θ0 ) is positive. Combining this observation with (123) and the fact that [Com(I + D• θ0 )]T = det[D• (I + θ0 )](I + D• θ0 )−1 coming from Lemmas 5.32 and 5.14, we obtain from the foregoing: Z Dθ20 FΩ (θ1 , θ2 ) = h(Hess f ) ◦ (I + θ0 ) [θ1 ] | θ2 i det [D• (I + θ0 )] Ω Z   + h∇f ◦ (I + θ0 ) | θ1 i trace D• θ2 [Com (I + D• θ0 )]T Ω Z   + h∇f ◦ (I + θ0 ) | θ2 i trace D• θ1 [Com (I + D• θ0 )]T Z Ω  
+ f ◦ (I + θ0 ) trace D• θ1 (I + D• θ0 )−1 trace D• θ2 [Com (I + D• θ0 )]T Ω Z   − f ◦ (I + θ0 ) trace D• θ1 (I + D• θ0 )−1 D• θ2 [Com (I + D• θ0 )]T . Ω

85

Therefore, we get from the triangle inequality: Z 2 2 h(Hess f ) ◦ (I + θ0 ) [θ1 ] | θ2 i (det [D• (I + θ0 )] − 1) |Dθ0 FΩ (θ1 , θ2 ) − D0 FΩ (θ1 , θ2 ) | 6 Ω Z h[(Hess f ) ◦ (I + θ0 ) − Hess f ] [θ1 ] | θ2 i + Ω Z n   X h∇f ◦ (I + θ0 ) | θ1 i ∂j (θ2 )i [Com (I + D• θ0 )]ij − Iij + Ω Z i,j=1 h∇f ◦ (I + θ0 ) − ∇f | θ1 i div (θ2 ) + Ω Z n   X h∇f ◦ (I + θ ) | θ i ∂j (θ1 )i [Com (I + D• θ0 )]ij − Iij 0 2 + Ω i,j=1 Z h∇f ◦ (I + θ0 ) − ∇f | θ2 i div (θ1 ) + Ω Z n  
X −1 f ◦ (I + θ ) trace D θ (I + D θ ) − I ∂j (θ2 )i [Com (I + D• θ0 )]ij 0 • 1 • 0 + Ω

i,j=1 n X

Z

  f ◦ (I + θ0 ) div (θ1 ) ∂j (θ2 )i [Com (I + θ0 )]ij − Iij Ω Z i,j=1 [f ◦ (I + θ0 ) − f ] div (θ1 ) div (θ2 ) + Ω     f ◦ (I + θ0 ) trace D• θ1 (I + D• θ0 )−1 − I D• θ2 [Com (I + D• θ0 )]T Z   f ◦ (I + θ0 ) trace D• θ1 D• θ2 [Com (I + D• θ0 ) − I]T + Z Ω [f ◦ (I + θ0 ) − f ] trace (D• θ1 D• θ2 ) , + +

+

Ω

and we obtain: n X

|Dθ20 FΩ (θ1 , θ2 ) − D02 FΩ (θ1 , θ2 ) | 6 kθ1 k∞ kθ2 k∞

2 2 k∂ij f ◦ (I + θ0 ) − ∂ij f kL1 (Rn ,R)

i,j=1

+ kdet [D• (I + θ0 )] − 1kL∞ (Rn ,R) kθ1 k∞ kθ2 k∞

n X

2 k∂ij f ◦ (I + θ0 ) kL1 (Rn ,R)

i,j=1

2

+ n (kθ1 k∞ kθ2 k0,1 + kθ2 k∞ kθ1 k0,1 )

n X

! k∂i f ◦ (I + θ0 ) kL1 (Rn ,R)

i=1

max ess sup | [Com (I + Dx θ0 )]ij − Iij |

16i,j6n

x∈Rn

+ n (kθ1 k∞ kθ2 k0,1 + kθ2 k∞ kθ1 k0,1 )

