Shape Optimization: Variations of Domains and Applications 9783111025438, 9783111025261

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Catherine Bandle, Alfred Wagner Shape Optimization

De Gruyter Series in Nonlinear Analysis and Applications

�

Editor in Chief Jürgen Appell, Würzburg, Germany Editors Catherine Bandle, Basel, Switzerland Manuel del Pino, Santiago de Chile, Chile Avner Friedman, Columbus, Ohio, USA Mikio Kato, Tokyo, Japan Wojciech Kryszewski, Torun, Poland Umberto Mosco, Worcester, Massachusetts, USA Vicenţiu D. Rădulescu, Krakow, Poland Simeon Reich, Haifa, Israel

Volume 42

Catherine Bandle, Alfred Wagner

Shape Optimization �

Variations of Domains and Applications

Mathematics Subject Classification 2020 Primary: 35Q93, 49Q10, 52A40; Secondary: 49R05, 53Z99 Authors Prof. Dr. Catherine Bandle University of Basel Department of Mathematics and Computer Science Spiegelgasse 1 4051 Basel Switzerland [email protected]

Priv. Doz. Dr. Alfred Wagner RWTH Aachen University Institute for Mathematics Templergraben 55 52062 Aachen Germany [email protected]

ISBN 978-3-11-102526-1 e-ISBN (PDF) 978-3-11-102543-8 e-ISBN (EPUB) 978-3-11-102581-0 ISSN 0941-813X Library of Congress Control Number: 2023932466 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2023 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

Preface The description of optimal forms plays an important role in mathematics and applications. In many areas of mathematics, pure and applied, it has recently received a lot of attention. In this book we consider domain functionals, like energies and eigenvalues of elliptic operators. Assuming the existence of an optimal domain, we are interested in its shape. Global methods like symmetrization and rearrangement inequalities are applicable in certain settings; however, for many interesting problems, no global approach is known. In the absence of a global approach, one can instead study the effect of local perturbations. In the spirit of calculus one considers one-parameter families of perturbations of a fixed domain. The first derivative with respect to this parameter can be used to derive a necessary condition for a domain to be optimal. The second derivative provides additional information to help to determine local maxima or minima. In general these derivatives are difficult to analyze. In particular the second derivative, which is crucial to understand stability, is complicated. We develop techniques that enable us to determine its sign in various cases of interest. The structure of this book can briefly be summarized as follows. We start with an overview of the main examples that will be treated in this book. They are selected in order to show the breadth of the applicability of our methods. Among them are energies of nonlinear boundary value problems, eigenvalues, and problems of fourth order. We then describe the class of admissible domain perturbations. It should be emphasized that our technique requires smoothness of the domains. The arguments will in general fail in the case of nonsmooth domains. Nonsmooth analysis for domain variations is not yet available. We pay special attention to volume and area preserving perturbations. In order to compare the family of perturbed domains, we introduce differential geometric tools. This allows us to compute variational formulas for volume and surface area. Our functionals are domain and boundary integrals. We present two methods to derive variational formulas. The first is the change of variables method. All quantities will be mapped onto a fixed domain. This requires only the chain rule. The second approach, due to Reynolds, is the moving surface method. It captures the shape by means of a boundary displacement and generally requires additional regularity. The resulting variational formulas coincide and are illustrated with some examples. Among them are problems of optimal control, convolutions, and weighted isoperimetric inequalities. It becomes apparent that there is a difference between variational formulas for purely geometrical functionals and for those functionals which depend on functions varying with the domain. https://doi.org/10.1515/9783111025438-201

VI � Preface The most involved variational formulas are derived for energies of semilinear elliptic problems. They depend only on the boundary data and the perturbations of the boundary. We discuss these formulas for different boundary value problems. To find precise estimates for the second variation, we expand it in a suitable basis of functions consisting of eigenfunctions of elliptic eigenvalue problems. In many examples we are able to determine the sign. This method applies to any critical domain. For the ball this system of eigenfunctions restricted to the sphere corresponds to the expansion with respect to spherical harmonics. The first and second variations allow us to derive some local isoperimetric inequalities. The well-known classical isoperimetric inequality states that among all domains of given volume the ball has the smallest perimeter. We will call an inequality isoperimetric if it relates geometrical or physical quantities defined on the same domain and if the equality sign is attained for some optimal domain. We extend this notion to a broader class of functionals such as energies and eigenvalues. Fourth order problems are studied at the end of the book by means of the moving surface method. For the buckling eigenvalue, we are able to use the variational formulas to prove uniqueness of the optimal domain. This book attempts to bridge the gap between analysis and geometry. Our exposition is example-driven and draws on problems that are the object of study in the current literature. We present them in a self-contained way, at times pointing out open questions that we hope will lead to further investigation and interesting discoveries. Domain variation has a long history, which can be traced back to Hadamard (1908). More detailed historical comments are given at the end of the relevant chapters. The book is aimed at readers with basic knowledge of analysis and geometry. We especially thank Simon Stingelin for producing all the illustrations in this book.

Contents Preface � V 1 1.1 1.2 1.2.1 1.2.2 1.3

Overview of the basic problems � 1 Semilinear elliptic boundary value problems � 1 Eigenvalue problems � 2 The problem of vibrating membranes � 2 Buckling problem for a clamped plate � 3 Green’s function, Hadamard’s formula � 4

2 2.1 2.2 2.2.1 2.2.2 2.2.3 2.3 2.3.1 2.3.2 2.3.3 2.3.4 2.3.5 2.4 2.4.1 2.4.2 2.4.3

Basic concepts � 7 Domain perturbations � 7 Geometry of hypersurfaces � 9 Preliminaries � 9 Projections � 12 Reduction to Hadamard perturbations � 13 First and second variation of the volume � 18 Discussion of 𝒱̈ (0) � 19 First and second variation of the surface area perimeter � 23 Area variations for the surface of the ball � 27 Hadamard perturbations � 30 Tangential perturbations � 31 Intrinsic versus extrinsic � 32 The distance function � 32 Shape derivatives � 34 Notes � 37

3 3.1 3.2 3.2.1 3.2.2 3.2.3 3.2.4 3.3 3.3.1 3.4 3.4.1 3.5

Spherical harmonics and eigenvalue problems � 38 Spherical harmonics � 38 The Steklov eigenvalue problem � 43 The ball � 43 The outer domain of a ball � 45 A Steklov–Dirichlet eigenvalue problem in annular domains � 45 Steklov eigenvalues in annular domains � 46 Membrane eigenvalue problems � 48 The generalized Steklov problem � 49 The buckling plate � 49 Fourth order Steklov eigenvalue problem � 50 Notes � 51

VIII � Contents 4 4.1 4.1.1 4.1.2 4.1.3 4.1.4 4.2 4.2.1 4.3

Variational formulas � 52 The change of variables method � 52 Introduction � 52 Volume integrals which depend on u only � 53 Boundary integrals which depend on u only � 55 The Dirichlet integral � 57 The moving surface method � 59 Second variation for special cases � 63 Notes � 69

5 5.1 5.2 5.2.1 5.2.2 5.3 5.3.1 5.3.2 5.4 5.4.1 5.5

Geometric inequalities, convolutions, cost functions � 71 Moment of inertia � 71 Isoperimetric inequalities in spaces of constant curvature � 75 Spherical space 𝕊n � 75 Hyperbolic space ℍn � 79 Convolutions � 81 ̈ for the ball � 86 The sign of 𝒞 (0) Gamov’s liquid model � 89 An optimal control problem � 92 The torsion problem � 93 Notes � 97

6 6.1 6.2 6.2.1 6.2.2 6.3 6.3.1 6.4 6.4.1 6.5 6.5.1 6.5.2 6.5.3

Domain variations for energies � 99 Energies and critical points � 99 Elliptic boundary value problems � 100 General setting, change of variables � 100 Boundary value problem for the shape derivative � 101 Differentiation of the energy � 105 First and second domain variation � 105 The first domain variation � 107 The first domain variation for volume preserving perturbations � 108 The second domain variation � 109 The transformations of ℱ1 (0), ℱ2 (0), and ℱ3 (0) � 109 Main results � 114 Notes � 118

7 7.1 7.2 7.3 7.4 7.4.1

Discussion of the main results � 119 Translations and rotations � 119 Hadamard perturbations � 121 Tangential perturbations υ = υτ � 123 Volume preserving perturbations � 125 Dirichlet energy � 125

Contents �

7.4.2 7.5 7.5.1

Robin energy � 126 Neumann energy � 128 Notes � 130

8 8.1 8.2 8.2.1 8.2.2 8.2.3 8.3

General strategy and applications � 131 A generalized Steklov eigenvalue problem � 131 Nearly spherical domains � 135 Robin energy � 135 The torsion problem with α > 0 � 143 Dirichlet energy � 144 Notes � 147

9 9.1 9.1.1 9.1.2 9.1.3 9.2 9.2.1 9.2.2 9.3

Eigenvalue problems � 148 Robin eigenvalue problem � 148 Domain variations for Robin eigenvalues with α > 0 � 149 ̇ � 151 The discussion of λ(0) ̈ Discussion of λ(0) for the ball � 153 Nonlinear eigenvalue problems � 157 Examples � 159 First variation of the turning point � 163 Notes � 166

10 10.1 10.2 10.2.1 10.3 10.3.1 10.3.2

Quantitative estimates � 167 Preliminary remark � 167 The Fraenkel asymmetry function � 168 Stability criterion � 169 Example � 170 The Robin energy for α > 0 � 170 Notes � 172

11 11.1 11.2 11.3

The Robin eigenvalues for α < 0 � 173 The interior eigenvalue problem � 173 The exterior eigenvalue problem � 178 Notes � 187

12 12.1 12.2 12.3 12.4

Problems with infinitely many positive and negative eigenvalues � 188 Eigenvalue problem with dynamical boundary conditions � 188 Known results � 189 Comparison with balls of the same volume � 191 Notes � 194

IX

X � Contents 13 13.1 13.2 13.2.1 13.3 13.3.1 13.3.2 13.3.3

The torsion problem for α < 0 � 195 General remarks � 195 Domain variations for nearly spherical domains � 197 Second variation � 197 An alternative approach: Robin versus Dirichlet torsion � 201 The functional 𝒥 (Ω) � 203 Domain variations of 𝒥 (t) � 203 The second variation of 𝒥 (Ωt ) in nearly spherical domains � 204

14 14.1 14.2 14.3 14.4

Problems in annular domains � 206 An eigenvalue problem related to trace inequalities � 206 Bernoulli’s problem � 209 The Robin torsion problem for α > 0 � 213 Notes � 216

15 15.1 15.2 15.3 15.4 15.5 15.6 15.7

The first buckling eigenvalue of a clamped plate � 217 The first eigenvalue � 217 Shape derivatives � 218 The first domain variation � 220 The second domain variation � 222 Minimization of the second domain variation � 226 The optimal domain is a ball � 232 Notes � 234

16 16.1 16.2 16.3 16.4 16.5 16.6

A fourth order Steklov problem � 236 The Steklov eigenvalue � 236 Shape derivatives � 237 First domain variation � 239 First variation for the ball � 242 Second variation for the ball � 243 Notes � 251

A A.1 A.2 A.2.1

General remarks � 253 Jacobi’s formula for determinants � 253 Hölder continuity � 253 Whitney’s extension theorem � 254

B B.1 B.1.1 B.1.2 B.2

Geometry � 257 Some concepts from differential geometry � 257 Curvilinear coordinates in ℝn � 258 Normal coordinates � 259 Implicit function theorem � 261

Contents �

C

Sobolev spaces and inequalities � 263

D D.1 D.2 D.2.1 D.2.2 D.2.3 D.2.4

Bilinear forms � 265 Abstract setting � 265 Examples � 266 Eigenvalues on the sphere � 266 Robin eigenvalues � 267 Buckling eigenvalue � 267 Steklov eigenvalue of fourth order � 268

Notation � 269 Bibliography � 271 Index � 277

XI

1 Overview of the basic problems In this chapter we describe briefly some selected classical problems to which we shall apply domain variations. The theory of domain variations started with Hadamard’s formula for the Green’s function. For historical reasons we review this formula at the end of this chapter.

1.1 Semilinear elliptic boundary value problems Let Ω ⊂ ℝn be a bounded, smooth domain with the outer normal ν and let g : ℝ → ℝ be a continuous function. The boundary value problems taken into consideration are Δu + g(u) = 0

in Ω,

subject to the boundary conditions u = 0,

Dirichlet,

𝜕ν u + αu = 0, 𝜕ν u = 0,

Robin,

Neumann,

where 𝜕ν denotes the derivative in the direction of the outer normal and α ∈ ℝ is a fixed number. The solutions are the Euler–Lagrange equations corresponding to the energy functional 2

2

ℰ (Ω, u) := ∫ |∇u| dx − 2 ∫ G(u) dx + α ∮ u dS. Ω

Ω

𝜕Ω

Here G is the primitive of g, i. e., G′ = g, and dS is the surface element of 𝜕Ω. For the Dirichlet and Neumann problem set α = 0. If Ω ⊂ ℝ2 is simply connected, the Dirichlet problem with g = 1 is a model for the stress of beams with constant cross-section. The torsional rigidity corresponds to −ℰ (Ω, u). Saint-Venant conjectured in 1856 that among all domains of given area, the circle has the largest torsional rigidity. This conjecture was later proved by Pólya in 1948. In this book we shall use the expression torsion problem for general boundary value problems with g(u) = 1. Special attention will be given to this particular problem with Robin boundary conditions. If α is negative, it is uniquely solvable if and only if α does not coincide with an eigenvalue of the Steklov problem Δφ = 0 in Ω,

𝜕ν φ = μφ on 𝜕Ω.

There is obviously no solution in the case of Neumann boundary conditions. https://doi.org/10.1515/9783111025438-001

2 � 1 Overview of the basic problems In many applications, a domain of prescribed volume is sought, for which the enrgy is a minimum or maximum. If u is a positive minimizer of ℰ (Ω, u) satisfying Dirichlet boundary conditions, rearrangement and symmetrizations are very powerful techniques to study optimal domains (see [98]). For more general problems, in particular for energies whose solutions change sign or satisfy a Robin boundary condition, they do not apply. A new approach to look for optimal domains with Robin boundary conditions is discussed in [29].

1.2 Eigenvalue problems 1.2.1 The problem of vibrating membranes Let Ω ⊂ ℝn be a bounded domain. Consider the eigenvalue problem Δϕ + λϕ = 0

in Ω,

subject to one of the following boundary conditions: ϕ = 0, +

𝜕ν ϕ + αϕ = 0, α ∈ ℝ , 𝜕ν ϕ = 0,

Dirichlet conditions, Robin boundary conditions, Neumann boundary conditions.

These problems possess a countable number of positive eigenvalues, 0 < λ1 < λ2 ≤ λ3 ≤ ⋅ ⋅ ⋅ λn → ∞. In the case of Neumann conditions, λ0 = 0 is an eigenvalue and the corresponding eigenfunction is ϕ0 = const. The fundamental frequency of the Dirichlet and Robin problems plays a central role in analysis and in physical applications. It can be characterized by the Rayleigh principle λD1 = min R(υ) := min W01,2 (Ω)

W01,2 (Ω)

∫Ω |∇υ|2 dx ∫Ω υ2 dx

for the Dirichlet eigenvalue problem and by λR1 := min R(α, υ) = min W 1,2 (Ω)

W 1,2 (Ω)

∫Ω |∇υ|2 dx + α ∮𝜕Ω υ2 ds ∫Ω υ2 dx

for the Robin eigenvalue problem. In both cases the minimum is achieved for the first eigenfunction. As a consequence of the Rayleigh principle together with Harnack’s inequality, the first eigenfunction is of constant sign and the lowest eigenvalue is simple.

1.2 Eigenvalue problems � 3

A physical model in two dimensions is a vibrating membrane which at rest covers the domain Ω. The deflection in the normal direction u(x, t) solves the wave equation utt = Δu. If we set u = eiωt ϕ(x), then ϕ is a solution of the eigenvalue problem described above with λ = ω2 . The Dirichlet conditions imply that the membrane is fixed on the boundary, the Robin conditions describe an elastic attachment, and in the case of Neumann conditions the membrane moves freely on the boundary (cf. Figure 1.1).

Figure 1.1: Different boundary conditions.

The eigenvalue λR1 (α) is a concave, monotone increasing function such that λR1 (0) = 0 and limα→∞ λR1 (α) = λD1 , where λD1 corresponds to the first eigenvalue with Dirichlet boundary conditions (see Figure 1.2).

Figure 1.2: Robin eigenvalue for the circle of area 1.

The well-known Rayleigh–Faber–Krahn inequality (cf. [99]) states that among all domains of given volume, λD1 (Ω) is minimal for the ball. Bossel [25] and Daners [41] have shown that this inequality also holds for λR1 (Ω) provided α is positive. Daners and Kennedy [42] proved that the ball is the unique minimal domain. 1.2.2 Buckling problem for a clamped plate In this subsection we are interested in the lowest eigenvalue of the problem Δ2 u + Λ(Ω)Δu = 0 u = 𝜕ν u = 0

in Ω, in 𝜕Ω.

4 � 1 Overview of the basic problems For its characterization we need the Rayleigh quotient ℛ(u, Ω) :=

∫Ω |Δu|2 dx

∫Ω |∇u|2 dx

.

Then Λ(Ω) := inf{ℛ(u, Ω) : u ∈ W02,2 (Ω)}.

(1.2.1)

The sign of the first eigenfunction may change depending on Ω. This is in contrast to the eigenfunctions described in the previous section. The quantity Λ(Ω) is called the buckling eigenvalue of Ω. It is well known that there is a discrete spectrum of positive eigenvalues of finite multiplicity and their only accumulation point is ∞. The corresponding eigenfunctions form an orthonormal basis of W02,2 (Ω). Some of these statements are found in [125]. In 1951, G. Pólya and G. Szegö formulated the following conjecture (see [98]): Among all domains Ω of given volume, the ball minimizes Λ(Ω). Partial results are known; however, the conjecture is still open for general domains. The most general result today is found in [115] and will be presented in this text. A crucial tool aside from the variational formulas is Payne’s inequality ([95], see also [56]). He showed that for any bounded domain λ2 (Ω) ≤ Λ(Ω), where λ2 (Ω) is the second eigenvalue of the Laplacian with Dirichlet boundary conditions, equality holds if and only if Ω is a ball.

1.3 Green’s function, Hadamard’s formula The modern theory of shape derivatives or domain variations started with the work of Hadamard [69]. The highlight was his variational formula for the Green’s function. The Green’s function corresponding to the Laplace operator in Ω with Dirichlet boundary conditions solves Δx G(x, y) = −δy (x) in Ω,

G(x, y) = 0 for x ∈ 𝜕Ω, y ∈ Ω.

It can be written in the form G(x, y) = h(x, y) + γ(|x − y|),

1

where γ(|x − y|) = { 2π

log |x − y|

1 |x (n−2)|𝜕B1 |

− y|

−2+n

if n = 2, if n > 2.

1.3 Green’s function, Hadamard’s formula

� 5

Here B1 is the unit ball in ℝn and h(x, y) is a harmonic function in Ω. We briefly describe Hadamard’s argument. Let Ωt be a family of domains, the boundary of which is obtained from 𝜕Ω by a shift tρ along its outer normal ν. The parameter t is close to zero and ρ is a smooth function defined on 𝜕Ω. The Green’s function in Ωt will be denoted by G(x, y : t). Hadamard studied its dependence on small |t| and showed that lim t −1 {G(x, y : t) − G(x, y)} = ∮ 𝜕ν G(x, z)𝜕ν G(z, y)ρ(z) dSz . t→0

𝜕Ω

His reasoning was as follows (see Figure 1.3).

Figure 1.3: Hadamard’s domain decomposition.

For x, y ∈ Ωt ∩ Ω, G(x, y : t) − G(x, y) = ∫ G(x, z : t)Δz G(z, y) − G(x, z)Δz G(z, y : t) dz Ωt ∩Ω

= ∫ G(x, z : t)𝜕ν G(z, y) dSz − ∫ G(x, z)𝜕ν̂ G(z, y : t) dSz , Γ−

Γ+

where Γ− = 𝜕Ω ∩ Ωt and Γ+ = 𝜕Ωt ∩ Ω. Here ν̂ stands for the outer normal of Ωt ∩ Ω. If the boundaries 𝜕Ω and 𝜕Ωt are sufficiently smooth, we have 0 = G(x, z : t) + t𝜕ν G(x, z : t)ρ(z) + o(t) on Γ− . A similar expansion holds for G(x, y) on Γ+ . Thus, G(x, y : t) − G(x, y) = − ∫ t𝜕ν G(x, z : t)ρ(z)𝜕ν G(z, y) dSz Γ−

+ ∫ t𝜕ν̂ G(x, z)ρ(z)𝜕ν̂ G(z, y : t) dSz + o(t 2 ), Γ+

Dividing by t and taking the limit as t → 0 we obtain Hadamard’s formula.

6 � 1 Overview of the basic problems In the same memoir Hadamard considered the Green’s function for Neumann boundary conditions as well as for the plate problem Δ2 G(x, y) = δy (x) in Ω, G(x, y) = 𝜕ν G(x, y) = 0 if x on 𝜕Ω and y in Ω. He showed that lim t −1 {G(x, y : t) − G(x, y)} = ∮ Δz G(x, z)Δz G(y, z)ρ(z) dS. t→0

𝜕Ω

Later Garabedian and Schiffer [63] gave a rigorous alternative proof of Hadamard’s formula. They also computed the second derivative of G(x, y : t) with respect to t. Generalizations to the Green’s function of general elliptic boundary value problems are found in [59] and for higher order problems in [96]. The question of smoothness has been studied in [117]. There it is shown that Hadamard’s formula holds for Ω ∈ C 1,1 , whereas for the second derivative Ω is required to be in C 2,α , 0 < α < 1. It should be pointed out that Schiffer [104] transformed Hadamard’s formula for plane domains such that it holds without any restriction on the boundary.

2 Basic concepts The concept of domain perturbation will be introduced. For a bounded smooth domain we define a one-parameter family of smooth diffeomorphisms. This leads to a family of smooth domains which are called “perturbed domains”. In the following we will consider special diffeomorphisms, for example those perturbing the initial domain only in the normal direction (Hadamard perturbations). We show that any domain perturbation can be chosen to be a Hadamard perturbation. An application of domain variations are the volume and surface functionals. The resulting variational formulas are frequently applied in this text. We review some differential geometric quantities of surface theory which later appear explicitly in variational formulas.

2.1 Domain perturbations We will use the following notation: x := (x1 , x2 , . . . , xn ): (x ⋅ y)

or

x ⋅ y: 1/2

|x| = (x ⋅ x) : n

dx = ∏ dxi : i=1

ei = (0, . . . , 0, 1, 0, . . . ),

i = 1, 2, . . . , n:

point in ℝn , expressed in Cartesian coordinates, Euclidean scalar product, length of a vector, volume element inℝn , coordinate axis.

→ 0 as t → 0 and O(t) means The symbol o(t) stands for expressions in t such that o(t) t | is bounded as t → 0. We shall always use the Einstein convention that the that | O(t) t repeated indices are being summed over. Throughout this text we assume that Ω ⊂ ℝn is a connected bounded smooth domain. The exact smoothness will be determined when needed. For given t0 > 0 and t ∈ (−t0 , t0 ), let {Ωt }|t| 0 be such that its inverse exists for all |t| < t0 and is of the form (DΦt )−1 = I − tDυ +

t2 (2(Dυ )2 − Dw ) + o(t 2 ), 2

(2.1.3)

where (Dυ )2 = 𝜕i υk 𝜕k υj . Thus, for t = 0 we obtain 󵄨󵄨 d 󵄨 (DΦt (x))ij 󵄨󵄨󵄨 = 𝜕i υj , 󵄨󵄨t=0 dt

󵄨 d −1 󵄨󵄨 (DΦt (x))ij 󵄨󵄨󵄨 = −𝜕i υj , 󵄨󵄨t=0 dt

(2.1.4)

󵄨 d2 −1 󵄨󵄨 󵄨󵄨 = 2𝜕i υk 𝜕k υj − 𝜕i wj . (D (x)) Φ 󵄨󵄨 ij t dt 2 󵄨t=0 After the change of coordinates the volume element dy has to be replaced by J(t) dx, where J(t) = det(DΦt ). By Jacobi’s formula (see Appendix A.1), we have for small t0 J(t) := det(I + tDυ + = 1 + t div υ +

t2 D + o(t 2 )) 2 w

t2 ((div υ)2 − Dυ : Dυ + div w) + o(t 2 ), 2

(2.1.5)

where Dυ : Dυ := 𝜕i υj 𝜕j υi . Observe that Dυ : Dυ is the trace of D2υ . Clearly, J(t) ≠ 0 if t0 is small and the map Φt : Ω → Ωt is therefore a local diffeomorphism by the inverse function theorem. Its inverse will be denoted by Φ−1 t . Next we will show that Φt : Ω → Ωt is a bijective map. For this purpose we introduce the function 1 Φ̂ t (x) = (Φt (x) − x). t

2.2 Geometry of hypersurfaces

� 9

Lemma 2.1. Assume that there exists a positive number c < 1/t0 such that ‖DΦ̂ ‖L∞ (Ω) ≤ c t

∀ |t| < t0 .

Then Φt (x) = Φt (z) implies that x = z. Proof. The proof is done by contradiction. We assume that for some fixed t with |t| < t0 there exist two points x, z ∈ Ω such that Φt (x) = Φt (z) and

|x − z| > 0.

Hence, 󵄨 󵄨 |x − z| = 󵄨󵄨󵄨−t(Φ̂ t (x) − Φ̂ t (z))󵄨󵄨󵄨 ≤ ct0 |x − z|.

The choice ct0 < 1 contradicts |x − y| > 0.

2.2 Geometry of hypersurfaces 2.2.1 Preliminaries The boundary 𝜕Ω will be represented by local coordinates as follows. Let V ⊂ ℝn be an open set such that ̃ V ∩ 𝜕Ω := {x(ξ) : ξ ∈ U ⊂ ℝn−1 }. The metric is given by the first fundamental form gij dξi dξj := (x̃ξi ⋅ x̃ξj ) dξi dξj . The inverse of the metric tensor gij will be denoted by g ij . The surface element of 𝜕Ω is dS = √det G dξ

with G = (gij )i,j=1,...,n−1 ,

where dξ = dξ1 ⋅ ⋅ ⋅ ⋅ ⋅ dξn−1 . The vectors x̃ξi , i = 1, 2, . . . , n − 1, span the tangent space in a ̃ and ν stands for the unit outer normal at a point x(ξ) ̃ point x(ξ) ∈ 𝜕Ω. In the Euclidean metric this is expressed as (ν ⋅ x̃ξi ) = 0 for i = 1, . . . , n − 1 and (ν ⋅ ν) = 1. We shall use the ̃ := ν(x(ξ)). ̃ notation ν(ξ) The integral of a given continuous function f : 𝜕Ω → ℝ over V ∩ 𝜕Ω is given by √det G(ξ) dξ = ∫ f ̃(ξ)√det G(ξ) dξ. ̃ ∮ f (x) dS = ∫ f (x(ξ)) V ∩𝜕Ω

U

U

10 � 2 Basic concepts i

i

0 0 Vi ∩ 𝜕Ω. Let such that 𝜕Ω ⊂ ∪i=1 Since 𝜕Ω is compact, there exists a finite covering {Vi }i=1

i

i0 , {pi }i=1

1. 2. 3.

0 , i. e., pi ∈ C0∞ (ℝn ) be a partition of unity subordinate to the covering {𝜕Ω ∩ Vi }i=0 n pi : ℝ → [0, 1], supp{pi } ⊂ Vi , i0 for any x ∈ Ω, ∑i=1 pi (x) ≡ 1.

For simplicity we shall write i0

∮ f (x) dS := ∑ ∮ pi (x)f (x) dS. 𝜕Ω

i=1 V ∩𝜕Ω i

̃ Next we measure the deviation of the surface from its tangent space. Assume that x(ξ) 2 is in C (𝜕Ω). Then 1 ̃ − x(0) ̃ x(ξ) = x̃ξi (0)ξi + x̃ξi ξj (0)ξi ξj + o(|ξ|2 ) 2 and 1 2 ̃ ̃ ̃ − x(0)) ̃ ((x(ξ) ⋅ ν(0)) = (x̃ξi ξj (0) ⋅ ν(0))ξ i ξj + o(|ξ| ) 2 The expression 1 Lij := −x̃ξi ξj ⋅ ν̃ = (ν̃ξj ⋅ x̃ξi + ν̃ξi ⋅ x̃ξj ) 2

(2.2.1)

is called the second fundamental form. The symmetric matrix ℒ with elements ik

ℒij = g Lkj

(2.2.2)

is called the Weingarten operator. Its eigenvalues are the principal curvatures κi of 𝜕Ω. Denote by T(ℒ) := g ik Lki the trace of ℒ. The mean curvature H is the arithmetic mean of its trace H :=

1 1 ik 1 n−1 T(ℒ) = g Lki = ∑κ. n−1 n−1 n − 1 i=1 i

(2.2.3)

Finally we like to state Weingarten’s equations: ν̃ξi = ℒik x̃ξk = g kj Lji x̃ξk .

(2.2.4)

Local orthonormal frame It is often convenient to use an orthonormal frame, the origin of which lies on the boundary, where the orthonormal axes {ei }n−1 i=1 span the tangent space and en points in the direction of the outer normal (see Figure 2.1). Then for any point p ∈ 𝜕Ω there exist a

2.2 Geometry of hypersurfaces

� 11

Figure 2.1: Local orthonormal frame.

neighborhood V = V (0) ⊂ ℝn , an open set U = U(0) ⊂ ℝn−1 , and a suitable translation and rotation such that 𝜕Ω ∩ V = {(x ′ , f (x ′ )) : x ′ := (x1 , x2 , . . . , xn−1 ) ⊂ U(0), f (0) = 0, 𝜕j f (0) = 0

for j ∈ {1, . . . , n − 1}}.

In this case g ij (0) = δij . Consequently, ̃ ′ )󵄨󵄨󵄨󵄨x ′ =0 = −𝜕i 𝜕j f (0)en , −𝜕i 𝜕j x(x

i, j ∈ {1, . . . , n − 1}.

Multiplication by ν = en together with the definition of the Weingarten operator ℒ (see (2.2.2)) yields ℒij = Lij = −𝜕i 𝜕j f (0). Example 2.1. In this example Lij and H will be computes for the sphere 𝜕BR . For a suit̃ able subset U ⊂ ℝn−1 we have |x(ξ)| = R if ξ ∈ U. Twice differentiation with respect to ξ results in ̃ + gij . 0 = (x̃ξi ξj ⋅ x)̃ + (x̃ξi ⋅ x̃ξj ) = (x̃ξi ξj ⋅ ν)R Thus, Lij = −(x̃ξi ξj ⋅ ν)̃ =

1 g R ij

(2.2.5)

and H=

1 ij 1 1 g Lji = g ij gji = . n−1 (n − 1)R R

(2.2.6)

Let f : 𝜕Ω → ℝ be a smooth function. Then ∇τ f ̃ := g ij

𝜕f ̃ x̃ 𝜕ξj ξi

̃ is called the tangential gradient of a function f ̃(ξ) := f (x(ξ)).

(2.2.7)

12 � 2 Basic concepts Any vector υ : 𝜕Ω → ℝn can be decomposed into υ = c0 ν + ∑n−1 i=1 ci x̃ξi . Multiplying successively by ν and x̃ξi we find υ = (υ ⋅ ν)ν + g ij (υ ⋅ x̃ξi )x̃ξj

and

(υ ⋅ w) = (υ ⋅ ν)(w ⋅ ν) + g ij (υ ⋅ x̃ξi )(w ⋅ x̃ξj ),

(2.2.8)

where w : 𝜕Ω → ℝn . As already emphasized before, we use the Einstein convention and sum over i, j = 1, 2, . . . , n − 1. For a smooth vector field υ : 𝜕Ω → ℝn , which is not necessarily tangent to 𝜕Ω, we define the tangential divergence by div𝜕Ω υ := g ij (υ̃ ξj ⋅ x̃ξi ),

̃ where υ̃ = υ(x(ξ)).

(2.2.9)

For υ = ν we get div𝜕Ω ν = (n − 1)H.

(2.2.10)

This follows immediately from (2.2.1), (2.2.3), and (2.2.9). In particular, for the ball BR of radius R centered at the origin we have H = R1 . 2.2.2 Projections Since 𝜕Ω is embedded in ℝn we can introduce the projection operator P : ℝn → Tx 𝜕Ω,

P(υ) := υ − (υ ⋅ ν)ν := υτ ,

υ = υ(x),

where Tx 𝜕Ω stands for the tangent space of 𝜕Ω in x. Consider a function f : 𝜕Ω → ℝ and let f ̂ be a C 1 -extension to Ω. Such an extension is not uniquely defined. Then the tangential gradient of a scalar function f on 𝜕Ω is given by ∇τ f := P(∇f ̂) = ∇x f ̂ − (ν ⋅ ∇x f ̂)ν.

(2.2.11)

Note that the tangential gradient is independent of the extension f ̂. This definition is equivalent to (2.2.7) in the following sense: We choose any point x0 ∈ 𝜕Ω and translate and rotate Ω such that x0 = 0

and

ν(0) = en = (0, . . . , 0, 1).

From (2.2.11) we deduce ∇τ f (0) = (𝜕1 f (0), . . . , 𝜕n−1 f (0), 0), which proves the claim.

(2.2.12)

2.2 Geometry of hypersurfaces

� 13

The decomposition (2.2.8) shows that ∇f ̂ in ℝn can be written as ∇f ̂ = (∇f ̂ ⋅ ν)ν + g ij (∇f ̂ ⋅ x̃ξi )x̃ξj . Thus, ∇τ f ̃ = g ij (0)(∇f ⋅ x̃ξi )x̃ξj ,

(2.2.13)

where we sum over i, j = 1, . . . , n − 1. For any smooth vector field υ : 𝜕Ω → ℝn we define div𝜕Ω υ̂ := div υ̂ − (ν ⋅ Dυ̂ ν) = 𝜕i υ̂ i − νj 𝜕j υ̂ i νi ,

(2.2.14)

where υ̂ : Ω → ℝn is any C 1 -extension of υ and υ̂ i is the i-th coordinate with respect to the coordinate system in ℝn . Note that tangential differential operators only depend on υ|𝜕Ω and not on the extension υ.̂ Definition (2.2.14) is equivalent to (2.2.9). The next identity will often be used. Let f : 𝜕Ω → ℝ. Then div𝜕Ω (fυ) = (υ ⋅ ∇τ f ) + f div𝜕Ω υ.

(2.2.15)

As an application of (2.2.15) we obtain div𝜕Ω [(υ ⋅ ν)ν] = ν ⋅ ∇τ (υ ⋅ ν) + (υ ⋅ ν) div𝜕Ω ν. By (2.2.10) this leads to div𝜕Ω [(υ ⋅ ν)ν] = (n − 1)(υ ⋅ ν)H.

(2.2.16)

The Laplace–Beltrami operator on 𝜕Ω is defined as Δ∗ := div𝜕Ω ∇τ .

(2.2.17)

Its representation in terms of local coordinates is given in Appendix B.1. We will frequently use integration by parts on 𝜕Ω. Assume f ∈ C 1 (𝜕Ω) and υ ∈ 0,1 C (𝜕Ω, ℝn ). Then the Gauss theorem on surfaces has the form ∮ f div𝜕Ω υ dS = − ∮(υ ⋅ ∇τ f ) dS + (n − 1) ∮ f (υ ⋅ ν)H dS. 𝜕Ω

𝜕Ω

(2.2.18)

𝜕Ω

2.2.3 Reduction to Hadamard perturbations In this section we first show that the boundary displacement under the perturbation Φt can be described by a vector field on 𝜕Ω whose first order approximation points in the normal direction.

14 � 2 Basic concepts Definition 2.1. Domain perturbations for which on 𝜕Ω the vector field υ points in the normal direction, i. e., υ = (υ ⋅ ν)ν, are called Hadamard perturbations or Hadamard variations. Note that no restriction is imposed on the second order perturbation w. To achieve our goal we will proceed as follows: For |t| < t0 let Φt be a smooth family of diffeomorphisms as described in Section 2.1. We denote by

–

Φ̃ t := Φt |𝜕Ω : 𝜕Ω → 𝜕Ωt ̃ is a local parametrization of 𝜕Ω. the restriction of Φt to 𝜕Ω. Furthermore, x(ξ) On 𝜕Ω we introduce new local coordinates ξ ̂ = φt (ξ) such that Ψ̃ t = Φ̃ t (ξ) satisfies

–

𝜕t Ψ̃ t |t=0 = (𝜕t Ψ̃ t |t=0 ⋅ ν)ν

on 𝜕Ω,

where Ψ̃ t : 𝜕Ω → 𝜕Ωt is a Hadamard perturbation. We then apply the extension theorem (Theorem A.2) and obtain a diffeomorphism Ψt : Ω → Ωt such that Ψt |𝜕Ω = Ψ̃ t .

–

Assume that 𝜕Ω ∈ C 2 and consider on 𝜕Ω a perturbation of the form t2 ̃ ̃ + w(ξ) ̃ + t υ(ξ) + o(t 2 ), Φ̃ t (ξ) = x(ξ) 2 where ξ ∈ U ⊂ ℝn−1 . Here we require υ̃ ∈ C 1 and w̃ ∈ C 0 . We decompose υ into its normal and tangential components. By (2.2.8) we have υ = (υ ⋅ ν)ν + υτ , where, in local coordinates, the tangential part is given by υτ = g ij (υ ⋅ x̃ξi )x̃ξj .

(2.2.19)

We denote by υτk the k-th coordinate of υτ in ℝn . Next we show that the parametrization φt can be chosen such that the perturbation Φ̃ t (ξ) can be transformed into a Hadamard perturbation Ψ̃ t . ̃ + Theorem 2.1. Let Φ̃ t : V ∩ 𝜕Ω → 𝜕Ωt be a diffeomorphism given locally by Φ̃ t (ξ) = x(ξ) t2 2 n−1 ̃ ̃ t υ(ξ) + 2 w(ξ) + o(t ) for all ξ ∈ U ⊂ ℝ and |t| < t0 . Then there exists a diffeomorphism φ : U → V of the form φ (ξ) = ξ + tη(ξ) =: ξ ̂ such that (see Figure 2.2) t

t

t2 ̂ ̂ ̂ ̂ Ψt (ξ)̂ := Φt ∘ φ−1 z(ξ) + o(t 2 ), t = x(ξ) + tρ(ξ)ν(ξ) + 2

ξ ̂ ∈ V ⊂ 𝜕Ω,

is a Hadamard perturbation. Moreover, the following identities hold: ηi (ξ) = g ij (ξ)υ̃ τj (ξ)

󵄨󵄨 ̃ ⋅ ν(ξ)) ̃ and ρ(̃ ξ)̂ = (υ(ξ) 󵄨󵄨ξ=φ−1 (ξ)̂

2.2 Geometry of hypersurfaces

� 15

Figure 2.2: Hadamard reduction.

and ik ̃ ̃ ⋅ ν(ξ)) ̃ ̃τ ̃ z(ξ)̂ = w(ξ) − x̃ξi ξj (ξ)g ik (ξ)υ̃ τk (ξ)g jl (ξ)υ̃ τl (ξ) − 2(υ(ξ) ξ g (ξ)υk (ξ)ν(ξ) i

̃ ⋅ ν(ξ))g ̃ − 2(υ(ξ)

ik

(ξ)υ̃ τk (ξ)ν̃ξi (ξ),

−1

̂ where ξ = φ (ξ).