n X

k∂i f ◦ (I + θ0 ) − ∂i f kL1 (Rn ,R)

i=1

+ n2 (n + 1)kθ1 k0,1 kθ2 k0,1 kf ◦ (I + θ0 ) kL1 (Rn ,R) k (I + D• θ0 )−1 − IkL∞ (Rn ,Rn2 ) max ess sup | [Com (I + Dx θ0 )]ij | 16i,j6n

x∈Rn

+ 2n2 kθ1 k0,1 kθ2 k0,1 kf ◦ (I + θ0 ) − f kL1 (Rn ,R) + 2n3 kθ1 k0,1 kθ2 k0,1 kf ◦ (I + θ0 ) kL1 (Rn ,R) max ess sup | [Com (I + Dx θ0 )]ij − Iij |. 16i,j6n

x∈Rn

2 Using successively (??) with θ0 , the fact that kθ0 k0,1 < 21 with (106) applied to f , ∂i f , and ∂ij f , for any (i, j) ∈ J1, nK2 , then (??)(??) with θ0 , and nally (??) with θ0 , we deduce that:

sup (θ1 ,θ2 )∈W 1,∞ ×W 1,∞ θ1 ,θ2 6=0

|Dθ20 FΩ (θ1 , θ2 ) − D02 FΩ (θ1 , θ2 )| kθ1 k1,∞ kθ2 k1,∞

6

2n+3 n3 (n!)2 kf k2,1 kθ0 k1,∞

+

n X

2 2 k∂ij f ◦ (I + θ0 ) − ∂ij f kL1 (Rn ,R)

i,j=1

+n

n X

k∂i f ◦ (I + θ0 ) − ∂i f kL1 (Rn ,R)

i=1 2

+ 2n kf ◦ (I + θ0 ) − f kL1 (Rn ,R) . 2 Since f ∈ W , we can apply Proposition 5.38 to f , ∂i f , and ∂ij f for any (i, j) ∈ J1, nK2 . By letting 2 correctly kθ0 k1,∞ → 0 in the previous inequality, we get |||Dθ0 FΩ − D02 FΩ ||| → 0 as kθ0 k1,∞ → 0. Hence, 2,1

86

we have proved that the map given by (126) is continuous at the origin. It remains to prove that (126) is continuous at any point of W 1,∞ ∩ B0,1 . The method is similar to the one used at the beginning of the proof to get the second-order dierentiability at any point of W 1,∞ ∩ B0,1 from the one at the origin. Let θ0 ∈ W 1,∞ be such that kθ0 k0,1 < 1. Proposition 5.31 ensures the Lipschitz continuous map I + θ0 has a Lipschitz continuous inverse. In particular, we deduce that Ω0 := (I + θ0 )(Ω) is measurable. From the foregoing, FΩ0 : θ ∈ W 1,∞ → F [(I + θ)(Ω0 )] ∈ R is twice continuously dierentiable at the origin. Let ε > 0 and set  := ε(1 − kθ0 k0,1 )2 > 0. There exists δ ∈]0, 1] such that for any θ ∈ W 1,∞ such that kθk1,∞ < δ , we have |||Dθ2 FΩ0 − D02 FΩ0 ||| 6 . Proceeding as in (110), we observe that for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθ0 k0,1 ), we have kθ0 + hk0,1 < 1 so we can write for any (θ1 , θ2 ) ∈ W 1,∞ × W 1,∞ :   Dθ20 +h FΩ (θ1 , θ2 ) − Dθ20 FΩ (θ1 , θ2 ) = Dθ2 FΩ0 θ1 ◦ (I + θ0 )−1 , θ2 ◦ (I + θ0 )−1  −D02 FΩ0 θ1 ◦ (I + θ0 )−1 , θ2 ◦ (I + θ0 )−1 , where we have set θ := h ◦ (I + θ0 )−1 . As in (111), we have kθk1,∞ 6 |Dθ20 +h FΩ (θ1 , θ2 ) − Dθ20 FΩ (θ1 , θ2 )|

khk1,∞ 1−kθ0 k0,1

< δ so we deduce that:

6

|||Dθ2 FΩ0 − D02 FΩ0 ||| kθ1 ◦ (I + θ0 )−1 k1,∞ kθ2 ◦ (I + θ0 )−1 k1,∞

6



kθ1 k1,∞ kθ2 k1,∞ = εkθ1 k1,∞ kθ2 k1,∞ . (1 − kθ0 k0,1 )2

Consequently, we obtain |||Dθ20 +h FΩ − Dθ20 FΩ ||| 6 ε for any h ∈ W 1,∞ such that khk1,∞ < δ(1 − kθ0 k0,1 ) and (126) is continuous at θ0 . We thus have proved that the map θ ∈ W 1,∞ 7→ F [(I + θ)(Ω)] ∈ R is twice continuously dierentiable at any point of W 1,∞ ∩ B0,1 and its dierential is well dened by (126). Finally, we now assume that Ω is an open bounded subset of Rn with a Lipschitz boundary. As in the proof of Theorem 5.30 (see (125) and below), let (θ1 , θ2 ) ∈ (W 1,∞ ∩ C 1 ) × (W 1,∞ ∩ C 1 ) and θ0 ∈ W 1,∞ ∩ C 1 be such that kθ0 k0,1 < 1. From Lemma 5.42, there exists two sequences (θ1k )k∈N and (θ2k )k∈N of maps in W 2,∞ converging respectively to θ1 ◦ (I + θ0 )−1 and θ2 ◦ (I + θ0 )−1 for the C 1 (Ω)-norm. Since the maps belong to W 2,∞ , we can apply (125) to θ1k and θ2k , then integrate it on Ω0 := (I + θ0 )(Ω). Using the Trace Theorem [17, Section 4.3], that we have proved to be valid for W 1,∞ -vector elds in the proof of Theorem 5.30 (see (119) and below), we obtain that (94) holds true with ∂Ω0 = (I + θ0 )(∂Ω), θ1k , and θ2k for any k ∈ N. Letting correctly k → +∞, we deduce from the uniform convergence on Ω, that (94) holds true with ∂Ω0 , θ1 ◦ (I + θ0 )−1 , and θ2 ◦ (I + θ0 )−1 . Hence, we have established that (127) is well dened by (94). If we now assume that Ω is an open bounded subset with a boundary of class C 1,1 , we can proceed exactly as in the end of the proof of Theorem 5.30 and show that (95) can be applied to ∂Ω0 , θ1 ◦ (I + θ0 )−1 , and θ2 ◦ (I + θ0 )−1 . We get that (127) is well dened by (95), concluding the proof of Proposition 5.44. If we look at the proofs of Theorems 5.295.30 and Propositions 5.39 or 5.435.44, we can notice that the arguments we use are very similar. In fact, the method is recursive and we can state the following general statement concerning the shape dierentiability of a volume integral.

Remark 5.45. To conclude this section, we can notice that Theorem 5.40 was stated with the less regularity assumptions as possible on Ω, θ, and f . For example, formulas (112)(113) are valid even for non-bounded very irregular (measurable) domains. The counterpart of this generality is the poor structure we get for the k-th order shape derivative. However, as in Theorem 5.295.30 and Propositions 5.435.44, if we assume that Ω is an open bounded subset of Rn with suciently smooth boundary (C k,1 in fact) and if we restrain the set of deformations to more regular ones (C k ∩ W k,∞ in fact), then we can obtain a better structure of the shape derivative (see e.g. [34]). More precisely, if the perturbations (θ1 , . . . , θk ) ∈ (W k,∞ ∩ C k )k are normal to the boundary, then one can check that there exists a function fΩ : ∂Ω → R, depending only on the geometry of the (hyper)-surface ∂Ω, the trace of f , and its derivatives up to order k, which is thus uniquely determined up to a set of zero A(• ∩ ∂Ω), such that: ! Z k Y k D0 FΩ (θ1 , . . . , θk ) = fΩ (x) hθj (x) | nΩ (x)i dA (x) . ∂Ω