Proof. 1. We look for ξ ̂ = φt of the form ξ + tη(ξ) + o(t). Let ηi (ξ) be the i-th component of η. Expansions with respect to t lead to t2 [x̃ (ξ)ηi (ξ)ηj (ξ)] + o(t 2 ), 2 ξi ξj ̃ + tηi (ξ)ρ̃ ξi (ξ) + o(t), ρ(φt ) = ρ(ξ) ̃ + tηi (ξ)x̃ξi (ξ) + x(φt ) = x(ξ)

̃ + tηi (ξ)ν̃ξi (ξ) + o(t) ν(φt ) = ν(ξ) and ̃ + tηi (ξ)z̃ξi (ξ) + o(t). ̃ + tη(ξ)) = z(ξ) z(φt ) = z(ξ 2. This leads to the following expansion of Ψ̃ t (ξ) = Ψt ∘ φt (ξ): ̃ + tηi (ξ)x̃ξi (ξ) + Ψt ∘ φt (ξ) = x(ξ)

t2 [x̃ (ξ)ηi (ξ)ηj (ξ)] 2 ξi ξj

̃ ν(ξ) ̃ + t[ηi (ξ)ρ̃ ξi (ξ)ν(ξ) ̃ + ρ(ξ)η ̃ + t{ρ(ξ) i (ξ)ν̃ ξi (ξ)]} +

t2 ̃ z(ξ). 2

For the identity Φ̃ t (ξ) = Ψ̃ t ∘ φt (ξ) + o(t 2 ) to hold, the following identities up to the order o(t 2 ) have to be satisfied: t: t2 : 2

̃ ̃ ν(ξ), ̃ υ(ξ) = ηi (ξ)x̃ξi (ξ) + ρ(ξ) ̃ ̃ + ρ(ξ)η ̃ ̃ + o(t 2 ). w(ξ) = x̃ξi ξj (ξ)ηi (ξ)ηj (ξ) + 2[ηi (ξ)ρ̃ ξi (ξ)ν(ξ) i (ξ)ν̃ ξi (ξ)] + z(ξ)

̃ ̃ ̃ ⋅ ν(ξ)). ̃ If we multiply the first equation by ν(ξ), we obtain ρ(ξ) = (υ(ξ) Then we multiply the same equation by x̃ξj (ξ) and find ηi (ξ) = g ij (ξ)υ̃ τj (ξ). Inserting these identities into the

16 � 2 Basic concepts ̂ The extension theorem (Theorem A.2) second equation we obtain the formula for z(ξ). then yields a diffeomorphism Ψt : Ω → Ωt which is a Hadamard perturbation. This establishes the theorem. ̃ Proposition 2.1. Let Φ̃ t : 𝜕Ω → 𝜕Ωt be a diffeomorphism given locally by Φt = x(ξ) + t2 ̃ ̃ t υ(ξ) + 2 w(ξ), where ξ ∈ U are local coordinates and |t| < t0 . Then there exist a 2

parametrization φt : U → U with φt (ξ) = ξ + tη(ξ) + t2 ϑ(ξ) and a Hadamard perturbation t2 ̃ t )ν(φ ̃ t ) + (z(φ ̃ t ))ν(φ ̃ t ) + o(t 2 ), ̃ t ) + t ρ(φ ̃ t ) ⋅ ν(φ Ψ̃ t (φt ) = x(φ 2

φt ∈ U,

such that Φ̃ t (ξ) = Ψ̃ t ∘ φt (ξ) + o(t 2 ) for all ξ ∈ U. As in Theorem 2.1 the following identities hold: ̃ ̃ ⋅ ν(ξ), ̃ ρ(ξ) = υ(ξ)

ηi (ξ) = g ij (ξ)υ̃ τj (ξ).

If the coordinates of ϑ are ̃ ϑi (ξ) = −g ij (ξ)x̃ξj ⋅ w(ξ) + g im (ξ)(x̃ξm (ξ) ⋅ x̃ξr ξs (ξ))g rk (ξ)υ̃ τk (ξ)g sl (ξ)υ̃ τl (ξ) lk ̃ ⋅ ν(ξ))g ̃ + 2(υ(ξ) (ξ)υ̃ τk (ξ)(νξl (ξ) ⋅ x̃ξm (ξ))g im (ξ),

̃ t ))ν(φ ̃ t ) = (z(ξ) ̃ ̃ + O(t), where ̃ t ) ⋅ ν(φ ̃ ⋅ ν(ξ)) then (z(φ ν(ξ) ̃ ̃ ̃ ̃ ⋅ ν(ξ)) (z(ξ) = (w(ξ) ⋅ ν(ξ)) + Lij (ξ)g ik (ξ)υ̃ τk (ξ)g jl (ξ)υ̃ τl (ξ) ̃ ⋅ ν(ξ)) ̃ ̃τ − 2g ik (ξ)(υ(ξ) ξ υk (ξ). i

Thus, an additional higher order term in the expansion of φt (ξ) yields a vector field z̃ which points in the normal direction. More generally the following remark holds. Remark 2.1. The reduction to Hadamard perturbation is valid for any order of t. This can be seen by applying an induction argument. Assume that N k t t N+1 ̃ + ̃ ̃ + ∑ ρ̃ (k) (ξ)ν(ξ) w(ξ) + o(t N+1 ) Φ̃ t (ξ) = x(ξ) k! (N + 1)! k=1

for some functions ρ̃ (k) . Then the parametrization φt (ξ) = ξ +

t N+1 (N+1) η (ξ), (N + 1)!

ξ ∈ U ⊂ ℝn−1 ,

yields a diffeomorphism N+1 k

Ψt (φ) = x(φ) + ∑

k=1

t (k) ρ (φ)ν(φ) + o(t N+1 ), k!

2.2 Geometry of hypersurfaces

� 17

where ̃ ̃ ρ̃ (N+1) = w(ξ) ⋅ ν(ξ)

and

η(N+1) (ξ) = g ij (ξ)w̃ jτ (ξ). i

Clearly, these computations require higher regularity of the boundary, e. g., 𝜕Ω ∈ C N+2 . The following example will play a major role in the sequel. Example 2.2. Let 𝜕Ω = 𝜕BR and let x̃ : U ⊂ ℝn−1 → ℝn be any local parametrization ̃ 2 = R2 for all ξ ∈ U. Then such that |x(ξ)| ̃ = ν(ξ)

1 ̃ x(ξ), R

gij (ξ) = x̃ξi (ξ) ⋅ x̃ξj (ξ),

̃ ⋅ x̃ξi ξj (ξ) = Lij (ξ) = −ν(ξ)

1 g (ξ). R ij

For small t0 we consider the diffeomorphism Φ̃ t : 𝜕BR → 𝜕Ωt given locally by its expan2 ̃ + t2 w(ξ) ̃ ̃ + t υ(ξ) sion Φ̃ t (ξ) = x(ξ) + o(t 2 ) for all ξ ∈ U. By Proposition 2.1 there exist a Hadamard variation Ψ̃ t and a parametrization φt such that Φ̃ t (ξ) = Ψt ∘ φt (ξ) + o(t 2 ) for all ξ ∈ U and t2 ̃ ν(ξ) ̃ + (z(ξ) ̃ ̃ ̃ ⋅ ν(ξ)) ̃ + t ρ(ξ) ν(ξ), Ψ̃ t (ξ) = x(ξ) 2

ξ ∈ U,

(2.2.20)

where ̃ ̃ ⋅ ν(ξ) ̃ ρ(ξ) = υ(ξ) and ̃ = (w(ξ) ̃ ̃ ̃ ⋅ ν(ξ) z(ξ) ⋅ ν(ξ)) +

1 kl g (ξ)υ̃ τk (ξ)υ̃ τl (ξ) − 2g jk (ξ)ρ̃ ξj (ξ)υ̃ τk (ξ). R

For later use we will express Ψ̃ t in terms of Cartesian coordinates. By means of (2.2.7) and (2.2.8) we obtain Ψt (x) = x + tρ(x)ν(x) + for x ∈ 𝜕BR .

t 2 1 󵄨󵄨 τ 󵄨󵄨2 ( 󵄨υ (x)󵄨󵄨 − 2υτ (x) ⋅ ∇τ ρ(x) + w(x) ⋅ ν(x))ν(x) + o(t 2 ) 2 R󵄨

(2.2.21)

18 � 2 Basic concepts

2.3 First and second variation of the volume We start with the volume 𝒱 (t) of Ωt for small t0 . After a change of variables x = Φ−1 t (y), it assumes the form 𝒱 (t) := |Ωt | = ∫ J(t) dx.

(2.3.1)

Ω

From (2.1.5) it follows that for small t 𝒱 (t) − 𝒱 (0) = t ∫ div υ dx + Ω

t2 ∫((div υ)2 − Dυ : Dυ + div w) dx + o(t 2 ). 2 Ω

This expansion enables us to compute the first variation limt→0 origin. In fact,

𝒱(t)−𝒱(0) t

of 𝒱 (t) at the

𝒱̇ (0) = ∫ div υ dx = ∮(υ ⋅ ν) dS. Ω

(2.3.2)

𝜕Ω

The second variation of 𝒱 (t) is 2

𝒱̈ (0) = ∫((div υ) − Dυ : Dυ + div w) dx.

(2.3.3)

Ω

Since ̈ = (div υ)2 − Dυ : Dυ + div w = 𝜕j (υj 𝜕i υi − υi 𝜕i υj + wj ), J(0)

(2.3.4)

integration by parts implies that 𝒱̈ (0) = ∮(υ ⋅ ν) div υ dS − ∮ υi 𝜕i υj νj dS + ∮(w ⋅ ν) dS. 𝜕Ω

𝜕Ω

(2.3.5)

𝜕Ω

By (2.2.14), (υ ⋅ ν) div υ − υi 𝜕i υj νj = (υ ⋅ ν) div𝜕Ω υ + (υ ⋅ ν)(ν ⋅ Dυ ν) − (υ ⋅ Dυ ν) = (υ ⋅ ν) div𝜕Ω υ − (υ − (υ ⋅ ν)ν) ⋅Dυ ν. ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ υτ

(2.3.6)

If we insert this identity into (2.3.5), we obtain τ

𝒱̈ (0) = ∮[(υ ⋅ ν) div𝜕Ω υ − (υ ⋅ Dυ ν) + (w ⋅ ν)] dS. 𝜕Ω

(2.3.7)

2.3 First and second variation of the volume

� 19

Note that in this second variation only the values of υ and w on the boundary and the tangential derivatives of υ appear. Therefore, V̈ (0) does not contain derivatives of the perturbation pointing inside the domain. Throughout this text we shall use the notion of volume preserving perturbation in a broader sense. Definition 2.2. The perturbation Φt (x) is called volume preserving if 𝒱̇ (0) = 0 and 𝒱̈ (0) = 0. In this text a domain perturbation will be called of first order if it is of the form Φt = x + tυ(x)

for |t| < t0 .

It will be called of second order if Φt = x + tυ(x) +

t2 w(x). 2

Remark 2.2. In the literature different concepts of volume preserving perturbations are considered: (i) the diffeomorphisms Φt for which |Ωt | = |Ω| for all |t| < t0 , (ii) first order diffeomorphisms such that |Ωt | = |Ω| + o(t), (iii) second order diffeomorphisms such that |Ωt | = |Ω| + o(t 2 ). Diffeomorphisms of type (iii) satisfy Definition 2.2. Clearly, a perturbation of type (i) is also a perturbation of type (ii) and (iii), but not necessarily vice versa.

2.3.1 Discussion of 𝒱̈ (0) Our goal is to express the second domain variation of the volume given in (2.3.7) in local coordinates. Using these coordinates we will see that 𝒱̈ (0) depends on the second fundamental form. From (2.2.18) it follows that the first term of 𝒱̈ (0) can be written as ∮[(υ ⋅ ν) div𝜕Ω υ] dS = − ∮(υ ⋅ ∇τ ρ) dS + (n − 1) ∮ ρ2 H dS, 𝜕Ω

𝜕Ω

ρ = (υ ⋅ ν).

𝜕Ω

The second term (υτ ⋅ Dυ ν) in (2.3.7) can be rewritten as (υτ ⋅ Dυ ν) = υτ ⋅ ∇(υ ⋅ ν) − (υτ ⋅ Dν υ) = υτ ⋅ ∇τ (υ ⋅ ν) − (υτ ⋅ Dν υτ ). The following formula will be derived in Section 2.4 (Dν )ij = 𝜕i νj = ℒkm (x̃ξm )i (x̃ξk )j .

(2.3.8)

20 � 2 Basic concepts This together with (2.2.19) yields (υτ ⋅ Dν υτ ) = (υ ⋅ x̃ξi )g ij Ljl (υ ⋅ x̃ξl ) = υτ ℒυτ . Thus, (υτ ⋅ Dυ ν) = υτ ⋅ ∇τ ρ − υτ ⋅ ℒυτ .

(2.3.9)

This leads to 2

τ

τ

τ

τ

𝒱̈ (0) = ∮[(n − 1)ρ H + (υ ⋅ ℒυ ) − 2(υ ⋅ ∇ ρ) + (w ⋅ ν)] dS.

(2.3.10)

𝜕Ω

Example 2.3. 1.

If Ω = BR the above formula (2.3.10) reads as τ

𝒱̈ (0) = −2 ∮ (υ ⋅ ∇ (υ ⋅ ν)) dS + 𝜕BR

n−1 ∮ (υ ⋅ ν)2 dS R 𝜕BR

1 󵄨 󵄨2 + ∮ 󵄨󵄨󵄨υτ 󵄨󵄨󵄨 dS + ∮ (w ⋅ ν) dS. R 𝜕BR

2.

(2.3.11)

𝜕BR

For the computation of (2.3.11) we applied the identity (Dν )ij = R1 (δij − νi νj ) on 𝜕BR . For Hadamard perturbations υτ = 0 we get 2

𝒱̈ (0) = (n − 1) ∮(υ ⋅ ν) H dS + ∮(w ⋅ ν) dS. 𝜕Ω

3.

(2.3.12)

𝜕Ω

For tangential perturbations Φt = x + tυτ (x) + τ

t2 w(x) 2

we have 𝒱̇ (0) = 0 and

τ

𝒱̈ (0) = ∮[υ ⋅ ℒυ + (w ⋅ ν)] dS.

(2.3.13)

𝜕Ω

4.

In convex domains the quadratic form υτ ℒυτ is positive. Let υ = (c1 x1 , c2 x2 , . . . , cn xn ) and w = 0. Then by (2.3.2) and (2.3.3), n

𝒱̇ (0) = |Ω| ∑ ci , i=1

5.

6.

n

2

n

2

𝒱̈ (0) = |Ω|((∑ ci ) − ∑ ci ). i=1

i=1

Consider in a domain in the plane the perturbation Φt = x+tυ(x), where υ = (−x2 , x1 ) 0 1 ). Consequently by (2.3.5), 𝒱̇ (0) = is a pure shear. Then div υ = 0 and Dυ = ( −1 0 0 and by (2.3.3), 𝒱̈ (0) = − ∮𝜕Ω (x ⋅ ν) dS = −2|Ω|. We replace Φt by a Hadamard perturbation (2.2.21) and compute 𝒱̈ (0) for the ball. In (2.3.12) we replace ρ(x) = (υ(x) ⋅ ν) by ρ(x)̂ and w(x) by

2.3 First and second variation of the volume

w(x)̂ + (

� 21

̂ 2 x̂ |υτ (x)| ̂ − 2υτ (x)̂ ⋅ ∇τ ρ(x)) . R R

Hence, 𝒱̈ (0) = ∮ ( 𝜕BR

7.

n−1 2 |υτ |2 ρ − 2(υτ ⋅ ∇τ ρ) + + (w ⋅ ν)) dS. R R

(2.3.14)

The variations of the volume and the area can be used to derive local Steiner formulas for parallel sets Ωρ which are known for convex bodies (see for instance [70] for an elementary introduction). For small |ρ| they read as |Ωρ | = |Ω| + ρ|𝜕Ω| + (n − 1)

ρ2 ∮ H dS + o(ρ2 ). 2 𝜕Ωρ

Remark 2.3. In general a first order perturbation for which 𝒱̇ (0) = 0 does not necessarily imply 𝒱̈ (0) = 0. It is easy to see that this can always be achieved by adding a term of the second order. In fact, consider the perturbations Φ(1) t = x + tυ(x)

and

Φ(2) t =x+

(1) Φt = Φ(2) t ∘ Φt = x + tυ(x) +

t2 w(x) + o(t 2 ), 2

t2 w + o(t 2 ). 2

(2) Clearly, Φ(1) t is a first order perturbation, whereas Φt and Φt are second order perturba(2) (2) tions. Let 𝒱 (1) (t) = |Φ(1) t (Ω)|, 𝒱 (t) = |Φt (Ω)|, and 𝒱 (t) = |Φt (Ω)|. The domain variations (1) (1) (2) of 𝒱 (t), 𝒱 (t) := 𝒱 (Ωt ), and 𝒱 (t) := 𝒱 (Ω(2) t ) are related as follows:

𝒱̇ (0) = 𝒱̇

(1)

(0)

and

𝒱̈ (0) = 𝒱̈

(1)

(0) + 𝒱̈ (2) (0).

By choosing w appropriately we can achieve 𝒱̈ (0) = 0. The role of w becomes apparent if we consider rotations. Consider a rotation in the (x1 , x2 )-plane. The polar angle θ1 (see the representation of x in spherical coordinates in Appendix B.1.1) is replaced by θ1 + t. Then for small |t|, x is transformed into x + 2 t(−x2 , x1 , 0, . . . , 0) + t2 (−x1 , −x2 , 0, . . . , 0) + o(t 2 ). In this case, Φ(1) t = x + t(−x2 , x1 , 0, . . . , 0), Φt = x + t(−x2 , x1 , 0, . . . , 0) +

Φ(2) t =x+

t2 (−x1 , −x2 , 0, . . . , 0) + o(t 2 ) and 2

t2 (−x1 , −x2 , 0, . . . , 0) + o(t 2 ). 2

Hence, div υ = 0 implies that 𝒱̇1 (0) = 𝒱̇2 (0) = 𝒱̇ (0) = 0.

22 � 2 Basic concepts Moreover, 0 −1 ( Dυ = ( 0 .. . (0

1 0 0 .. . 0

0 0 ⋅⋅⋅ .. . ⋅⋅⋅

⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ .. . ⋅⋅⋅

0 0 0) ). .. . 0)

From (2.3.5) it follows that 𝒱̈1 (0) = 2 ∮(ν1 x1 + ν2 x2 ) dS = −𝒱̈2 (0) ≠ 0. 𝜕Ω

Obviously, 𝒱̈ (0) = 0. In conclusion, the second order variation of the rotated domain vanishes only if in the expansion of the rotation, second order terms are taken into account. As an example consider the ellipse E = (4 cos(θ1 ), 2 sin(θ1 )). The transformations of E under the perturbations Φ(1) t and Φt for t = 0.5 are pictured in Figure 2.3.

Figure 2.3: Ellipses under first and second order rotation.

This example allows a generalization. Proposition 2.2. Let Ω be convex and let Φt (x) = x + tυτ (x) be a first order tangential perturbation. Then 𝒱̇ (0) = 0 and 𝒱̈ (0) > 0. Proof. The first statement is immediate (see (2.3.2)). Clearly, if Φt = x + tυτ , then for convex domains Ω ⊂ Ωt and |Ωt | > |Ω|. Because ℒ is positive for convex domains this is in accordance with formula (2.3.13). Note that Proposition 2.2 is not necessarily true if Φt (x) = x + tυτ +

t2 w. 2

2.3 First and second variation of the volume

� 23

2.3.2 First and second variation of the surface area perimeter We now compute the perimeter or the surface area 𝒮 (t) := |𝜕Ωt | of Ωt . If the boundary ̃ 𝜕Ω is represented locally by x(ξ), then in view of (2.1.1), a local parametrization of 𝜕Ωt is given by ̃ + ̃ = x(ξ) ̃ + t υ(ξ) y(ξ)

t2 ̃ w(ξ) + o(t 2 ). 2

̃ ̃ ̃ ̃ As before we write υ(ξ) = υ(x(ξ)) and w(ξ) = w(x(ξ)). In order to compute the metric on 𝜕Ωt we introduce the following notation: gij := (x̃ξi ⋅ x̃ξj ),

G := (gij )i,j=1,2,...,n−1 ,

aij := (x̃ξi ⋅ υ̃ ξj ) + (x̃ξj ⋅ υ̃ ξi ),

A := (aij )i,j=1,...,n−1 ,

bij := 2(υ̃ ξi ⋅ υ̃ ξj ) + (w̃ ξi ⋅ x̃ξj ) + (w̃ ξj ⋅ x̃ξi ),

B := (bij )i,j=1,...,n−1 .

Then the metric gijt on 𝜕Ωt is given by gijt := (yξi ⋅ yξj ) = gij + taij +

t2 b + o(t 2 ), 2 ij

Gt := (gijt )i,j=1,...,n .

(2.3.15)

The surface element on 𝜕Ωt is 1

dS(t) =

√det Gt dξ

2 t2 = √det G{det(I + tG A + G−1 B + o(t 2 ))} dξ, 2

−1

where dξ = dξ1 . . . dξn−1 . It can therefore be written as dS(t) = m(t)dS,

(2.3.16)

where dS is the surface element of 𝜕Ω. Let us write for short T(C) := trace(C). By Jacobi’s formula (see Section A.1), det(I + tG−1 A +

t 2 −1 G B + o(t 2 )) 2

= 1 + tT(G−1 A) +

t2 2 [(T(G−1 A)) − G−1 A : G−1 A + T(G−1 B)] + o(t 2 ). 2

Then for small t we get m(t) = √det(I + tG−1 A +

t 2 −1 G B) + o(t 2 ) 2

t t 2 (T(G−1 A))2 G−1 A : G−1 A T(G−1 B) = 1 + T(G−1 A) + ( − + ) 2 2 4 2 2 + o(t 2 ).

(2.3.17)

24 � 2 Basic concepts Let us write for short 𝜕i∗ υ̃ := g ij υ̃ ξj .

(2.3.18)

With this notation we have by (2.2.9) and (2.3.15) div𝜕Ω υ̃ = (𝜕i∗ υ̃ ⋅ x̃ξi ),

g ij ajk = (𝜕i∗ x̃ ⋅ υ̃ ξk ) + (𝜕i∗ υ̃ ⋅ x̃ξk ).

A straightforward computation leads to T(G−1 A) = 2 div𝜕Ω υ,̃

G−1 A : G−1 A = g ij ajk g kl ali = (𝜕i∗ υ̃ ⋅ x̃ξk )(𝜕k∗ υ̃ ⋅ x̃ξi ) + 2(𝜕i∗ υ̃ ⋅ x̃ξk )(υ̃ ξi ⋅ 𝜕k∗ x) + (υ̃ ξk ⋅ 𝜕i∗ x)(υ̃ ξi ⋅ 𝜕k∗ x) ̃ = 2(𝜕i∗ υ̃ ⋅ x̃ξk )(𝜕k∗ υ̃ ⋅ x̃ξi ) + 2(𝜕i∗ υ̃ ⋅ x̃ξk )(υ̃ ξi ⋅ 𝜕k∗ x), T(G−1 B) = 2(υ̃ ξi ⋅ 𝜕i∗ υ)̃ + 2 div𝜕Ω w.̃ Decomposing υ = υτ + (υ ⋅ ν)ν and applying (2.2.10), we obtain ̇ m(0) = div𝜕Ω υτ + (n − 1)H(υ ⋅ ν).

(2.3.19)

Consequently, from (2.3.16) and (2.3.19) it follows that at t = 0 the first variation of the surface area 𝒮 (t) of 𝜕Ωt is

̇ = ∮ m(0) ̇ 𝒮 (0) dS = (n − 1) ∮ H(υ ⋅ ν) dS. 𝜕Ω

(2.3.20)

𝜕Ω

This together with the first variation of the volume implies the following lemma. ̇ = Lemma 2.2. For volume preserving perturbations the surface area is critical, i. e., 𝒮 (0)

0, if and only if H is constant. Proof. Let

H :=

1 ∮ H dS |𝜕Ω| 𝜕Ω

be the mean value of H. From (2.3.2) we have ∮𝜕Ω (υ ⋅ ν) dS = 0. Hence, 0 = ∮{H − H}(υ ⋅ ν) dS. 𝜕Ω

Choose (υ ⋅ ν) = H − H. Then ∮(H − H)2 dS = 0. 𝜕Ω

The assertion is now obvious.

2.3 First and second variation of the volume

� 25

By Alexandrov’s uniqueness theorem [1] any closed connected C 2 surface of constant mean curvature is a sphere. We mention an identity which will frequently be used in the sequel. From (2.3.19) and (2.2.18) it follows that ̇ dS = − ∮(υτ ⋅ ∇f ) dS + (n − 1) ∮ fH(υ ⋅ ν) dS. ∮ f (x)m(0) 𝜕Ω

𝜕Ω

(2.3.21)

𝜕Ω

̈ Next we shall compute m(0). From (2.3.17) and (2.3.18) we get ̈ m(0) = (div𝜕Ω υ)̃ 2 − (𝜕i∗ υ̃ ⋅ x̃ξk )(𝜕k∗ υ̃ ⋅ x̃ξi )

− (𝜕i∗ υ̃ ⋅ x̃ξk )(υ̃ ξi ⋅ 𝜕k∗ x)̃ + (𝜕j∗ υ̃ ⋅ υ̃ ξj ) + div𝜕Ω w.̃

(2.3.22)

Hence, (2.3.16) implies that the second variation of the surface area perimeter at t = 0 is (cf. Section 2.2.1) i0

̈ = ∮ m(0) ̈ ̈ dS = ∑ ∮ m(0) dS 𝒮 (0) i=1 V ∩𝜕Ω

𝜕Ω

i

i0

̃ 2 − (𝜕j∗ υ̃ ⋅ x̃ξk )(𝜕k∗ υ̃ ⋅ x̃ξj ) − (𝜕j∗ υ̃ ⋅ x̃ξk )(υ̃ ξj ⋅ 𝜕k∗ x)]dS ̃ ̃ = ∑ ∫ pi (x)[(div 𝜕Ω υ) i=1 U

i

i0

i0

i=1 U

i=1 U

̃ dS. ̃ j∗ υ̃ ⋅ υ̃ ξj ) dS + (n − 1) ∑ ∫ pi (x)(w̃ ⋅ ν)H + ∑ ∫ pi (x)(𝜕 i

(2.3.23)

i

In short, we have 2

∗

∗

∗

∗

∗

̈ = ∮(div𝜕Ω υ)̃ − (𝜕 υ̃ ⋅ x̃ξ )(𝜕 υ̃ ⋅ x̃ξ ) − (𝜕 υ̃ ⋅ x̃ξ )(υ̃ ξ ⋅ 𝜕 x)̃ + (𝜕 υ̃ ⋅ υ̃ ξ ) dS 𝒮 (0) i k i k j k i k i j 𝜕Ω

̃ dS. + ∮(div𝜕Ω w) 𝜕Ω

̈ if 𝜕Ω is the sphere 𝜕BR and Φt is Example 2.4. 1. We start with the computation of 𝒮 (0) a Hadamard perturbation such that υ = ρν. Then the following formulas are valid: n−1 ρ, R υ̃ ξi ⋅ 𝜕k∗ x̃ = g kl ((ρξi ν̃ + ρν̃ξi ) ⋅ x̃ξl ) = ρg kl (ν̃ξi ⋅ x̃ξl ) = −ρg kl (ν̃ ⋅ x̃ξl ξi ) ρ ρ = ρg kl Lli = g kl gli = δki , R R ρ 𝜕i∗ υ̃ ⋅ x̃ξk = g ij (υ̃ ξj ⋅ x̃ξk ) = ρg ij (ν̃ξj ⋅ x̃ξk ) = −ρg ij (ν̃ ⋅ x̃ξk ξj ) = δik . R div𝜕Ω υ̃ =

26 � 2 Basic concepts Moreover, 𝜕j∗ υ̃ ⋅ υ̃ ξj = g jk υ̃ ξk ⋅ υ̃ ξj = g jk (ρξk ν̃ + ρν̃ξk ) ⋅ (ρξj ν̃ + ρν̃ξj ) = g jk ρξk ρξj + ρ2 g jk ν̃ξk ⋅ ν̃ξj . For the last term we use (2.2.8), take into account that (ν̃ξk ⋅ ν)̃ = 0, and apply (2.2.5): ̃ ν̃ξj ⋅ ν)̃ + ρ2 g jk g mn (ν̃ξk ⋅ x̃ξm )(ν̃ξj ⋅ x̃ξn ) ρ2 g jk ν̃ξk ⋅ ν̃ξj = ρ2 g jk (ν̃ξk ⋅ ν)( = ρ2 g jk g mn (ν̃ξk ⋅ x̃ξm )(ν̃ξj ⋅ x̃ξn ) = ρ2 g jk g mn Lkm Ljn = = (n − 1)

ρ2 . R2

ρ2 ρ2 jk mn g g gkm gjn = 2 δkn δkn 2 R R

Hence, 𝜕j∗ υ̃ ⋅ υ̃ ξj = g ik ρξk ρξj + (n − 1)

ρ2 . R2

Then (2.3.22) takes the form ̈ m(0) = g jk ρξk ρξj + (n − 1)(n − 2)

ρ2 + div𝜕Ω w.̃ R2

Hence, ij

̈ = ∮ (g ρξ ρξ + 𝒮 (0) i j 𝜕BR

2.

(n − 1)(n − 2) 2 n − 1 ρ + (w ⋅ ν)) dS. R R2

(2.3.24)

Let Ω be a smooth domain in the plane. Its boundary curve (x1 (s), x2 (s)), s ∈ I, is ̇ satisfies |x(s)| ̇ parametrized by arc length. Then the tangent vector x(s) = 1; thus, ̈ g 11 = 1. Moreover, κ := x(s) = H is the curvature of the curve in x(s). For any ̃ := υ(x(s)). We need to compute the terms in (2.3.22). smooth vector field υ we set υ(s) From (2.2.9) we deduce 2

2

̃ ̃ ⋅ x(s)) ̇ (div𝜕Ω υ(x(s))) = (𝜕s υ(s) . Since ̇̃ = 𝜕 υ(s) ̇̃ ̃ ⋅ x(s) 𝜕i∗ υ̃ ⋅ x̃ξk = g 11 (s)𝜕s υ(s) ⋅ x(s), s ̃ ̇̃ ̃ ⋅ x(s), υ̃ ⋅ 𝜕∗ x = 𝜕 υ(s) ξi

k

s

and 2

̃ 𝜕j∗ υ̃ ⋅ υ̃ ξj = (𝜕s υ(s)) ,

2.3 First and second variation of the volume

� 27

we obtain 2

2

̈ = ∫(𝜕s υ(s)) ̃ ̃ ⋅ x(s)) ̃ ⋅ ν(s)) ̃ ̃ ̇ 𝒮 (0) − (𝜕s υ(s) ds + ∫ κ(s)( w(s) ds. I

I

̇ into its normal and tangential components, we If we decompose υ = ρν + υτ x(s) obtain ̃ = 𝜕s ρ(s) ̃ ν(s) ̃ + ρ(s)𝜕 ̃ ̃ + 𝜕s υ̃ τ (s)x(s) ̇ + υ̃ τ (s)x(s). ̈ 𝜕s υ(s) s ν(s) By Frenet’s formula, ̃ = κ(s) ̃ x(s) ̇ 𝜕s ν(s)

and

̃ ̃ ν(s). ̈ = −κ(s) x(s)

Hence, ̃ ̃ ν(s) ̃ + κ(s) ̃ x(s) ̃ = 𝜕s ρ(s) ̃ υ̃ τ (s)ν(s). ̃ ρ(s) ̇ − κ(s) ̇ + 𝜕s υ̃ τ (s)x(s) 𝜕s υ(s) ̈ can be written as The first integrand for 𝒮 (0) 2

2

2

̃ − κ(s) ̃ ⋅ x(s)) ̃ ̃ υ̃ τ (s)) . ̇ = (𝜕s ρ(s) − (𝜕s υ(s) (𝜕s υ(s)) Hence, τ

2

̈ = ∫(𝜕s ρ(s) ̃ − κ(s) ̃ ⋅ ν(s)) ̃ ̃ υ̃ (s)) ds + ∫ κ(s)( ̃ 𝒮 (0) w(s) ds. I

3.

I

1 Consider a stretch in one direction υ = (x1 , 0, . . . , 0). Then div𝜕Ω = g ij 𝜕x 𝜕ξ

x1 2g ij 𝜕ξ ij

j

𝜕x1 . 𝜕ξk

j

𝜕x1 , 𝜕ξj

g ij ajk =

Without loss of generality we may assume that at a fixed point on 𝜕Ω,

g = δij ,

̈ m(0) = 2 ∑( i=j̸

2

2

𝜕x 𝜕x1 ) ( 1) . 𝜕ξi 𝜕ξj

2.3.3 Area variations for the surface of the ball In (2.3.24) we computed the second variation for the area for Hadamard perturbations. We now consider more general perturbations υ = ρ(x)ν(x) + υτ (x) on 𝜕BR which contain also tangential components. According to Theorem 2.1 and in particular (2.2.21), the perturbation Φt (x) = x + t[ρ(x)ν(x) + υτ (x)] +

t2 w(x) 2

28 � 2 Basic concepts can be replaced by the Hadamard perturbation ̂ 2 t 2 |υτ (x)| ̂ Φ(̂ x)̂ = x̂ + tρ(x)̂ + { − 2υτ (x)̂ ⋅ ∇τ ρ(x)̂ + w(x)}ν( x)̂ + o(t 2 ). 2 R We can apply (2.3.24) provided (w ⋅ ν) is replaced by ω := (w ⋅ ν) +

̂ 2 |υτ (x)| ̂ − 2(υτ (x)̂ ⋅ ∇τ ρ(x)). R

Thus, the second variation of the perimeter for general perturbations reads as ij

̈ = ∮ (g ρξ ρξ + 𝒮 (0) i j 𝜕BR

(n − 1)(n − 2) 2 n − 1 ω) dS. ρ + R R2

(2.3.25)

Recall that (see (2.3.14)) 𝒱̈ (0) = ∮ ( 𝜕BR

|υτ |2 n−1 2 n−1 2 ρ − 2(υτ ⋅ ∇τ ρ) + + (w ⋅ ν)) dS = ∮ [ ρ + ω] dS. R R R 𝜕BR

Consequently, the second variation of the area is ij

̈ = ∮ g ρξ ρξ dS − 𝒮 (0) i j

n−1 ̈ n−1 𝒱 (0). ∮ ρ2 dS + 2 R R

(2.3.26)

𝜕BR

𝜕BR

Recall that g ij ρξi ρξj = |∇τ (υ ⋅ ν)|2 . For later use we set 󵄨2

τ

𝒮0̈ (0) := ∫ 󵄨󵄨󵄨∇ (υ ⋅ ν)󵄨󵄨󵄨 dS −

󵄨

𝜕BR

n−1 ∫ (υ ⋅ ν)2 dS. R2

(2.3.27)

𝜕BR

Thus, ̈ = 𝒮0̈ (0) + 𝒮 (0)

n−1 ̈ 𝒱 (0). R

(2.3.28)

̈ Consider now volume preservation in the sense of Definition 2.2. Then 𝒮 (0) = 𝒮0̈ (0). From 𝒱̇ (0) = 0 it follows that ∮ B ρ dS = 0. Denote R

1

𝒦1 := {ρ ∈ C (𝜕BR ) : ∮ ρ dS = 0}. 𝜕BR

Then (cf. Appendix D.2.1) min 𝒦1

∮𝜕B g ij ρξi ρξj dS R

∮𝜕B

R

ρ2 dS

=

n−1 . R2

2.3 First and second variation of the volume

� 29

Equality holds if and only if ρ is a spherical harmonic of degree 1. It belongs to the x n-dimensional eigenspace ℋ1 . This eigenspace is spanned by { Ri }ni=1 . Consequently, ̈ = 0 for ρ ∈ ℋ1 . In particular we proved the following lemma. 𝒮 (0) Lemma 2.3. Let Φt : BR → Ωt be a perturbation which is volume preserving, i. e., 𝒱̇ (0) = ̇ = 0, 𝒮 (0) ̈ ≥ 0, and equality holds if and only if ρ = (υ ⋅ ν) ∈ ℋ1 . 𝒱̈ (0) = 0. Then 𝒮 (0) ̈ if 𝒮 (0) ̇ = 𝒮 (0) ̈ = 0. Definition 2.3. A perturbation is said to be in the kernel of 𝒮 (0) x

The perturbation Φt = x + t Ri ei + 𝒱̈ (0) = ∮ ( 𝜕BR

t2 w, 2

where w is such that

n−1 (xi )2 + (w ⋅ ν)) dS = 0, R3

is in the kernel of 𝒮 ̈ of the ball. We now restrict to the following class of volume preserving perturbations in order to obtain a strictly positive 𝒮 .̈ Let 1

𝒦2 := {ρ ∈ C (𝜕BR ) : ∮ ρ dS = 0, ∮ xρ(x) dS = 0},

min

𝜕BR

ij

∮𝜕B g ρξi ρξj dS

𝒦2

R

∮𝜕B ρ2 dS R

=

𝜕BR

2n . R2

The eigenspace of functions for this eigenvalue is denoted by ℋ2 (cf. Appendix D.2.1). Thus, for any ρ ∈ ℋ2 we have 2n 󵄨 󵄨2 ∫ 󵄨󵄨󵄨∇τ ρ󵄨󵄨󵄨 dS ≥ 2 ∮ ρ2 dS. R

(2.3.29)

𝜕BR

𝜕BR

Consequently, ̈ ≥ 𝒮 (0)

n+1 ∮ ρ2 dS. R2 𝜕BR

Definition 2.4. For any domain Ω ⊂ ℝn the barycenter is defined as b :=

1 ∫ x dx. |Ω| Ω

We consider volume preserving perturbations which leave the barycenter unchanged in first order: bt :=

1 1 ∫ y dy = ∫ x dx + o(t) = b + o(t). |Ωt | |Ω| Ωt

Ω

30 � 2 Basic concepts This condition is equivalent to 1 ∫(x + tυ(x))(1 + t div υ) dx = b + o(t). |Ω| Ω

A straightforward computation yields ∮𝜕Ω (υ ⋅ ν)x dS = 0. Definition 2.5. A perturbation Φt satisfies the barycenter condition if ∮𝜕Ω (υ ⋅ ν)x dS = 0. If the barycenter condition is satisfied, a stronger estimate in Lemma 2.3 is available. Lemma 2.4. Let Φt : BR → Ωt be a perturbation which is volume preserving, i. e., 𝒱̇ (0) = ̇ = 0 and 𝒱̈ (0) = 0. Moreover, let Φt satisfy the barycenter condition. Then 𝒮 (0) ̈ ≥ 𝒮 (0)

n+1 ∮ ρ2 dS. R2 𝜕BR

2.3.4 Hadamard perturbations In this subsection we evaluate the second variation of the area (2.3.23) for Hadamard perturbations. For the computations we will use Weingarten’s formula (see (2.2.4)): ν̃ξi = g jk Lji x̃ξk . Thus, 𝜕i∗ υ̃ = g ij

𝜕 (ρ̃ ν)̃ = g ij (ρ̃ ξj ν̃ + ρ̃ ν̃ξj ) 𝜕ξj

and

𝜕i∗ x = g ij xξj .

Hence, ̃ ij (ν̃ξj ⋅ x̃ξi ) = ρg ̃ ij Lij , div𝜕Ω υ̃ = (𝜕i∗ υ̃ ⋅ x̃ξi ) = ρg

̃ ij g sm Ljs (x̃ξm ⋅ x̃ξk ) = ρg ̃ ij Ljk , (𝜕i∗ υ̃ ⋅ x̃ξk ) = ρg ̃ sm Lis g kl gml = ρg ̃ sk Lis . (υ̃ ξi ⋅ 𝜕k∗ x)̃ = ρg

Let ℒ denote the Weingarten operator (see (2.2.2)) and T(ℒ) its trace. Then ̃ ℒ), div𝜕Ω υ̃ = ρT( (𝜕i∗ υ̃

⋅

x̃ξk )(𝜕k∗ υ̃

⋅ x̃ξi ) = (𝜕i∗ υ̃ ⋅ x̃ξk )(υ̃ ξi ⋅ 𝜕k∗ x) = ρ̃ 2 T(ℒ2 ).

Moreover, (𝜕j∗ υ̃ ⋅ υ̃ ξj ) = g js ρ̃ ξs ρ̃ ξj + ρ̃ 2 (g js νξs ⋅ νξj ) = g js ρ̃ ξs ρ̃ ξj + ρ̃ 2 T(ℒ2 ).

2.3 First and second variation of the volume

� 31

Inserting these relations into (2.3.22), we get 2

̈ m(0) = g js ρ̃ ξs ρ̃ ξj + ρ̃ 2 [(T(ℒ)) − ℒ : ℒ] + div𝜕Ω w.̃

(2.3.30)

We are now able to write down the second variation of area: i0

ij

2

2

̈ = ∑ ∫{g ρ̃ ξ ρ̃ ξ + ρ̃ [(T(ℒ)) − ℒ : ℒ] + (n − 1)H(w̃ ⋅ ν)} ̃ dS. 𝒮 (0) i j i=1 U

(2.3.31)

i

The eigenvalues of ℒ are the principal curvatures κ1 , κ2 , . . . , κn−1 of 𝜕Ω. The traces of the matrices ℒ and ℒ2 can be expresses as follows: T(ℒ) = κ1 + κ2 + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + κn−1 = (n − 1)H, 2

ℒ : ℒ = T(ℒ ) =

κ12

+

κ22

+ ⋅⋅⋅ + ⋅⋅⋅ +

(2.3.32)

2 κn−1 .

Consequently, ̈ m(0) = g sj ρ̃ ξs ρ̃ ξj + 2ρ̃ 2 ∑ κ̃ i κ̃ j + div𝜕Ω w.̃ i 0 the first eigenfunction is a nonconstant radial function. ̃ If we set κ := RR for some fixed R̃ > 0, then in the limit R → ∞ we obtain the outer domain Ω = ℝn \ BR̃ . For the eigenvalue μk , k > 0, we obtain the asymptotic formula limR→∞ μk = k+n−2 , which is the k-th eigenvalue of the Steklov problem in the exterior R̃ of the ball BR̃ . A slight variant is the eigenvalue problem Δϕ = 0 in BR \ BκR ,

ϕ = 0 on 𝜕BR ,

and

𝜕ν ϕ + αϕ = μϕ on 𝜕BκR .

(3.2.7)

In this case ϕk,i = (r −k−n+2 − R−2k−n+2 r k )Yk,i (ξ)

(3.2.8)

and μk :=

καR + κ2k+n−2 (k − καR) + k + n − 2 . κR(1 − κ2k+n−2 )

(3.2.9)

3.2.4 Steklov eigenvalues in annular domains In the annular domain Ω = BR \ BκR (0),

R > 0 and 0 < κ < 1,

the Steklov eigenvalue problem reads as Δϕ = 0 in BR \ BκR ,

𝜕ν ϕ + αϕ = μϕ on 𝜕BR ∪ 𝜕BκR .

(3.2.10)

Clearly, μ0 = α with ϕ = const. is the lowest eigenvalue. Separation of variables leads to

3.2 The Steklov eigenvalue problem � 47

ϕ±k,i (r, ξ) = (r k + A± r −k−n+2 )Yk,i (ξ), with A± := (κR)n+2k−2

καR − μ±k κR − k καR − μ±k κR + k + n − 2

and μ±k = α +

√Z (k + n − 2)(1 + κ2k+n−1 ) κk(1 + κ2k+n−2 ) + ± . 2k+n−2 2k+n−2 2κR(1 − κ ) 2κR(1 − κ ) 2κR(1 − κ2k+n−2 )

Here, 2

2

Z := [(1 − κ)k + n − 2] + [(1 − κ)k − κ(n − 2)] κ4k+2n−4 + 2κ2k+n−2 [(κ2 + 6κ + 1)k(k + n − 2) + κ(n − 2)2 ].

Figure 3.1 shows μ±k as a function of k for α = −4, n = 3, and R = 1. In Figure 3.1a we choose κ = 0.1, while in Figure 3.1b κ = 0.9.

Figure 3.1: Steklov eigenvalues in annular domains.

Remark 3.2. The following asymptotic behavior can be shown. 1. For κ → 0 we have limκ→0 μ+k (κ) = ∞ and limκ→0 μ−k (κ) = μk = α + Rk , which is the k-th Steklov eigenvalue of the ball (see Remark 3.1). On the other hand, limκ→1 μ+k (κ) = ∞ and limκ→1 μ−k (κ) = 0. ̃ 2. Let κ = R for some fixed R̃ > 0. Then R

lim μ+k = α +

R→∞

k+n−2 R̃

and

lim μ−k = 0.

R→∞

Thus, μ+k (R) converges for R → ∞ to the k-th Steklov eigenvalue of the outer ball domain ℝn \ BR̃ (see Section 3.2.2).

48 � 3 Spherical harmonics and eigenvalue problems

3.3 Membrane eigenvalue problems In this section we consider the Helmholtz equation Δu + λu = 0

in BR ,

(3.3.1)

subject to one of the following boundary conditions: – Dirichlet, u = 0; – Neumann, 𝜕ν u = 0; – Robin, 𝜕ν u + αu = 0, α ∈ ℝ. Separation of variables u = r γ J(r)Yk (θ), where Yk (ξ) is a spherical harmonic of degree k, leads to J ′′ (r) + J ′ (r)(

γ(γ + n − 2) − k(k + n − 2) 2γ + n − 1 ) + J(r)( + λ) = 0. r r2

If we set γ = − n−2 , then 2 r 2 J ′′ (r) + rJ ′ (r) + (λr 2 − ν2 )J(r) = 0,

ν=k+

n−2 . 2

The solutions which are regular at the origin are the Bessel function Jν (√λr) if λ > 0 and the modified Bessel functions Iν (√|λ|r) if λ < 0. The boundary conditions give rise to a countable number of eigenfunctions and eigenvalues λ1 < λ2 ≤ λ3 ≤ ⋅ ⋅ ⋅ tending to infinity. We shall write r −(n−2)/2 𝒥ν (√|λ|r)Yk (ξ),

ν = k + (n − 2)/2,

where 𝒥ν stands for either the Bessel function or the modified Bessel function. The lowest eigenvalues λD , λN , and λR corresponding to Dirichlet, Neumann, and Robin boundary conditions, respectively, are λD1 =

j02 , R2

where j0 is the first zero of J(n−2)/2 ,

j2 λN1 = 0 and λN2 = λ̃N2 = ⋅ ⋅ ⋅ = λ̃Nn+1 = 12 , where j1 is the first zero of J1+(n−2)/2 , R λR1 = lowest solution of 2−n ′ 𝒥 n−2 (√|λ|R) + √|λ|R𝒥 n−2 (√|λ|R) + αR𝒥 n−2 (√|λ|R) = 0, 2 2 2 2 where 𝒥 = J or I depending on the sign of λ.

3.4 The buckling plate

� 49

3.3.1 The generalized Steklov problem We are looking for an eigenvalue μ such that for fixed λ the problem Δϕ + λϕ = 0 in BR ,

𝜕ν ϕ = μϕ on 𝜕BR

(3.3.2)

has a nontrivial solution. In contrast to the classical Steklov problem, μ = 0 is not the lowest eigenvalue anymore. If we separate variables ϕk (r, ξ) = ck (r)Yk (ξ), we obtain where ν = k +

ck (r) = r −(n−2)/2 𝒥ν (√|λ|r),

n−2 . 2

The eigenvalue μk is then determined by 2−n ′ 𝒥k+ n−2 (√|λ|R) + √|λ|R𝒥k+ n−2 (√|λ|R) = μk R𝒥k+ n−2 (√|λ|R). 2 2 2 2 The lowest eigenvalue corresponds to the lowest root for k = 0. It is simple and the eigenfunction is radial. Note that ∮ ϕk ϕi dS = 0

if k ≠ i.

𝜕BR

3.4 The buckling plate The eigenvalue problem of the buckling plate is Δ2 u + ΛΔu = 0 in BR ,

u = 𝜕ν u = 0 on 𝜕BR .

The eigenvalues are positive. This is easily seen by testing this equation with u. The differential equation can be written as (Δ + Λ)Δu = 0, and it therefore breaks into two equations, namely Δu = υ and Δυ + Λυ = 0. Consequently, Δ(u +

υ ) = 0. Λ

Hence, u + Λυ is a harmonic function. Every solution of Δ2 u + ΛΔu = 0 is therefore of the form u = h + ϕ, where Δh = 0 and Δϕ + Λϕ = 0. It can be written in the form ∞

u(r, ξ) = ∑ (hk r k + ck r −(n−2)/2 Jνk (√Λr))Yk (ξ), k=0

50 � 3 Spherical harmonics and eigenvalue problems where νk = k + (n − 2)/2. The boundary conditions together with ∮𝜕B Ym Yk dS = δkm lead 1 to a linear system for the unknowns hk and ck : hk Rk + ck R−(n−2)/2 Jνk (√ΛR) = 0,

khk Rk−1 − ck

n − 2 −n/2 √ R Jνk ( ΛR) + ck R−(n−2)/2 √ΛJν′k (√ΛR) = 0. 2

It has a nontrivial solution provided the corresponding determinant vanishes. Thus, Rk−(n−2)/2 √ΛJν′k (√ΛR) − Rk−n/2 νk Jνk (√ΛR) = 0. Replacing Jν′ (z) by νz Jν (z) − Jν+1 , we get −√ΛRk−(n−2)/2 Jνk +1 (√ΛR) = 0. Obviously, Λ = 0 cannot be an eigenvalue. Therefore, √ΛR = jνk +1 (ℓ), where jν (ℓ) is the ℓ-th zero of Jν . An argument based essentially on the completeness of the spherical harmonics on 𝜕B1 reveals that we have found all eigenvalues. Hence, Λ1 (BR ) = (

j n (1)

u1 (r) = r −

2

R

n−2 2

2

(3.4.1)

),

R J n−2 (√Λr) − R− 2

n−2 2

r J n−2 (√ΛR). 2

Consequently, the first eigenvalue is simple and its eigenfunction is radial and monotone. Note that in the ball Λ1 coincides with the second eigenvalue of the membrane with Dirichlet boundary conditions.

3.4.1 Fourth order Steklov eigenvalue problem Consider the eigenvalue problem Δ2 φ = 0 in BR ,

φ = 0 on 𝜕BR ,

Δφ = d𝜕ν φ on 𝜕BR .

(3.4.2)

From Δ2 = ΔΔ = 0 it follows that Δφ = h, where h is harmonic. The sequence {hk }∞ 0 with k hk = r Yk (ξ) is a basis for the harmonic functions in BR . The solutions of Δuk = ak r k Yk (ξ) can be obtained by separation of variables, uk = Ak (r)Yk (ξ). The function Ak is a regular solution of A′′ k + It is of the form Ak (r) =

n − 1 ′ k(n − 2 + k) Ak − Ak = ak r k . r r2

ak r k+2 2(n+2k)

+ bk r k . Hence,

3.5 Notes

∞

φ(r, ξ) = ∑ (bk r k + k=0

�

51

ak r k+2 )Yk (ξ). 2(n + 2k)

The boundary conditions lead to φ(R, ξ) = 0

󳨐⇒

bk = −

ak R2 2(n + 2k)

and Δφ = d𝜕ν φ

󳨐⇒

ak Rk = d(kbk Rk−1 + ak

(k + 2)Rk+1 ). 2(n + 2k)

Consequently, the eigenvalues of (3.4.2) are dk =

n + 2k , R

k = 0, 1, 2, . . . ,

and the corresponding eigenfunctions are φk = ak (r k −

r k+2 )Yk (ξ). R2

The first eigenvalue is simple and its eigenfunction is radial and of constant sign.

3.5 Notes References to spherical harmonics and to the Funk–Hecke formula are for instance the Lecture Notes of C. Müller [91].

4 Variational formulas We present two methods to derive variational formulas for domain functionals. The change of variables method maps the regions via a diffeomorphism into a fixed domain and expresses the functionals over this fixed domain. In this case one perturbs the coefficients. Its implementation requires only the chain rule. The moving surface method captures the change in shape by means of a boundary displacement. This displacement is expressed by a one-parameter family of perturbations. It is used to understand the evolution of a given domain functional as a function of the parameter.

4.1 The change of variables method 4.1.1 Introduction Let {Ωt }|t| 0, λ1 < 0, then |λ1 | > |λk |, for k = 2, . . ., implies that ̈ ≥ 0. In this case the ball is a local minimum. 𝒞 (0)

5.3.2 Gamov’s liquid model In Gamov’s liquid model the energy of a water drop is given by the surface energy and the Coulomb force. After suitable normalization it takes the form ℰ (Ω) = |𝜕Ω| +

1 1 dx dy, ∫∫ 8π |x − y| Ω Ω

where Ω is a bounded domain in ℝ3 . Let Ω = BR and consider the family of volume ̇ = 0, and by (2.3.26) and the previous computations, preserving perturbations. Then ℰ (0) ̈ = 𝒮0̈ (0) + ℰ (0)

2 1 ̈ 󵄨 τ 󵄨2 2 𝒞 (0) = ∮ 󵄨󵄨󵄨∇ ρ󵄨󵄨󵄨 dS − 2 ∮ ρ dS 8π R 𝜕BR

𝜕BR

1 − ∮ ∮ ρ(x)[ρ(y) − ρ(x)x ⋅ y]|x − y|−1 dSx dSy . 8π 𝜕BR 𝜕BR

The sign is not clear because the second variation of the surface area is positive and that of the convolution is negative. A more subtle analysis is needed to determine which one dominates. In [32] the author gave a detailed analysis of volume preserving perturbations of the first and the second order. We consider the special case n = 3 and F(u) := Then F > 0 and F ′ ≤ 0 and

1 . √u

(5.3.18)

90 � 5 Applications 1

λ1 = |𝜕B1 | ∫ F(2R2 (1 − t))t(1 − t 2 )

n−3 2

dt

−1 1

=

|𝜕B1 | t 8π dt = . ∫ √2 3R 2 √ R (1 − t) −1

(5.3.19)

By (5.3.18) it follows ∮𝜕B |F(R2 |ξ − η|2 ) dSη < ∞. Hence, (5.3.17) applies and thus 1

̈ = 𝒮0̈ (0) + ℰ (0)

R4 ∞ ∑ (λ − λ1 )ck2 . 8π k=2 k

Since the perturbations is volume preserving and since the barycenter condition holds, we get the lower bound ̈ ≥ ℰ (0)

4 R4 ∞ ∑ (λ − λ1 )ck2 , ∮ ρ2 dS + 2 8π k=2 k R 𝜕BR

where ∞

ρ = ρ(Rξ) = ∑ ck Yk (ξ). k=2

Since |λk | ≤ λ1 for all k ≥ 2, ∞

2

̈ ≥ 4 ∑ c − 2λ1 ℰ (0) k k=2

R4 ∞ 2 ∑c . 8π k=2 k

We apply (5.3.19) and obtain ∞

2

2 3

∞

3

2

̈ ≥4∑c − R ∑c . ℰ (0) k k k=2

k=2

1

Hence, for R < 6 3 ∼ 1.8 the ball is a local minimizer for the Gamov energy. For large R such a statement cannot hold. As an example we consider perturbations such that ρ = Yk ,

k ≥ 2.

Then ρ is volume preserving up to first order (since k ≥ 2) and w can be chosen such that (Φt )t is volume preserving up to second order. As a consequence we have ci = δik . Hence, 4

̈ = R (λk − λ1 ). 𝒞 (0)

5.3 Convolutions

� 91

Since n = 3 the surface variation reads as 󵄨

τ

󵄨2

4

2

𝒮0̈ (0) = ∮ 󵄨󵄨󵄨∇ ρ󵄨󵄨󵄨 − 2 ρ dS = k(k + 1) − 4, R

for k ≥ 2.

𝜕BR

Hence, the Gamov energy becomes ̈ = k(k + 1) − 4 + ℰ (0)

R4 (λ − λ1 ), 8π k

k ≥ 2.

From (5.3.16) and (5.3.19) we deduce λ1 =

8π . 3R

Moreover (again applying (5.3.16)), a direct calculation yields λ2 =

8π . 5R

Note that |λk | − λ2 ≤ 0 for k ≥ 2. Then, for k ≥ 2 we have R4 R4 16π (λ2 − λ1 ) = k(k + 1) − 4 − ( ) 8π 8π 15R 2 = k(k + 1) − 4 − R3 . 15

̈ ≤ k(k + 1) − 4 + ℰ (0)

̈ < 0 if Clearly, ℰ (0) 1

R > R∗ := (

3 15 k(k + 1) − 30) . 2

(5.3.20)

We summarize our results. Proposition 5.5. Let Ω be a bounded smooth domain in ℝ3 . The Gamov energy is given by ℰ (Ω) = |𝜕Ω| +

1 1 dx dy. ∫∫ 8π |x − y| Ω Ω

Then among all domains of given volume the ball is a critical point of the energy. Moreover, the following statements hold true: 1 – If R < 6 3 ∼ 1.8, the ball is a local minimizer. – For large volumes (resp. R) the ball is not a local minimizer. Moreover, (5.3.20) gives a ̈ < 0 for R > R∗ . lower bound R∗ such that ℰ (0) 1 – No result is known for 6 3 < R < R∗ . This result corresponds to an earlier work [79].

92 � 5 Applications

5.4 An optimal control problem We compute the domain variations of the cost function 𝒞 (t) = ∫ C(u(y, t)) dy, Ωt

where u is a solution of the boundary value problem Δu+g(u) = 0 in Ωt and u = 0 on 𝜕Ωt . We assume that Ωt is obtained from BR by means of the diffeomorphism Φt described in the previous chapters. From formula (4.1.6) we obtain ′

′

̇ = ∫ C (u)u dx + C(0) ∮ (υ ⋅ ν) dS, 𝒞 (0) BR

𝜕BR

and from (4.1.10) it follows that ′

′′

′2

′′

′

′

̈ = ∫ C (u)u dx + ∫ C (u)u dx + 2C (0) ∮ u (υ ⋅ ν) dS 𝒞 (0) BR

BR

𝜕BR

′

+ C (0) ∮ (υ ⋅ ∇u)(υ ⋅ ν) dS + C(0)𝒱̈ (0). 𝜕BR

Suppose that u = u(r) is radial. Here r = |x| and ur (r) = satisfies Δu′ + g ′ (u)u′ = 0 in BR ,

d u(r). The first shape derivative dr

u′ = −(∇u ⋅ υ) = −ur (R)(υ ⋅ ν) on 𝜕BR .

(5.4.1)

The second shape derivative u′′ solves the equation Δu′′ + g ′′ (u)u′ 2 + g ′ (u)u′′ = 0

in BR .

Formula (4.1.8) implies that on the boundary, u′′ = −(υ ⋅ D2 uυ) − 2(υ ⋅ ∇u′ ) − (w ⋅ ∇u) = − urr (R)(υ ⋅ ν)2 − ur (R)

|υτ |2 − 2υ ⋅ ∇u′ − ur (R)(w ⋅ ν). R

The fact that the differential equation for u′′ depends also on u′ makes the discussion more difficult. This is not the case in the next example.

5.4 An optimal control problem

� 93

5.4.1 The torsion problem We consider the special case g(u) = 1. Then u(x) =

1 2 (R − |x|2 ). 2n

The first and second shape derivatives, u′ and u′′ , are harmonic functions. For Hadamard perturbations they satisfy the boundary conditions u′ =

R (υ ⋅ ν) and n

u′′ = −2(υ ⋅ ν)𝜕ν u′ +

1 R (υ ⋅ ν)2 + (w ⋅ ν). n n

In the boundary condition for u′′ , we replace (υ ⋅ ν) by u′ : u′′ = −

2n ′ ′ n ′ 2 R u 𝜕ν u + 2 u + (w ⋅ ν) R n R

on 𝜕BR .

(5.4.2)

For the first variation we obtain ′

′

̇ = ∫ C (u(r))u dx + C(0)𝒱̇ (0). 𝒞 (0) BR

Since u′ is harmonic, the mean value theorem implies 1 1 ∮ u′ dS = n−1 ∮ u′ dS r n−1 R 𝜕Br

(5.4.3)

𝜕BR

for all 0 ≤ r ≤ R. Hence, R

R

∫ C ′ (u(r))u′ dx = ∫ C ′ (u(r)) ∮ u′ dSr dr = ∫ C ′ (u(r)) BR

0

= ∫ C ′ (u(r)) 0

=

0

𝜕Br

R

r n−1 ∮ u′ dSr dr r n−1

r n−1 ∮ u′ dSR dr Rn−1 𝜕BR

1

R

Rn−1

∮ u′ dSR ∫ C ′ (u(r))r n−1 dr. 0

𝜕BR

Together with the boundary conditions of u′ this leads to R

̇ ={ 𝒞 (0)

1 ∫ C ′ (u)r n−1 dr + C(0)}𝒱̇ (0). nRn−2 0

𝜕Br

94 � 5 Applications If 𝒱̇ (0) = 0, the first variation vanishes. Thus, the ball is a critical point of the cost functionals for all volume preserving perturbations. The second variation is of the form ′

′′

′2

′′

̈ = ∫ C (u)u dx + ∫ C (u)u dx + 𝒞 (0) BR

BR

nC ′ (0) ∮ u′ 2 dS + C(0)𝒱̈ (0). R

(5.4.4)

𝜕BR

We will determine the sign under the assumptions 𝒱̇ (0) = 𝒱̈ (0) = 0

and C ′ (u), C ′′ (u) ≥ 0.

̈ will be discussed separately. The first two terms of 𝒞 (0) 1. ∫B C ′ (u)u′′ dx R

Since u′′ is harmonic, the mean value theorem (see (5.4.3)) applies. Moreover, u′′ satisfies the boundary conditions (5.4.2). Hence, ′

1−n

′′

∫ C (u)u dx = R BR

R

∮ u dS ∫ C ′ (u(r))r n−1 dr ′′

0

𝜕BR R

= R1−n ∫ C ′ (u(r))r n−1 dr ∮ (− 0

𝜕BR

2n ′ ′ n ′ 2 R u 𝜕ν u + 2 u + (w ⋅ ν)) dS. R n R

From (2.3.12) and the boundary condition for u′ we obtain ∮ (w ⋅ ν) dS = −

n2 (n − 1) n−1 ∮ (υ ⋅ ν)2 dS = − ∮ u′ 2 dS. R R3 𝜕BR

𝜕BR

𝜕BR

Consequently, ′

1−n

′′

∫ C (u)u dx = R BR

R

∮ u dS ∫ C ′ (u(r))r n−1 dr ′′

𝜕BR

0

R

= R1−n ∫ C ′ (u(r))r n−1 dr ∮ (− 0

𝜕BR

2n ′ ′ n(n − 2) ′ 2 u 𝜕ν u + u ) dS. R R2

(5.4.5)

2. ∫B C ′′ (u)u′ 2 dx R As before we introduce polar coordinates and write R

∫ C ′′ (u(r))u′ 2 dx = ∫ C ′′ (u(r)) ∮ u′ 2 dS dr. BR

0

𝜕Br

(5.4.6)

5.4 An optimal control problem

� 95

Since u′ 2 is subharmonic, the mean value theorem does not apply. We therefore pursue a different scheme. Let {ϕk }∞ k=0 be the orthonormal basis of Steklov eigenfunctions (see Section 3.2.1) Δϕk = 0 in BR ,

𝜕ν ϕk = μk ϕk on 𝜕BR .

Expressed in polar coordinates, the eigenfunctions are of the form ϕ0 (r, ξ) =

1

ϕk (r, ξ) =

, |𝜕BR |1/2

rk

R(n−1+2k)/2

Yk,i (ξ),

(5.4.7)

where Yk,i is a normalized spherical harmonic such that 2 (ξ) dS = 1 ∮ Yk,i

and

𝜕B1

∮ Yk,i (ξ)Ys,j (ξ) dS = 0

if (k, i) ≠ (s, j).

𝜕B1

In order to estimate the integral ∮𝜕B u′ 2 dS we expand u′ into the Fourier series r

∞

u′ = ∑ ck ϕk . k=0

Since 𝒱̇ (0) =

n R

∮𝜕B u′ dS = 0, it follows that c0 = 0. Hence, R

∞

∮ u′ 2 dS = ∑ ck2 .

(5.4.8)

k=1

𝜕BR

For r ≤ R we obtain ∮ u′ 2 dS = r n−1 ∮ u′ 2 (r, ξ) dS. 𝜕Br

𝜕B1

Consequently, r n−1+2k 2 c . n−1+2k k k=1 R ∞

∮ u′ 2 (R, ξ) dS = r n−1 ∮ u′ 2 (r, ξ) dS = ∑ 𝜕Br

𝜕B1

Since r ≤ R and k ≥ 1, r n+1 2 r n+1 c = n+1 ∮ u′ 2 dS. n+1 k R R k=2 ∞

∮ u′ 2 (r, ξ) dS ≤ ∑ 𝜕Br

Then by (5.4.6)

𝜕BR

96 � 5 Applications

′2

′′

∫ C (u(r))u dx ≤ BR

1

R

′2

∮ u dS ∫ C ′′ (u(r))r n+1 dr.

Rn+1

0

𝜕BR

d ′ Furthermore, since C ′′ (u) = − nr dr C (u), we have R

′′

∫ C (u)r

n+1

n

2

R

dr = −nC (0)R + n ∫ C ′ (u)r n−1 dr. ′

0

0

This leads to the final estimate ′2

′′

∫ C (u(r))u dx ≤ BR

1

Rn+1

′2

n

2

R

∮ u dS(−nC (0)R + n ∫ C ′ (u)r n−1 dr). ′

(5.4.9)

0

𝜕BR

̈ These two steps enable us to derive an upper bound for 𝒞 (0). If we insert (5.4.5) and (5.4.9) into (5.4.4) and use Green’s theorem 󵄨 󵄨2 ∫ 󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 dx = ∮ u′ 𝜕ν u′ dS, BR

𝜕BR

we obtain the estimate 1−n

̈ ≤R 𝒞 (0)

R

∫ C ′ (u)r n−1 dr(−2

n 󵄨󵄨 ′ 󵄨󵄨2 n(n − 2) ∮ u′ 2 dS) ∫ 󵄨∇u 󵄨 dx − R 󵄨 󵄨 R2 BR

0

𝜕BR

R

+

1 ∮ u′ 2 dS(n2 ∫ C ′ (u)r n−1 dr − nC ′ (0)Rn ) Rn+1 𝜕BR

n + C ′ (0) ∮ u′ 2 dS. R

0

𝜕BR

Since μ1 =

1 R

and 𝒱̇ (0) = 0, we have 1 󵄨 󵄨2 ∫ 󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 dx ≥ ∮ u′ 2 dS. R

BR

𝜕BR

Finally we obtain R

̈ ≤ 2 ∫ C ′ (u)r n−1 dr{R−1 ∮ u′ 2 dS − ∫ 󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨2 dx} ≤ 0. 𝒞 (0) 󵄨 󵄨 Rn 0

𝜕BR

BR

(5.4.10)

5.5 Notes

�

97

The fact that the second variation is nonpositive is compatible with the isoperimetric inequality (see [9]), which states that among all domains of fixed volume, 𝒞 is maximal for the ball. A special case for 𝒞 Let C(u) = up , p > 1, and consider the functional 1/p

𝒰 (t) = 𝒞 (t)

.

From 1/p

𝒞 (t)

p

1 u p ) dy)}, = exp{ log(umax ∫( p umax Ωt

it follows that max u(y, t) = lim 𝒰 (t). y∈Ωt

p→∞

2

R Consequently, for nearly spherical domains of given volume, maxΩt u(x, t) ≤ 2n . By means of symmetrization it can be shown that this result holds globally for any domain with the same volume as BR [9, 119].

Problem 5.2. Are the results of this section also true for torsion problems with Robin boundary conditions? Is it possible to discuss Payne–Rayner type inequalities [94] for the first eigenfunctions of the membrane problem?

5.5 Notes Isoperimetric inequalities for functionals of the type 𝒢 (t) considered in Section (5.1) have been dealt with the techniques of rearrangement and symmetrization. They go back to Hardy, Littlewood, and Pólya [72] and Pólya and Szegö [98]. An excellent presentation of the essential ingredients is given in [87]. For further developments and an extensive bibliography we refer to [28]. Global inequalities for weighted isoperimetric inequalities are derived in [22]. By means of symmetrization it is shown that under certain conditions, among all domains of given weighted volume, the weighted surface area attains its minimum for the ball. The classical isoperimetric inequalities for the spherical and hyperbolic spaces were first proved by E. Schmidt in 1940 and 1943. Proofs and further references are found in [34].

98 � 5 Applications Convolutions play an important role in potential theory. The effect of symmetrization on convolutions is completely explained by Riesz’ inequality [103]; see also [72] and [87]. In this case, among all domains of given area, 𝒞 (t) achieves its maximum for the ball. A survey of the existence and nonexistence of global minimizers in the Gamov liquid problem is given in [37]. For an overview of control problems we refer to [120].

6 Domain variations for energies This chapter deals with domain functionals (energies) depending on the solutions of semilinear elliptic boundary value problems. In contrast to optimal control problems these solutions are critical points of the energies in the sense of Fréchet. The Euler– Lagrange equations allow to eliminate the shape derivative in the first domain variation. Consequently, the second domain variation depends only on first order shape derivatives. In this chapter we use the change of variables method.

6.1 Energies and critical points In this section we show that the first variation of a functional governed by a function satisfying the Euler–Lagrange equation does not contain the shape derivative of the first order. Let ℰ (t) = ∫ G(y, u(y, t), ∇u(y, t)) dy, Ωt

where G(y, u, p) : ℝn × ℝ × ℝn → ℝ is a continuously differentiable function in all 2n + 1 variables. Suppose that u satisfies the Euler–Lagrange equation Gu (y, u, ∇u) −

𝜕 G (y, u, ∇u) = 0 in Ωt , 𝜕yi pi

u = 0 on 𝜕Ωt .

(6.1.1)

If Reynolds’ theorem (Theorem 4.1) applies, then t

̇ = ∫ 𝜕t G dy + ∮ (𝜕t Φt ⋅ ν )G dS. ℰ (t) Ωt

𝜕Ωt

In this case, 𝜕t G = Gu 𝜕t u + Gpi 𝜕y𝜕 𝜕t u. The divergence theorem implies that i

∫ 𝜕t G dx = ∫{Gu − Ωt

Ωt

𝜕 G }u dx + ∮ νit Gpi 𝜕t u dS. 𝜕yi pi 𝜕Ωt

The boundary conditions u(y, t) = 0 on 𝜕Ωt yield (cf. Section 2.4.2) 0 = 𝜕t Φt ⋅ ∇y u(y, t) + 𝜕t u(y, t). Finally we get t

t

̇ = − ∮ ν Gp (∇y u ⋅ 𝜕t Φt ) dSt + ∮ G(𝜕t Φt ⋅ ν ) dSt . ℰ (t) i i 𝜕Ωt https://doi.org/10.1515/9783111025438-006

𝜕Ωt

(6.1.2)

100 � 6 Domain variations for energies In particular, ̇ = − ∮ νi Gp (∇u ⋅ υ) dS + ∮ G (υ ⋅ ν) dS. ℰ (0) i 𝜕Ω

𝜕Ω

Consider now the more general case ℰ (t) = ∫ G(y, u(y, t), ∇u(y, t)) dy + ∮ B(u(y, t)) dSt . Ωt

𝜕Ωt

If u(y, t) is a critical point of ℰ (t), it satisfies the Euler–Lagrange equation (6.1.1) and the boundary conditions νit Gpi + Bu (u) = 0. Then by Reynolds’ theorem (Theorem 4.1), t

t

̇ = ∮ [G(y, u, ∇u) + Bu (u)𝜕νt u + (n − 1)H B(u)](𝜕t Φt ⋅ ν ) dSt . ℰ (t) 𝜕Ωt

Two remarks stand out: 1. If u satisfies the Euler–Lagrange equations corresponding to ℰ (t), then its first variation contains only boundary integrals. 2. If in addition u satisfies the boundary conditions associated to the Euler–Lagrange ̇ is independent of the shape derivative u′ . equation, ℰ (t)

6.2 Elliptic boundary value problems 6.2.1 General setting, change of variables Let {Ωt }|t| 0 in supp Z + and (υ ⋅ ν) < 0 in supp Z − . In this case we get ̇ ℰ (0) > 0, which is obviously a contradiction. The first assertion is proved in the same way. ̇ (0) = 0. Example 6.2. If Ω = BR and u(x) = u(|x|), then ℰD/R By means of the moving plane method – proposed by Serrin in [107] – the existence of positive solutions with u = 0 and |∇u| = const. on 𝜕Ω implies that Ω is a ball. This method does not seem to apply to Robin boundary conditions. Problem 6.1. Is the overdetermined problem Δu + g(u) = 0 in Ω, 𝜕ν u + αu = 0 on 𝜕Ω, and |∇u|2 − 2G(u) − 2α2 u2 + α(n − 1)u2 H = const.

on 𝜕Ω

only solvable in a ball?

6.5 The second domain variation The aim of this section is to find a suitable form of the second variation of the energy in order to determine its sign. By (6.3.7) we have ℰD̈ (0) = ℱ1 + ℱ2 ,

where 󵄨2

󵄨

̇̃ 󵄨󵄨 dx, ℱ1 (0) := ∫ Ä ij (0)𝜕i u𝜕j u dx − 2 ∫󵄨󵄨󵄨∇u(0) 󵄨 Ω

Ω

̈ dx + 2 ∫ g ′ (u)̃ u̇̃ 2 (0) dx. ℱ2 (0) := −2 ∫ G(u)J(0) Ω

Ω

By (6.3.8), the Robin energy is ℰR̈ (0) = ℱ1 + ℱ2 + ℱ3 ,

where 2

2

̈ ℱ3 (0) := α ∮ u m(0) dS − 2α ∮ u̇̃ dS. 𝜕Ω

𝜕Ω

6.5.1 The transformations of ℱ1 (0), ℱ2 (0), and ℱ3 (0) Step (i). We start with ℱ1 (0). From Lemma 4.1 we have

110 � 6 Domain variations for energies Ä ij (0)𝜕i u𝜕j u = [(div υ)2 − Dυ : Dυ ]|∇u|2 + 4(∇uDυ ⋅ Dυ ∇u) + 2Dυ ∇u ⋅ Dυ ∇u − 4 div υ(∇u ⋅ Dυ ∇u)

+ div w|∇u|2 − 2(∇u ⋅ Dw ∇u).

(6.5.1)

From (4.1.4) it follows that ̇̃ ∇u(0) = ∇(υ ⋅ ∇u) + ∇u′ = Dυ ∇u + D2 uυ + ∇u′ . Thus, 󵄨 ̇̃ 󵄨󵄨2 2 2 2 −2󵄨󵄨󵄨∇u(0) 󵄨󵄨 = −2(Dυ ∇u ⋅ Dυ ∇u) − 4(Dυ ∇u ⋅ D uυ) − 2(υD u ⋅ υD u) 󵄨 󵄨2 − 4(∇u′ ⋅ Dυ ∇u) − 4(υ ⋅ D2 u∇u′ ) − 2󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 . Clearly, −4(∇u′ ⋅ Dυ ∇u) − 4(υ ⋅ D2 u∇u′ ) = −4 div[∇u′ (υ ⋅ ∇u)] + 4Δu′ (υ ⋅ ∇u) + 4(∇u′ ⋅ D2 uυ) − 4(υ ⋅ D2 u∇u′ )

= −4 div[∇u′ (υ ⋅ ∇u)] + 4Δu′ (υ ⋅ ∇u). Thus, 󵄨 ̇̃ 󵄨󵄨2 2 2 2 −2󵄨󵄨󵄨∇u(0) 󵄨󵄨 = −2(Dυ ∇u ⋅ Dυ ∇u) − 4(Dυ ∇u ⋅ D uυ) − 2(υD u ⋅ υD u) 󵄨 󵄨2 − 4 div[(υ ⋅ ∇u)∇u′ ] + 4(υ ⋅ ∇u)Δu′ − 2󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 . Inserting this expression and (6.5.1) in ℱ1 (0) we find ℱ1 (0) = ∫(E0 + E1 + E2 ) dx, Ω

where E0 := ((div υ)2 − Dυ : Dυ )|∇u|2 +4(∇uD ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ υ ⋅ Dυ ∇u) (1)

(∗)

−4 div υ(∇u ⋅ Dυ ∇u) − 4(Dυ ∇u ⋅ D2 uυ) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ −2(υD2 u ⋅ υD2 u) . ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (2)

(3)

The terms containing w are collected in E1 and partially expressed in divergence form: E1 := div w|∇u|2 − 2(∇u ⋅ Dw ∇u)

= div[w|∇u|2 − 2(w ⋅ D2 u∇u) − 2 div[∇u(w ⋅ ∇u)] + 2Δu(w ⋅ ∇u) + 2(w ⋅ D2 u∇u)

= div[w|∇u|2 − 2∇u(w ⋅ ∇u)] + 2Δu(w ⋅ ∇u).

6.5 The second domain variation

� 111

The term E2 collects all terms containing u′ : 󵄨 󵄨2 E2 = −4 div[(υ ⋅ ∇u)∇u′ ] + 4(υ ⋅ ∇u)Δu′ − 2󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 . The goal is to write as many terms of E0 as possible in divergence form so that the domain integrals of the remaining terms are canceled out. With the help of (2.3.4) the first term in the expression of E0 can be rewritten as ((div υ)2 − Dυ : Dυ )|∇u|2 = div[(υ div υ − υDυ )]|∇u|2

= div[(υ div υ − υDυ )|∇u|2 ] − 2 div υ(υ ⋅ D2 u∇u) + 2(υDυ ⋅ D2 u∇u).

Moreover, − 4 div υ(∇u ⋅ Dυ ∇u) − 4(Dυ ∇u ⋅ D2 uυ)

= 4 div[∇u(υ ⋅ Dυ ∇u) − υ(∇u ⋅ Dυ ∇u)] + 4(υD2 u ⋅ ∇uDυ ) 2 −4(∇uD ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ υ ⋅ Dυ ∇u) −4Δu(υ ⋅ Dυ ∇u) − 4(υDυ ⋅ D u∇u). (∗)

This is best seen by calculating the first term on the right-hand side. The terms underbracketed with a (∗) in the above formula and in E0 are canceled out. Hence, E0 = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ div[(υ div υ − υDυ )|∇u|2 ] −2 div υ(υ ⋅ D2 u∇u) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (1)

(1)

2

+4 div[∇u(υ ⋅ Dυ ∇u) − υ(∇u ⋅ Dυ ∇u)] −2(υDυ ⋅ D u∇u) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (1),(2)

(2)

2

+4(υD u ⋅ ∇uDυ ) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ −4Δu(υ ⋅ Dυ ∇u) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ −2(υD2 u ⋅ υD2 u) . ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (2)

(2)

(3)

The four terms in E0 which are neither in divergence form nor contain the Δ operator will be considered separately separately. Let us sum up the first three terms: − 2(υDυ ⋅ D2 u∇u) − 2(υD2 u ⋅ υD2 u) − 2 div υ(υD2 u ⋅ ∇u) = −2 div[υ(υD2 u ⋅ ∇u)] ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ +2υi υj 𝜕i 𝜕j 𝜕k u𝜕k u . (∗)

Furthermore, 4(υD2 u ⋅ ∇uDυ ) = 4 div[∇u(υD2 u ⋅ υ)] − 4Δu(υ ⋅ D2 uυ)

−4υi υj 𝜕i 𝜕j 𝜕k u𝜕k u − 4(υD2 u ⋅ ∇uDυ ).

Note that the term on the left side also appears on the right side with reversed sign. Consequently,

112 � 6 Domain variations for energies 4(∇uDυ ⋅ D2 uυ) = 2 div[∇u(υD2 u ⋅ υ)] − 2Δu(υ ⋅ D2 uυ) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ −2υi υj 𝜕i 𝜕j⏟𝜕⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ k u𝜕k u . (∗)

The four terms underbracketed with (∗) are canceled out when we sum them all. Inserting them into E0 we obtain E0 = div[(υ div υ − υ ⋅ Dυ )|∇u|2 ] + 4 div[∇u(υ ⋅ Dυ ∇u) − υ(∇u ⋅ Dυ ∇u)]

− 4Δu(υ ⋅ Dυ ∇u) − 2 div[υ(υ ⋅ D2 u∇u) − ∇u(υ ⋅ D2 uυ)] − 2Δu(υ ⋅ D2 uυ).

(6.5.2)

The divergence theorem then implies the following proposition. Proposition 6.1. We have 2

ℱ1 (0) = ∮[υ ⋅ ν div υ − (υ ⋅ Dυ ν) + (w ⋅ ν)]|∇u| dS 𝜕Ω

+ 4 ∮[𝜕ν u(υ ⋅ Dυ ∇u) − (υ ⋅ ν)(∇u ⋅ Dυ ∇u)] dS 𝜕Ω

− 2 ∮[(υ ⋅ ν)(υ ⋅ D2 u∇u) − 𝜕ν u(υ ⋅ D2 uυ) + 𝜕ν u(w ⋅ ∇u)] dS 𝜕Ω

− 2 ∫ Δu[2(υ ⋅ Dυ ∇u) + (υ ⋅ D2 uυ) − (w ⋅ ∇u)] dx Ω

󵄨 󵄨2 − 2 ∫[󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 − 2Δu′ (υ ⋅ ∇u)] dx − 4 ∮(υ ⋅ ∇u)𝜕ν u′ dS. Ω

𝜕Ω

Step (ii). Consider now ℱ2 (0). The integral ℱ2 (0) is transformed similarly to ℱ1 (0). From (2.1.5) and the definition ̇̃ we have of the material derivative u(0) ′

2

̈ dx + 2 ∫ g (u)u̇̃ (0) dx ̃ ℱ2 (0) = −2 ∫ G(u(0)) J(0) Ω

Ω 2

= −2 ∫ G(u(x))((div υ) − Dυ : Dυ + div w) dx Ω

+ 2 ∫ g ′ (u)[u′ 2 + 2u′ (υ ⋅ ∇u)] dx + ∫ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 2g ′ (u)(υ ⋅ ∇u)2 dx. Ω

Ω

(∗)

We apply (2.3.4) and obtain ̈ = (div υ)2 − Dυ : Dυ + div w = div(υ div υ − υDυ + w). J(0) Since g(u) = G′ (u), ̈ = −2 div[G(u)(υ div υ − υDυ + w)] + 2(∇G(u) ⋅ [υ div υ − υDυ + w]) −2G(u)J(0)

6.5 The second domain variation

� 113

= −2 div[G(u)(υ div υ − υDυ + w)]

+ 2g(u)[(υ ⋅ ∇u) div υ − (υ ⋅ Dυ ∇u) + (w ⋅ ∇u)].