j=1

Since fΩ does not depend on (θ1 , . . . , θk ), it can be interpreted as the k-th order partial derivative of the k map (89) with respect to the domain Ω. For this reason, it is often denoted ∂∂ΩFk (Ω) by abuse of notation. For example, in the particular case k = 3, the third-order shape derivative of (89) takes the following form for any open bounded subset Ω of Rn with a boundary of class C 2,1 and for any (θ1 , θ2 , θ3 ) ∈ (W 2,∞ ∩ C 2 )3 which are normal to the boundary ∂Ω i.e. such that θi = (θi )n nΩ on ∂Ω for any i ∈ J1, 3K: Z  D03 FΩ (θ1 , θ2 , θ3 ) = hHess f (nΩ ) | nΩ i + 2HΩ h∇f | nΩ i ∂Ω

+f

87



2 HΩ − trace D∂Ω n2Ω




 (θ1 )n (θ2 )n (θ3 )n dA,

where the notation have already been introduced in Denition 5.28 and Theorems 5.295.30. Hence, the third-order partial derivative of (89) with respect to any open bounded subset Ω of Rn with C 2,1 -boundary is given by: 
 ∂3F (Ω) = hHess f (nΩ ) | nΩ i + 2HΩ h∇f | nΩ i + f HΩ (x)2 − trace D∂Ω n2Ω , 3 ∂Ω where the boundary values of f ∈ W 3,1 (Rn , R), ∇f , and Hess f have to be understood in the sense of trace.

5.2.9 About Lipschitz boundaries, C 1,1 domains, and complex valued functionals First, we briey recall the denition of a Lipschitz open subset of Rn . In the context of bounded domains, this denition is equivalent to require the uniform cone property on the boundary [26, Theorem 1.2.2.2].

Denition 5.46.

Let n > 2 and Ω be an open subset of Rn . We say that Ω is Lipschitz or has a Lipschitz boundary if for any point x0 ∈ ∂Ω, there exists a direct orthonormal frame centred at x0 such that in this local frame, there exists an L-Lipschitz continuous map ϕ : Dr (00 ) →] − a, a[ with a > 0 and L > 0, satisfying ϕ(00 ) = 0, and such that:  0 0 0 0 0  ∂Ω ∩ (Dr (0 ) ×] − a, a[) = {(x , ϕ(x )) , x ∈ Dr (0 )} Ω ∩ (Dr (00 ) ×] − a, a[)



=

{(x0 , xn ),

x0 ∈ Dr (00 ) and − a < xn < ϕ(x0 )} ,

where Dr (00 ) = {x0 ∈ Rn−1 , |x0 | < r} is the open ball of Rn−1 centred at the origin 00 and of radius r > 0. Then, we briey recall the denition of C 1,1 -regularity for open subsets of Rn . In the context of bounded domains, there is also two equivalent characterizations of the C 1,1 -regularity, that are well-known and that were studied in details in [14, Theorems 1.6-1.8]. Roughly speaking, the following denition is equivalent to require a uniform ball condition on the boundary, which is itself equivalent to assume that the boundary has a positive reach i.e. that there exists an open tubular neighbourhood of the boundary in which every point has a unique projection on the boundary. Finally, it is also equivalent to require the C 1,1 -regularity of the oriented distance function x ∈ Rn 7→ d(x, Ω) − d(x, Rn \Ω) ∈ R (see e.g. [16])

Denition 5.47.