The term 2g(u)(υ ⋅ ∇u) div υ can be written in divergence form with some additional terms: 2g(u)(υ ⋅ ∇u) div υ = 2 div[g(u)(υ ⋅ ∇u)υ] − 2υ ⋅ ∇[g(u)(υ ⋅ ∇u)]

= 2 div[g(u)(υ ⋅ ∇u)υ] − ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 2g ′ (u)(υ ⋅ ∇u)2 −2g(u)(υ ⋅ D2 uυ) (∗)

− 2g(u)(υ ⋅ Dυ ∇u). The underbracketed terms are canceled out; hence,

ℱ2 (0) = −2 ∮ G(u)[(υ ⋅ ν) div υ − (υ ⋅ Dυ ν) + (w ⋅ ν)] dS 𝜕Ω

+ 2 ∮ g(u)(υ ⋅ ν)(υ ⋅ ∇u) dS 𝜕Ω

− 2 ∫ g(u)[2(υ ⋅ Dυ ∇u) + (υ ⋅ D2 uυ) − (w ⋅ ∇u)] dx Ω

+ 2 ∫ g ′ (u)[u′ 2 + 2(υ ⋅ ∇u)u′ ] dx. Ω

This together with Proposition 6.1 yields the following proposition. Proposition 6.2. We have 2

ℱ1 (0) + ℱ2 (0) = ∮[(υ ⋅ ν) div υ − υ ⋅ (Dυ ν) + (w ⋅ ν)](|∇u| − 2G(u)) dS 𝜕Ω

+ 2 ∮ g(u)(υ ⋅ ν)(υ ⋅ ∇u) dS + 4 ∮[𝜕ν u(υ ⋅ Dυ ∇u) − (υ ⋅ ν)(∇u ⋅ Dυ ∇u)] dS 𝜕Ω

𝜕Ω 2

− 2 ∮[(υ ⋅ ν)(υ ⋅ D u∇u) − 𝜕ν u(υ ⋅ D2 uυ) + 𝜕ν u(w ⋅ ∇u)] dS 𝜕Ω

− 2 ∫(Δu + g(u))[2(υ ⋅ Dυ ∇u) + (υ ⋅ D2 uυ) − (∇u ⋅ w)] dx Ω

󵄨 󵄨2 − 2 ∫[󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 − g ′ (u)u′ 2 ] dx Ω

+ 4 ∫(Δu′ + g ′ (u)u′ )(υ ⋅ ∇u) dx − 4 ∮(υ ⋅ ∇u)𝜕ν u′ dS. Ω

𝜕Ω

114 � 6 Domain variations for energies Step (iii). We continue with the transformation of ℱ3 (0). ̇̃ by (υ ⋅ ∇u) + u′ and obtain In ℱ3 (0) = α ∮𝜕Ω [u2 m̈ − 2u̇̃ 2 (0)] dS we replace u(0) 2

2

̈ ℱ3 (0) = α ∮[u (x)m(0) − 2(υ ⋅ ∇u) − 2u

′2

− 4(υ ⋅ ∇u)u′ ] dS.

(6.5.3)

𝜕Ω

6.5.2 Main results We now collect the previous transformations to express the second variation of the energy. Dirichlet energy In this case, ℰD̈ (0) = ℱ1 (0) + ℱ2 (0). In view of equation (6.2.1), the boundary conditions (6.2.2), and ∇u = 𝜕ν uν, Proposition 6.2 implies 2

ℰD̈ (0) = ∮[(υ ⋅ ν) div υ − υ ⋅ (Dυ ν) +(w ⋅ ν)](|∇u| − 2G(0)) dS ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ =:I1

𝜕Ω

+ 2g(0) ∮ 𝜕ν u(υ ⋅ ν)2 dS + 4 ∮(𝜕ν u)2 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (υ − (υ ⋅ ν)ν) ⋅Dυ ν) dS 𝜕Ω

=υτ

𝜕Ω

− 2 ∮[(υ ⋅ ν)(υ ⋅ D2 u∇u) − 𝜕ν u(υ ⋅ D2 uυ) +(𝜕ν u)2 (w ⋅ ν)] dS ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ =:I2

𝜕Ω ′

+ R(u ).

(6.5.4)

Here R(u′ ) stands for the two integrals in Proposition 6.2 which contain u′ . According to (2.2.14) we have div υ = div𝜕Ω υ − (ν ⋅ Dυ ν) on 𝜕Ω. Hence, I1 = (υ ⋅ ν) div𝜕Ω υ − (υτ ⋅ Dυ ν).

(6.5.5)

Decomposing υ = (υ ⋅ ν)ν + υτ and ∇u = 𝜕ν uν + ∇τ u = 𝜕ν ν, we obtain I2 = −𝜕ν u(υ ⋅ D2 uυτ ) = −𝜕ν u(υτ ⋅ D2 uυ). By Proposition 6.2 we have 󵄨 󵄨2 R(u′ ) = −2 ∫[󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 − g ′ (u)u′ 2 ] dx + 4 ∫(Δu′ + g ′ (u)u′ )(υ ⋅ ∇u) dx Ω

Ω ′

− 4 ∮(υ ⋅ ∇u)𝜕ν u dS. 𝜕Ω

Since Δu′ + g ′ (u)u′ = 0 in Ω and u′ = −(υ ⋅ ∇u) on 𝜕Ω, it follows that

(6.5.6)

6.5 The second domain variation

� 115

󵄨 󵄨2 R(u′ ) = −2 ∫[󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 − g ′ (u)u′ 2 ] dx + 4 ∮ u′ 𝜕ν u′ dS. Ω

𝜕Ω

For the last integral, the divergence theorem together with Δu′ + g ′ (u)u′ = 0 yields 󵄨 󵄨2 4 ∮ u′ 𝜕ν u′ dS = 4 ∫[󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 − g ′ (u)u′ 2 ] dx. Ω

𝜕Ω

Hence, R(u′ ) = 2Q0 (u′ ), where 󵄨 󵄨2 Q0 (u′ ) := ∫{󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 − g ′ (u)u′ 2 } dx.

(6.5.7)

Ω

Inserting these identities into (6.5.4) we are led to the final form of ℰD̈ (0). Theorem 6.2. Suppose that Δu +g(u) = 0 in Ω and u = 0 on 𝜕Ω. Then the second variation ℰD̈ (0) assumes the form τ

2

ℰD̈ (0) = ∮[(υ ⋅ ν) div𝜕Ω υ − (υ ⋅ Dυ ν) + (w ⋅ ν)](|∇u| − 2G(0)) dS 𝜕Ω

+ 2g(0) ∮ 𝜕ν u(υ ⋅ ν)2 dS + 4 ∮(𝜕ν u)2 (υτ ⋅ Dυ ν) dS 𝜕Ω

𝜕Ω τ

2

+ 2 ∮ 𝜕ν u(υ ⋅ D uυ) dS − 2 ∮(𝜕ν u)2 (w ⋅ ν) dS + 2Q0 (u′ ). 𝜕Ω

𝜕Ω

Discussion We now transform various terms in ℰD̈ (0) to get a better insight in its structure. As before we write υ = υτ + ρν. By (2.3.9), (υτ ⋅ Dυ ν) = (υτ ⋅ ∇τ ρ) − υτ ℒυτ . Hence, 4 ∮(𝜕ν u)2 (υτ ⋅ Dυ ν) dS = 4 ∮(𝜕ν u)2 [(υτ ⋅ ∇τ ρ) − υτ ℒυτ ] dS. 𝜕Ω

𝜕Ω

Let P be any fixed point on 𝜕Ω which we choose to be the origin of the Cartesian coordinate system such that e1 , . . . , en−1 are in the tangent space and en points in the direction of the outer normal. Since u = 0 on 𝜕Ω, we have 𝜕i 𝜕j u = 0 for i, j = 1, . . . , n − 1. Therefore,

116 � 6 Domain variations for energies 0 0 (υτ ⋅ D2 uυ) = (υτ1 , υτ2 , . . . , υτn−1 , 0) ( . .. 𝜕n 𝜕1 u τ = ρυi 𝜕i 𝜕ν u.

0 0 .. . 𝜕n 𝜕2

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

0 0 .. . 𝜕n 𝜕n−1 u

𝜕1 𝜕n u υτ1 𝜕2 𝜕n u υτ2 .. ) ( .. ) . . 𝜕n 𝜕n u ρ

Since this expression is independent of the special Cartesian coordinate system, it follows that 2 ∮ 𝜕ν u(υτ ⋅ D2 uυ) dS = ∮ ρ(υτ ⋅ ∇τ (𝜕ν u)2 ) dS. 𝜕Ω

𝜕Ω

These transformations lead to the following observation. Corollary 6.1. The second variation ℰD̈ (0) depends only on the boundary values of υ, w, 𝜕ν u and their tangential derivatives. It is independent of the extension of the perturbation inside Ω. Robin energy In this case, ℰr̈ (0) = ℱ1 (0) + ℱ2 (0) + ℱ3 (0). Adding up the contributions in Proposition 6.1 and (6.5.3), we get 2

ℰR̈ (0) = ∮[(υ ⋅ ν) div υ − (υ ⋅ Dυ ν) +(w ⋅ ν)](|∇u| − 2G(u)) dS ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ I1

𝜕Ω

+ 4 ∮(𝜕⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ν u(υ ⋅ Dυ ∇u) − (υ ⋅ ν)(∇u ⋅ Dυ ∇u)) dS I2

𝜕Ω

2 2 + 2 ∮(𝜕⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ν u(υ ⋅ D uυ) − (υ ⋅ ν)(υ ⋅ D u∇u)) dS I3

𝜕Ω

+ 2 ∮ g(u)(υ ⋅ ν)(υ ⋅ ∇u) dS − 4 ∮(υ ⋅ ∇u)(𝜕ν u′ + αu′ ) dS 𝜕Ω

𝜕Ω 2

̈ − 2α ∮(υ ⋅ ∇u) dS − 2 ∮(w ⋅ ∇u)𝜕ν u dS + α ∮ u2 (x)m(0) dS 𝜕Ω

𝜕Ω

𝜕Ω

− 4 ∫(Δu + g(u))(υ ⋅ Dυ ∇u) dx − 2 ∫(Δu + g(u))(υ ⋅ D2 uυ) dx Ω

Ω

+ 2 ∫(Δu + g(u))(w ⋅ ∇u) dx + 4 ∫(Δu′ + g ′ (u)u′ )(υ ⋅ ∇u) dx − 2Qg (u′ ), Ω

(6.5.8)

Ω

where 󵄨 󵄨2 Qg (u′ ) := ∫󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 dx − ∫ g ′ (u)u′ 2 dx + α ∮ u′ 2 dS Ω

Ω ′

′

𝜕Ω ′

= ∮ u (𝜕ν u + αu ) dS 𝜕Ω

(6.5.9)

6.5 The second domain variation

� 117

is a quadratic form in u′ . Formula (6.5.8) can further be simplified. The last four integrals vanish because Δu+ g(u) = 0 and Δu′ + g ′ (u)u′ = 0 in Ω. We apply (6.5.5), and we obtain I1 = (υ ⋅ ν) div𝜕Ω υ − (υτ ⋅ Dυ ν). We decompose υ and ∇u in their normal and tangential parts: I2 = 𝜕ν u(υτ ⋅ Dυ ∇u) + 𝜕ν u(υ ⋅ ν)(ν ⋅ Dυ ∇u)

− (υ ⋅ ν)(∇τ u ⋅ Dυ ∇u) − (υ ⋅ ν)𝜕ν u(ν ⋅ Dυ ∇u)

= 𝜕ν u(υτ ⋅ Dυ ∇u) − (υ ⋅ ν)(∇τ u ⋅ Dυ ∇u). The Robin boundary condition then implies

I2 = −αu(υτ ⋅ Dυ ∇u) − (υ ⋅ ν)(∇τ u ⋅ Dυ ∇u). Similarly, we obtain I3 = 𝜕ν u(υ ⋅ D2 uυ) − (υ ⋅ ν)(υ ⋅ D2 u∇u)

= 𝜕ν u(υ ⋅ D2 uυτ ) − (υ ⋅ ν)(υ ⋅ D2 u∇τ u)

= −αu(υ ⋅ D2 uυτ ) − (υ ⋅ ν)(υ ⋅ D2 u∇τ u).

Inserting all these expressions into (6.5.4), we obtain the final result. Theorem 6.3. Suppose that Δu + g(u) = 0 in Ω and 𝜕ν u + αu = 0 on 𝜕Ω. Then the second variation ℰR̈ (0) assumes the form τ

2

ℰR̈ (0) = ∮[(υ ⋅ ν) div𝜕Ω υ − (υ ⋅ Dυ ν) + (w ⋅ ν)](|∇u| − 2G(u)) dS 𝜕Ω

− 4 ∮{αu(υτ ⋅ Dυ ∇u) + (υ ⋅ ν)(∇τ u ⋅ Dυ ∇u)} dS 𝜕Ω

− 2 ∮{αu(υ ⋅ D2 uυτ ) + (υ ⋅ ν)(υ ⋅ D2 u∇τ u)} dS 𝜕Ω

+ 2 ∮ g(u)(υ ⋅ ν)(υ ⋅ ∇u) dS − 4 ∮(υ ⋅ ∇u)(𝜕ν u′ + αu′ ) dS 𝜕Ω

𝜕Ω 2

− 2α ∮(υ ⋅ ∇u) dS + 2α ∮(w ⋅ ∇u)u dS 𝜕Ω

𝜕Ω 2

̈ + α ∮ u m(0) dS − 2Qg (u′ ), 𝜕Ω

where Qg (u′ ) is given by (6.5.9).

118 � 6 Domain variations for energies Discussion We are mainly interested in the terms on 𝜕Ω which contain derivatives in the normal direction. Let {ei }∞ 1 be the same coordinate system as in the discussion of Theorem 6.3, i. e., en points in the direction of the outer normal ν. Then on 𝜕Ω we have υ = υτi ei + ρen and ∇u = ∇τ u − αen , i = 1, . . . , n − 1. Consequently, (υ ⋅ ∇u) = (υτ ⋅ ∇τ u) − αρ. Set Dυτ = (𝜕i υτj ). A straightforward computation yields −4 ∮{αu(υτ ⋅ Dυ ∇u) + (υ ⋅ ν)(∇τ u ⋅ Dυ ∇u)} dS 𝜕Ω

= −4 ∮ (αuυτ + ρ∇τ u) ⋅ Dυτ ∇τ u dS + 4 ∮(α2 u2 + αuρ)(∇τ u ⋅ ∇τ ρ) dS. ±Ω

𝜕Ω

On the boundary the matrix D2 u is of the form 𝜕1 𝜕1 u 𝜕2 𝜕1 u D2 u = ( . .. −α𝜕1 u

𝜕1 𝜕2 u . . . 𝜕2 𝜕2 u . . . .. .... −α𝜕2 u . . .

𝜕1 𝜕n−1 u 𝜕2 𝜕n−1 u .. . −α𝜕n−1 u

−α𝜕1 u −α𝜕2 u .. ) . . 2 αu

Thus, D2 u does not contain derivatives in the normal direction. By Lemma 6.1, 𝜕ν u′ + αu′ depends on u and υ on 𝜕Ω only. The same is true for Q(u′ ) = ∮𝜕Ω u′ (𝜕ν u′ + αu′ ) dS. Hence, we have proved the following property of the second variation. Corollary 6.2. The second variation ℰR̈ (0) depends only on the boundary values of υ, w, u and their tangential derivatives. It is independent of the extension of the perturbation Φt into Ω.

6.5.3 Notes The computation of the second variation of the energy was first carried out in [13] and [14]. In [75] the second variation of ℰ is presented in an abstract form.

7 Discussion of the main results In the first section some simple variations are presented which lead to integral identities for boundary value problems. The rest of this chapter is devoted to the first and second domain variation of energies under Hadamard and tangential perturbations. Special emphasis is given to volume preserving perturbations. It turns out that tangential perturbations have an effect on the second domain variation only.

7.1 Translations and rotations Translations The perturbation Φt (x) = x + ta where a is a constant vector in ℝn corresponds to a displacement of the domain Ω in the direction a. In this case obviously the volume and the surface area remain unchanged. Clearly, the energies ℰD (t) = const. and ℰR (t) = const. Therefore, all shape derivatives vanish. By (6.4.2), 2

ℰḊ (0) = − ∮(a ⋅ ν)|∇u| dS = 0. 𝜕Ω

This identity can also be checked directly. Indeed, − ∮(a ⋅ ν)|∇u|2 dS = − ∫ ai 𝜕i (𝜕j u𝜕j u) dx = −2 ∫ ai 𝜕i 𝜕j u𝜕j u dx Ω

𝜕Ω

Ω

= −2 ∫ ai u𝜕i Δu dx = 2 ∫ ai u𝜕i g(u) dx = 2 ∫ ai 𝜕i (g(u)u − G(u)) dx Ω

Ω

Ω

= −2G(0) ∮(a ⋅ ν) dS = 0. 𝜕Ω

Similarly, the invariance of ℰR (t) with respect to translations and (6.4.4) imply that for any domain Ω, 2

2 2

2

ℰṘ (0) = ∮(a ⋅ ν)[|∇u| − 2G(u) − 2α u + α(n − 1)Hu ] dS = 0. 𝜕Ω

Obviously an additional second order term affects the second domain variation. Rotations Let Φt be a rotation of the (x1 , x2 )-plane considered in Remark 2.3. Expansion up to the second order yields t2 2 Φt (x) = x + t (−x ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 2 , x1 , 0, . . . , 0) + (−x1 , −x2 , 0, . . . , 0) + o(t ). 2 υ https://doi.org/10.1515/9783111025438-007

120 � 7 Discussion of the main results The first variation of the Dirichlet energy is then by (6.4.2) 2

ℰḊ (0) = ∮(−x2 ν1 + x1 ν2 )|∇u| dS = 0. 𝜕Ω

Similarly, by (6.4.4) 2

2 2

2

ℰṘ (0) = ∮(−x2 ν1 + x1 ν2 )[|∇u| − 2G(u) − 2α u + α(n − 1)Hu ] dS = 0. 𝜕Ω

This identity holds for any Ω and for any solution of the Robin boundary value problem. Pohozaev’s identity In 1965 Pohozaev [97] published an identity which he used to prove nonexistence of positive solutions of some Dirichlet boundary value problems in star-shaped domains. This identity is related to the first variation ℰḊ (0) of the Dirichlet energy. From (6.3.2) and Lemma 4.1 it follows that 2

ℰḊ (0) = ∫ div υ|∇u| dx − 2 ∫(∇u ⋅ Dυ ∇u) dx − 2 ∫ G(u) div υ dx, Ω

Ω

Ω

and from (6.4.2) and G(0) = 0 we have 2

ℰḊ (0) = − ∮(υ ⋅ ν)|∇u| dS. 𝜕Ω

If we equate these two terms, we obtain ∫ div υ|∇u|2 dx − 2 ∫(∇u ⋅ Dυ ∇u) dx − 2 ∫ G(u) div υ dx = − ∮(υ ⋅ ν)|∇u|2 dS. Ω

Ω

Ω

𝜕Ω

Choosing υ = x and keeping in mind that ∫Ω |∇u|2 dx = ∫Ω f (u)u dx, we are led to Pohozaev’s identity ∮(x ⋅ ν)|∇u|2 dS = (2 − n) ∫ f (u)u dx + 2n ∫ G(u) dx. 𝜕Ω

Ω

(7.1.1)

Ω

In star-shaped domains the left-hand side is nonnegative, whereas for supercritical nonn+2 linearities such as f (u) = up with p ≥ n−2 it is negative. This is impossible and therefore no positive solution exists.

7.2 Hadamard perturbations

� 121

7.2 Hadamard perturbations This section focuses on Hadamard perturbations υ = (υ ⋅ ν)ν = ρν. The first variation is given in Section 6.4 and depends only on ρ. For the computation of the second domain variations we use Theorems 6.2 and 6.3. We start with the Dirichlet energy (see Section 6.2.1). From Theorem 6.2 we obtain 2

ℰD̈ (0) = ∮[(υ ⋅ ν) div𝜕Ω υ + (w ⋅ ν)](|∇u| − 2G(0)) dS 𝜕Ω

+ 2g(0) ∮ 𝜕ν uρ2 dS − 2 ∮(w ⋅ ν)(𝜕ν u)2 dS + 2Q0 (u′ ). 𝜕Ω

𝜕Ω

Since by (2.2.16) (υ ⋅ ν) div𝜕Ω υ = ρ div𝜕Ω (ρν) = (n − 1)ρH, it follows that 2

2

ℰD̈ (0) = ∮[(n − 1)ρ H + (w ⋅ ν)](|∇u| − 2G(0)) dS 𝜕Ω

+ 2g(0) ∮ 𝜕ν uρ2 dS − 2 ∮(w ⋅ ν)(𝜕ν u)2 dS + 2Q0 (u′ ). 𝜕Ω

(7.2.1)

𝜕Ω

For the second variation of the Robin energy we use the equations for u and u′ . Theorem 6.3 then implies 2

ℰR̈ (0) = ∮[(υ ⋅ ν) div𝜕Ω υ + (w ⋅ ν)](|∇u| − 2G(u)) dS 𝜕Ω

− 4 ∮(υ ⋅ ν)(∇τ u ⋅ Dυ ∇u) dS − 2 ∮(υ ⋅ ν)(υ ⋅ D2 u∇τ u) dS 𝜕Ω

𝜕Ω

+ 2 ∮ g(u)(υ ⋅ ν)(υ ⋅ ∇u) dS − 4 ∮(υ ⋅ ∇u)(𝜕ν u′ + αu′ ) dS 𝜕Ω

𝜕Ω 2

− 2α ∮(υ ⋅ ∇u) dS + 2α ∮(w ⋅ ∇u)u dS 𝜕Ω

𝜕Ω

̈ + α ∮ u2 m(0) dS − 2Qg (u′ ), 𝜕Ω

where Qg (u′ ) is defined in (6.5.9). We consider the various integrals separately. The first integral is the same as (7.2.1). In the second integral we set (Dυ )ij = 𝜕i υj = 𝜕i (ρνj ) = 𝜕i ρνj + ρ𝜕i νj

and

∇u = (𝜕ν u)ν + ∇τ u.

(7.2.2)

122 � 7 Discussion of the main results Herewith we calculate ∇τ u ⋅ Dυ ∇u. Keeping in mind that by (2.4.3), ∇iτ 𝜕i νj νj = 0, we get −4 ∮(υ ⋅ ν)(∇τ u ⋅ Dυ ∇u) dS 𝜕Ω

= −4 ∮ ρ𝜕ν u(∇τ u ⋅ ∇τ ρ) dS + 4 ∮ ρ2 (∇τ u ⋅ Dν ∇τ u) dS. 𝜕Ω

𝜕Ω

Replacing υ by ρν we obtain for the third integral −2 ∮(υ ⋅ ν)(υ ⋅ D2 u∇τ u) dS = −2 ∮ ρ2 (∇τ u ⋅ D2 uν) dS. 𝜕Ω

𝜕Ω

In view of (6.2.12) the shape derivative u′ solves Δu′ + g ′ (u)u′ = 0 in Ω. By (6.2.14) it satisfies for Hadamard perturbations the boundary condition 𝜕ν u′ + αu′ = −(υ ⋅ D2 uν) + (∇τ u ⋅ Dυ ν) − α(υ ⋅ ∇u) = −ρ𝜕ν2 u + ∇iτ u𝜕i υj ∇jτ ρ − α2 ρ𝜕ν u

on 𝜕Ω.

For the second term in the last sum we apply (7.2.2). This implies ∇iτ u𝜕i υj ∇jτ ρ = ∇iτ u(𝜕i ρνj + ρ𝜕i νj )∇jτ ρ = ρ∇iτ u𝜕i νj ∇jτ ρ = ρ(∇τ u ⋅ Dν ∇τ ρ).

Hence, −4 ∮(υ ⋅ ∇u)(𝜕ν u′ + αu′ ) dS 𝜕Ω

= 4 ∮ ρ2 𝜕ν2 u𝜕ν u dS − 4 ∮ ρ2 𝜕ν u(∇τ u ⋅ Dν ∇τ ρ) dS − 4α2 ∮ ρ2 𝜕ν uu dS. 𝜕Ω

𝜕Ω

𝜕Ω

In conclusion, we obtain the following theorem. Theorem 7.1. Assume Δu + g(u) = 0 in Ω. Let Φt be a Hadamard perturbation and let ρ = (υ ⋅ ν). 1. If u = 0 on 𝜕Ω, then 2

2

ℰD̈ (0) = ∮[(n − 1)ρ H + (w ⋅ ν)](|∇u| − 2G(0)) dS 𝜕Ω

+ 2g(0) ∮ 𝜕ν uρ2 dS − 2 ∮(w ⋅ ν)(𝜕ν u)2 dS + 2Q0 (u′ ), 𝜕Ω

where Q0 (u′ ) = ∫Ω [|∇u′ |2 − 2g ′ (u)u′ 2 ] dx.

𝜕Ω

7.3 Tangential perturbations υ = υτ

2.

� 123

If 𝜕ν u + αu = 0 on 𝜕Ω, then 2

2

ℰR̈ (0) = ∮[(n − 1)ρ H + w ⋅ ν](|∇u| − 2G(u)) dS 𝜕Ω

+ 4α ∮ ρu(∇τ u ⋅ ∇τ ρ) dS + 4 ∮ ρ2 (∇τ u ⋅ Dν ∇τ u) dS 𝜕Ω

𝜕Ω 2

τ

2

− 2 ∮ ρ (∇ u ⋅ D uν) dS − 2α ∮ g(u)ρ2 u dS 𝜕Ω

+

𝜕Ω

4 ∮ ρ2 𝜕ν2 u𝜕ν u dS

2

+ 4α ∮ ρ u(∇τ u ⋅ Dν ∇τ ρ) dS

𝜕Ω

𝜕Ω

+ 2α3 ∮ ρ2 u2 dS + 2α ∮(w ⋅ ∇u)u dS 𝜕Ω

𝜕Ω

2

̈ + α ∮ u m(0) dS − 2Qg (u′ ), 𝜕Ω

̈ where Qg (u′ ) = ∫Ω |∇u′ |2 − g ′ (u)u′ 2 dx + α ∮𝜕Ω u′ 2 dS (see (6.5.9)) and m(0) is given in (2.3.30). These formulas simplify if we assume volume or area preserving perturbations, or if we consider special geometries like the ball.

7.3 Tangential perturbations υ = υτ 2

In the case of purely tangential perturbations Φt = x +tυτ + t2 w, the first variations of the volume and the surface area vanish. This follows from (2.3.2) and (2.3.20). The tangential perturbations have an effect only in the second variations (see (2.3.13) and (2.3.36)). A similar phenomenon holds for the energies. By (6.4.2) and (6.4.4) the first variation of the Dirichlet and the Robin energy vanishes for all tangential perturbations. Theorem 6.2 and the subsequent discussion imply that τ

τ

2

2

τ

τ

ℰD̈ (0) = ∮[υ ℒυ + (w ⋅ ν)](|∇u| − 2G(0)) dS − 2 ∮ |∇u| (2υ ℒυ + (w ⋅ ν)) dS 𝜕Ω

𝜕Ω ′

+ 2Q0 (u ). By (6.2.13) the shape derivative satisfies u′ = −(υ ⋅ ν)𝜕ν u = 0 on 𝜕Ω. Consequently, (6.5.7) implies that u′ = 0 in Ω. Thus, Q0 (u′ ) = 0 and

124 � 7 Discussion of the main results τ

τ

2

ℰD̈ (0) = ∮[υ ℒυ + (w ⋅ ν)][(𝜕ν u) − 2G(0)] dS 𝜕Ω

− 2 ∮(𝜕ν u)2 [2υτ ℒυτ + (w ⋅ ν)] dS.

(7.3.1)

𝜕Ω

In convex domains ℒ is positive. This leads to the following observation. Corollary 7.1. Let Φt = x + tυτ be a first order tangential perturbation, let G(0) = 0, and let Ω be convex. Then and ℰD̈ (0) = − ∮(𝜕ν u)2 υτ ℒυτ dS ≤ 0.

ℰḊ (0) = 0

𝜕Ω

In the case of Robin boundary conditions, Lemma 6.1 implies that 𝜕ν u′ + αu′ = 0 on 𝜕Ω, and thus by (6.5.9) Qg (u′ ) = 0. By Theorem 2.1 and Proposition 2.1 we can replace the tangential perturbation x + tυτ + 2 t2 w locally by the Hadamard perturbation x + t2 w,̂ where 2 ŵ = υτ ℒυτ ν + w. If we introduce ŵ into Theorem 7.1, we get τ

τ

2

2 2

2

̈ dS. ℰR̈ (0) = ∮(υ ℒυ + w ⋅ ν)(|∇u| − 2G(u) − 2α u ) + αu m(0) 𝜕Ω

̈ By (2.3.35) it follows that m(0) = div𝜕Ω w.̂ Gauss’ theorem (2.2.18) then implies that ̈ α ∮ u2 m(0) dS = −α ∮(ŵ ⋅ ∇τ u2 ) dS + α(n − 1) ∮ u2 (ŵ ⋅ ν)H dS. 𝜕Ω

𝜕Ω

𝜕Ω

Since ŵ ⋅ ∇τ u2 = w ⋅ ∇τ u2 we obtain τ

τ

2

2 2

ℰR̈ (0) = ∮(υ ℒυ + w ⋅ ν)(|∇u| − 2G(u) − 2α u ) dS 𝜕Ω

− α ∮(w ⋅ ∇τ u2 ) dS + α(n − 1) ∮ u2 (ŵ ⋅ ν)H dS. 𝜕Ω

(7.3.2)

𝜕Ω

In conclusion, the tangential perturbations have an effect on the second variation of the energy.

7.4 Volume preserving perturbations

� 125

7.4 Volume preserving perturbations The effect of volume preserving perturbations in the sense of Definition 2.2 for the first variation of the energy has been discussed in Section 6.4.1. There only 𝒱̇ (0) = 0 played a role. In this section we therefore restrict ourselves to the second variation. In addition to 𝒱̇ (0) = 0, also 𝒱̈ (0) = 0 will be assumed.

7.4.1 Dirichlet energy Assume that Ω is a critical domain of the Dirichlet energy, i. e., ℰḊ (0) = 0. By Theorem 6.1, 𝜕ν u = const. on 𝜕Ω. Under the assumption τ

𝒱̈ (0) = ∮[(υ ⋅ ν) div𝜕Ω υ − (υ ⋅ Dυ ν) + (w ⋅ ν)] dS = 0, 𝜕Ω

if we take into account that g(u) = g(0) and G(u) = G(0) on 𝜕Ω, Theorem 6.2 reads as 2

2

τ

ℰD̈ (0) = 2g(0) ∮ 𝜕ν u(υ ⋅ ν) dS + 4 ∮(𝜕ν u) (υ ⋅ Dυ ν) dS 𝜕Ω

𝜕Ω τ

2

+ 2 ∮ 𝜕ν u(υ ⋅ D uυ) dS − 2 ∮(𝜕ν u)2 (w ⋅ ν) dS + 2Q0 (u′ ). 𝜕Ω

𝜕Ω

This expression can be simplified. In fact, 𝜕ν u = const. and 𝒱̈ (0) = 0 can be read as an integral equation for ∮𝜕Ω (w ⋅ ν) dS. We obtain 4 ∮(𝜕ν u)2 (υτ ⋅ Dυ ν) dS − 2 ∮(𝜕ν u)2 (w ⋅ ν) dS 𝜕Ω

𝜕Ω 2

τ

= 4(𝜕ν u) ∮(υ ⋅ Dυ ν) dS − 2(𝜕ν u)2 ∮(w ⋅ ν) dS 𝜕Ω 2

𝜕Ω τ

= 2(𝜕ν u) ∮[(υ ⋅ Dυ ν) + ρ div𝜕Ω υ] dS. 𝜕Ω

Moreover, on 𝜕Ω we have (υτ ⋅ D2 uυ) = 0. This is easily seen by choosing an orthonormal frame such that the origin lies in an arbitrary, but fixed boundary point. The en -axis points in the direction of the outer normal and ℝn−1 is the tangent space. At the origin we have 𝜕i 𝜕j u = 0 for i, j = 1, 2, . . . , n −1. Thus,

126 � 7 Discussion of the main results 0 0 (υτ ⋅ D2 uυ) = (υτ1 , υτ2 , . . . , υτn−1 , 0) ( . .. 0

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

υτ1 .. ) ( . ) = ρ(υτ ⋅ ∇τ 𝜕ν u). 𝜕n−1 𝜕n u υτn−1 𝜕n 𝜕n u ρ 𝜕1 𝜕n u 𝜕2 𝜕n u

The claim now follows from the boundary condition 𝜕ν u = const. Gauss’ theorem (2.2.18) then leads to the following proposition. Proposition 7.1. Let u be a solution of Δu + g(u) = 0 in Ω and u = 0 on 𝜕Ω and let u′ be the ̇ corresponding shape derivative solving (6.2.12)–(6.2.13). We assume ℰ (0) = 0. Then the ̈ for volume preserving perturbations can be expressed in the form second variation ℰ (0) g(0) 󵄨 ′ 󵄨2 ′ ′2 ℰD̈ (0) = 2 ∫[󵄨󵄨󵄨∇u 󵄨󵄨󵄨 − g (u)u ] dx + 2 ∮ u′ 2 dS + 2(n − 1) ∮ u′ 2 H dS. 𝜕ν u

Ω

𝜕Ω

𝜕Ω

In particular, 𝜕ν u = const. ≠ 0 on 𝜕Ω. If υ = υτ is tangential to 𝜕Ω, then u′ = 0 on 𝜕Ω and thus 󵄨 󵄨2 ∫[󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 − g ′ (u)u′ 2 ] dx = ∮ u′ 𝜕ν u′ dS = 0.

Ω

𝜕Ω

Consequently, we have the following corollary. Corollary 7.2. The tangential perturbations υ = υτ are in the kernel of ℰD̈ (0), i. e., ℰḊ (0) = ℰD̈ (0) = 0. 7.4.2 Robin energy For the Robin boundary condition a domain Ω is critical for volume preserving perturbations if ℰṘ = 0. This results in an overdetermined boundary value problem given in Theorem 6.1, where the additional boundary |∇u|2 − 2G(u) − 2α2 u2 + (n − 1)u2 H = const.

on 𝜕Ω

was derived. We denote this constant by C. The second domain derivative ℰR̈ (0) was computed in Theorem 7.1. With the additional boundary condition the first integral is I := ∮[(n − 1)ρ2 H + w ⋅ ν](|∇u|2 − 2G(u)) dS 𝜕Ω

= ∮[(n − 1)ρ2 H + w ⋅ ν](C + 2α2 u2 − (n − 1)u2 H) dS. 𝜕Ω

From (2.3.12) we deduce

7.4 Volume preserving perturbations

�

127

I = ∮[(n − 1)ρ2 H + w ⋅ ν](2α2 u2 − (n − 1)u2 H) dS. 𝜕Ω

The other integrals in ℰR̈ (0) in Theorem 7.1 remain unchanged. Next we restrict ourselves to volume preserving tangential perturbations. We assume again that Ω is a critical domain for ℰR . Thus, the overdetermined boundary con̈ dition holds in this case as well. Moreover, by (2.3.35) we have m(0) = div𝜕Ω w,̂ where ŵ = (υτ ℒυτ )ν + w. Then (7.3.2) together with (2.2.18) implies that 2

2

ℰR̈ (0) = C ∮(ŵ ⋅ ν) dS − (n − 1)α ∮ u H(ŵ ⋅ ν) dS + α ∮ u div𝜕Ω ŵ dS. 𝜕Ω

𝜕Ω

𝜕Ω

Moreover, we have (ŵ ⋅ ∇τ u2 ) = (w ⋅ ∇τ u2 ) and α ∮ u2 div𝜕Ω ŵ dS = α ∮[−(ŵ ⋅ ∇τ u2 ) + (n − 1)u2 (ŵ ⋅ ν)H] dS. 𝜕Ω

𝜕Ω

With these observations we get τ 2

ℰR̈ (0) = C ∮(ŵ ⋅ ν) dS − α ∮(w ⋅ ∇ u ) dS. 𝜕Ω

(7.4.1)

𝜕Ω

From (2.3.13) we get the additional condition 0 = ∮ υτ ℒυτ + (w ⋅ ν) dS = ∮ ŵ ⋅ ν dS. 𝜕Ω

𝜕Ω

Hence, τ 2

ℰR̈ (0) = −α ∮(w ⋅ ∇ u ) dS.

(7.4.2)

𝜕Ω

In summary, we have the following proposition. Proposition 7.2. Let Ω be a critical domain for the energy ℰR w. r. t. volume preserving 2 perturbations. Then for tangential volume preserving perturbations Φt = x + tυτ + t2 w, under the additional condition wτ = 0 we have ℰR̈ (0) = 0.

In accordance with Definition 2.3 we now have the following corollary. Corollary 7.3. The second order volume preserving tangential perturbations with wτ = 0 are in the kernel of ℰR̈ (0).

128 � 7 Discussion of the main results

7.5 Neumann energy Consider the Neumann problem Δu + g(u) = 0

in Ω,

𝜕ν u = 0

in 𝜕Ω.

(7.5.1)

For the existence of a solution the compatibility condition ∮ g(u) dS = 0 𝜕Ω

has to be satisfied. The Neumann energy is given by 2

ℰN (Ω, u) := ∫ |∇u| dy − 2 ∫ G(u) dx. Ω

(7.5.2)

Ω

From (6.4.4) we get the first variation 2

ℰṄ (0) = ∮(υ ⋅ ν){|∇u| − 2G(u)} dS. 𝜕Ω

According to Theorem 6.1, ℰṄ (0) = 0 for all volume preserving perturbations if and only if u satisfies in addition to (7.5.1) the boundary condition |∇u|2 − 2G(u) = C = const.

on 𝜕Ω.

(7.5.3)

For the second variation we use Theorem 6.3 with α = 0. Then in consideration of (7.5.3) this yields τ

2

τ

ℰN̈ (0) = C 𝒱̈ (0) − 4 ∮ ρ(∇ u ⋅ Dυ ∇u) dS − 2 ∮ ρ(υ ⋅ D u∇ u) dS 𝜕Ω

𝜕Ω

+ 2 ∮ ρ(υ ⋅ ∇u)g(u) dS − 4 ∮(υ ⋅ ∇u)𝜕ν u′ dS − 2QgN (u′ ), 𝜕Ω

𝜕Ω

where 󵄨 󵄨2 QgN (u′ ) := ∫󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 dx − ∫ g ′ (u)u′ 2 dx. Ω

Ω

The shape derivative satisfies (6.2.12) and (6.2.14) with α = 0. Hence, Δu′ + g ′ (u)u′ = 0 By Lemma 6.1

in Ω,

𝜕ν u′ = −(υ ⋅ D2 uν) + (∇τ u ⋅ Dυ ν).

(7.5.4)

7.5 Neumann energy �

129

𝜕ν u′ = ρ[g(u) + Δ∗ u] + (∇τ u ⋅ ∇ρ). The following Lemma will allow us to simplify the expression for ℰN̈ (0). Lemma 7.1. We have −4 ∮(υ ⋅ ∇τ u)𝜕ν u′ dS = −4 ∮(υ ⋅ ∇τ u)g(u)ρ dS 𝜕Ω

𝜕Ω

+ 4 ∮ ρ(υ ⋅ ∇τ u)(∇τ u ⋅ ∇τ υ) dS + 4 ∮ ρ(∇τ u ⋅ Dυ ∇τ u) dS 𝜕Ω

𝜕Ω τ

2

+ 4 ∮ ρ(∇ u ⋅ D uυ) dS − 4(n − 1) ∮ ρ(∇u ⋅ υ)H dS. 𝜕Ω

𝜕Ω

Proof. In view of the boundary condition for u′ mentioned above we have −4 ∮(υ ⋅ ∇τ u)𝜕ν u′ dS = −4 ∮ g(u)ρ(υ ⋅ ∇τ u) dS 𝜕Ω

𝜕Ω

− 4 ∮(υ ⋅ ∇τ u)ρΔ∗ u dS − 4 ∮(υ ⋅ ∇τ u)(∇τ u ⋅ ∇ρ) dS. 𝜕Ω

𝜕Ω

Since Δ∗ u = div𝜕Ω ∇τ u we can apply the Gauss theorem (2.2.18) and get −4 ∮(υ ⋅ ∇τ u)ρΔ∗ u dS = 4 ∮(υ ⋅ ∇u)(∇τ u ⋅ ∇τ ρ) dS 𝜕Ω

𝜕Ω

+ 4 ∮ ρ∇τ u ⋅ ∇τ (υ ⋅ ∇τ u) dS − 4(n − 1) ∮ ρ(υ ⋅ ∇u)H dS. 𝜕Ω

𝜕Ω

Moreover, 4 ∮ ρ∇τ u ⋅ ∇τ (υ ⋅ ∇τ u) dS = 4 ∮ ρ(∇τ u ⋅ Dυ ∇τ u) dS + 4 ∮ ρ(∇τ u ⋅ D2 uυ) dS. 𝜕Ω

𝜕Ω

𝜕Ω

The assertion now follows. This lemma then implies 2

τ

ℰN̈ (0) = C 𝒱̈ (0) + 2 ∮ ρυD u∇ u dS − 2 ∮ ρ(υ ⋅ ∇u)g(u) dS 𝜕Ω

𝜕Ω τ

+ 4 ∮ ρ(υ ⋅ ∇ u)(∇ u ⋅ ∇ υ) dS − 4(n − 1) ∮ ρ(∇τ u ⋅ υ)H dS − 2QgN (u′ ). (7.5.5) 𝜕Ω

τ

τ

𝜕Ω

This formula has been derived under the assumption that 𝒱̇ (0) = 0 and therefore |∇u|2 − 2G(u) = const. on 𝜕Ω. If we impose in addition that 𝒱̈ (0) = 0, then

130 � 7 Discussion of the main results 2

τ

ℰN̈ (0) = 2 ∮ ρ(υ ⋅ D u∇ u) dS − 2 ∮ ρ(υ ⋅ ∇u)g(u) dS 𝜕Ω

𝜕Ω

+ 4 ∮ ρ(υ ⋅ ∇τ u)(∇τ u ⋅ ∇τ υ) dS − 4(n − 1) ∮ ρ(∇τ u ⋅ υ)H dS −

𝜕Ω

𝜕Ω 2QgN (u′ ).

For Hadamard perturbations this leads to 2

τ

N

′

ℰN̈ (0) = 2 ∮ ρ(υ ⋅ D u∇ u) dS − 2Qg (u ). 𝜕Ω

The first term can be written in a more accessible form. The Hadamard perturbation υ = ρν implies ρ(υ ⋅ D2 u∇uτ ) = ρ2 (∇uτ ⋅ D2 uν). Furthermore, by 𝜕ν u = 0 (∇τ u ⋅ D2 uν) = (∇τ u ⋅ ∇𝜕ν u) − (∇τ u ⋅ Dν ∇τ u) = −(∇τ u ⋅ Dν ∇τ u). Consequently, 2

τ

τ

N

′

ℰN̈ (0) = −2 ∮ ρ (∇ u ⋅ Dν ∇ u) dS − 2Qg (u ). 𝜕Ω

In summary, we obtain the following proposition. Proposition 7.3. Let u be a solution of Δu + g(u) = 0 in Ω and 𝜕ν u = 0 on 𝜕Ω and let ℰN (t) be the corresponding Neumann energy (see (7.5.2)). Assume that Φt is a Hadamard, volume preserving perturbation. Under the assumption that ℰṄ (0) the second variation is 2

τ

τ

N

′

ℰN̈ (0) = −2 ∮ ρ (∇ u ⋅ Dν ∇ u) dS − 2Qg (u ), 𝜕Ω

where QgN (u′ ) is given by (7.5.4). Note that by (2.3.8) the term Dν contains the second fundamental form. 7.5.1 Notes The relation between the first domain variation and Pohozaev’s identity was discussed in [123]. This paper also contains a proof that the overdetermined torsion problem is solvable in balls only. A related interpretation of Pohozaev’s identity is found in [100]. This text studies the influence of transformation groups to critical points for abstract functionals. Pohozaev’s identity follows there as a special case.

8 General strategy and applications In this chapter we present a strategy to determine the sign of the second variation. It is based on a generalized Steklov eigenvalue problem associated to the linearization of the elliptic equation. Our goal is to express the shape derivative as a Fourier series of the corresponding Steklov eigenfunctions. The lower eigenvalues turn out to be essential for the sign of the second variation of the energy. We start with a general discussion of this eigenvalue problem. The strategy of Fourier series expansions is then applied to domain functionals defined on nearly spherical domains. We obtain a lower bound for the second variation of the Robin energy for α > 0. In the special case of the torsion energy this lower bound establishes the positivity of the second variation. Finally, we apply the same method to the Dirichlet energy and obtain a lower bound as well.

8.1 A generalized Steklov eigenvalue problem It was shown in Section 6.2.2 that the shape derivatives of the boundary value problem Δu + g(u) = 0 in Ω with Robin boundary conditions satisfy Δu′ + g ′ (u)u′ = 0 in Ω and 𝜕ν u′ + αu′ = kR (u, x) on 𝜕Ω. The solutions of such problems will be computed by means of the eigenfunctions of the Steklov type eigenvalue problem Δϕ + σ(x)ϕ = 0 in Ω,

𝜕ν ϕ + αϕ = μϕ on 𝜕Ω,

where σ(x) = g ′ (u).

(8.1.1)

In order to write it in a weak form we introduce the bilinear forms defined in the Hilbert spaces W 1,2 (Ω) and L2 (𝜕Ω), namely aσ (ϕ, υ) := ∫(∇ϕ ⋅ ∇υ) dx − ∫ σ(x)ϕυ dx Ω

Ω

and b(ϕ, υ) := ∮ ϕυ dS. 𝜕Ω

Since 𝜕Ω is Lipschitz, the embedding W 1,2 (Ω) 󳨅→ L2 (𝜕Ω) is a compact linear operator. The space W01,2 (Ω) is the kernel of the trace operator. We write ℋ for the quotient W 1,2 (Ω)/W01,2 (Ω). The weak formulation of (8.1.1) is: find ϕ ∈ ℋ such that aσ (ϕ, υ) = (μ − α) b(ϕ, υ)

for all υ ∈ ℋ.

(8.1.2)

If σ(x) ∈ C 0 (Ω), aσ (ϕ, ϕ) ≥ ‖∇ϕ‖2L2 (Ω) − σ∞ ‖ϕ‖2L2 (Ω) , https://doi.org/10.1515/9783111025438-008

where σ∞ := max{σ(x), 0}. Ω

132 � 8 General strategy and applications From (C.0.5) it follows that σ∞ ∫ ϕ2 dx ≤ Ω

1 ∫ |∇ϕ|2 dx + c1/2 ∮ ϕ2 dS. 2 Ω

𝜕Ω

Hence, the corresponding Rayleigh quotient is bounded from below by RS (ϕ) :=

aσ (ϕ, ϕ) ≥ −c1/2 . b(ϕ, ϕ)

Consequently, infℋ RS (ϕ) =: μ0 − α > −∞. Lemma 8.1. The Steklov eigenvalue problem (8.1.1), resp. (8.1.2), possesses: – countably many eigenvalues of finite multiplicity μ0 < μ̃ 1 ≤ μ̃ 2 ≤ ⋅ ⋅ ⋅ with limn→∞ μ̃ n = ∞, – a system of eigenvalues {ϕi }∞ 0 which is orthonormal with respect to the inner product b(ϕi , ϕj ) and complete in ℋ. Moreover: – The eigenvalues are characterized by the Rayleigh principle μi − α = min RS (υ), υ∈ℋ

–

where b(υ, ϕj ) = 0 for j = 0, 1, 2, . . . , i − 1.

The eigenspace corresponding to the lowest eigenvalue is one-dimensional and the eigenfunction is of constant sign.

Proof. We prove the existence of μ0 and show that the corresponding eigenspace is onedimensional. For the proof of the existence of higher eigenvalues we refer to the classical literature on compact operators in Hilbert spaces. Let {φn }∞ 0 be a minimizing sequence in ℋ such that limn→∞ RS (φn ) = μ0 − α. It can be chosen such that RS (φn ) ∈ (μ0 − α − ϵ0 , μ0 − α + ϵ0 ) for all n > 1. Observe that RS (φ) does not change if φ is replaced by αφ. Consequently, we can assume that b(φn , φn ) = 1. The modified Friedrich inequality (C.0.5) implies that ∫ |∇φn |2 dx + ∫ σ − φ2n dx ≤ μ − α + ϵ0 + ∫ σ + φ2n dx

Ω

Ω

Ω

≤ μ − α + ϵ0 + σ∞ [ϵ ∫ |∇φn |2 dx + cϵ ], Ω

where σ ± = max{o, ±ϕ}. If we choose ϵ =

1 , 2σ∞

then

1 ∫ |∇φn |2 dx ≤ C, 2 Ω

8.1 A generalized Steklov eigenvalue problem

� 133

for some constant C which is independent of n. By Friedrich’s inequality (C.0.4) and the normalization of φn it follows that also ‖φn ‖L2 (Ω) is uniformly bounded. Thus, {φn }∞ n=1 is a bounded sequence in W 1,2 (Ω). There exists a subsequence which converges weakly to a function ϕ0 ∈ ℋ. Consequently, the infimum of RS is attained in ℋ. If σ(x) = 0, then μ − α = 0 and ϕ0 = constant. Next we prove that the minimizer is a solution of (8.1.2). Set for short Tσ (ϕ0 , υ) := aσ (ϕ0 , υ) − (μ0 − α)b(ϕ0 , υ). Since ϕ0 is a minimizer, we have 0 ≤ Tσ (ϕ0 + tυ, ϕ0 + tυ) = Tσ (ϕ0 , ϕ0 ) + t 2 Tσ (υ, υ) + 2tTσ (ϕ0 , υ). If Tσ (υ, υ) > 0, the minimum is attained for t = −

Tσ (ϕ0 ,υ) . Tσ (υ,υ)

Consequently,

2

(Tσ (ϕ0 , υ)) ≤ Tσ (υ, υ)Tσ (ϕ0 , ϕ0 ). Since Tσ (ϕ0 , ϕ0 ) vanishes, we have Tσ (υ, ϕ0 ) = 0. Hence, ϕ0 is the eigenfunction of the Steklov problem (8.1.1) and μ0 is the corresponding eigenvalue. If the eigenfunction ϕ0 changes sign, then |ϕ0 | is also a minimizer. It is therefore a weak solution of (8.1.1). By the regularity theory – in particular Harnack’s inequality – |ϕ0 | is a classical solution; hence, |ϕ0 | = ϕ0 . Suppose that there exist two different eigenfunctions ϕ and ψ to the lowest eigenvalue μ. Then any linear combination ζ = c1 ϕ + c2 ψ is again an eigenfunction to μ. If ϕ and ψ are different, we can find constants c1 and c2 such that ζ changes sign. This is a contradiction to the previous observation. Notation According to Lemma 8.1, any function u ∈ ℋ can be written as a Fourier series. We denote by ϕk,1 , . . . , ϕk,dk the eigenfunctions corresponding to μk , and dk denotes its multiplicity. Then ∞ dk

u = ∑ ∑ ck,i ϕk,i . k=0 i=1

The eigenfunctions {ϕk,i }, where k = 0, 1, . . . and i = 1, 2, . . . , dk , can always be chosen such that ∮ ϕk,i ϕk,j dS = δij

for i, j = 1, . . . , dk

𝜕Ω

and ∮ ϕk,i ϕℓ,j dS = δkℓ 𝜕Ω

for k, ℓ = 0, 1, 2, . . . .

134 � 8 General strategy and applications Then ck,i = ∮ uϕk,i dS. 𝜕Ω

To simplify notation we will often set ϕk =

dk

−1/2 d k

2 {∑ ck,i } i=1

∑ ck,i ϕk,i i=1

and ck =

1/2

dk

2 {∑ ck,i } i=1

.

(8.1.3)

In this notation u can be written as ∞

u = ∑ ck ϕk .

(8.1.4)

k=0

Observe that ∮ ϕk ϕℓ dS = δkℓ . 𝜕Ω

Representation of Qg (u′ ) We apply the expansion described above to the quadratic form 2 󵄨 󵄨2 Qg (u′ ) = ∫[󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 − g ′ (u)u′ 2 ] dx + α ∮ u′ dS, Ω

𝜕Ω

which appears in the second variation of the Robin energy (cf. Theorem 6.3). From the partial differential equation Δu′ + g ′ (u)u′ = 0 in Ω it follows that Qg (u′ ) = ∮(u′ 𝜕ν u′ + αu′ 2 ) dS. 𝜕Ω ′ We now use (8.1.4) and insert u′ = ∑∞ i=0 ci ϕi into Qg (u ). Since ϕi is a Steklov eigenfunction corresponding to μi we get

Qg (u′ ) = ∑ ci cj ∮(ϕi 𝜕ν ϕj + αϕi ϕj ) dS = ∑ μj ci cj ∮ ϕi ϕj dS. i,j

i,j

𝜕Ω

𝜕Ω

From the orthonormality of {ϕi }∞ i=0 it then follows that ∞

∞

di

i=0

i=0

j=1

2 Qg (u′ ) = ∑ μi ci2 = ∑ μi ∑ ci,j .

(8.1.5)

8.2 Nearly spherical domains

� 135

The Fourier coefficients are obtained from the boundary condition 𝜕ν u′ + αu′ =: K(u, ρ), derived in Lemma 6.1. Multiplication with ϕi and integration yields ∮(𝜕ν u′ ϕi + αu′ ϕi ) dS = ∮ K(u, ρ)ϕi dS. 𝜕Ω

𝜕Ω

Replacing u′ by its expansion and using the orthonormality of {ϕi }∞ i=0 we find that ci =

1 ∮ K(u, ρ)ϕi (x) dS. μi

(8.1.6)

𝜕Ω

8.2 Nearly spherical domains We call a domain Ω nearly spherical if there are a constant c > 0 and a C 1,1 -diffeomorphism Φ : BR → Ω such that ‖Φ‖C 1,1 ≤ c. As a consequence there exists a radius R0 = R0 (c) such that Ω ⊂ BR0 . With regard to Section 2.1, Ωt = {y = x + tυ(x) +

t2 w(x) + o(t 2 ) : x ∈ BR , |t| < t0 } 2

describes a family of nearly spherical domains for t0 sufficiently small (see Lemma 2.1).

8.2.1 Robin energy Our goal is to discuss the first and second domain variation of the energy ℰR (t) (cf. (6.2.5)) corresponding to the boundary value problem (6.2.1) with Robin boundary conditions. We will assume that the solution u = u(r) of problem (6.2.1), (6.2.3) in BR is radial. Then according to (6.4.4) the first variation is of the form 2

ℰṘ (0) = −(2G(u(R)) + [α − α

n−1 2 ]u (R)) ∮ (υ ⋅ ν) dS. R

(8.2.1)

𝜕BR

This together with (2.3.2) and (2.3.20) leads to the following simple conclusions: 1. ℰṘ (0) = 0 if 𝒱̇ (0) = 0, ̇ = 0, 2. ℰṘ (0) = 0 if 𝒮 (0) 3. ℰṘ (0) = 0 if υ = υτ is a purely tangential perturbation. We now proceed to the calculation of ℰR̈ (0). To this end we need the shape derivative which by Lemma 6.1 solves

136 � 8 General strategy and applications Δu′ + g ′ (u)u′ = 0 ′

′

in BR ,

(8.2.2)

𝜕ν u + αu = kg (u(R))(υ ⋅ ν)

on 𝜕BR ,

(8.2.3)

where kg (u(R)) = g(u(R)) −

α(n − 1) u(R) + α2 u(R). R

(8.2.4)

In order to determine ℰR̈ (0) for a general perturbation Φt we use Theorem 6.3. Since Ω = BR and u = u(|x|), the formula in Theorem 6.3 simplifies. Step 1. Since u is radial, tangential derivatives of u in ℰR̈ (0) are zero: ∇u|𝜕BR = ur (R)ν

and ∇τ u|𝜕BR = 0.

Moreover, 𝜕i 𝜕j u(x)|𝜕BR = 𝜕i 𝜕j u(|R|) = urr (R)νi νj +

ur (R) (δij − νi νj ), R

where νi =

xi . R

The last identity implies (υ ⋅ D2 uυτ ) = υi 𝜕i 𝜕j uυτj =

ur (R) 󵄨󵄨 τ 󵄨󵄨2 󵄨υ 󵄨 . R 󵄨 󵄨

We put this information into ℰR̈ (0) and write down the integrals in the same order as in Theorem 6.3. Taking innto account the boundary condition ur (R) + αu(R) = 0 we have 2

2 2

τ

ℰR̈ (0) = [ur (R) − 2G(u(R))]𝒱̈ (0) + 4α u (R) ∮ (υ ⋅ Dυ ν) dS 𝜕BR

u(R)ur (R) 󵄨󵄨 τ 󵄨󵄨2 ∮ 󵄨󵄨υ 󵄨󵄨 dS −2αu(R)g(u(R)) ∮ (υ ⋅ ν)2 dS R 𝜕BR 𝜕BR ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ −2

(1)

+ 4αu(R) ∮ (υ ⋅ ν)(𝜕ν u′ + αu′ ) dS − 2α3 u2 (R) ∮ (υ ⋅ ν)2 dS 𝜕BR

𝜕BR

̈ − 2Qg (u′ ). −2α2 u2 (R) ∮ (w ⋅ ν) dS +αu2 (R)𝒮 (0) 𝜕BR ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (2)

8.2 Nearly spherical domains

� 137

Step 2. For the first integral on the right side we recall (2.3.11). We have 4ur2 (R) ∮ (υτ ⋅ Dυ ν) dS = −2ur2 (R)𝒱̈ (0) + 𝜕BR

2(n − 1) 2 ur (R) ∮ (υ ⋅ ν)2 dS R 𝜕BR

2ur2 (R)

󵄨 󵄨2 ∮ 󵄨󵄨󵄨υτ 󵄨󵄨󵄨 dS +2ur2 (R) ∮ (w ⋅ ν) dS . R 𝜕BR 𝜕BR ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

+

(1)

(2)

Since ur (R) = −αu(R), the underbracketed integrals are canceled out. This implies 2 2

ℰR̈ (0) = −[α u (R) + 2G(u(R))]𝒱̈ (0)

+(

2(n − 1) 2 ur (R) − 2αu(R)g(u(R)) − 2α3 u2 (R)) ∮ (υ ⋅ ν)2 dS R 𝜕BR

+ 4αu(R) ∮ (υ ⋅ ν)(𝜕ν u′ + αu′ ) dS 𝜕BR

2

̈ − 2Qg (u′ ). + αu (R)𝒮 (0) Step 3. We insert the boundary condition (8.2.3) for u′ . With (8.2.4) we obtain 2 2

2

ℰR̈ (0) = −[α u (R) + 2G(u(R))]𝒱̈ (0) + 2αu(R)kg (u(R)) ∮ (υ ⋅ ν) dS 𝜕BR

2

̈ − 2Qg (u′ ). + αu (R)𝒮 (0) Theorem 8.1. Let {Ωt }|t| qk+m (r) for any m ∈ ℝ+ , Sturm’s first comparison theorem [71, Theorem 3.1] applies and yields μk < μk+m . The eigenvalues μk are computed from the boundary condition at r = R, i. e., μk =

d a (R) dr k

ak (R)

+ α.

Hence, the dimension of the eigenspace of μk is the dimension of the space spanned by the spherical harmonics of degree k. Thus, μk,1 = μk,2 = ⋅ ⋅ ⋅ = μk,dk = μk . (iii) Completeness. To show that the system {ϕk,i }, k ≥ 0, i = 1, . . . , dk , is complete, we assume that there exists an eigenvalue μ ≠ μi,k for all i and k. If ϕ is the corresponding eigenfunction, then (μk,i − α) ∮ ϕk,i ϕ dS = ∫ [(∇ϕk,i ⋅ ∇ϕ) − g ′ (u)ϕk,i ϕ + BR

𝜕BR

(μ − α) ∮ ϕk,i ϕ dS = ∫ [(∇ϕk,i ⋅ ∇ϕ) − g ′ (u)ϕk,i ϕ + BR

𝜕BR

k(n − 2 + k) ϕk,i ϕ] dx, r2 k(n − 2 + k) ϕk,i ϕ] dx. r2

Hence, ∮ Yk,i ϕ dS = 0,

for all k = 0, . . . and i = 1, 2, 3, . . . , dk .

𝜕B1

The completeness of the spherical harmonics implies that ϕ ≡ 0. Hence, no other eigenvalues and eigenfunctions exist. If g ′ (u(r)) is continuous in BR , we approximate it by analytic functions and use the continuous dependence of the differential equation on its data. This establishes the existence of a solution to (8.2). As in (8.1.4) we set ∞

∞

u′ = ∑ ci ϕi

and (υ ⋅ ν) = ∑ bi ϕi ,

i=0

i=0

where {ϕi }i≥0 is an orthonormal basis of eigenfunctions of the Steklov problem (8.2.9). By (8.1.6) and the boundary condition of u′ (cf. (8.2.3)), ci = In view of (8.1.5) we have

kg (u(R)) μi

∮ (υ ⋅ ν)ϕi dS = 𝜕BR

kg (u(R)) μi

bi .

(8.2.11)

8.2 Nearly spherical domains

� 141

∞

Qg (u′ ) = ∑ μi ci2 . i=0

Let us write for short 2

′

ℱ := −2Qg (u ) + 2αu(R)kg (u(R)) ∮ (υ ⋅ ν) dS. 𝜕BR

The boundary condition 𝜕ν u′ + αu′ = kg (u(R))(υ ⋅ ν) and ϕ0 (R) = const. imply that ∮ (𝜕ν u′ + αu′ )ϕ0 dS = kg (u(R))ϕ0 (R) ∮ (υ ⋅ ν) dS. 𝜕BR

𝜕BR

For volume and area preserving perturbations it follows from (2.3.2) and (2.3.20) that ∮ (υ ⋅ ν) dS = 0. 𝜕BR

Hence, ∞

0 = ∮ (𝜕ν u′ + αu′ )ϕ0 dS = ∮ ∑ ci (𝜕ν ϕi + αϕi )ϕ0 dS = c0 μ0 . 𝜕BR i=0

𝜕BR

This implies b0 = 0. ∞ 2 2 2 In ℱ we replace Qg (u′ ) by ∑∞ i=1 μi ci and ∮𝜕B (υ ⋅ ν) dS by ∑i=1 bi , and we obtain R

∞

2 2

ℱ = 2 ∑ ci μi [ i=1

1 αu(R) − ] kg (u(R)) μi

∞

= 2 ∑ b2i [αu(R)kg (u(R)) −

kg2 (u(R))

i=1

μi

].

(8.2.12)

From (8.2.7) we then obtain the following proposition. Proposition 8.1. Let Φt be a volume preserving perturbation. Then 2

ℰR̈ (0) = αu (R)𝒮0̈ (0) + ℱ .

For tangential perturbations we have (υ ⋅ ν) = 0 and therefore bi = 0 for all i. This implies ℰṘ (0) = 0 and by (2.3.27), ℰR̈ (0) = 0. Consequently, the volume preserving tangential perturbations are in the kernel of ℰR (0). From (8.2.8) we get the following proposition.

142 � 8 General strategy and applications Proposition 8.2. Let Φt be an area preserving perturbation. Then ℰR̈ (0) =

R (k (u(R))u(R) + 2G(u(R)) − u(R)g(u(R)))𝒮0̈ (0) + ℱ . n−1 g

With (8.2.4) this reads as ℰR̈ (0) =

n−1 2 R (α2 u2 (R) − αu (R) + 2G(u(R)))𝒮0̈ (0) + ℱ . n−1 R

Note that also in this case area preserving tangential perturbations belong to the kernel of ℰR̈ (0). Lower bounds for ℰR̈ (0) The sign of ℰR̈ (0) depends on the solution and the data of the Robin boundary value problem. We consider volume or area preserving perturbations. We now impose the barycenter condition ∮ x(υ(x) ⋅ ν(x)) dS = 0. 𝜕BR

Recall that ϕ1 = a1 (r)Y1 (ξ), where ξ = implies

x |x|

and x ∈ ℋ1 . Then the orthonormality of {ϕi }∞ i=1 ∞

0 = ∮ ϕ1 (R, ξ)(∑ bi ϕi (R, ξ)) dS = b1 . i=1

𝜕BR

Consequently, ∞

2

ℱ = 2 ∑ bi [αu(R)kg (u(R)) −

kg2 (u(R))

i=2

μi

].

(8.2.13)

Let μp be the smallest positive eigenvalue of (8.2.9). Then ℱ ≥ [αu(R)kg (u(R)) −

kg2 (u(R)) μp

] ∮ ρ2 dS. 𝜕BR

We recall that 𝒮0̈ (0) is nonnegative (cf. Section 2.3.3). By Lemma 2.4 the barycenter condition leads to the stronger estimate ̈ ≥ 𝒮 (0)

n+1 ∮ (υ ⋅ ν)2 dS. R2 𝜕BR

This together with the estimates for ℱ leads to the following theorem.

8.2 Nearly spherical domains

� 143

Theorem 8.2. Let {Φt }|t| 0 We apply the previous arguments to the torsion problem Δu + 1 = 0 in Ωt ,

𝜕ν u + αu = 0 on 𝜕Ωt ,

α > 0,

in nearly spherical domains. If α > 0 the problem has a unique positive solution. In the R R2 r2 ball it is therefore radial and of the form u(r) = αn + 2n − 2n . In this case we have u(R) =

R αn

and k1 (u(R)) =

1 + αR . n

The corresponding Steklov problem is Δϕ = 0 in BR ,

𝜕ν ϕ + αϕ = μϕ on 𝜕BR ,

where by Remark 3.1, μi = Ri + α. From (8.2.1) the first variation is ℰṘ (0) = −

R (αR + n + 1) ∮ (υ ⋅ ν) dS. αn2

(8.2.14)

𝜕BR

Clearly, it vanishes for volume or area preserving perturbations. It is negative, resp. positive, if |Ωt | > |BR ], resp. |Ωt | < |BR ].

144 � 8 General strategy and applications For volume preserving perturbations Proposition 8.1 gives ℰR̈ (0) =

∞ R2 ̈ 2 R(1 + αR)(i − 1) . 𝒮0 (0) + 2 ∑ bi 2 αn n2 (i + αR) i=1

(8.2.15)

Consequently, for volume preserving perturbations, ℰR̈ (0) ≥ 0. If (υ ⋅ ν) = b1 ϕ1 , then 𝒮0̈ (0) = 0 and thus ℰR̈ (0) = 0. If we impose the barycenter condition, then b1 = 0. Consequently, ℰR̈ (0) =

∞ R2 ̈ R(1 + αR)(i − 1) 𝒮 (0) + 2 b2i . ∑ 0 2 αn n2 (i + αR) i=2

In view of Theorem 8.2 we obtain ℰR̈ (0) ≥ {

n + 1 2(1 + αR)R + 2 } ∮ (υ ⋅ ν)2 dS > 0, αn2 n (2 + αR)

(8.2.16)

𝜕BR

provided Φt is not a purely tangential perturbation. In this case the ball is among all nearly spherical domains of prescribed volume a local minimizer of the Robin energy. This is in accordance with the result of Buçur and Giacomini [29], who showed that the ball has the minimal energy among all domains of given volume. In the case of area preserving transformations, Proposition 8.2 yields ℰR̈ (0) =

∞ 1 3 2 ̈0 (0) + 2 ∑ b2 R(1 + αR)(i − 1) (αR + R (n + 1)) 𝒮 i α(n − 1)n2 n2 (i + αR) i=1

≥{

2(1 + αR) (αR + n + 1)(n + 1) + } ∮ ρ2 dS. α(n − 1)n2 (2 + αR)n2 𝜕BR

Clearly, ℰR̈ (0) is nonnegative. Imposing the barycenter condition we obtain similar lower estimates as for volume preserving perturbations. Hence, the ball is a local minimizer for area preserving perturbations. Problem 8.1. If there exists an optimal domain, must it be a ball?

8.2.3 Dirichlet energy Let ℰD (t) be the Dirichlet energy (cf. Section 6.2.1) defined in the nearly spherical domain Ωt . Assume that the corresponding Dirichlet problem has a solution u(y, t) which is differentiable in t. In addition, we require that in the ball u is radially symmetric. From (6.4.2) we get

8.2 Nearly spherical domains

� 145

2

ℰḊ (0) = −(ur (R) + 2G(0)) ∮ (υ ⋅ ν) dS. 𝜕BR

̇ = 0, or υ = υτ . It vanishes whenever 𝒱̇ (0) = 0, 𝒮 (0) From Theorem 6.2 it follows that 2

2

ℰD̈ (0) = (ur (R) − G(0))𝒱̈ (0) + 2g(0)ur (R) ∮ (υ ⋅ ν) dS 𝜕BR

+ ur2 (R) ∮ {4(υτ ⋅ Dυ ν) − 2(w ⋅ ν)} dS 𝜕BR

+ 2ur (0) ∮ (υτ ⋅ D2 uυ) dS + 2Q0 (u′ ). 𝜕BR

By (2.3.7), −2 ∮ (w ⋅ ν) dS = −2𝒱̈ (0) + 2 ∮ (υ ⋅ ν) div𝜕BR υ dS − 2 ∮ (υτ ⋅ Dυ ν) dS. 𝜕BR

𝜕BR

𝜕BR

For radial solutions u we have from Subsection 8.2.1 (υτ ⋅ D2 uυ) = 2

ur (R) τ 2 |υ | . R

Hence,

2

ℰD̈ (0) = −(ur (R) + G(0))𝒱̈ (0) + 2g(0)ur (R) ∮ (υ ⋅ ν) dS 𝜕BR

+ 2ur2 (R) ∮ {(υτ ⋅ Dυ ν) + (υ ⋅ ν) div𝜕BR υ + 𝜕BR

1 󵄨󵄨 τ 󵄨󵄨2 󵄨υ 󵄨 } dS R󵄨 󵄨

+ 2Q0 (u′ ). On the sphere we have (υτ ⋅ Dυ ν) = υτ ⋅ ∇τ (υ ⋅ ν) − |υτ |2 /R. By Gauss’ theorem (2.2.18), 2

2

ℰD̈ (0) = − (ur (R) + G(0))𝒱̈ (0) + 2g(0)ur (R) ∮ (υ ⋅ ν) dS 𝜕BR

(n − 1) + 2ur2 (R) ∮ (υ ⋅ ν)2 dS + 2Q(u′ ). R

(8.2.17)

𝜕BR

For volume preserving perturbations, taking into account the boundary condition u′ = −ur (R)(υ ⋅ ν) on 𝜕BR , we get ℰD̈ (0) = 2

g(0) (n − 1) ∮ u′ 2 dS + 2 ∮ u′ 2 dS + 2Q(u′ ). ur (R) R 𝜕BR

This is the formula given in Proposition 7.1.

𝜕BR

(8.2.18)

146 � 8 General strategy and applications

lem

The strategy used for the Robin energy requires eigenfunctions of the Steklov probΔϕ + g ′ (u)ϕ = 0 in BR ,

𝜕ν ϕ = μϕ in 𝜕BR .

Set u′ = ∑∞ i=0 ci ϕi (cf. (8.1.4)). Then ∞

𝜕ν u′ = ∑ μi ci ϕi = −(∇u ⋅ υ) = −ur (R)(υ ⋅ ν). i=0

Consequently, ∞ μi ci ϕi =: ∑ bi ϕi u (R) i=0 i=0 r ∞

(υ ⋅ ν) = − ∑

and

∞

∮ (υ ⋅ ν)2 dS = ∑ b2i . i=0

𝜕BR

By Lemma 8.1, ϕ0 is of constant sign. The condition 𝒱̇ (0) = 0 implies that 0 = ∮ (υ ⋅ ν) dS = b0 . 𝜕BR

Furthermore, ∞

∞

i=1

i=1

Q0 (u′ ) = ∑ μi ci2 = ur2 (R) ∑

b2i . μi

Observe that μi ci = 0 implies bi = 0. Thus volume preserving perturbations yield ℰD̈ (0) = 2{g(0)ur (R) +

2

∞ b n−1 2 ur (R)} ∮ (υ ⋅ ν)2 dS + 2ur2 (R) ∑ i . R μ i=1 i

(8.2.19)

𝜕BR

We set μp := min{μk : μk ≥ 0, k ≥ 1} and take into account the boundary condition u′ = −ur (R)(υ ⋅ ν) on 𝜕BR . Then ℰD̈ (0) ≥ 2{

g(0) n−1 2 + − } ∮ u′ 2 dS. ur (R) R μp

(8.2.20)

𝜕BR

Remark 8.1. 1. If u(y, t) is a positive minimizer of ℰD (t), then Schwarz symmetrization applies. It shows that among all domains of given volume the ball yields the smallest energy. Moreover, the ball is the unique domain with such a property (see [119] and [9]). 2. Formula (8.2.19) is valid for sign changing solutions and for any critical point of ℰD (0).

8.3 Notes

� 147

Problem 8.2. Does ℰḊ (0) = 0 and ℰD̈ (0) > 0 imply that the ball is the only optimal domain?

8.3 Notes A detailed discussion of related Steklov eigenvalue problems is found in the work of Auchmuty; see for instance [6], [7], and the references cited therein. A discussion of the sign of the second variation in nearly spherical domains was first given in [13] and [14].

9 Eigenvalue problems This chapter is devoted to domain variations of membrane eigenvalue problems. The computations rely strongly on the calculations in Chapter 6. It is shown that among all nearly spherical domains of given volume the ball has the smallest first eigenvalue. The second part of this chapter deals with nonlinear eigenvalue problems. The peculiarity of these problems is that there exists a finite interval of eigenvalues for which solutions exist. The endpoint (turning point) of this interval where the solution branch bends is of particular interest. It can be expressed by means of a Rayleigh quotient depending on two functions. An important example is the Gelfand problem. We discuss the one-dimensional and the radial two-dimensional case analytically and compute the solution branches numerically for Dirichlet and Robin boundary conditions. At the end the first variation of the turning point of general nonlinear eigenvalue problems is derived for volume preserving perturbations.

9.1 Robin eigenvalue problem Let {Ωt }|t| 0.

̃ := u(Φt (x), t) If we transform this problem back onto Ω, the transformed solution u(t) solves (see Section 6.2.1) ̃ + λ(t)u(t)J(t) ̃ =0 LA(t) u(t)

in Ω.

(9.1.1)

Here we set λ(t) := λ(Ωt ) and λ(0) = λ(Ω). On 𝜕Ω we obtain ̃ + αm(t)u(t) ̃ = 0. 𝜕νA u(t)

(9.1.2)

̃ ∈ W 1,2 (Ω). The solutions of (9.1.1) satisfy We assume that u(t) ̃ ij (t)𝜕j ψ dx + α ∮ uψm(t) ̃ ̃ dS = λ(t) ∫ uψJ(t) dx, ∫ 𝜕i uA Ω

𝜕Ω

https://doi.org/10.1515/9783111025438-009

Ω

∀ ψ ∈ W 1,2 (Ω).

9.1 Robin eigenvalue problem

� 149

For ψ = ũ we obtain λ(t) =

̃ ̃ dx + α ∮ ũ 2 (t)m(t) dS ∫Ω Aij (t)𝜕i u(t)𝜕 j u(t) 𝜕Ω ∫Ω ũ 2 (t)J(t) dx

.

(9.1.3)

Observe that ̃ ij (t)𝜕j ψ dx + α ∮ uψm(t) ̃ dS : W 𝒢ũ (t)[ψ] := ∫ 𝜕i uA Ω

1,2

(Ω) → ℝ

𝜕Ω

is a self-adjoint linear operator, analytic in t for small |t|. The same is true for the linear operator 2

̃ ℬũ (t)[ψ] := ∫ uψJ(t) dx : L (Ω) → ℝ. Ω

The following result holds true for the eigenvalue problem 𝒢ũ = λ(t)ℬũ . Theorem 9.1. Let λ(0) be an eigenvalue of multiplicity d. Then there exists δ > 0 such that for |t| < δ there exist exactly d eigenvalues λ1 (t), λ2 (t), . . . , λd (t). After a possible renaming these eigenvalues and the corresponding eigenfunctions are analytic in t. A proof of this result is given for example in the book of Chow and Hale [38, Theorem 5.5]. Figure 9.1a illustrates the branches and Figure 9.1b shows that the order of the eigenvalues can change at the bifurcation point.

Figure 9.1: Bifurcations.

9.1.1 Domain variations for Robin eigenvalues with α > 0 Let λ(t) be a differentiable branch of eigenvalues in Ωt with α > 0. We use the same arguments as for the energy in Section 6.3 to compute the derivatives of λ(t).

150 � 9 Eigenvalue problems Differentiation of (9.1.3) and the normalization ∫ ũ 2 (t)J(t) dx = 1

(9.1.4)

Ω

lead to ̇ = 2 ∫(∇u(t) ̇̃ ⋅ A(t)∇u(t)) ̇ ̃ ̃ ⋅ A(t)∇ ̃ λ(t) dx + ∫(∇u(t) u(t)) dx Ω

Ω

̇̃ u(t)m(t) ̇ dS. ̃ + 2α ∮ u(t) dS + α ∮ ũ 2 (t)m(t) 𝜕Ω

𝜕Ω

The expressions containing u̇̃ can be eliminated. If we test (9.1.1) with u,̇̃ we get ̇̃ ̇̃ ⋅ A(t)∇u(t)) ̇̃ u(t)m(t) ̃ ̃ λ(t) ∫ u(t)u(t)J(t) dx = ∫(∇u(t) dx + α ∮ u(t) dS. Ω

Ω

𝜕Ω

The normalization (9.1.4) implies d ̇̃ ̇ ̃ u(t)J(t) ̃ 2 J(t)] + u(t) dx = 0 ∫ ũ 2 (t)J(t) dx = ∫[2u(t) dt Ω

(9.1.5)

Ω

for all t ∈ (−t0 , t0 ). Hence, ̇ = ∫(∇u(t) ̇ ̇ dx ̃ ⋅ A(t)∇ ̃ λ(t) u(t)) dx − λ(t) ∫ ũ 2 (t)J(t) Ω

Ω 2

̇ dS. + α ∮ ũ (t)m(t)

(9.1.6)

𝜕Ω

Next we compute the second derivative of λ(t): ̈ = ∫(∇u(t) ̇ ∫ ũ 2 (t)J(t) ̇̃ ⋅ A(t)∇ ̈ ̇ ̇ dx ̃ ⋅ A(t)∇ ̃ ̃ λ(t) u(t)) dx + 2 ∫(∇u(t) u(t)) dx − λ(t) Ω

Ω

Ω

̇̃ u(t) ̇ dx − λ(t) ∫ ũ 2 (t)J(t) ̈ dx ̃ J(t) − 2λ(t) ∫ u(t) Ω

Ω

̇̃ m(t) ̇ dS + α ∮ u(t) ̈ dS. ̃ u(t) ̃ 2 m(t) + 2α ∮ u(t) 𝜕Ω

𝜕Ω

As for the energy in Section 6.3 we can eliminate the domain integrals which contain both ũ and u.̇̃ For this purpose we differentiate (9.1.1) with respect to t, multiply it by 2u,̇̃ and integrate over Ω. We obtain

9.1 Robin eigenvalue problem

� 151

̇̃ ̇̃ ⋅ A(t)∇u(t)) ̇̃ ̇ ̃ ⋅ A(t)∇ 0 = −2 ∫(∇u(t) u(t)) dx − 2 ∫(∇u(t) dx Ω

Ω

̇̃ dS + 2 ∮(ν ⋅ A(t)∇u(t)) ̇̃ u(t) ̇̃ dS ̇ ̃ u(t) + 2 ∮(ν ⋅ A(t)∇ u(t)) 𝜕Ω

𝜕Ω

̇ ∫ u(t) ̇̃ ̇̃ J(t) ̇ dx ̃ u(t) ̃ u(t)J(t) dx + 2λ(t) ∫ u(t) + 2λ(t) Ω

Ω

̇2

+ 2λ(t) ∫ ũ (t)J(t) dx. Ω

̈ we obtain Adding this expression to λ(t) ̈ = ∫(∇u(t) ̇̃ ⋅ A(t)∇u(t)) ̇̃ ̈ ̃ ⋅ A(t)∇ ̃ λ(t) u(t)) dx − 2 ∫(∇u(t) dx Ω

Ω

̇̃ ̇̃ ̇ ̇ ̃ ̃ m(t)) + 2 ∮ u(t)((ν ⋅ A(t)∇ u(t)) + (ν ⋅ A(t)∇u(t)) + αu(t) dS 𝜕Ω

̇ ∫ 2u(t) ̇̃ ̇ dx ̈ dS + λ(t) ̃ u(t)J(t) − ũ 2 (t)J(t) + α ∮ ũ 2 (t)m(t) Ω

𝜕Ω

̈ dx. + λ(t) ∫ 2u̇̃ 2 (t)J(t) − ũ 2 (t)J(t) Ω

From the boundary condition (9.1.2) it follows that νi Ȧ ij 𝜕j ũ + νi Aij 𝜕j ũ + αṁ ũ + αmu̇̃ = 0

on 𝜕Ω.

Hence, ̈ = ∫(∇u(t) ̇̃ ⋅ A(t)∇u(t)) ̇̃ ̈ ̃ ⋅ A(t)∇ ̃ λ(t) u(t)) dx − 2 ∫(∇u(t) dx Ω

Ω

̈ dS −2α ∮ u̇̃ 2 (t)m(t) dS + α ∮ ũ 2 (t)m(t) 𝜕Ω

𝜕Ω

̇2

̇ ∫ ũ 2 (t)J(t) ̈ dx − 2λ(t) ̇ dx. + λ(t) ∫ 2ũ (t)J(t) − ũ (t)J(t) Ω

2

(9.1.7)

Ω

̇ 9.1.2 The discussion of λ(0) 2

The same arguments as in Section 6.4 with G(u) replaced by λ(0) u2 imply that (cf. (6.4.4)) ̇ λ(0) = ∮(υ ⋅ ν)[|∇u|2 − λ(0)u2 − 2α2 u2 + α(n − 1)Hu2 ] dS. 𝜕Ω

(9.1.8)

152 � 9 Eigenvalue problems ̇ This expression characterizes the critical domains for which λ(0) = 0. We will write λ := λ(0) for short. In analogy to Theorem 6.1 we have the following. Theorem 9.2. Let Ωt = Φt (Ω) be a family of volume preserving perturbations of Ω. Then ̇ = 0, if and only if Ω is a critical domain for the eigenvalue λ(t), i. e., λ(0) |∇u|2 − λu2 − 2α2 u2 + α(n − 1)u2 H = const.

on 𝜕Ω.