Let n > 2 and Ω be an open subset of Rn . We say that Ω is a C 1,1 -domain or has a boundary of class C 1,1 if for any point x0 ∈ ∂Ω, there exists a direct orthonormal frame centred at x0 such that in this local frame, there exists a map ϕ : Dr (00 ) →] − a, a[ continuously dierentiable with a > 0, such that ϕ and its gradient ∇ϕ are L-Lipschitz continuous with L > 0, satisfying ϕ(00 ) = 0, ∇ϕ(00 ) = 00 , and also:  0 0 0 0 0  ∂Ω ∩ (Dr (0 ) ×] − a, a[) = {(x , ϕ(x )) , x ∈ Dr (0 )} Ω ∩ (Dr (00 ) ×] − a, a[)



where Dr (0 ) = {x ∈ R 0

0

n−1

=

{(x0 , xn ),

x0 ∈ Dr (00 ) and − a < xn < ϕ(x0 )} ,

, kx k < r} is the open ball of Rn−1 centred at the origin 00 and of radius r > 0. 0

Finally, we recall this classical result which can interpreted as the divergence theorem for C 1,1 -surfaces.

Lemma 5.48.

Let Ω be an open bounded subset of Rn with a C 1,1 -boundary as in Denition 5.47. Then, for any v ∈ W (∂Ω, Rn ) such that hv | nΩ i = 0, we have: Z div∂Ω (v)dA = 0. 1,1

∂Ω

Proof. Consider any open bounded subset of Rn with a C 1,1 -boundary. Let x0 ∈ ∂Ω. Denition 5.47 ensures the existence of a local C 1,1 -parametrization X : x0 ∈ D(x0 ) 7→ (x0 , ϕ(x0 )) ∈ ∂Ω ∩ C(x0 ), where D(x0 ) is an open hyper-disk centred at x0 , and where C(x0 ) is an open cylinder of Rn centred at x0 . First, we assume that v : ∂Ω → Rn is a smooth map with compact support in ∂Ω ∩ C(x0 ). We can decompose v ◦ X in the basis (∂1 X, . . . , ∂n−1 X, nΩ ). Since hv | nΩ i = 0, we have: v◦X =

n−1 X

g ij hv ◦ X | ∂j Xi∂i X + hv ◦ X | nΩ ◦ Xi (nΩ ◦ X) =

i,j=1

n−1 X

g ij vj ∂i X,

i,j=1

where we set vj = hv ◦ X | ∂j Xi and where (g )16i,j6n−1 is the inverse matrix of the rst fundamental form dened for any (i, j) ∈ J1, n − 1K2 by gij := h∂i X | ∂j Xi. In this decomposition, note that v ◦ X is a Lipschitz continuous map so from Rademacher's Theorem [17, Section 3.1.2], it is dierentiable almost everywhere and we can compute: * ! + n−1 n−1 n−1 X kl X kl X ij div∂Ω (v) ◦ X = g h∂l (v ◦ X) | ∂k Xi = g ∂l g vj ∂i X | ∂k X ij

k,l=1

=

n−1 X k,l=1

g kl

n−1 X i,j=1



i,j=1

k,l=1



∂l g ij vj h∂i X | ∂k Xi

! +

n−1 X

n−1 X

i=1

j=1

88

! g ij vj

n−1 X

 k,l=1

 g kl h∂i (∂l X) | ∂k Xi .

Since X is a C 1,1 -map, it is twice-dierentiable almost everywhere and at the point where it is the case, we have ∂l (∂i X) = ∂i (∂l X). Moreover, the matrix (g kl ) is symmetric so we deduce that:  ! n−1 n−1 !  n−1 n−1 X kl X kl n−1 X  ij  X X ij g h∂l (∂i X) | ∂k Xi div∂Ω (v) ◦ X = g ∂l g vj gik + g vj  i,j=1