(9.1.9)

This additional condition leads to an overdetermined boundary value problem for u. The radial solutions in the ball clearly satisfy this condition. Problem 9.1. Is the ball the unique domain for which this overdetermined problem has a solution? Example 9.1. Let Ω = BR and let λ be the lowest eigenvalue. The corresponding eigenfunction is radial and of constant sign. Then n−1 ̇ λ(0) = −u2 (R)(α2 − α + λ) ∮ (υ ⋅ ν) dS. R

(9.1.10)

𝜕BR

The sign of K := α2 − α

n−1 +λ R

can be determined as follows. Since u(r) is of constant sign, the function z = defined in (r0 , R) and satisfies the Riccati equation dz n−1 + z2 + z+λ=0 dr r

ur u

is well-

in (0, R).

At the endpoint z(R) = −α, z(0) = 0, and zr (R) = −K. Assume α > 0 and let u(r) > 0 in (0, R). If zr (R) > 0, then there exists a number ρ ∈ (0, R) such that zr (ρ) = 0, z(ρ) < 0, and zrr (ρ) ≥ 0. From the equation we get zrr (ρ) = n−1 z < 0, which leads to a contradiction. ρ2 Hence, K >0

if α > 0.

(9.1.11)

K 0, ̇ λ(0) 0. R

̈ for the ball 9.1.3 Discussion of λ(0) From (9.1.7) we obtain for t = 0 and λ = λ(0) 󵄨 ̇̃ 󵄨󵄨2 2 2 ̈ ̈ = ∫(A(0)∇u ̈ λ(0) ⋅ ∇u) dx − 2 ∫󵄨󵄨󵄨∇u(0) dx 󵄨󵄨 dx + λ ∫[2u̇̃ (0) − u J(0)] Ω

Ω

Ω 2

̇ ∫ u J(0) ̇ dx + α ∮ u m(0) ̈ dS − 2α ∮ u̇̃ 2 (0) dS. − 2λ(0) 2

Ω

𝜕Ω

𝜕Ω

Next we discuss the case where Ω is a ball BR and the eigenfunction u corresponding to ̈ ̈ we can proceed exactly λ(0) is radial. Because of the similarity between λ(0) and ℰ (0) as in Section 8.2. We consider volume preserving perturbations from now on; hence, ̇ λ(0) = 0 for the ball. By (6.2.12) and Lemma 6.1 the shape derivative u′ solves Δu′ + λu′ = 0

in BR ,

𝜕ν u + αu = kλu (u(R))(υ ⋅ ν) ′

′

(9.1.13) on 𝜕BR ,

where kλu (u(R)) = (λ −

α(n − 1) + α2 )u(R). R

Then by (8.2.7) ̈ = −2Q (u′ ) − u2 (R)(λ + α2 )𝒱̈ (0) λ(0) λ ̈ + 2αu(R)kλu ∮ (υ ⋅ ν)2 dS + αu2 (R)𝒮 (0).

(9.1.14)

𝜕BR

Here 󵄨 󵄨2 Qλ (u′ ) = ∫ 󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 dx + α ∮ u′ 2 dS − λ ∫ u′ 2 dx. BR

𝜕BR

BR

By the observation in Example 9.1 we have αu(R)kλu = αu2 (R)K ≥ 0 independently of the sign of α.

154 � 9 Eigenvalue problems ̈ for volume preserving perturbations The sign of λ(0) In this section we shall assume that λ(t) is the lowest eigenvalue. By the Bossel–Daners inequality the principal eigenvalue λ(Ω) is minimal for all domains Ω such that |Ω| = |B|. We will show that the local result follows directly from ̈ the positivity λ(0). ̇ We assume 𝒱̇ (0) = 𝒱̈ (0) = 0. In particular, 𝒱̇ (0) = 0 implies λ(0) = 0. We shall show ̈ that the strategy developed in Section 8 leads to λ(0) > 0. Under the assumptions stated above we have by (9.1.14) ̈ λ(0) = −2Qλ (u′ ) + 2αu(R)kλu ∮ (υ ⋅ ν)2 dS + αu2 (R)𝒮0̈ (0). 𝜕BR

Following the strategy of Chapter 8 we shall express the quadratic form Qλu in terms of the Steklov eigenfunctions Δϕ + λϕ = 0

in BR ,

(9.1.15)

𝜕ν ϕ + αϕ = μϕ in 𝜕BR . Let {ϕi }∞ i=0 be normalized such that ∮ ϕi ϕj dS = δij 󳨐⇒ ∫ [(∇ϕi ⋅ ∇ϕj ) − λϕi ϕj ] dx = (μi − α)δij . BR

𝜕BR

The first eigenfunction of the Robin eigenvalue problem in the ball is radial and is of the form u(r) = Cn J n−2 (√λ1 r)r − 2

n−2 2

,

(9.1.16)

where Cn is determined by the normalization ∫B u2 dx = 1 (see Section 3.3.1). The funcR tion u(r) is also a solution of the Steklov problem (9.1.15) with μ = 0. Since it is of constant sign, μ = 0 is the lowest eigenvalue. ∞ Consequently, 0 = μ0 < μ̃ 1 ≤ μ̃ 2 ≤ ⋅ ⋅ ⋅ . Let u′ = ∑∞ i=0 ci ϕi and (υ ⋅ ν) = ∑i=0 bi ϕi (see Section 8.1 for the notation). By the orthonormality of the Steklov eigenfunctions we obtain cj = ∮ u′ ϕj dS 𝜕BR

and

bj = ∮ (υ ⋅ ν)ϕj dS. 𝜕BR

Since ϕ0 is radial, b0 = 0 for volume preserving perturbations. The relations between bj and cj for j ≥ 1 are obtained from (9.1.13) and (9.1.15). In fact,

9.1 Robin eigenvalue problem

� 155

0 = ∫ (Δu′ + λu′ )ϕj dx = ∮ (ϕj 𝜕ν u′ − u′ 𝜕ν ϕj ) dS BR

𝜕BR

= ∮ (ϕj kλu (υ ⋅ ν) − μj u′ ϕj ) dS. 𝜕BR

Thus, kλu bj = μj cj ,

j ≥ 1.

2 Recall that Qλ (u′ ) = ∑∞ i=1 μi ci . If we insert this expression and ∞

∞

i=1

i=1

∮ (υ ⋅ ν)2 dS = ∑ b2i = ∑( 𝜕BR

2

μi ci ) kλu

̈ into λ(0), we obtain ∞ αu(R) 1 ̈ λ(0) = 2 ∑ ci2 μ2i [ − ] + αu2 (R)𝒮0̈ (0). k (u(R)) μ λu i i=1

Replacing μi ci by kλu bi and kλu by Ku(R), we get ∞ α 1 ̈ λ(0) = 2u2 (R)K 2 ∑ b2i [ − ] + αu2 (R)𝒮0̈ (0). K μ i i=1

(9.1.17)

Next we shall show that the term inside brackets is nonnegative. Since μi > μ1 for i > 1, α α 1 1 > − . − K μi K μ1 In Example 9.1 it was shown that K and α are positive. The inequality equivalent to L := μ1 − α +

α K

−

1 μ1

> 0 is

n−1 λ − ≥ 0. R α

To establish this inequality we have to compute μ1 . By Lemma 8.2 the eigenfunctions of μ1 are of the form ϕ(x) = ∑ ci a1 (r)Y1,i (ξ), i

ξ ∈ S n−1 ,

where Y1,i (ξ), i = 1, . . . , n, denotes the i-th spherical harmonics of the first order and a1 (r) = r

2−n 2

J n (√λr). 2

156 � 9 Eigenvalue problems The boundary condition in (9.1.15) implies that a1′ (R) = (μ1 − α)a1 (R). From ur (R) + αu(R) = 0, (9.1.16), and the well-known Bessel identity (z−ν Jν (z))z = −r −ν Jν+1 (z), it follows that α = √λ

Jn/2 (√λR) . J(n−2)/2 (√λR)

(9.1.18)

The same argument and the boundary condition for a1 implies Jn/2+1 (√λR) 1 + α − √λ = μ1 . R Jn/2 (√λR) Introducing these two expressions in L, we obtain L=

√λ n − (J n (√λR) + J n −1 (√λR)). 2 R J n (√λR) 2 +1 2

The recurrence relation for Bessel functions Jν−1 (z) + Jν+1 (z) =

2ν J z ν

(9.1.19)

implies that L = 0 and K = αμ1 . Consequently, α 1 αL − = = 0. K μ1 Kμ1 Then (9.1.17) becomes ∞ α 1 ̈ λ(0) = 2u2 (R)K 2 ∑ b2i [ − ] + αu2 (R)𝒮0̈ (0). K μi i=2

Since 0 < μi < μi+1 for i ≥ 1,

α K

−

1 μi

≥

α K

−

1 μi+1

and we get

α 1 ∞ ̈ λ(0) ≥ 2K 2 u2 (R)[ − ] ∑ b2i + αu2 (R)𝒮0̈ (0) ≥ 0. K μ2 i=2 By imposing the barycenter condition (cf. Definition 2.5), Lemma 2.4 yields α 1 n+1 ̈ λ(0) ≥ (2K 2 [ − ] + α 2 )u2 (R) ∮ (υ ⋅ ν)2 dS > 0. K μ2 R 𝜕BR

(9.1.20)

9.2 Nonlinear eigenvalue problems

� 157

Theorem 9.3. Let λ be the principal eigenvalue of the Robin eigenvalue problem with α > 0. Then among all nearly spherical domains of equal volume the ball yields the lowest eigenvalue.

9.2 Nonlinear eigenvalue problems In this chapter we discuss nonlinear eigenvalue problems of the form Δu + λf (u) = 0 in Ω,

𝜕ν u + αu = 0 on 𝜕Ω, α > 0,

(9.2.1)

where λ > 0 and f is a positive twice differentiable function such that f (0) > 0 and f (u)/u → ∞ as u → ∞. We assume that there exists a turning point 0 < λ∗ < ∞ such that: – for 0 < λ < λ∗ there exists a classical minimal solution, – for λ = λ∗ there is a unique classical solution, – for λ > λ∗ there are no solutions. For convex and monotone functions f the existence of such a turning point has been proved for instance in [39, 16]. Problems of this type arise in a variety of models such as combustion, thermal explosions, and gravitational equilibrium of polytropic stars (cf. Crandall and Rabinowitz [39]). The turning point is characterized by two different boundary value problems, namely original problem: linearized problem:

Δu + λ∗ f (u) = 0 in Ω, Δϕ + λ f (u)ϕ = 0 in Ω, ∗ ′

𝜕ν u + αu = 0 on 𝜕Ω,

(9.2.2)

𝜕ν ϕ + αϕ = 0 on 𝜕Ω,

(9.2.3)

where λ∗ is a simple eigenvalue of (9.2.3). It can be expressed in terms of the Rayleigh quotient λ∗ =

∫Ω (∇u ⋅ ∇ϕ) dx + α ∮𝜕Ω uϕ dS ∫Ω f (u)ϕ dx

=

∫Ω (∇u ⋅ ∇ϕ) dx + α ∮𝜕Ω uϕ dS ∫Ω f ′ (u)uϕ dx

,

where u and ϕ are solutions of (9.2.2) and (9.2.3). Let {Ωt }, |t| ≤ t0 , be a family of small perturbations of Ω as considered throughout this text. Next we will show that problem (9.2.1) has a turning point λ(t) in Ωt . The next lemma and its proof are taken from Henry’s book [76]. Lemma 9.1. Let u∗ and ϕ∗ be solutions of (9.2.2) and (9.2.3). Assume that λ∗ is simple. Furthermore, let f ′ (u) > 0 and 2

∫ f ′′ (u)(ϕ∗ ) dx ≠ 0. Ω

(9.2.4)

158 � 9 Eigenvalue problems Then in each Ωt there exists a turning point λ∗ (t) such that λ∗ (0) = λ∗ . Moreover, it is twice continuously differentiable for small |t|. Proof. We shall use the notation of Section 9.1 and denote ℬA := 𝜕νA +αm and ℬ0 := 𝜕ν +α for short. Choose p > n/2 and set G:

(λ, υ, ϕ, t) → (LA υ + λf (υ)J, ℬA υ, LA ϕ + λf ′ (υ)Jϕ, ℬA ϕ, ∫ ϕ2 J dx), Ω

where G:

ℝ × W 2,p (Ω) × W 2,2 (Ω) × (−t0 , t0 )

→ C 0 (Ω) × C 0 (𝜕Ω) × L2 (Ω) × L2 (𝜕Ω) × ℝ.

If ϕ∗ is normalized such that ∫Ω (ϕ∗ )2 dx = 1, then clearly G(λ∗ , u∗ , ϕ∗ , 0) = (0, 0, 0, 0, 1). We are looking for a solution G(λ, u, ϕ, t) = (0, 0, 0, 0, 1) in a neighborhood of (λ∗ , u∗ , ϕ∗ , 0). We will apply the implicit function theorem. The derivative of G(λ, υ, ϕ, t) with respect to t evaluated at t = 0 is the quintuple (T1 , T2 , T3 , T4 , T5 ), where T1 := Δυ̇ + λḟ (u∗ ) + λ∗ f ′ (u∗ )υ,̇ T2 := ℬ0 υ,̇ ̇ ∗, T := Δϕ̇ + λḟ ′ (u∗ )ϕ∗ + λ∗ f ′ (u∗ )ϕ̇ + λ∗ f ′′ (u∗ )υϕ 3

T4 := ℬ0 ϕ,̇

and

T5 := 2 ∫ ϕ∗ ϕ̇ dx. Ω

We have to show that it vanishes if and only if (υ,̇ ϕ,̇ λ)̇ = (0, 0, 0). From (9.2.3) and Δυ̇ + λḟ (u∗ ) + λ∗ f ′ (u∗ )υ̇ = 0 in Ω,

𝜕ν υ̇ + αυ̇ = 0 on 𝜕Ω,

it follows that λ̇ ∫Ω f (u∗ )ϕ∗ dx = 0. Our assumption f ′ (u) > 0 implies that ϕ∗ does not change sign and thus ∫ f (u∗ )ϕ∗ dx ≠ 0. Ω

Hence, λ̇ = 0. Since λ∗ is simple it follows that υ̇ = cϕ∗ . The equation 2

Δϕ̇ + λ∗ f ′ (u∗ )ϕ̇ + cλ∗ f ′′ (u∗ )(ϕ∗ ) = 0 together with (9.2.3) and (9.2.4) implies that c = 0. It follows now immediately that ϕ̇ = bϕ∗ and the normalization yields b = 0. Hence, by the implicit function theorem G = (0, 0, 0, 0, 1) has for |t| < t0 a unique solution (λ(t), u(t), ϕ(t), t) such that (λ(0), u(0), ϕ(0)) = (λ∗ , u∗ , ϕ∗ ).

9.2 Nonlinear eigenvalue problems � 159

In addition to the existence of a turning point the implicit function theorem implies that for small |t|, u(t), ϕ(t), λ(t) are twice continuously differentiable with respect to t.

9.2.1 Examples A classical example is the Gelfand problem Δu + λeu = 0 [64] for λ > 0. The 1 − d Gelfand equation Let Ω be the interval (−a, a). By scaling, the differential equation u′′ + λeu = 0 transforms to the equation u′′ + a2 λeu = 0 on (−1, 1). We set λa := a2 λ. Without specifying the boundary conditions the solution is u(x) := ln(

1 − tanh2 ( x+d ) 2c 2c2 λa

).

(9.2.5)

We consider only symmetric solutions and set therefore d = 0. In particular, u′ (0) = 0. For Dirichlet data u(±1) = 0, (9.2.5) yields 1=

1 − tanh2 ( 2c1 ) 2c2 λa

.

For a given λa this equation gives a condition on c ≥ 0. It is equivalent to ZD (c, λa ) := hD (c) − λa = 0,

where hD (c) :=

1 1 (1 − tanh2 ( )), 2 2c 2c

for all c ≥ 0. Clearly, h ≥ 0 for c ≥ 0 and h(0) = limc→∞ h(c) = 0 (see Figure 9.2, where we set a = 1). There is a critical c∗ such that the function ZD has exactly two zeros c1 and

Figure 9.2: The function hD .

160 � 9 Eigenvalue problems c2 for 0 < λa < hD (c∗ ). Consequently, ui (x) = ln(

1 − tanh2 ( 2cx ) 2ci2 λa

i

for i = 1, 2

)

both solve the Dirichlet boundary value problem. Note that for i = 1, 2 ‖ui ‖L∞ (−1,1) = ui (0) = (

1 ). 2ci2 λa

Figure 9.3a shows the values of u(0) in dependence of λa for a = 1. For the Robin boundary condition ±u′ (±1) + αau(±1) = 0 with α > 0, the constant c must satisfy ZR (c, λa , α) := hR (c, α) − λa = 0. Here hR (c, α) :=

1 −2 A(c,α) 2 (1 + e c ) e B(c,α) 2 c

with 1

A(c, α) = 1 + α + (α − 1)e c

and

1

B(c, α) = αc(1 + e c ).

Note that limα→∞ hR (c, α) = hD (c).

Figure 9.3: The function hR .

Figure 9.3b shows hR as a function of c for different values of α and a = 1. As before, there is a critical value c∗ such that there are two zeros c1 and c2 of ZR for 0 < λa < hR (c∗ ). The two values c1 and c2 give two different solutions u1 and u2 .

9.2 Nonlinear eigenvalue problems

� 161

The turning point λ∗ for the Dirichlet (resp. Robin) boundary value problem is the largest value such that ZD (resp. ZR ) has two solutions for 0 < λa < λ∗ . Thus, λ∗ = hD (c∗ ) (resp. λ∗ = hR (c∗ )). Remark 9.1. The solution of the one-dimensional Gelfand equation such that u′ (0) = 0 without taking into account any boundary conditions can be calculated by multiplying u′′ + λa eu = 0 by u′ and then integrating it from 0 to x ∈ (−a, a). This leads to u′ 2 (x) + 2λa (eu(x) − eu(0) ) = 0.

(9.2.6)

Because of the equation u′′ = −λa eu < 0 for λa > 0, u is concave on (−a, a). Hence, (9.2.6) implies u′ (x) = −√2λa [eu(0) − eu(x) ]. Separation of variables and integration yield u(0)

λ e u(x) = u(0) − 2 log cosh(√ a 2 u′ (x) = −2√

x),

λa eu(0) λ eu(0) tanh(√ a x). 2 2

The Robin boundary condition at x = ±1 leads to the following relations between λ and u(0) = ‖u‖L∞ : u(0)

λ e 2√ a 2

u(0)

λ e tanh(√ a 2

u(0)

λ e ) = α[u(0) − 2 ln cosh(√ a 2

)].

We write this equation as 2√z tanh(√z) = α[ln(

2z ) − 2 ln(cosh(√z))], λa

z :=

λa eu(0) , 2

and solve it for λa : λa =

8z

(1 + e2√z )2

eA(z,α) := h̃ R (z, α),

A(z, α) =

2√z[e2√z α − e2√z + α + 1] α(1 + e2√z )

.

There is a maximal range of values for λa for which there exist two values of u(0), i. e., two solutions of the 1 − d Gelfand equation.

162 � 9 Eigenvalue problems The radial Gelfand equation in the disc The Gelfand problem Δu+λeu = 0 in the disc BR can be transformed after scaling r → r/R into the problem Δu + λR eu = 0 in B1 , where λR = λR2 . The radial solutions are of the form 8β , λR (r 2 + β)2

u(r) = ln

λR := R2 λ.

In the case of Dirichlet data, u(1) = 0. By the maximum principle the solution is positive. There are two values of β for which the boundary condition is satisfied, namely β1 =

1 (4 − λR + 2√4 − 2λR ) and λ

β2 =

1 (4 − λR − 2√4 − 2λR ). λ

The corresponding solutions ui (r) (for β = βi ) have a unique maximum in r = 0; thus, ‖ui ‖L∞ (B1 ) = ui (0). Obviously, there is no solution if λR > 2. Consequently, the turning point for BR is λ∗ = R22 . After scaling, the Robin boundary conditions 𝜕ν u + αu = 0 on 𝜕BR become ur (1) + αRu(1) = 0 with α > 0. The parameter β is then determined by the zeros of Zrad :=

8β − 4 e (1+β)αR − λR . (1 + β)2

Set hrad (β) :=

4 8β − (1+β)αR e . (1 + β)2

8β Note that limα→∞ hrad = (1+β) 2 corresponds to the case of Dirichlet data. Clearly, hrad ≥ 0 for β ≥ 0 and limβ→0 hrad (β) = limβ→∞ hrad (β) = 0. This is true for all α > 0. The λR corresponding to the turning point is given by

λ∗R = hrad (β∗ ) = max{ β>0

The maximum is attained for β∗ =

λ∗R

∗ 2

=λ R =

2 γ

8β − 4 e αR(1+β) }. 2 (1 + β)

+ √ γ42 + 1, where γ = αR. This leads to

8γ(2 + √γ2 + 4) exp(

−4

γ+2+√γ2 +4

(γ − 2 + √γ2 + 4)2

) .

Figure 9.4 shows that λ∗ tends to 2 as γ tends to ∞. Hence, for fixed R > 0 the turning point for the Robin problem converges to the turning point of the Dirichlet problem as α → ∞. The three dots in Figure 9.4 (a) show the different values of α for which the graph of λ → u(0) is plotted in Figure 9.4 (b) for R = 1.

9.2 Nonlinear eigenvalue problems � 163

Figure 9.4: Results for the ball.

9.2.2 First variation of the turning point In this section we compute the first variation of the turning point of the nonlinear problem (9.2.1). For simplicity we shall write λ(t) instead of λ∗ (t). We will use the same notation as in Chapter 6, that is, ũ is the solution of (9.2.2) and ϕ̃ is the solution of (9.2.3) in Ωt after pullback to Ω. It follows from the general discussion in Section 9.2 that λ(t) can be represented with a Rayleigh quotient. After the change of variables it reads as λ(t) =

̃ ̃ ̃ ⋅ A(t)∇ϕ(t)) ̃ ϕ(t)m(t) dx + α ∮𝜕Ω u(t) dS ∫Ω (∇u(t) . ̃ ̃ ϕ(t)J(t) dx ∫ f (u(t)) Ω

Differentiation in t = 0 together with the fact that J(0) = m(0) = 1 and Aij (0) = δij yields ̇ ∫ f (u)ϕ dx + λ ∫[f ′ (u)uϕ ̇ ̇ + f (u)ϕ̇ + f (u)ϕJ(0)] λ(0) dx Ω

Ω

̇ dx ̇ = ∫[(∇u̇ ⋅ ∇ϕ) + (∇u ⋅ A∇ϕ) + (∇u ⋅ ∇ϕ)] Ω

̇ ̇ + uϕ̇ + uϕm(0)] + α ∮[uϕ dS.

(9.2.7)

𝜕Ω

̇ Testing Δu + λf (u) = 0 in Ω and 𝜕ν u + αu = 0 on 𝜕Ω with ϕ(0) we get ̇ dS. λ ∫ f (u)ϕ̇ dx = ∫(∇ϕ̇ ⋅ ∇u) dx + α ∮ ϕu Ω

Ω

𝜕Ω

̇ we get Testing Δϕ + λf ′ (u)ϕ = 0 in Ω and 𝜕ν ϕ + αϕ = 0 on 𝜕Ω with u(0)

164 � 9 Eigenvalue problems λ ∫ f ′ (u)ϕu̇ dx = ∫(∇ϕ ⋅ ∇u)̇ dx + α ∮ ϕu̇ dS. Ω

Ω

𝜕Ω

Introducing these expressions in (9.2.7) we obtain ̇ dx + α ∮ uϕṁ dS. λ̇ ∫ f (u)ϕ dx + λ ∫ f (u)ϕJ ̇ dx = ∫(∇u ⋅ A∇ϕ) Ω

Ω

Ω

(9.2.8)

𝜕Ω

In the sequel we shall write u and ϕ for the functions u and ϕ corresponding to λ∗ . In ∗

∗

t = 0 we apply (2.1.5), (4.1.16), (2.3.19), and (2.3.21). Hence, ∫ f (u)ϕ dx = ∫ f (u∗ )ϕ∗ dx, Ω

Ω

∫ f (u)ϕJ ̇ dx = ∮ f (u∗ )ϕ∗ (υ ⋅ ν) dS − ∫[f ′ (u∗ )ϕ∗ (∇u∗ ⋅ υ) dx Ω

Ω

𝜕Ω

− ∫ f (u )(∇ϕ ⋅ υ)] dx, ∗

∗

Ω

̇ dx = ∮[(υ ⋅ ν)(∇u∗ ⋅ ∇ϕ∗ ) + αϕ∗ (∇u∗ ⋅ υ) + αu∗ (∇ϕ∗ ⋅ υ)] dS ∫(∇u ⋅ A∇ϕ) Ω

𝜕Ω

− λ∗ ∫[f ′ (u∗ )ϕ∗ (∇u ⋅ υ) + f (u∗ )(∇ϕ∗ ⋅ υ)] dx, Ω

and ̇ dS = ∮[(n − 1)H(υ ⋅ ν)u∗ ϕ∗ − (υτ ⋅ ∇τ (u∗ ϕ∗ ))] dS. ∮ uϕm(0) 𝜕Ω

𝜕Ω

Consequently, ̇ ∫ f (u∗ )ϕ∗ dx = α ∮[(υ ⋅ ∇u∗ )ϕ∗ + (υ ⋅ ∇ϕ∗ )u∗ ] dS λ(0) Ω

𝜕Ω

+ ∮(υ ⋅ ν)[(∇u∗ ⋅ ∇ϕ∗ ) − λ∗ f (u∗ )ϕ∗ ] dS 𝜕Ω

+ α ∮[(n − 1)H(υ ⋅ ν)u∗ ϕ∗ − (υτ ⋅ ∇τ (u∗ ϕ∗ ))] dS. 𝜕Ω

For ρ = (υ ⋅ ν) this becomes ̇ ∫ f (u∗ )ϕ∗ dx = −2α2 ∮ ρu∗ ϕ∗ dS + α(n − 1) ∮ Hρu∗ ϕ∗ dS λ(0) Ω

𝜕Ω

𝜕Ω

+ ∮ ρ[(∇u ⋅ ∇ϕ ) − λ f (u )ϕ∗ ] dS. ∗

𝜕Ω

∗

∗

∗

(9.2.9)

9.2 Nonlinear eigenvalue problems � 165

On the boundary ∇u∗ and ∇ϕ∗ can be decomposed in their normal and tangential components. Thus, in view of the Robin boundary conditions (∇u∗ ⋅ ∇ϕ∗ ) = α2 u∗ ϕ∗ + (∇τ u∗ ⋅ ∇τ ϕ∗ ). Finally we get for Hadamard perturbations ̇ ∫ f (u∗ )ϕ∗ dx λ(0) Ω

= ∮ ρ[(∇τ u∗ ⋅ ∇τ ϕ∗ ) − α2 u∗ ϕ∗ − λ∗ f (u∗ )ϕ∗ + α(n − 1)Hu∗ ϕ∗ ] dS.

(9.2.10)

𝜕Ω

For volume preserving perturbations we get an overdetermined boundary value problem for the bilinear form in u∗ and ϕ∗ . Example 9.2. Let Ω be the ball BR . In this case u := u∗ and ϕ := ϕ∗ are radial. The boundary conditions ∇u(R) = −αu(R)ν and ∇ϕ(R) = −αϕ(R)ν and (9.2.10) implies ̇ ∫ f (u)ϕ dx = αϕ(R)u(R) ∮ m(0) ̇ λ(0) dS + α2 u(R)ϕ(R) ∮ (υ ⋅ ν) dS BR

𝜕BR

𝜕BR

− λf (u(R))ϕ(R) ∮ (υ ⋅ ν) dS. 𝜕BR

̇ Since by (2.3.19), m(0) = div𝜕Ω υτ +

n−1 (υ R

⋅ ν), we obtain

̇ ∫ f (u)ϕ dx = [α n − 1 − α2 − λ f (u(R)) ]ϕ(R)u(R) ∮ (υ ⋅ ν) dS. λ(0) R u(R) BR

𝜕BR

̇ vanishes for volume preserving perturbations. This formula reveals that λ(0) that

In the case of Dirichlet boundary conditions, (9.2.8) together with (4.1.16) implies λ̇∗ (0) ∫ f (u∗ )ϕ∗ dx = − ∮ ρ(∇u∗ ⋅ ∇ϕ∗ ) dS. Ω

(9.2.11)

𝜕Ω

This result was first obtained in [88]. Problem 9.2. Does the overdetermined boundary condition (9.2.11) for volume preserving perturbations imply that Ω is a ball? A first step in this direction would be to show that for a critical domain 𝜕ν u = const. and 𝜕ν ϕ = const. on the boundary. Problem 9.3. The second variation is rather involved because it depends on two functions u∗ and ϕ∗ and their shape derivatives. The determination of its sign remains open.

166 � 9 Eigenvalue problems

9.3 Notes The first and the second domain variation of the first Dirichlet eigenvalue of the Laplace operator were derived in [63]. In this paper the second domain variation is expressed in terms of the material derivatives, which makes it difficult to discuss their sign. The Rayleigh–Faber–Krahn inequality implies that the ball is not only a local but also a global minimizer of the first eigenvalue among all domains of given volume [98]. This is also true for the first Robin eigenvalue (see [25, 41]). For further studies see [74]. Daners and Kennedy proved [42] that the ball is the only minimizer. A different approach which requires less regularity has been taken by Buçur and Giacomini [29]. Problems with large α are studied in [85]. The nonlinear eigenvalue problem was studied in [16]. In [17] and for more general problems in [18] it was shown that in the case of Dirichlet boundary conditions the ball has the smallest turning point among all domains of given volume. The Gelfand problem has applications in several fields, like astrophysics and the theory of combustion. It is the stationary solution of the Frank-Kamenetskii model in combustion. In 1973 Joseph and Lundgren [78] completely characterized the structure of the radial solutions of the Gelfand equation with Dirichlet data for all dimensions. A detailed discussion of this equation is found in [118]. Mignot, Murat, and Puel [88] studied the variation of the turning point of the Gelfand problem in the case of Dirichlet data. Henry [76] extended their result to more general nonlinearities.

10 Quantitative estimates The stability of an optimal domain plays an important role in applications. By stability we mean the existence of a neighborhood of the optimal domain in which there is no other optimal domain. A quantitative inequality gives an estimate for the size of this neighborhood. We focus mainly on nearly spherical domains. A rather general criterion based on the second variation is derived and then applied to various examples.

10.1 Preliminary remark In the calculus of variations it is well known that a critical point with a positive second variation may not be a strict (isolated) minimum. This is shown by an example taken from [35]. Let J : L2 (0, 1) → ℝ be given by 1

J(u) := ∫ sin(u(s)) ds. 0

Clearly, u∗ (s) := − π2 is a global minimizer. It is easily checked that: (1) J ′ (u∗ )[φ] = 0 and (2) J ′′ (u∗ )[φ, φ] = ‖φ‖L2 (0,1) . Property (2) is often referred to as “strict coercivity” with respect to L2 . The family of functions −π uϵ (s) := { 3π2 2

if s ∈ [0, 1 − ϵ], if s ∈ [1 − ϵ, 1]

also satisfies J(uϵ ) = J(u∗ ) = 0 for all 0 < ϵ < 1. Moreover, ‖u∗ − uϵ ‖L2 = 2π √ϵ. Thus, in any L2 -neighborhood of u∗ there are infinitely many different global minimizers. The reason for this phenomenon is a lack of differentiability of the functional J. While it is twice differentiable on L∞ (0, 1) it is not on the larger space L2 (0, 1). One might be tempted to restrict the functional J to the smaller space L∞ (0, 1). However, then property (2) is not satisfied: There does not exist any δ > 0 such that J ′′ (u∗ )[φ, φ] ≥ ‖φ‖L∞ (0,1) . This is called the “norm discrepancy phenomenon.” For the domain functionals the situation is similar. They are differentiable in the class of smooth domains; however, the second domain variation typically provides an in̈ ≥ c‖υ‖L2 . We will show that the norm discrepancy phenomenon equality of the type ℰ (0) cannot occur because of the smoothness of the domain perturbations. Thus, any optimal domain is isolated in a sense that will be specified in the next section. https://doi.org/10.1515/9783111025438-010

168 � 10 Quantitative estimates

10.2 The Fraenkel asymmetry function Let {Ωt }|t| 0 and ℱ (0) < 0. We define δ(t) := −

δℱ (t) . δℐ (t)

Expansion with respect to t yields δ(t) = −

ℐ (0) ℱ̈ (0) + o(1). ℱ (0) ℐ̈ (0)

Assume there is a γ− > 0 such that γ− ≤ δ(t) = −

ℐ (0) ℱ (t) − ℱ (0) . ℱ (0) ℐ (t) − ℐ (0)

(10.2.4)

Then we deduce 󵄨󵄨

󵄨󵄨

− 󵄨 ℱ (0) 󵄨󵄨 ℱ (t) − ℱ (0) ≥ γ 󵄨󵄨󵄨 󵄨(ℐ (t) − ℐ (0)). 󵄨󵄨 ℐ (0) 󵄨󵄨󵄨

Clearly, the same result holds if ℐ (0) < 0 and ℱ (0) > 0. Next we will apply (10.2.3) and (10.2.5) for different domain functions.

10.3 Example 10.3.1 The Robin energy for α > 0 Let Ω = BR , α > 0 and consider the Robin energy 2

2

ℰ (u, Ω) = ∫ |∇u| − 2G(u) dx + α ∮ u dS, Ω

𝜕Ω

where u solves Δu + g(u) = 0 𝜕ν u + αu = 0

in Ω, in 𝜕Ω.

(10.2.5)

10.3 Example

� 171

We assume ∫ ug(u) − 2G(u) dx < 0. Ω

This ensures ℰ (0) < 0. Let {Ωt }|t| 0 and ℰ (0) < 0 we are in the case of Section 10.2.1. We like to verify (10.2.4). Note that −

̈ ℐ (0) ℰ (0) 𝒮 (0) ℱ =− (αu2 (R) + ). ̈ ℰ (0) ℐ̈ (0) ℰ (0) 𝒮 (0)

For the torsion problem g = 1 and α > 0 it follows that ℱ ≥ 0. In this case, lim inf δ(t) = lim inf t→0

t→0

δ(ℰ (t)) S(0) 2 = αu (R) > 0. δ(𝒮 (t)) ℰ (0)

Thus, there exists a t0 > 0 such that for all |t| < t0 we have δ(t) > following proposition holds.

S(0) αu2 (R). 2ℰ(0)

Thus the

Proposition 10.1. Assume there exists a t0 > 0 such that the isoperimetric deficit δS(t) > c0 for all |t| < t0 . Assume ℱ is nonnegative. Then we have ℰ (t) − ℰ (0) ≥

1 2 αu (R)(S(t) − S(0)). 2

Hence, the ball is a strict local minimizer for the Robin energy. Robin eigenvalue problem Next we discuss the stability of the lowest eigenvalue λ(t) for nearly spherical domains. For this purpose we assume that λ(t) is three times continuously differentiable or equivalently υ and w are in C 3,1 . We set δλ(t) :=

λ(t) δλ(t) − 1 and δ(t) := . λ(0) δ𝒮 (t)

̈ leads Then expansion of λ(t) around t = 0 together with the lower bound (9.1.20) for λ(0) to

172 � 10 Quantitative estimates

δ(t) =

̈ 𝒮 (0)λ(0)

̈ λ(0)𝒮 (0)

≥α Recall that

α K

−

1 μ2

+ o(1) ≥

2

𝒮 (0)u (R)

λ(0)

𝒮 (0)u2 (R)

λ(0)

{

2K 2 α 1 ∞ [ − ] ∑ b2i + α} ̈ K μ2 i=2 𝒮 (0)

.

> 0. Hence, λ(t) − λ(0) ≥ α𝒮 (0)u2 (R).

Another example of a quantitative estimate is the first Steklov eigenvalue for the biLaplace operator (see Chapter 16).

10.3.2 Notes Quantitative estimates for optimal domains attracted recently a lot of attention. The first results date back to the first half of the nineteenth century, when Bernstein and Bonnesen proved quantitative estimates for the isoperimetric inequality. Fuglede [58] extended their results to higher dimensions. In the last decades there has been growing interest in sharpening the classical isoperimetric inequality in terms of an asymmetric functional. For this purpose Fusco et al. [60] used the Fraenkel asymmetry measure. We refer to [61] and to [62] for a review on this topic. Subsequently, Brasco et al. [26] studied improvements of isoperimetric inequalities related to eigenvalues. More recent results can be found in [27].

11 The Robin eigenvalues for α < 0 Eigenvalue problems with negative α are of interest for bounded smooth domains Ω and for its complement ℝn \ Ω. Domain variations lead to different results than the problems with positive α. It will be shown that the ball and the outer ball are local maximizers for the first eigenvalue λ(Ω) among all nearly spherical and exterior spherical domains of equal volume.

11.1 The interior eigenvalue problem The equation Δu + λ(Ω)u = 0 in Ω, 𝜕ν u + αu = 0

(11.1.1)

on 𝜕Ω

with α < 0 was studied by Bareket [20]. The trace inequality (see Section D.1) implies λ1 (Ω) > −∞. By the direct method of the calculus of variations there exists a minimizer u ∈ W 1,2 (Ω) which is a solution to (11.1.1). As for α > 0 this problem also has a countable number of eigenvalues tending to infinity. The minimizer is of constant sign and the eigenspace of the smallest eigenvalue is one-dimensional (see Theorem D.1 and Section D.2.2). Hence, λ(Ω) < λ2 (Ω) ≤ ⋅ ⋅ ⋅ ≤ λn (Ω) ≤ ⋅ ⋅ ⋅ . The variational characterization of the first eigenvalue λ(Ω) is given by λ(Ω) = inf{∫ |∇υ|2 dx + α ∮ υ2 dS : υ ∈ W 1,2 (Ω), ∫ υ2 dx = 1}. Ω

Let υ =

1 . √|Ω|

(11.1.2)

Ω

𝜕Ω

Then λ(Ω) ≤ α

|𝜕Ω| . |Ω|

Thus, λ(Ω) is negative and limα→−∞ λ(Ω) = −∞. By Poincaré’s principle (see Theorem D.1), any higher eigenfunction ui satisfies ∫ uui dx = 0. Ω

Since the first eigenfunction u is of constant sign, ui for i > 1 has to change sign. https://doi.org/10.1515/9783111025438-011

174 � 11 The Robin eigenvalues for α < 0 Under the assumption that the eigenfunction of the perturbed domain Ωt , u(y, t), is normalized by ∫Ω u(y, t)2 dy = 1, the first variation of a simple eigenvalue λ(Ω) is the t same as for positive α, namely (see (9.1.8)) ̇ λ(0) = ∮(υ ⋅ ν)[|∇u|2 − u2 (λ(0) + 2α2 − α(n − 1)H)] dS. 𝜕Ω

Domain variations for nearly spherical domains Let {Ωt }|t| |BR | for small |t|. Then ̇ λ(0) > 0. This improves the local monotonicity property derived in [19]. Note that this observation extends partly the result of Giorgi and Smits [67], who proved that λ(Ω) > λ(BR ) for any BR ⊂ Ω. Next we want to discuss the second variation of the lowest eigenvalue in balls. The computations of Section 9.1.3 apply. The only difference is that λ is negative and therefore the Bessel functions in u and ϕ have to be replaced by modified Bessel functions.

11.1 The interior eigenvalue problem

� 175

We recall the Steklov problem (9.1.15) associated to the Robin eigenvalue problem with negative λ and α: Δϕ + λϕ = 0

in BR ,

𝜕ν ϕ + αϕ = μϕ on 𝜕BR .

(11.1.6)

It was discussed in Section 8.2.1, where g ′ (u(r)) = λ < 0 in this section. Let μ0 < μ1 < ⋅ ⋅ ⋅ denote eigenvalues of the eigenvalue equation (11.1.6) without counting their multiplicity (see Section 9.1.3). Since the first eigenfunction of the Robin eigenvalue problem u1 is an eigenfunction of the Steklov problem corresponding to μ = 0 and it does not change sign, we obtain μ = μ0 = 0. In Lemma 8.2 the eigenfunctions ϕk,i were expanded with respect to the spherical harmonics. The radial coefficients of this expansion with negative λ are given by as (r) = an r

2−n 2

Is+ n −1 (√|λ|r). 2

Hence for volume preserving perturbations and by (9.1.17) we obtain ∞ α 1 ̈ λ(0) = 2u2 (R)K 2 ∑ b2i [ − ] + αu2 (R)𝒮0̈ (0), K μi i=1

(11.1.7)

where K = α2 −

n−1 α + λ < 0. R

(11.1.8)

The sum starts with i = 1 because the perturbations are volume preserving. Since (9.1.19) holds also for the modified Bessel functions [

1 α − ] = 0. K μ1

From (11.1.7) and μi > 0 we obtain the following upper bound: ∞

̈ λ(0) ≤ 2u2 (R)Kα ∑ b2i + αu2 (R)𝒮0̈ (0). i=2

(11.1.9)

In Lemma 2.1 it was shown that the barycenter condition implies 𝒮0̈ (0) ≥

n+1 ∮ (υ ⋅ ν)2 dS. R2 𝜕BR

Thus, n+1 ̈ λ(0) ≤ αu2 (R){2K + 2 } ∮ (υ ⋅ ν)2 dS. R 𝜕BR

(11.1.10)

176 � 11 The Robin eigenvalues for α < 0 Proposition 11.1. Assume α < 0. Then n+1 ̈ λ(0) ≤ αu2 (R){2K + 2 } ∮ (υ ⋅ ν)2 dS < 0 R 𝜕BR

for all volume preserving perturbations satisfying the barycenter condition. Proof. We will show the positivity of the bracket {2K + this is the case if

n+1 }. R2

From (11.1.8) and (11.1.10)

n−1 n+1 |α| + > −λ. R 2R2

α2 +

(11.1.11)

Step 1. The value λ(BR ) is implicitly given as a solution of u′ (R) + αu(R) = 0, where u(r) is given in (11.1.5). Thus, (−

n−2 + αR)I n−2 (√−λR) + √−λRI ′n−2 (√−λR) = 0. 2 2 2

Next we use the rule for derivatives Iν′ (z) =

ν I (z) + Iν+1 (z), z ν

where Iν′ (z) = 𝜕z Iν (z). This implies αI n−2 (√−λR) + √−λI n (√−λR) = 0. 2

2

We apply the recurrence identity Iν (z) = for ν =

n−2 . 2

2(ν + 1) Iν+1 (z) + Iν+2 (z) z

This gives αR√−λI n +1 (√−λR) − (λR − nα)I n (√−λR) = 0. 2

2

(11.1.12)

This equation needs to be satisfied by some λ for which also (11.1.11) holds. Step 2. For s ∈ [0, 1] we set ℓ(s) := s(α2 + |α|

n−1 n+1 + ). R 2R2

We want to show that there exists an s∗ ∈ [0, 1] such that −ℓ(s∗ ) = λ.

(11.1.13)

11.1 The interior eigenvalue problem � 177

Step 3. We reformulate the problem. We set y := |α|R > 0 and z(s) := √s(y2 + (n − 1)y +

n+1 ). 2

Then −ℓ(s∗ ) is a solution of (11.1.12) if F(s∗ ) := −I n +1 (z(s∗ ))yz(s∗ ) + I n (z(s∗ ))(z2 (s∗ ) − ny) = 0. 2

2

(11.1.14)

We want to show that F(s) < 0 for s sufficiently small and F(1) > 0. The intermediate value theorem implies the existence of an s∗ such that (11.1.14) holds. Step 4. For s small enough such that 0 < z(s) < √ n2 + 1 we have the asymptotic expansion F(s) ∼ −nI n (z(s))y < 0. 2

This implies F(s) < 0 for small s. Step 5. For s = 1 we set z := z(1) and show that F(1) = −I n +1 (z)yz + I n (z)(z2 − ny) > 0, 2

2

(11.1.15)

where z = √y2 + (n − 1)y +

n+1 . 2

We write (11.1.15) in the equivalent form, using the fact that z2 − ny > 0 for all n and all y > 0, namely F1 :=

zI n (z) 2

I n +1 (z) 2

−

yz2 > 0. z2 − ny

Step 6. The Simpson–Spector inequality states (see, e. g., [109, Theorem 2] or [128, formula (1.9)]) that for all z > 0 and all n ≥ 2 there holds zI n (z) 2

I n +1 (z) 2

≥

n+1 √ 2 n+1n+3 + z + . 2 2 2

This lower bound reduces the proof of F1 > 0 to 2 n+1 √z2 + n + 1 n + 3 > yz − . 2 2 2 z2 − ny

178 � 11 The Robin eigenvalues for α < 0 Note that for small y the right-hand side becomes negative. We will show that for all y ≥ 0, z2 +

2

yz2 n+1 n+1n+3 >( 2 − ). 2 2 2 z − ny

This is equivalent to 1 1 1 y4 − 2y3 + (2n3 + 6n + 8)y2 + (2n3 − 6n − 4)y + (n + 1)3 > 0. 4 4 4 Note that the sum of the last two terms in parentheses is always positive. For the other terms we note that 1 1 y4 − 2y3 + (2n3 + 6n + 8)y2 = y2 (y2 − 2y + (2n3 + 6n + 8)) > 0. 4 4 This proves F(1) > 0 and hence the existence of an s∗ such that (11.1.14) holds. Thus, among all domains of equal volume the ball is a local maximizer. Problem 11.1. Is the statement in Proposition 11.1 true for all volume preserving perturbations even without requiring the barycenter condition? Problem 11.2. Is it true that the ball has the largest first eigenvalue in the class of simply connected domains of prescribed volume?

11.2 The exterior eigenvalue problem Let Ω be a nearly spherical domain in the sense of Section 8.2. We will write D := ℝn \ Ω and call D a spherical exterior domain (see Figure 11.1). For α < 0 we consider the eigenvalue problem λ(D) := inf{∫ |∇u|2 dx + α ∮ u2 dS : u ∈ W 1,2 (D), ∫ u2 dx = 1}. D

𝜕Ω

(11.2.1)

D

Lemma 11.1. Let D be a spherical exterior domain. Then there exists a positive constant c = c(α) such that λ(D) ≥ −c(α). Proof. It was shown in [108] that there exists a linear and continuous trace operator 1 T : W 1,2 (D) → W 2 ,2 (𝜕Ω) such that T(u) = u|𝜕Ω for u ∈ C 1 (D). For any Lipschitz boundary 1 𝜕Ω the imbedding W 2 ,2 (𝜕Ω) 󳨅→ L2 (𝜕Ω) is compact. Hence, the same argument as for bounded domains applies and proves the claim.

11.2 The exterior eigenvalue problem

� 179

Figure 11.1: Exterior domain.

In [81] the following result for exterior domains D was shown. Proposition 11.2. There exists a number α⋆ (D) ≤ 0 such that λ(D) is a negative discrete eigenvalue if and only if α < α⋆ (D). Moreover, α⋆ (D) = 0 for n = 2 and α⋆ (D) < 0 for n ≥ 3. . In particular, for the exterior of a ball of radius R there holds α⋆ (ℝn \ BR ) = − n−2 R It is well known that under the assumptions of Proposition 11.2 there exists an infinite sequence of eigenvalues λ1 (D) < λ2 (D) ≤ ⋅ ⋅ ⋅ with ∞ as the only accumulation point. For the rest of this section we assume α < α⋆ (D) and we set DR := ℝn \ BR . From [65, Theorem 8.38] we know that λ(D) := λ1 (D) is simple and the corresponding eigenfunction is of constant sign. The first eigenfunction in DR is radial. It solves u′′ (r) +

n−1 ′ u (r) + λ(DR )u(r) = 0 r

in (R, ∞)

(11.2.2)

with the boundary conditions u′ (R) − αu(R) = 0

and

lim u(r) = 0.

r→∞

(11.2.3)

The solution is u(r) = Cn r

2−n 2

󵄨 󵄨 K n−2 (√󵄨󵄨󵄨λ(DR )󵄨󵄨󵄨r), 2

(11.2.4)

where K n−2 denotes the modified Bessel functions of second kind of order n−2 . The con2 2 stant Cn is determined by the normalization, while the eigenvalue is implicitly given by the boundary condition. For given t0 and |t| < t0 let {Ωt }|t| 0, dr 2 ρ

which is a contradiction. Therefore it follows K ≤ 0. The case K = 0 can be excluded in a similar way. For the exterior domain we also obtain the same upper bound as for the interior problem. n+1 ̈ λ(0) ≤ αu2 (R){2K + 2 } ∮ (υ ⋅ ν)2 dS. R

(11.2.7)

𝜕BR

Note that in this inequality the solution u(R) is expressed in terms of the Bessel functions Kμ rather than Iμ . The following result is analogous to Proposition 11.1.

182 � 11 The Robin eigenvalues for α < 0 Proposition 11.3. Assume α < α⋆ (BR ) for 2 ≤ n ≤ 5. Then n+1 ̈ λ(0) ≤ αu2 (R){2K + 2 } ∮ (υ ⋅ ν)2 dS < 0 R 𝜕BR

for all volume preserving perturbations which also satisfy the barycenter condition. Proof. From (11.2.7) we deduce that the proposition is proven if we can show that {2K + n+1 } is negative. Here K is given in (11.2.5). This is equivalent to R2 2α2 + 2

n−1 n+1 α + 2 + 2λ < 0, R R

(11.2.8)

where λ solves u′ (R) − αu(R) = 0. Step 1. We argue as in the proof of Proposition 11.1. Thus, the boundary condition for u(r) becomes (

2−n − αR)K n−2 (√−λR) + √−λRK ′n−2 (√−λR) = 0. 2 2 2

The differentiation rule Kν′ (z) =

ν K (z) − Kν+1 (z), z ν

where Kν′ (z) = 𝜕z Kν (z), implies √−λRK n (√−λR) + αRK n−2 (√−λR) = 0. 2

2

Finally we apply the recurrence identity Kν (z) = − for ν =

n−2 . 2

2(ν + 1) Kν+1 (z) + Kν+2 (z) z

This gives αR2 √−λK n +1 (√−λR) − K n (√−λR)(R2 λ + nαR) = 0. 2

2

(11.2.9)

Step 2. For s ∈ [0, 1] we set ℓ(s) := s(α2 +

n−1 n+1 α+ ) > 0. R 2R2

(11.2.10)

Note that the assumptions on α ensure the positivity of α2 + n−1 α + n+1 . As in the proof R 2R2 ∗ of Proposition 11.1 it remains to show that there exists an s ∈ [0, 1] such that the corresponding −ℓ(s∗ ) is a solution to (11.2.9).

11.2 The exterior eigenvalue problem

� 183

Step 3. In this step we reformulate the problem. Set y := |α|R and z(s) := √s(y2 − (n − 1)y +

n+1 ). 2

Since α < α∗ for 2 ≤ n ≤ 5 it follows from (11.2.10) that y2 − (n − 1)y + n+1 > 0. Then (11.2.9) 2 takes the form F(s) := K n +1 (z(s))yz(s) − K n (z(s))(z2 (s) + ny) = 0. 2

2

(11.2.11)

We claim that F(s) > 0 for s sufficiently small and F(1) < 0. Hence, the existence of an s∗ such that (11.2.11) holds is established. Step 4. In the sequel the Turan type inequality (see [106, Theorems 1 and 5 and Corollary 3]) ν 2

2

+ √ ν4 + z2 z


0 for sufficiently small s. Step 5. For s = 1 we set z := z(1) and want to show that for 2 ≤ n ≤ 5 and y > n − 2 F(1) = K n +1 (z)yz − K n (z)(z2 + ny) < 0, 2

2

where z = √y2 − (n − 1)y +

n+1 2

and

The cases n = 2, 3, 4, 5 will be discussed separately.

y ≥ n − 2.

(11.2.13)

184 � 11 The Robin eigenvalues for α < 0 Step 6. If n = 2 and y ≥ 0, then (11.2.13) reads as 3 F(y) := F(1) = K2 (z)yz − K1 (z)(y + y2 + ) < 0 2

(11.2.14)

where 3 z = √y2 − y + . 2 We will show that F is an increasing function in y and then we will use asymptotic behavior of K1 and K2 . We first compute F ′ (y) using the differentiation rule and the recurrence identity for the modified Bessel functions of second kind: (−8y4 + 4y3 − 4y2 − 10y + 6)K1 (z) + 4K0 (z)(y3 + 21 y2 +

F ′ (y) =

8y2

− 8y + 12

3 4

+ y)z

.

Note that the denominator of F ′ (y) is positive. We apply (11.2.12) for ν = 0. Hence, K0 (z) ≥

1 2

K1 (z)

+ √ 41 + z2

.

We thus obtain the lower bound ′

F (y) ≥ K1 (z)(

(−2y2 − y + 1)√4y2 − 4y + 7 + 4y3 + y + 4 2 + 2√4y2 − 4y + 7

).

It is easy to verify that the right-hand side is positive. Thus, F(y) is strictly increasing in y. To show that F(y) < 0 for all y we apply the asymptotic formula Kν (z) ∼ √

π −z 4ν2 − 1 1 e (1 + + o( )) 2z 8z z

as z → ∞. From (11.2.14) we thus get π −z 15 1 e (1 + + o( ))yz 2z 8z z π 3 1 3 − √ e−z (1 + + o( ))(y2 + y + ). 2z 8z z 2

F(y) ∼ √

Thus, F(y) ∼ −√πe−z

2z(8y2 − 7y + 12) − 16y3 + 22y2 − 18y + 9 + o( z1 )(yz − y2 − y − 32 ) 3

8(4y2 − 4y + 6) 4

as y → ∞. From this formula it also follows that limy→∞ F(y) = 0.

11.2 The exterior eigenvalue problem

� 185

Step 7. Let n = 3. In this case we use the explicit representation, K 3 (z) =

√πe−z (1 + z1 ) √2z

2

K 3 +1 (z) =

√πe−z (1 +

√2z

2

3 z

,

+

3 ) z2

.

For y ≥ 1, F(1) reads as F(y) := F(1) =

1 √π 2 √ 2 (y − 2y + 2) 4 (y − 1 − √y2 − 2y + 2)e− y −2y+2 . √2

Clearly, F(y) < 0 and consequently, (11.2.13) holds. Step 8. We consider the case n = 4. To prove (11.2.13) we set z=

√4y2 − 12y + 10 2

.

For y ≥ 2 we get F(y) := F(1) = 2

y2 − 3y +

5 2

((y − 2)K1 (z) − zK0 (z)),

√4y2 − 12y + 10

where again the differentiation rule for the modified Bessel functions of second kind has been used. From (11.2.12) we deduce K1 (z)
3. Clearly, F(y) < 0 for y > 3; thus, (11.2.13) holds. Remark 11.1. The situation is more delicate if n > 5. Inequality (11.2.10) holds if and only if α < α∗ := −

1 (n − 1 + √n2 − 4n − 1). 2R

(11.2.15)

It is easy to check that α∗ ≤ α∗ (see Proposition 11.2), with equality if and only if n = 5. The condition α < α∗ seems to be too weak to guarantee (11.2.13). Figure 11.2 shows F(1) as a function of y := |α|R, where α satisfies (11.2.15). Thus, for n > 5 our result is weaker than in [81].

Figure 11.2: The graph of F(1) as a function of y for n = 6.

Remark 11.2. L. Bundrock [33] proved Proposition 11.3 for all n ≥ 2. Krejcirik [80] has shown that for all α < 0 and in the plane the ball is a maximizer among all smooth convex domains of equal volume. The same result holds true if we compare the ball with smooth convex domains of the same perimeter. The situation changes considerably if n ≥ 3: We have λ(ℝn \ Ω) = 0 for α = α∗ (ℝn \ Ω). For the ball α∗ (ℝn \ BR ) = − n−2 . R

11.3 Notes

� 187

11.3 Notes Eigenvalue problems with Robin boundary conditions with negative α appear in acoustic analysis. Bareket [20] showed that for nearly circular domains of given area in the plane the circle has the largest first eigenvalue λ for α < 0. Her proof was based on the construction of suitable trial functions in the variational characterization. This result was extended to higher dimensions for nearly spherical domains by Ferone, Nitsch, and Trombetti [48]. The question whether or not the ball is optimal for all domains of the same volume remained open until Freitas and Krejcirik in [57] showed that annuli have a larger eigenvalue than the ball with the same volume for sufficiently small α < 0 (see also [4]). Further properties of this problem are found in [19]. The exterior problem is discussed in a series of papers [80, 81, 33].

12 Problems with infinitely many positive and negative eigenvalues This chapter deals with an eigenvalue problem which possesses two sequences of eigenvalues, one positive and one negative. The peculiarity of the problem is that the eigenvalues appear both in the equation in the domain and in the boundary condition. The smallest positive and the largest negative eigenvalue share some properties with the lowest Robin eigenvalue. Under certain conditions the ball is a global minimizer for the smallest positive eigenvalue and a local minimizer for the largest negative eigenvalue.

12.1 Eigenvalue problem with dynamical boundary conditions A reaction–diffusion problem with dynamical boundary conditions gives rise to the eigenvalue problem Δu + λu = 0 in Ω,

𝜕ν u = λσu in 𝜕Ω,

(12.1.1)

where σ, λ ∈ ℝ and Ω ⊂ ℝn is a bounded domain with smooth boundary. The special feature of this problem is that the eigenvalue appears in the domain and on the boundary. Except for λ0 = 0 they are critical values of the functional Jσ (υ) :=

∫Ω υ2 dx + σ ∫𝜕Ω υ2 dS ∫Ω |∇υ|2 dx

,

υ ∈ W 1,2 (Ω).

For positive σ, Jσ (υ) is positive and the classical theory for symmetric operators applies. François [55] showed that in this case the spectrum consists of countably many eigenvalues, which are bounded from below and tend to infinity. The lowest eigenvalue is zero and the corresponding eigenfunction is the constant. In this section we are interested in the problem with negative σ. It has been studied in [12]. A more general approach is found in [45]. It is known that in addition to λ = 0 there exist two sequences of eigenvalues, one tending to +∞ and the other tending to −∞. The eigenfunctions are complete in W 1,2 (Ω) except in the resonance case |Ω| + σ|𝜕Ω| = 0. The motivation to study this problem comes from the parabolic equations with dynamical boundary conditions 𝜕t u − Δu = 0

σ𝜕t u + 𝜕ν u = 0

in (0, ∞) × Ω,

in (0, ∞) × 𝜕Ω,

u(x, 0) = u0 (x)

https://doi.org/10.1515/9783111025438-012

in Ω.

12.2 Known results

� 189

It is known that they are well posed for positive σ in the space C([0, T], W 1,2 (Ω)) in the sense of Hadamard. Moreover, there exists a smooth solution globally in time, whereas for σ < 0 this is not the case in dimensions n ≥ 2.

12.2 Known results Some results for the case σ < 0 are summarized below. The results presented here are taken from [12]. For u, υ ∈ W 1,2 (Ω) let a(u, υ) := ∫ uυ dx + σ ∫ uυ dS Ω

𝜕Ω

be an inner product on L2 (Ω) ⊕ L2 (𝜕Ω). We define 𝒦 := {u ∈ W

1,2

(Ω) : ∫ |∇u|2 dx = 1, a(u, 1) = 0}. Ω

Moreover, let (u, υ) := ∫Ω (∇u ⋅ ∇υ) dx. We consider critical points of the functional 𝒥σ on

𝒦. Any critical point solves (12.1.1) in the weak sense:

(u, υ) = λa(u, υ) for all υ ∈ W 1,2 (Ω). In [12] the authors showed the existence of two infinite sequences of eigenvalues. One sequence consists of negative eigenvalues {λ−k }k and the other of positive eigenvalues {λk }k . The corresponding eigenfunctions {u±k }k ∈ 𝒦 solve (12.1.1). The eigenvalues are ordered like ⋅ ⋅ ⋅ ≤ λ−k ≤ ⋅ ⋅ ⋅ ≤ λ−1 < λ0 = 0 < λ1 ≤ λ2 ≤ ⋅ ⋅ ⋅ ≤ λk ≤ ⋅ ⋅ ⋅ . Note that they depend on σ. For the case n > 1 one calculates the following asymptotes: lim λ−k = −∞

k→∞

and

lim λk = ∞.

k→∞

In case n = 1 there are infinitely many positive eigenvalues but only finitely many negative ones. We are interested in the bottoms of the positive and the negative spectra, λ1 and λ−1 , for which the following theorem holds. Theorem 12.1. 1.

|Ω| If σ < σ0 = − |𝜕Ω| , then λ1 is simple and

{ } 1 = sup ∫ υ2 dx + σ ∮ υ2 dS } λ1 (σ) υ∈𝒦 { {Ω } 𝜕Ω

190 � 12 Two spectra 2.

If σ0 < σ < 0, then λ−1 is simple and { } 1 = inf {∫ υ2 dx + σ ∮ υ2 dS } . λ−1 (σ) υ∈𝒦 {Ω } 𝜕Ω

Lemma 12.1. The eigenvalue λ1 (σ) is for σ < σ0 and λ−1 (σ) is for 0 > σ > σ0 monotone decreasing in σ. Proof. We assume σ1 > σ2 . We distinguish two cases. Case 1. From the characterization of λ1 we get 1 ≥ ∫ u2 dx + σ1 ∫ u2 dS ≥ ∫ u2 dx + σ2 ∫ u2 dS. λ1 (σ1 ) Ω

Ω

𝜕Ω

𝜕Ω

For u we choose the eigenfunction of λ1 (σ2 ) and we obtain 1 1 ≥ . λ1 (σ1 ) λ1 (σ2 ) This gives λ1 (σ1 ) ≤ λ1 (σ2 ). Case 2. From the characterization of λ−1 we get 1 ≤ ∫ u2 dx + σ2 ∫ u2 dS ≤ ∫ u2 dx + σ1 ∫ u2 dS. λ−1 (σ2 ) Ω

Ω

𝜕Ω

𝜕Ω

In this case we choose u as the eigenfunction of λ−1 (σ1 ) and we obtain 1 1 ≤ . λ−1 (σ2 ) λ−1 (σ1 ) Since λ−1 (σ) < 0, we have λ−1 (σ1 ) ≤ λ−1 (σ2 ). Remark 12.1. 1.

In [12] the authors considered the map σ → λ(σ) given by λ1 (σ) if σ < σ0 , { { { λ(σ) = {0 if σ = σ0 , { { {λ−1 (σ) if 0 > σ > σ0 .

They proved that it is a smooth curve satisfying lim λ(σ) = λD1

σ→−∞

and

lim λ(σ) = −∞,

σ→0

where λD1 is the first Dirichlet eigenvalue for the Laplacian.

(12.2.1)

12.3 Comparison with balls of the same volume

2.

� 191

For 0 > σ > σ0 , λ1 (σ) is strictly decreasing and limσ→0 λ1 (σ) = λN1 (Ω), where λN1 (Ω) is the first nontrivial Neumann eigenvalue of the Laplacian. The different eigenvalues as functions of σ are shown in Figure 12.1.

Figure 12.1: Eigenvalues as functions of σ.

We are interested in the domain dependence of λ±1 (σ, Ω). Note that σ0 = σ0 (Ω) depends on Ω as well. In particular for domains of given volume and for a ball BR with the same volume the isoperimetric inequality gives σ0 (Ω) = −

|B | |B | |Ω| = − R ≥ − R = σ0 (BR ). |𝜕Ω| |𝜕Ω| |𝜕BR |

(12.2.2)

12.3 Comparison with balls of the same volume By exploiting previous results on the Robin eigenvalue problem we can prove the following isoperimetric inequality. Theorem 12.2. Let |Ω| = |BR | and assume σ < σ0 (BR ) < 0. Then λ1 (σ, Ω) ≥ λ1 (σ, BR ). Equality holds if and only if Ω = BR . Proof. Let λR (α, Ω), α > 0, be the lowest Robin eigenvalue defined in (11.1.1). The variational characterization (11.1.2) implies that λR (α, Ω) is concave and increasing as a function of α and λR (α, Ω) ≤ λD1 (Ω). We claim that lim λR (α, Ω) = λD1 (Ω),

(12.3.1)

α→∞

where λD1 (Ω) denotes the first Dirichlet eigenvalue of the Laplacian. This can be seen as follows. For any sequence {αi }i∈ℕ with limi→∞ αi = ∞ there exists a subsequence {αik }k∈ℕ such that the sequence {λR (α , Ω)} converges to some 0 < ζ ̂ ≤ λD (Ω). By scaling we ik

k∈ℕ

1

192 � 12 Two spectra may assume that the corresponding eigenfunctions are normalized, i. e., ‖φk ‖L2 (Ω) = 1. Since φk solves ∫(∇φk ⋅ ∇υ) dx + αik ∫ φk υ dS = λR (αik , Ω) ∫ φk υ dx

Ω

∀ υ ∈ W 1,2 (Ω),

(12.3.2)

Ω

𝜕Ω

we easily get the following convergence results: (i) The sequence {φk }k is bounded in W 1,2 (Ω); hence, there exists φ̂ ∈ W 1,2 (Ω) such that φk → φ̂ as k → ∞ weakly in W 1,2 (Ω) and thus strongly in L2 (Ω). In particular, ‖φ‖̂ L2 (Ω) = 1. (ii) Since ∫ |φk |2 dS ≤ 𝜕Ω

λR (αik , Ω) αik

→0

as k → ∞,

the compactness of the trace operator gives φ̂ = 0 a. e. on 𝜕Ω. Hence, υ̂ ∈ W01,2 (Ω). (iii) We can pass to the limit k → ∞ in (12.3.2) and get ̂ dx ∫ ∇φ̂ ⋅ ∇υ dx = ζ ̂ ∫ φυ Ω

Ω

∀ υ ∈ W01,2 (Ω).

(12.3.3)

From φ̂ ∈ W01,2 (Ω) and the fact that φ̂ solves (12.3.3) we deduce that φ̂ is an eigenfunction of the Dirichlet Laplace operator and ζ ̂ is a Dirichlet eigenvalue. Since 0 < ζ ̂ ≤ λD1 (Ω) we necessarily get ζ ̂ = λD1 (Ω). This establishes (12.3.1). The eigenvalue λ1 (σ, Ω) coincides with λR (α, Ω) for α = −σλ1 (σ, Ω). Hence, it solves the equation λR (|σ|λ, Ω) = λ or equivalently λR (α, Ω) =

α . |σ|

This equation has a solution if and only if (see Figure 12.2) 󵄨󵄨 1 d R |Ω| 󵄨 < λ (α, Ω)󵄨󵄨󵄨 . = 󵄨󵄨α=0 |𝜕Ω| |σ| dα Consequently, we must have σ < σ0 (Ω). From the Bossel–Daners inequality it follows α that λR (α, Ω) ≥ λR (α, BR ). Since by assumption σ < σ0 (BR ), the equation |σ| = λR (α, BR )

has a solution α1 which is smaller than the solution α2 of now follows from

α |σ|

α1 = −σλ1 (σ, BR ) ≤ α2 = −σλ1 (σ, Ω).

= λR (α, Ω). The assertion

12.3 Comparison with balls of the same volume

� 193

Figure 12.2: Construction of λ1 (α, Ω).

The claim of Theorem 12.2 cannot be true for all 0 < σ < σ0 because by Remark 12.1 λ1 (σ, Ω) is close to λN1 (Ω). The Szegö–Weinberger inequality [124] states that λN1 (Ω) ≤ λN1 (BR ), which suggests that for small |σ|, λ1 (σ, Ω) is also smaller than λ1 (σ, BR ). Next we consider the negative eigenvalue λ−1 (σ, Ωt ) in σ0 < σ < 0, where Ωt is a nearly spherical domain. As mentioned before, λ−1 (σ, Ωt ) is simple. A similar argument as for the previous theorem leads to the following theorem. Theorem 12.3. Assume σ0 (Ωt ) < σ < 0 and let Ωt be a nearly spherical domain of the same volume as BR . Then 0 > λ−1 (σ, Ωt ) ≥ λ−1 (σ, BR ). Proof. Since σλ−1 (σ, Ωt ) is positive, λ−1 (σ, Ωt ) coincides with λR (−α, Ωt ), where α = σλ−1 (σ, Ωt ) (see Figure 12.3). Hence, λ−1 (σ, Ωt ) satisfies λR (−σλ−1 (σ, Ωt ), Ωt ) = λ−1 (σ, Ωt ).

Figure 12.3: Construction of λ−1 (σ, Ω).

194 � 12 Two spectra For negative α we have by Proposition 11.1, λR (α, Ωt ) ≤ λR (α, BR ). If σ > σ0 (Ωt ) > σ(BR ), then the equations λR (−σZ, Ωt ) = Z

and

λR (−σz, BR ) = z

have respectively a unique solution Z ≥ z. This establishes the claim. In [85] it is shown that for α → −∞, λR (α, Ω) = −α2 + o(α−2 ). This implies that lim σ 2 λ−1 (σ, Ω) = −1.

σ→0

12.4 Notes The parabolic problem has been studied by various authors (cf. for instance [46, 77], and [122]). A discussion of eigenvalue problems with eigenvalues in the interior and on the boundary is found in [45] (cf. also [121]). Generalizations to elliptic operators are treated in [11]. Monotonicity properties and isoperimetric inequalities for λ1 and λ−1 are discussed in [11].

13 The torsion problem for α < 0 In contrast to the problem with positive α, a solution of the torsion problem for negative α exists only under an additional compatibility condition. Special attention is paid to the energy of nearly spherical domains of equal volume. The stability of the ball is affected. The energy does not have a local minimum for the ball anymore. The sign of the second variation depends not only on the size of α but also on the dimension. In the last part we split u = h + z, where h is a harmonic function satisfying a nonhomogeneous Robin boundary condition and z is the torsion function with homogeneous Dirichlet boundary conditions. This splitting leads to an alternative discussion of the stability analysis of the energy. At the end we discuss a related functional for which domain variations lead to a global result.

13.1 General remarks We study the torsion problem Δu + 1 = 0 in Ω,

𝜕ν u + αu = 0 on 𝜕Ω,

(13.1.1)

where α < 0. Integration of this equation leads to |Ω| = α ∮ u dS. 𝜕Ω

In contrast to the case where α > 0, the solution must therefore take on negative values. The associated Steklov eigenvalue problem (8.1.1) is Δϕ = 0 in Ω,

𝜕ν ϕ + αϕ = μϕ in 𝜕Ω,

with the eigenvalues α = μ0 < μ1 < ⋅ ⋅ ⋅ . Testing problem (13.1.1) with the corresponding eigenfunctions ϕi and i ≥ 1, we obtain − ∫ ϕi dx = ∫[ϕi Δu − uΔϕi ] dx = ∮[ϕi 𝜕ν u − u𝜕ν ϕi ] dS Ω

Ω

𝜕Ω

= −μi ∮ uϕi dS. 𝜕Ω

This is a compatibility condition for u. If https://doi.org/10.1515/9783111025438-013

(13.1.2)

196 � 13 The torsion problem for α < 0 ∫ ϕi dx = 0

for some i ∈ ℕ,

(13.1.3)

Ω

then either μi = 0 or ∮𝜕Ω uϕi dS = 0. If it is satisfied, then for all β ∈ ℝ, u + βϕi is also a solution of (13.1.1). The solution is therefore not unique. Any solution of (13.1.1) is a critical point of the energy 2

2

ℰR (u, Ω) := ∫ |∇u| dx − 2 ∫ u dx + α ∮ u ds Ω

Ω

(13.1.4)

𝜕Ω

for u ∈ W 1,2 (Ω). If μi = 0 and if there exists a solution of the form u + βϕi , then ℰR (u + βϕi , Ω) = ℰR (u, Ω).

Example 13.1. Let Ω = BR . The Steklov eigenfunctions for the ball are of the form ck r k Yk (ξ), where ξ ∈ 𝜕B1 , k ∈ ℕ ∪ {0}, and Yk (ξ) are the spherical harmonics of degree k. The corresponding eigenvalues are μk = Rk + α, ∀k, without counting multiplicity. The compatibility condition (13.1.3) is satisfied for all ϕk , k ≥ 1. Consequently (13.1.1) has a solution for all α ≠ 0 in the ball . It is of the form 2

R { 2n + u={ 2 R { 2n +

R αn R αn

− −

r2 2n 2

r 2n

if 0 ≠ μj , j = 0, 1, 2, . . . , +ψ

if 0 = μj ,

where ψ is any function in the eigenspace of μj . In both cases we get ℰR = − ∫ u dx = −|BR |( BR

R2 R + ). n(n + 2) αn

In Figure 13.1 radial solutions are shown for R = 1 and different values of α.

Figure 13.1: Radial solutions for n = 2.

(13.1.5)

13.2 Domain variations for nearly spherical domains

� 197

13.2 Domain variations for nearly spherical domains We now consider the torsion problem (13.1.1) in nearly spherical domains {Ωt }|t| |BR |, then ℰṘ (0) > 0; otherwise, if αR + n + 1 < 0, then ℰṘ (0) < 0. Note that −αR = n + 1 is equivalent to μn+1 (BR ) = 0. 13.2.1 Second variation The second variation for volume preserving perturbations was computed in (8.2.7) for problems with a nonlinearity g(u). The formal computations are independent of the sign of α. The torsion problem for α > 0 was treated in Section 8.2.2. Recall that u′ stands for the shape derivative satisfying Δu′ = 0

in BR ,

1 + αR 𝜕ν u′ + αu′ = ( )(υ ⋅ ν) n

on 𝜕BR .

By (8.2.7) we have ′

ℰR̈ (0) = −2Q(u ) +

2R R2 ̈ (1 + αR) ∫ (υ ⋅ ν)2 dS + 2 𝒮 (0). 2 n αn

(13.2.1)

𝜕BR

In this case Q(u′ ) := ∫B |∇u′ |2 dx + α ∫𝜕B u′ 2 dS. Testing the equation for u′ with ϕk we R R get

198 � 13 The torsion problem for α < 0

(

k 1 + αR ) ∮ ϕk (υ ⋅ ν) dS = μk ∮ ϕk u′ dS = ( + α) ∮ ϕk u′ dS. n R 𝜕BR

𝜕BR

𝜕BR

If μk (BR ) ≠ 0, the shape derivative exists, whereas if μk = 0, it exists only under the assumption ∮ (υ ⋅ ν)ϕk dS = 0. 𝜕BR

As in Section 8.2 we set ∞

u′ (x) = ∑ ck ϕk

and

k=0

∞

(υ ⋅ ν) = ∑ bk ϕk , k=0

where by (8.2.11) nck μk 1 + αR

bk =

for k = 0, 1, 2, . . . .

Since the perturbation is volume preserving, we deduce that b0 = c0 = 0. By (8.2.15), ∞ R2 ̈ R(1 + αR)(k − 1) 𝒮 (0) + 2 b2k , ∑ 0 2 αn n2 (k + αR) k=1

ℰR̈ (0) =

where 𝒮0̈ (0) is given in (2.3.27). Lemma 13.1. The expansion of 𝒮0̈ (0) into a series of Steklov eigenvalues leads to 󵄨

τ

󵄨2

𝒮0̈ (0) := ∮ [󵄨󵄨󵄨∇ ρ󵄨󵄨󵄨 − 𝜕BR

∞ n−1 2 −2 (k(n − 2 + k) − n + 1)b2k . ρ ] dS = R ∑ R2 k=0

Proof. The normalized Steklov eigenfunctions are of the form ϕk =

1

R

n−1 +k 2

r k Yk (ξ).

Then by the orthonormality of the spherical harmonics, ∞

∞

󵄨 󵄨2 ∮ 󵄨󵄨󵄨∇τ ρ󵄨󵄨󵄨 dS = ∑ R−2 ∮ (∇τ Yk ⋅ ∇τ Yi ) dS = ∑ R−2 Λk b2k , 𝜕BR

k,i=0

k=0

𝜕B1

where by (3.1.2), Λk = k(n − 2 + k). Moreover, ∞

∮ ρ2 dS = ∑ b2k . k=0

This establishes the assertion.

13.2 Domain variations for nearly spherical domains

� 199

We insert the expansion for 𝒮0̈ (0) into ℰR̈ (0) and obtain b2k αR αR − ) + k(k + n − 2) − n + 1} . {2(1 + αR)2 ( 2 1 + αR k + αR αn ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ k=1 ∞

ℰR̈ (0) = ∑

ℓk

For the analysis of ℓk (ξ), k = 1, 2, 3, . . . , we set ξ := −αR > 0. Hence, 2ξ(ξ − 1)(k − 1) + k(k + n − 2) − n + 1 k−ξ 2ξ(ξ − 1) + k + n − 1]. = (k − 1)[ k−ξ

ℓk = ℓk (ξ) =

(13.2.2)

Figure 13.2 shows ℓk as a function of ξ for k = n = 3.

Figure 13.2: The function ℓk (ξ) with k, n = 3.

Clearly, ℓ1 (ξ) = 0 for all ξ. For k > 1 note that ℓk (ξ) > 0 if 0 < ξ < k. If ξ > k, the sign of ℓk (ξ) is more involved. We have lim ℓk (ξ) = −∞ ξ↘k

and

lim ℓk (ξ) = −∞.

ξ→∞

The maximum of ℓk (ξ) for ξ > k is achieved at ξm = k + √k 2 − k and takes the value ℓ(ξm ) = n + 1 − 3k − 4√k 2 − k. Since min(3k + 4√k 2 − k − 1) = 10.65685425, k≥2

we have for all k ≥ 2 and ξ > k ℓk (ξ, n) < 0

provided n ≤ 9.

200 � 13 The torsion problem for α < 0 On the other hand, if n > 10 we can find for any given k0 ≥ 2 an integer n0 such that ℓk0 (ξ, n0 ) > 0 for ξ in a neighborhood of ξm (k0 , n0 ). ̈ These observations lead to the following result concerning the sign of ℰ (0). Theorem 13.1. Let ℰ be the torsion energy given in (13.1.4) and let α < 0. Then for volume preserving perturbations the following statements hold: ̈ ≤ 0. (i) Let 0 < −αR < 2, −α ≠ R1 , and n ≤ 10. Then ℰ (0) ̈ ≥ 0. (ii) Assume k < −αR < k + 1, k ≥ 2, and n ≤ 9. If bj = 0 for j = 2, .., k − 1, then ℰ (0) ̂ ̂ (iii) If n > 10, there exists a value ξ such that ℓ (ξ) > 0 for some k ≥ 2. k

The last assertion implies that ℰ (0) has a saddle point. Example 13.2. Let Ωt ⊂ ℝ2 be the ellipse whose boundary 𝜕Ω is given by {

R cos(θ) , (1 + t)R sin(θ)}, 1+t

where (r, θ) are the polar coordinates in the plane. This ellipse has the same area as the 2 circle BR . For small |t| we have y = x + t(−x1 , x2 ) + t2 (x1 , 0) + o(t 2 ). The eigenvalues and eigenfunctions of the Steklov eigenvalue problem (8.1.1) in BR are μk = where ck =

1 √πR2k+1

k +α R

ck cos(kθ),

ϕk,i = r k {

and

ck sin(kθ),

is the normalization constant. We have (υ ⋅ ν) = −R cos(2θ) = b2 ϕ2

and 󵄨

󵄨2

τ

̈ = ∮ (󵄨󵄨∇ (υ ⋅ ν)󵄨󵄨 − 𝒮 (0) 󵄨 󵄨 𝜕BR

(υ ⋅ ν)2 ) dS = 3πR. R2

A straightforward computation yields ̈ =[ ℰ (0)

R(1 + αR) 3 + ] ∮ (υ ⋅ ν)2 dS, 4α 2(2 + αR) 𝜕BR

with ∮𝜕B (υ ⋅ ν)2 dS = πR3 . From this expression it follows immediately that R

>0

if −αR > 2,

0 and therefore ℰ + = 0. Consequently, ℰR (Ω) = ℰD (Ω) + ℰ − . 13.3.1 The functional 𝒥 (Ω) Assume that − R1 < α < 0. With μ0 = α and ϕ0 = +

ℰ =−

1 , √|𝜕Ω|

(∮𝜕Ω ϕ0 𝜕ν z dS)2 α

we get

=−

|Ω|2 . α|𝜕Ω|

From Theorem 13.2 it follows that ℰR (Ω) ≤ ℰD (Ω) −

|Ω|2 . α|𝜕Ω|

Define 𝒥 (Ω) := ℰD (Ω) −

|Ω|2 , α|𝜕Ω|

α < 0.

If |Ω| = |BR |, then by Schwarz symmetrization and

ℰD (Ω) ≥ ℰD (BR )

|B |2 |Ω|2 ≤ R . |𝜕Ω| |𝜕BR |

2

|Ω| In 𝒥 (Ω) the functionals ℰD (Ω) and − α|𝜕Ω| compete. It is not clear from the start which one prevails. In order to tackle this problem we compute its first and second domain variation.

13.3.2 Domain variations of 𝒥 (t) Let Φt : Ω → Ωt be a volume preserving perturbation. Set 𝒥 (t) := 𝒥 (Ωt ) and 𝒮 (t) = |𝜕Ωt |. Then the first variation of 𝒥 (t) is 𝒥 ̇ (0) = ℰḊ (0) +

|Ω|2 ̇ 𝒮0 (0), α|𝜕Ω|2

where by (6.4.2) and (2.3.20) 2

ℰḊ (0) = − ∫ |∇z| (υ ⋅ ν) dS, 𝜕Ω

𝒮0̇ (0) = (n − 1) ∫ (υ ⋅ ν)H dS. 𝜕Ω

204 � 13 The torsion problem for α < 0 Consequently, 2

𝒥 ̇ (0) = − ∮(|∇z| − 𝜕Ω

|Ω|2 (n − 1)H)(υ ⋅ ν) dS. α|𝜕Ω|2

(13.3.4)

̇ = 0 for all volume preserving perturbations, then in If Ω is a critical domain, that is J(0) addition to z = 0 on 𝜕Ω the condition |∇z|2 −

|Ω|2 (n − 1)H = const. α|𝜕Ω|2

on 𝜕Ω

(13.3.5)

must hold. By [107, Theorem 3] concerning overdetermined boundary value problems, the ball is the only domain for which z is constant and |∇z| = c(H) for a nonincreasing function c on 𝜕Ω. Consequently, we have the following lemma. Lemma 13.2. Among all domains of equal volume, the ball is the only critical domain for the functional 𝒥 (Ω).