k,l=1

=

n−1 X

n−1 X

∂l

i=1

! ij

g vj

j=1

i,l=1

n−1 X

! gik g

kl

+

j=1

 !  n−1 X kl 1 g vj  g ∂i (glk ) . 2 j=1

n−1 X

n−1 X

i=1

k=1

k,l=1

ij

k,l=1

Then, we observe that the rst term has a simplication since (g ij ) is the inverse matrix of g := (gij ) and the second term is the dierential of a determinant by Proposition 5.24. Hence, we obtain: ! !  n−1 n−1 n−1 X X ij X n−1 X ij
1 Trace ∂i (g)g −1 div∂Ω (v) ◦ X = ∂l g vj δil + g vj 2 j=1 i=1 j=1 i,l=1 =

n−1 X

∂i

i=1

=

n−1 X

! ij

g vj

+

j=1

n−1 X 1 p ∂i det(g) i=1

n−1 X

n−1 X

i=1

j=1

! ij

g vj

n−1 X ij p det(g) g vj

∂i (det(g)) 2det(g)

! .

j=1

Since v has compact support in ∂Ω ∩ C(x0 ), so does vj = hv ◦ X | ∂j Xi on D(x0 ) and we get: ! Z Z n−1 n−1 XZ X ij p p div∂Ω (v)dA = (div∂Ω (v) ◦ X) det(g) = det(g) g vj = 0. ∂i ∂Ω

D(x0 )

i=1

D(x0 )

j=1

The result of Lemma 5.48 is thus established if v : ∂Ω → Rn is a smooth map with compact support in ∂Ω ∩ C(x0 ) for any x0 ∈ ∂Ω. Then, we assume that v ∈ C ∞ (∂Ω, Rn ). Since ∂Ω is compact, there exists a nite number K > 1 of points denoted by x1 , . . . , xK such that ∂Ω ⊂ ∪K k=1 C(xk ). We can build a partition of unity onPthis set. There exists K smooth maps ξk : Rn → [0, 1] with compact support in C(xk ), and such that K k=1 ξk = 1 on ∂Ω. Hence, we have successively: ! Z Z K K Z X X div∂Ω (v)dA = div∂Ω ξk v dA = div∂Ω (ξk v)dA = 0, ∂Ω

∂Ω

k=1

k=1

∂Ω

where the last equality comes from the previous case because ξk v is a smooth map with compact support in C(xk ) for any k = 1, . . . , K . The result of Lemma 5.48 holds for any v ∈ C ∞ (∂Ω, Rn ). Finally, we assume that v ∈ W 1,1 (∂Ω, Rn ). By density, there exists a sequence of smooth maps vm ∈ C ∞ (∂Ω, Rn ) such that vm − v tends to zero in W 1,1 (∂Ω, Rn )(combine e.g. [26, Denition 1.3.3.2] and [7, Theorem 9.2]). We can apply the previous case on each vm and we get: Z Z n Z X div∂Ω (v)dA = div∂Ω (v − vm )dA 6 |∇∂Ω (vi ) − ∇∂Ω (vim ) |dA −→ 0. ∂Ω

To conclude, we proved

∂Ω

R ∂Ω

i=1

∂Ω

m→+∞

div∂Ω (v)dA = 0 for any map v ∈ W 1,1 (∂Ω, Rn ) such that hv | nΩ i = 0.

Remark 5.49. To conclude, all the results of Section 5.2 still holds true with a complex-valued map f in (89) instead of real-valued one. More precisely, Theorems 5.295.30 and 5.40, and Propositions 5.39, 5.43, and 5.44 are still valid for a complex valued map f in (89). Indeed, decomposing the real and imaginary parts of such an f , on which we can apply the dierentiability results, we obtain by linearity the results for f . More generally, if the map f in (89) is valued in a real or complex nite dimensional space, the results of Section 5.2 remain valid. Indeed, since being dierentiable for a multi-valued map is equivalent to require the dierentiability of its components, we can apply the results of Section 5.2 to each components (or to the real and imaginary parts in the complex case). We deduce that Theorems 5.295.30 and 5.40, and Propositions 5.39 and 5.435.44 hold true for any map f in (89) which is valued in a real or complex 2 2 nite dimensional space. In particular, its is true if such an f is valued in the matrix spaces Rn and Cn .

89

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