13.3.3 The second variation of 𝒥 (Ωt ) in nearly spherical domains The second variation for volume preserving perturbations for nearly spherical domains is 𝒥 ̈ (0) = ℰD̈ (0) − 2

In the ball we have z(r) =

R2 2n

−

r2 2n

Δz′ = 0 in BR ,

|BR |2 ̈ |BR |2 2̇ 𝒮 (0) + 𝒮0 (0). 0 α|𝜕BR |3 α|𝜕BR |2

and its shape derivative satisfies z′ = −(υ ⋅ ∇z) = (υ ⋅ ν)|∇z| on 𝜕BR .

From (8.2.18) it follows that ℰD̈ (0) = −

2 󵄨 󵄨2 ∮ z′ 2 dS + 2 ∫ 󵄨󵄨󵄨∇z′ 󵄨󵄨󵄨 dx. R 𝜕BR

BR

For the ball BR we have 𝒮0̇ (0) = 0

󵄨 󵄨2 n − 1 and 𝒮0̈ (0) = ∮ (󵄨󵄨󵄨∇τ (υ ⋅ ν)󵄨󵄨󵄨 − 2 (υ ⋅ ν)2 ) dS ≥ 0. R 𝜕BR

Then

13.3 An alternative approach: Robin versus Dirichlet torsion

𝒥 ̈ (0) = −

�

205

R2 ̈ 2 󵄨 󵄨2 ∫ z′ 2 dS + 2 ∫ 󵄨󵄨󵄨∇z′ 󵄨󵄨󵄨 dx − 2 𝒮 (0). R αn BR

𝜕BR

If on 𝜕BR we replace (υ ⋅ ν) by Rn z′ , then 𝒥 ̈ (0) depends only on z′ . It is of the form 2 1 󵄨 ′ 󵄨2 󵄨 󵄨2 n − 1 𝒥 ̈ (0) = 2 ∫ 󵄨󵄨󵄨∇z 󵄨󵄨󵄨 dx − ∫ z′ 2 dS − ∫ (󵄨󵄨󵄨∇τ z′ 󵄨󵄨󵄨 − 2 z′ 2 ) dS. R α R BR

𝜕BR

𝜕BR

Sign of 𝒥 ̈ (0) The volume constraint ∮𝜕B z′ dS = 0 implies that R

1 󵄨 󵄨2 ∫ 󵄨󵄨󵄨∇z′ 󵄨󵄨󵄨 dx ≥ ∮ z′ 2 dS. R

BR

𝜕BR

We get the lower estimate 𝒥 ̈ (0) ≥ −

1 󵄨 󵄨2 n − 1 ∫ (󵄨󵄨󵄨∇τ z′ 󵄨󵄨󵄨 − 2 z′ 2 ) dS ≥ 0. α R

(13.3.6)

𝜕BR

Since 𝒥 (Ω) is bounded from below by ℰD (BR ) for any domain of fixed volume |BR | and the ball is the only optimal domain, we have the following proposition. Proposition 13.2. Among all domains of prescribed volume, the ball minimizes 𝒥 (Ω). Some results in this chapter are published in [15].

14 Problems in annular domains In this chapter we will discuss some examples of the previous chapters for domains with holes. To be as explicit as possible we will concentrate on small perturbations of annular domains. In general the spherical ring is a critical domain of the energy; however, in contrast to the ball, the sign of the second variation depends on the perturbation.

14.1 An eigenvalue problem related to trace inequalities Let Ω and K ⊂ Ω be two bounded smooth domains, and let WK1,2 (Ω) := {u ∈ W 1,2 (Ω) : u ≡ 0 on K}. In [24] the authors considered the following eigenvalue problem: λ := λ(Ω \ K) = inf{

∫Ω (|∇u|2 + u2 ) dx ∮𝜕Ω u2 dS

: u ∈ WK1,2 (Ω)}.

Any minimizer satisfies Δu = u in Ω \ K,

𝜕ν u = λu on 𝜕Ω,

u = 0 in 𝜕K.

(14.1.1)

A typical situation is shown in Figure 14.1.

Figure 14.1: The domain Ω \ K .

We are interested in the change of λ when K is perturbed and Ω remains unchanged. In this case the family of diffeomorphisms {Φt }|t| 0

� 213

Figure 14.4: ℓ(12, k, κ).

14.3 The Robin torsion problem for α > 0 For α > 0 we consider the torsion problem Δu + 1 = 0 𝜕ν u + αu = 0

in BR \ BκR ,

(14.3.1)

on 𝜕(BR \ BκR ).

There exists a unique solution which must be radial. It is of the form 2

r 2−n {− 2n + cr + d u(r) = { 2 r {− 4 + c ln r + d

if n ≥ 3,

(14.3.2)

if n = 2.

The constants c and d are determined by the boundary conditions. In particular, for n = 3 and R = 1 we obtain u(κ) =

2 + α(1 − 3κ2 + 2κ3 ) − 2κ3 . 6α[1 + κ2 + ακ(1 − κ)]

(14.3.3)

Let Ωt be a volume preserving perturbation of BR \ BκR and let the outer boundary 𝜕BR be unperturbed. Consider the torsion energy 2

2

ℰR (t) = ∫ (|∇u(y)| − 2u(y)) dy + α ∮ u (y) dS(y) = − ∫ u(y) dy. Ωt

𝜕Ωt

Ωt

Since u is radial we can apply the results in Section 6.4.1, and get ̇ = 0. ℰ (0) ̈ For the computation of the second derivative ℰ (0) we need the shape derivative u′ . From (6.2.12) and Lemma 6.1 it follows that

214 � 14 Problems in annular domains Δu′ = 0 in BR \ BκR with the boundary conditions ur′ + α u′ = 0 on 𝜕BR

− ur′ + α u′ = k1 (u(κR))ρ on 𝜕BκR ,

and

where ur′ denotes the derivative with respect to r and ρ = (υ ⋅ ν) on 𝜕BκR . Note that 𝜕ν u′ = −ur′ on 𝜕BκR and k1 (u(κR)) = 1 + α

n−1 u(κR) + α2 u(κR). κR

(14.3.4)

From (6.5.9) it follows that Q(u′ ) =

󵄨 󵄨2 ∫ 󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 + α BR \BκR

u′ 2 dS = k1 (u(κR)) ∮ ρu′ dS.

∮ 𝜕(BR \BκR )

𝜕BκR

For volume preserving Hadamard perturbations we can apply (8.2.7) and obtain ℰR̈ (0) = 2αu(κR)(1 + α

n−1 u(κR) + α2 u(κR)) ∮ ρ2 dS κR 𝜕BκR

2

′

+ αu (κR)𝒮0̈ (0) − 2Q(u ).

(14.3.5)

In summary, we get the following statement. Lemma 14.1. For 0 < κ < 1 let BR \ BκR be an annulus in ℝn , n ≥ 2. Then the first domain variation for volume preserving Hadamard perturbations, leaving the outer boundary 𝜕BR unperturbed, is zero. The corresponding second domain variation is given in (14.3.5). Based on the strategy developed in Section 8.1, we discuss the sign of ℰR̈ (0) for the special case n = 3 and R = 1. For this purpose we will expand u′ into a series of eigenfunctions of the Steklov problem Δϕ = 0 in B1 \ Bκ , with ϕr + αϕ = 0 on 𝜕B1 and −ϕr + αϕ = μϕ on 𝜕Bκ . For ξ ∈ 𝜕B1 , ϕk (r, ξ) = (r k +

k + α −k−1 r )Yk (ξ) k+1−α

and μk :=

α(k + 1 − α)κ2k+2 − k(k + 1 − α)κ2k+1 + (k + α)(k + 1 + ακ) . κ[(k − α + 1)κ2k+1 + α + k]

For k = 0 we obtain μ0 =

α(1 − α)κ2 + α(ακ + 1) κ[(1 − α)κ + α]

14.3 The Robin torsion problem for α > 0

� 215

and 0 < μ0 < μ1 < ⋅ ⋅ ⋅ < μk < μk+1 < ⋅ ⋅ ⋅ . Let the eigenfunctions {ϕk }k≥1 be normalized such that ‖ϕk ‖L2 (𝜕B1 ) = 1. An expansion on 𝜕Bκ with respect to {ϕk }k≥1 which takes into account the volume preserving property and the barycenter condition of the perturbation yields ∞

u′ (x) = ∑ ck ϕk k=2

and

∞

ρ = ∑ bk ϕk . k=2

This implies the following identities: 1)

∮ ρ2 dS = 𝜕Bκ

∞ 1 c 2 μ2 , ∑ k12 (u(κ)) k=2 k k

∞ 1 (k(k + 1) − 2)ck2 μ2k ∑ κ2 k12 (u(κ)) k=2

2)

𝒮0̈ (0) =

3)

−2Q(u′ ) = −2k1 (u(κ)) ∮ ρu′ dS = −2 ∑ ck2 μk ,

(see (2.3.27)),

∞

k=2

𝜕Bκ

where u(κ) and k1 are given in (14.3.3) and (14.3.4). The second domain variation then becomes ∞

αu(κ) αu2 (κ) 2 + 2 2 (k(k + 1) − 2) − )ck2 μ2k . k (u(κ)) μ κ k (u(κ)) 1 k ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 1

̈ = ∑ (2 ℰ (0) k=2

ℓ(k)

Let α = 2.5. Then Figure 14.5a (resp. Figure 14.5b) shows ℓ(k) for κ = 0.1 (resp. κ = 0.9) as a function of k ≥ 2.

Figure 14.5: ℓ(k) for n = 3, R = 1, and α = 2.5.

216 � 14 Problems in annular domains As in the first example (see Section 14.1), we see that for κ close to 1 the coefficient ℓ(k) becomes negative for small k. Remark 14.1. We have discussed the case where only the inner sphere is perturbed. Alternatively, we could also perturb the outer sphere. In that case, the second variation is positive. Problem 14.1. The case α < 0 is more difficult. Numerical experiments show that there are values of α for which there is a unique solution. The solution of (14.3.1) is not unique if α coincides with an eigenvalue of the Steklov eigenvalue problem Δϕ = 0 in BR \ BκR and 𝜕ν ϕ = μϕ on 𝜕(BR \ BκR ). A complete analysis is not yet available.

14.4 Notes For further reading about the mathematical background of problem (14.1.1) and its relation to trace inequalities we refer to [24, Section 1]. Information about applications and the solutions of the Bernoulli problem are found in [52]. Some important questions of optimal shapes related to the Bernoulli problem are discussed in [105].

15 The first buckling eigenvalue of a clamped plate The first and second domain variation of the first buckling eigenvalue are computed by means of the moving surface method. It turns out that for optimal domains the second variation is a quadratic form in the shape derivative only. The positivity of this form together with Payne’s inequality implies that among all domains of given volume the ball is the only optimal domain. This establishes Pólya and Szegö’s conjecture for smooth domains.

15.1 The first eigenvalue Let Ω ⊂ ℝn be a bounded connected domain and let ℛ(u, Ω) :=

∫Ω |Δu|2 dx

∫Ω |∇u|2 dx

for u ∈ W02,2 (Ω). We set ℛ(u, Ω) = ∞ if the denominator vanishes. Then the lowest buckling eigenvalue of Ω is given by Λ(Ω) := inf{ℛ(u, Ω) : u ∈ W02,2 (Ω), ‖∇u‖L2 (Ω) = 1}. The infimum is attained by the first eigenfunction u which solves the Euler–Lagrange equation Δ2 u + Λ(Ω)Δu = 0 u = 𝜕ν u = 0

in Ω,

(15.1.1)

on 𝜕Ω.

(15.1.2)

The corresponding weak form is ∫ ΔuΔψ dx = Λ(Ω) ∫(∇u ⋅ ∇ψ) dx, Ω

Ω

∀ ψ ∈ C0∞ (Ω).

Problem (15.1.1)–(15.1.2) possesses countable many eigenvalues tending to infinity. The corresponding eigenfunctions form a basis in W02,2 (Ω) which is orthonormal with respect to the inner product ∫Ω ΔuΔυ dx. In contrast to the membrane, the sign of the first eigenfunction may change depending on Ω. For the ball the first eigenvalue is simple and the first eigenfunction is radial; see Section 3.4 where the eigenfunctions have been computed explicitly. https://doi.org/10.1515/9783111025438-015

218 � 15 Buckling eigenvalue

15.2 Shape derivatives Let Ω be a bounded smooth domain in ℝn (at least C 2,α ). For given t0 > 0 and t ∈ (−t0 , t0 ) let {Ωt := Φt (Ω)}|t| 0 for y ∈ 𝜕Ωt , then Ω0 ⊂ Ωt for small t. Thus, ̇ is negative, which implies that the first buckling eigenvalue is decreasing under set Λ(t) inclusion. This is in accordance with the inequality Λ(A) ≥ Λ(B) for any eigenvalue of the buckling plate whenever A ⊂ B.

For t = 0 Theorem 15.1 implies that ̇ Λ(0) = − ∮ |Δu|2 (υ ⋅ ν) dS. 𝜕Ω

Assume that Ω is a critical point of Λ(t) for all volume preserving perturbations. From (2.3.2) it follows that |Δu| = c0 for some constant c0 . This constant can be determined if we use the following representation of Λ(Ω). If we multiply (15.1.1) by (x ⋅ ∇u) and integrate by parts, we obtain Λ(Ω) =

1 ∫ |Δu|2 (x ⋅ ν) dS, 2

∫ |∇u|2 dx = 1.

(15.3.3)

Ω

𝜕Ω

Consequently, c0 := √

2Λ(0) . |Ω|

From now on we assume that 𝜕Ω consists only of one connected component. Corollary 15.1. Let Ωt be a family of volume preserving perturbations of Ω. Then Ω is a ̇ critical point of the energy Λ(t), i. e., Λ(0) = 0, if and only if |Δu| = √

2Λ(0) |Ω|

on 𝜕Ω.

(15.3.4)

Proof. In this case u is a solution of the overdetermined boundary value problem (15.1.1)–(15.1.2) with Δu = const. on 𝜕Ω. This implies that Δu + Λ(Ω)u = √

2Λ(0) |Ω|

in Ω.

(15.3.5)

222 � 15 Buckling eigenvalue In fact, if we set U := Δu + Λ(Ω)u, (15.1.1)–(15.1.2) imply ΔU = 0 in Ω

and U = √

2Λ(0) on 𝜕Ω. |Ω|

Hence, U = const. in Ω. From (15.3.5) we get in addition 𝜕ν Δu = 0

in 𝜕Ω.

(15.3.6)

Remark 15.2. There exists a k ∈ ℕ such that Λ(0) = μk , where μk is the k-th eigenvalue of the membrane with Neumann boundary conditions. This follows immediately from 2 the fact that u − √ |Ω|λ(0) is an eigenfunction corresponding to μk . The value of k is not known. In [127] it was shown that the overdetermined boundary value problem for n = 2 has a solution u if and only if Ω is a ball.

15.4 The second domain variation ̈ We now compute Λ(0) for volume preserving domain perturbations Φt and under the ̇ assumption that Λ(0) = 0, i. e., that Ω is a critical domain. It was observed previously that the last assumption implies that u satisfies the overdetermined boundary value problem. Consequently, by Lemma 15.2, the shape derivative is a solution of Δ2 u′ + Λ(Ω)Δu′ = 0 ′

u =0

in Ω,

(15.4.1)

on 𝜕Ω,

(15.4.2)

𝜕ν u′ = −c0 (υ ⋅ ν) on 𝜕Ω,

c0 := √

2Λ(0) . |Ω|

(15.4.3)

The normalization (15.3.1) and the fact that we restrict ourselves to volume preserving perturbations imply ∫(∇u ⋅ ∇u′ ) dx = 0

and

Ω

∮ 𝜕ν u′ dS = 0. 𝜕Ω

In view of Lemma 15.1, (15.3.2) assumes the form ̇ = ∮ Δut (νt ⋅ ∇(𝜕t ut (y))) dSt . Λ(t) 𝜕Ωt

(15.4.4)

15.4 The second domain variation

� 223

Next we apply Reynolds’ theorem (Theorem 4.1). Let B(ut ) := Δut (νt ⋅ ∇(𝜕t ut (y))). The following quantities appear in Reynolds formula: 𝜕t B(ut )|t=0 = Δu′ 𝜕ν u′ + Δu(ν′ ⋅ ∇u′ ) + Δu𝜕ν u′′ ,

𝜕νt B(ut )|t=0 = (υ ⋅ ν)𝜕ν (Δu𝜕ν u′ ). Then (4.2.2) implies

̈ Λ(0) = ∮ Δu′ 𝜕ν u′ dS + ∮ Δu(ν′ ⋅ ∇u′ ) dS + ∮ Δu𝜕ν u′′ dS 𝜕Ω

𝜕Ω

𝜕Ω ′

+ ∮(υ ⋅ ν)𝜕ν (Δu𝜕ν u ) dS + (n − 1) ∮(υ ⋅ ν)Δu𝜕ν u′ H dS. 𝜕Ω

(15.4.5)

𝜕Ω

Recall that ν′ is the shape derivative of ν defined in (2.4.9). By (15.4.2) we have ∇u′ = 𝜕ν u′ ν and by (2.4.12) we obtain ∮ Δu(ν′ ⋅ ∇u′ ) dS = 0. 𝜕Ω

For the fourth integral we apply (15.3.6) and (15.3.4). Then ∮(υ ⋅ ν)𝜕ν (Δu𝜕ν u′ ) dS = ∮(υ ⋅ ν)𝜕ν Δu𝜕ν u′ dS + ∮(υ ⋅ ν)Δu𝜕ν2 u′ dS 𝜕Ω

𝜕Ω

𝜕Ω

= 0 + c0 ∮(υ ⋅

ν)𝜕ν2 u′ dS.

𝜕Ω

With the help of (15.4.2) (15.2.5) we obtain 𝜕ν2 u′ = Δu′ − (n − 1)𝜕ν u′ H.

(15.4.6)

Hence, (15.4.3) yields ∮(υ ⋅ ν)𝜕ν (Δu𝜕ν u′ ) dS = c0 ∮(υ ⋅ ν)Δu′ dS − c0 (n − 1) ∮(υ ⋅ ν)𝜕ν u′ H dS. 𝜕Ω

𝜕Ω

𝜕Ω

With these computations, (15.4.5) simplifies to ̈ Λ(0) = ∮ Δu′ 𝜕ν u′ dS + ∮ Δu𝜕ν u′′ dS + c0 ∮(υ ⋅ ν)Δu′ dS. 𝜕Ω

𝜕Ω

𝜕Ω

224 � 15 Buckling eigenvalue In the first integral on the right-hand side we use (15.4.3) and apply Corollary 15.1 to the second integral. ̈ Λ(0) = c0 ∮ 𝜕ν u′′ dS.

(15.4.7)

𝜕Ω

̈ can be expressed in terms of the first shape derivative u′ . Next it will be shown that Λ(0) Differentiation of the equation in Lemma 15.1 with respect to t and evaluation at t = 0 implies by (15.3.6) and Δu = c0 that (ν′ ⋅ ∇u′ ) + (υ ⋅ Dν ∇u′ ) + 𝜕ν u′′ + (ν ⋅ D2 u′ υ) = −Δu′ (υ ⋅ ν) − c0 (υ ⋅ ν′ ) − c0 (υ ⋅ Dν υ) − c0 (w ⋅ ν). Recall that (ν′ ⋅ ∇u′ ) = 0 on 𝜕Ω. Moreover, by (15.4.3) (υ ⋅ Dν ∇u′ ) = −c0 (υ ⋅ Dν ν)(υ ⋅ ν) = 0. The last equality follows from (2.4.3). Thus, ̈ Λ(0) = −c0 ∮(υ ⋅ ν)Δu′ dS − c0 ∮(ν ⋅ D2 u′ υ) dS 𝜕Ω

𝜕Ω

− c02 ∮(υ ⋅ ν′ ) dS − c02 ∮(υ ⋅ Dν υ) dS − c02 ∮(w ⋅ ν) dS. 𝜕Ω

𝜕Ω

𝜕Ω

For the first integral we use (15.4.3) and then (15.4.2). We obtain −c0 ∮(υ ⋅ ν)Δu′ dS = ∮ Δu′ 𝜕ν u′ dS − ∮ u′ 𝜕ν Δu′ dS. 𝜕Ω

𝜕Ω

𝜕Ω

Gauss’ theorem, partial integration, and equation (15.4.1) for u′ lead to 󵄨 󵄨2 󵄨 󵄨2 −c0 ∮(υ ⋅ ν)Δu′ dS = ∫󵄨󵄨󵄨Δu′ 󵄨󵄨󵄨 dx − Λ(Ω) ∫󵄨󵄨󵄨∇u′ 󵄨󵄨󵄨 dx. Ω

𝜕Ω

Ω

τ

For the second integral we set υ = υ − (υ ⋅ ν)ν. Then (15.4.6) yields − c0 ∮(υ ⋅ D2 u′ ν) dS = −c0 ∮(υτ ⋅ D2 u′ ν) − (υ ⋅ ν)(Δu′ − (n − 1)𝜕ν u′ H) dS. 𝜕Ω

𝜕Ω

The first integral on the right-hand side can further be simplified: (D2 u′ ν)i = 𝜕i 𝜕j u′ νj = 𝜕i (𝜕j u′ νj ) − 𝜕j u′ 𝜕i νj . Because u′ = 0 on 𝜕Ω, ∇u′ = 𝜕ν u′ ν must necessarily hold. Hence, (D2 u′ ν)i = 𝜕i 𝜕j u′ νj = 𝜕i (𝜕j u′ νj ) − 𝜕k u′ νk νj 𝜕i νj .

(15.4.8)

15.4 The second domain variation

� 225

The last term is zero by (2.4.1). The boundary condition (15.4.3) then implies −c0 ∮ υτ ⋅ D(𝜕ν u′ ν) ⋅ ν dS = c02 ∮ υτ ⋅ ∇τ (υ ⋅ ν) dS. 𝜕Ω

𝜕Ω

For the third integral in (15.4.8) we recall Lemma 2.6, where it was shown that ν′ := 𝜕t ν(t, y)|t=0 = −∇τ (υ ⋅ ν)

(ν ⋅ ν′ ) = 0

and

on 𝜕Ω.

Consequently, −c02 ∮(υ ⋅ ν′ ) dS = c02 ∮ υ ⋅ ∇τ (υ ⋅ ν) dS = c02 ∮ υτ ⋅ ∇τ (υ ⋅ ν) dS. 𝜕Ω

𝜕Ω

𝜕Ω

We define 󵄨 ′ 󵄨2 󵄨 ′ 󵄨2 ′ ℰ (u ) := ∫󵄨󵄨󵄨Δu 󵄨󵄨󵄨 dx − Λ(Ω) ∫󵄨󵄨󵄨∇u 󵄨󵄨󵄨 dx. Ω

(15.4.9)

Ω

With the simplifications derived above, (15.4.8) assumes the form ̈ Λ(0) = 2ℰ (u′ ) + 2c02 ∮ υτ ⋅ ∇τ (υ ⋅ ν) dS 𝜕Ω

−

c02 (n

− 1) ∮(υ ⋅ ν)2 H dS − c02 ∮(υ ⋅ Dν υ) dS 𝜕Ω

𝜕Ω

− c02 ∮(w ⋅ ν) dS.

(15.4.10)

𝜕Ω

Next we use the volume constraint (2.3.7) and obtain −c02 ∮(w ⋅ ν) dS = c02 ∮(υ ⋅ ν) div𝜕Ω υ dS − c02 ∮(υτ ⋅ Dτυ ν) dS. 𝜕Ω

𝜕Ω

𝜕Ω

By Gauss’ theorem (2.2.18) −c02 ∮(w ⋅ ν) dS = −c02 ∮ υτ ⋅ ∇τ (υ ⋅ ν) dS + c02 (n − 1) ∮(υ ⋅ ν)2 H dS 𝜕Ω

𝜕Ω

−

c02

𝜕Ω τ

∮(υ ⋅

Dτυ ν) dS.

𝜕Ω

Thus, (15.4.10) becomes ̈ Λ(0) = 2 ∮ 𝜕ν u′ Δu′ dS + c02 ∮ υτ ⋅ ∇τ (υ ⋅ ν) dS − c02 ∮(υτ ⋅ Dτυ ν) dS 𝜕Ω

−

c02

𝜕Ω

∮(υ ⋅ Dν υ) dS. 𝜕Ω

𝜕Ω

226 � 15 Buckling eigenvalue The integrals which contain the vector field υ cancel. To see this recall that υτ ⋅ Dτυ ν = υτ ⋅ ∇τ (υ ⋅ ν) − υτ ⋅ Dτν υ and υτ ⋅ Dτν υ = υτ ⋅ Dτν υτ + ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (υ ⋅ ν)υτ ⋅ Dτν ν . =0

Moreover (2.4.1) implies υ ⋅ Dν υ = υτ ⋅ Dν υτ . ̈ Inserting these terms in the formula for Λ(0) we obtain the following result. ̇ Theorem 15.2. Let Φt be a volume preserving perturbation. Assume Λ(0) = 0. Then ̈ Λ(0) = 2ℰ (u′ ), where ℰ (u′ ) is defined in (15.4.9) and the shape derivative u′ is a solution of the boundary value problem (15.4.1)–(15.4.3).

15.5 Minimization of the second domain variation We consider the quadratic functional 2

2

ℰ (φ) := ∫ |Δφ| dx − Λ(Ω) ∫ |∇φ| dx Ω

(15.5.1)

Ω

for φ ∈ W01,2 ∩ W 2,2 (Ω). By the definition of Λ we have minW 2,2 (Ω) ℰ (φ) = 0. In this case the minimizer is a 0 buckling eigenfunction. If the set of admissible functions φ is enlarged, ℰ (φ) need not be bounded from below. A suitable subset of W01,2 ∩ W 2,2 (Ω), containing the shape derivatives defined in (15.4.1)–(15.4.3), will be constructed. It will be convenient to work with an alternative representation of ℰ . For φ ∈ W01,2 ∩ 2,2 W (Ω) partial integration of the first integral leads to 2

󵄨2

2

2

ℰ (φ) = ∫󵄨󵄨󵄨D φ󵄨󵄨󵄨 − Λ(Ω)|∇φ| dx + ∮ Δφ𝜕ν φ − φ ⋅ (D φν) dS.

󵄨

Ω

𝜕Ω

Taking into account (15.2.5), we obtain Δφ𝜕ν φ − ∇φ ⋅ (D2 φν) = 𝜕ν2 φ𝜕ν φ + (n − 1)(𝜕ν φ)2 H − ∇φ ⋅ (D2 φν)

= ν ⋅ D2 φ ⋅ ν(ν ⋅ ∇φ) + (n − 1)(𝜕ν φ)2 H − ∇φ ⋅ (D2 φν)

= (n − 1)(𝜕ν φ)2 H.

15.5 Minimization of the second domain variation

� 227

Consequently, we get 󵄨2

2

2

2

ℰ (φ) = ∫󵄨󵄨󵄨D φ󵄨󵄨󵄨 dx − Λ(Ω) ∫ |∇φ| dx + (n − 1) ∮(𝜕ν φ) H dS.

󵄨

Ω

Ω

(15.5.2)

𝜕Ω

Remark 15.3. The functional ℰ is lower semicontinuous with respect to weak convergence in W01,2 ∩ W 2,2 (Ω). ̈ In the sequel we assume Λ(0) ≥ 0. We know from Theorem 15.2 that this is equivalent to ℰ (φ) ≥ 0 for all φ which are shape derivatives of u. We set 1,2

𝒵 := {φ ∈ W0 ∩ W

2,2

(Ω) : ∮ 𝜕ν φ dS = 0, ∮(𝜕ν φ)2 dS > 0, ∫ ∇u ⋅ ∇φ dx = 0}. 𝜕Ω

Ω

𝜕Ω

All shape derivatives are in 𝒵 since they satisfy (15.4.4). However, 𝒵 also contains elements which are not shape derivatives. The next lemma ensures that 𝒵 is not empty. Lemma 15.3. For each 1 ≤ k ≤ n the directional derivative 𝜕k u belongs to 𝒵 . Furthermore, ℰ (𝜕k u) = 0. Proof. First we show that 𝜕k u is a shape derivative. Let 1 ≤ k ≤ n. Due to (15.1.1) and (15.1.2), 𝜕k u satisfies Δ2 𝜕k u + Λ(Ω)Δ𝜕k u = 0 𝜕k u = 0

in Ω,

(15.5.3)

in 𝜕Ω.

According to (15.2.6) we have 𝜕ν 𝜕k u = c0 νk on 𝜕Ω. Hence, ∮ 𝜕ν 𝜕k u dS = c0 ∮ νk dS = 0. 𝜕Ω

𝜕Ω

In addition, we find that ∫ ∇u ⋅ ∇𝜕k u dx = Ω

1 ∮ |∇u|2 νk dS = 0. 2 𝜕Ω

We obtain that 𝜕ν 𝜕k u does not vanish identically on 𝜕Ω. Thus, 𝜕k u ∈ 𝒵 . Moreover, (15.3.6) and (15.5.3) imply 2

ℰ (𝜕k u) = ∫(Δ 𝜕k u + Λ(Ω)Δ𝜕k u)𝜕k u dx + ∮ 𝜕k Δu𝜕ν 𝜕k u dS = 0. Ω

This proves the lemma.

𝜕Ω

228 � 15 Buckling eigenvalue It will be shown that ℰ |𝒵 ≥ 0.

For this purpose consider ̃ ℰ (φ) :=

ℰ (φ)

∮𝜕Ω (𝜕ν φ)2 dS

,

where φ ∈ 𝒵 and we set ℰ ̃ = ∞ if ∮𝜕Ω (𝜕ν φ)2 dS = 0. By the scaling invariance we may assume ∮(𝜕ν φ)2 dS = 1. 𝜕Ω

The following statement holds. Theorem 15.3. The infimum of the functional ℰ ̃ in 𝒵 is finite. Proof. We argue by contradiction. Let us assume that inf𝒵 ℰ ̃ = −∞ and consider a sequence {ŵ k }k ⊂ 𝒵 such that ∮(𝜕ν ŵ k )2 dS = 1 𝜕Ω

and lim ℰ (̃ ŵ k ) = lim ℰ (ŵ k ) = −∞.

k→∞

k→∞

Let H∞ := max𝜕Ω |H|. Then 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 2 󵄨󵄨∮ H(𝜕ν ŵ k ) dS 󵄨󵄨󵄨 ≤ H∞ < ∞. 󵄨󵄨 󵄨󵄨 𝜕Ω

From (15.5.2) it follows that ℰ (ŵ k ) = ℰ (̃ ŵ k ) ≥ −Λ(Ω) ∫ |∇ŵ k | dx − (n − 1)H∞ . Ω

The assumption limk→∞ ℰ (̃ wk̂ ) = −∞ implies k→∞

∫ |∇ŵ k |2 dx 󳨀→ ∞. Ω

We define

(15.5.4)

15.5 Minimization of the second domain variation

wk :=

�

229

1 ŵ . ̂ ‖∇wk ‖L2 (Ω) k

Then we have ‖∇wk ‖L2 (Ω) = 1

and

k→∞

∮(𝜕ν wk )2 dS 󳨀→ 0.

(15.5.5)

𝜕Ω

Since ℰ (ŵ k ) = ℰ (̃ ŵ k ) for each k ∈ ℕ the estimate (15.5.4) implies ℰ (wk ) =

ℰ (ŵ k )

‖∇ŵ k ‖2L2 (Ω)

=

ℰ (̃ ŵ k )

‖∇ŵ k ‖2L2 (Ω)

≥ −Λ(0) − C :

The infimum of ℰ taken over M := {wk : k ∈ ℕ} is finite. Therefore, there exists a subsequence of {wk }k – again denoted by {wk }k – such that lim ℰ (wk ) = inf ℰ . M

k→∞

Poincaré’s inequality and the previous estimates imply 󵄨 󵄨2 ‖wk ‖2W 2,2 (Ω) = ∫󵄨󵄨󵄨D2 wk 󵄨󵄨󵄨 + |∇wk |2 + wk2 dx Ω

≤ ℰ (wk ) + C ∫ |∇wk |2 dx + (n − 1) ∮ |H|(𝜕ν wk )2 dS ≤ C.

Ω

𝜕Ω

(15.5.6)

Thus, the sequence (wk )k is uniformly bounded in W 2,2 (Ω). There exists a w ∈ W 2,2 (Ω) such that (wk )k converges weakly to w. In view of (15.5.5), the limit function w satisfies ‖∇w‖L2 (Ω) = 1 and w = 𝜕ν w = 0 on 𝜕Ω. Hence, w ∈ W02,2 (Ω). Since by our assumption ℰ (̃ ŵ k ) converges to −∞, there exists a k0 ∈ ℕ such that ℰ (wk ) =

ℰ (̃ ŵ k )

‖∇ŵ k ‖2L2 (Ω)

0 the sequence (φk )k is uniformly bounded in W 2,2 (Ω) and φk converges weakly to a φ∗ ∈ W 2,2 (Ω). We deduce that φ∗ ∈ 𝒵 and that it satisfies ̃ ∗ ) = inf𝒵 ℰ .̃ In addition, we have ℰ (φ 2

∮(𝜕ν φ∗ ) dS = 1. 𝜕Ω

Hence, φ∗ minimizes ℰ ̃ in 𝒵 . Next we compute the first variation under the assumption υ ∈ W01,2 ∩ W 2,2 (Ω). Since ∗ φ is a minimizer it satisfies the constraints in 𝒵 . 󵄨󵄨 ℰ (φ∗ + tυ) d 󵄨󵄨 󵄨 ∗ 2 dt ∮ (𝜕ν (φ + tυ)) dS 󵄨󵄨󵄨t=0 𝜕Ω + μ1

󵄨󵄨 d d 󵄨 ∮ 𝜕ν (φ∗ + tυ) dS 󵄨󵄨󵄨 + μ2 ∫ ∇u ⋅ ∇(φ∗ + tυ) dx|t=0 = 0, 󵄨󵄨t=0 dt dt Ω

𝜕Ω

where μ1 and μ2 are the Lagrange parameters for the two constraints ∮ 𝜕ν φ dS = 0,

∫ ∇u ⋅ ∇φ dx = 0.

𝜕Ω

Ω

Since υ ∈ 𝒵 was chosen arbitrarily, φ∗ satisfies the Euler–Lagrange equality Δ2 φ∗ + Λ(Ω)Δφ∗ = μ2 Δu ∗

∗

Δφ − ℓ𝜕ν φ = const.

in

Ω,

in 𝜕Ω,

where ℓ := min𝒵 ℰ .̃ It is now easy to see that μ2 = 0. In fact, if we multiply the first equation by u, integrate over Ω, and take into account u ∈ W02,2 (Ω) and (15.3.6), we get

15.5 Minimization of the second domain variation

�

231

∫ φ(Δu2 + Λ(Ω)Δu) dx = −μ2 ∫ |∇u|2 dx. Ω

Ω

The fact that u is the eigenfunction associated to Λ(Ω) and ‖∇u‖L2 (Ω) = 1 implies μ2 = 0. The following theorem collects the previous results. ̃ ∗ ) = min𝒵 ℰ .̃ Furthermore, Theorem 15.4. There exists a function φ∗ ∈ 𝒵 such that ℰ (φ ∗ any minimizer φ ∈ 𝒵 satisfies Δ2 φ∗ + Λ(Ω)Δφ∗ = 0 in Ω, Δφ∗ − ℓ𝜕ν φ∗ = const.

(15.5.7) and

φ∗ = 0 in 𝜕Ω,

(15.5.8)

where ℓ = min𝒵 ℰ .̃ The next theorem shows ℓ = 0. Theorem 15.5. Suppose φ∗ ∈ 𝒵 is a minimizer of ℰ .̃ Hence, φ∗ satisfies equations ̃ ∗ ) = 0. In particular, ℰ ≥ 0 in 𝒵 . (15.5.7)–(15.5.8). Then we have ℰ (φ Proof. Since φ∗ satisfies equations (15.5.7)–(15.5.8) and 𝜕Ω is smooth, φ∗ is a smooth function on Ω. Hence, we define a volume preserving perturbation Φt of Ω such that 𝜕ν u′ (x) = 𝜕ν φ∗ (x)

for x ∈ 𝜕Ω.

Note that this can be achieved by setting υ = −c0−1 ∇φ∗ on 𝜕Ω. In this way, each minimizer φ∗ implies the existence of vector fields υ and w in the sense of Section 2.1. We define ψ := u′ − φ∗ . Then ψ ∈ W02,2 (Ω) and Δ2 ψ + Λ(Ω)Δψ = 0

in Ω.

The uniqueness of u implies ψ = αu for an α ∈ ℝ. Since φ∗ ∈ 𝒵 , equation (15.4.4) yields 0 = ∫ ∇u ⋅ ∇u′ dx − ∫ ∇u ⋅ ∇φ∗ dx = ∫ ∇u ⋅ ∇ψ dx = α. Ω

Ω

Ω

̃ ∗ ) ≥ 0. Consequently u′ ≡ φ∗ . Thus, φ∗ is a shape derivative. Since Ω is optimal, ℰ (φ Finally we apply Lemma 15.3 and find ̃ ∗ ) = min ℰ ̃ ≤ ℰ (𝜕 ̃ k u) = 0. 0 ≤ ℰ (φ 𝒵

This proves the claim.

232 � 15 Buckling eigenvalue

15.6 The optimal domain is a ball We will apply an inequality due to L. E. Payne to show that the optimal domain Ω is a ball. Let λ2 denote the second Dirichlet eigenfunction of the Laplacian. Payne’s inequality (see [95] and [56]) states that for any open bounded domain G λ2 (G) ≤ Λ(G). Equality holds if and only if G is a ball. Note that we do not require any smoothness of G. We denote by u1 and u2 the first and the second Dirichlet eigenfunction for the Laplacian in Ω. Thus, for k = 1, 2 we have Δuk + λk (Ω)uk = 0 uk = 0

in Ω, in 𝜕Ω,

where λk (Ω) is the k-th Dirichlet eigenvalue for the Laplacian in Ω. Note that 0 < λ1 (Ω) < λ2 (Ω). For short we will write λk instead of λk (Ω) and Λ instead of Λ(Ω). In addition we assume ‖uk ‖L2 (Ω) = 1 and use the fact that ∫ u1 u2 dx = 0. Ω

Without loss of generality we may assume that ∫ u1 dx > 0

and

Ω

∫ u2 dx ≤ 0. Ω

Consequently, there exists a t ∈ (0, 1] such that ∫(1 − t)λ1 u1 + tλ2 u2 dx = 0.

(15.6.1)

Ω

Lemma 15.4. Set ψ(x) := (1 − t)u1 (x) + tu2 (x) + cu(x)

for x ∈ Ω,

where u is the first buckling eigenfunction in Ω and c is defined as c :=

1 ∫[(1 − t)λ1 u1 + tλ2 u2 ]Δu dx. Λ Ω

Then ψ ∈ 𝒵 .

15.6 The optimal domain is a ball

� 233

Proof. 1. Note that ψ ∈ W01,2 ∩ W 2,2 (Ω). The boundary condition 𝜕ν u = 0, the equations for u1 and u2 , and (15.6.1) imply ∮ 𝜕ν ψ dS = ∫(1 − t)Δu1 + tΔu2 dx Ω

𝜕Ω

= − ∫(1 − t)λ1 u1 + tλ2 u2 dx = 0.

(15.6.2)

Ω

By classical arguments and the unique continuation principle, 𝜕ν ψ does not vanish identically in 𝜕Ω. Thus, to show that ψ ∈ 𝒵 it remains to prove that ∫ ∇u ⋅ ∇ψ dx = 0.

(15.6.3)

Ω

We recall that Δu = c0 in 𝜕Ω. Since ψ ∈ W 2,2 ∩ W01,2 (Ω), 0 = ∫(Δ2 u + ΛΔu)ψ dx Ω

= ∫ ΔuΔψ dx − Λ ∫(∇u ⋅ ∇ψ) dx + c0 ∮ 𝜕ν ψ dS. Ω

Ω

𝜕Ω

The last integral vanishes because of (15.6.2). We compute Δψ in the first integral and obtain 0 = − ∫[(1 − t)λ1 u1 + tλ2 u2 ]Δu dx + c ∫ |Δu|2 dx − Λ ∫(∇u ⋅ ∇ψ) dx. Ω

Ω

Ω

Since ‖∇u‖L2 (Ω) = 1, the second integral is equal to Λ. Thus, the definition of c implies (15.6.3). Unless t = 1 and Ω is a ball ψ doesn’t satisfy (15.4.1). Therefore, it is not a shape derivative. ̃ Since ψ ∈ 𝒵 Theorem 15.5 implies ℰ (ψ) ≥ 0, 2

2

ℰ (ψ) = ∫ |Δψ| − Λ|∇ψ| dx ≥ 0. Ω

We write ψ = ψ̃ + cu. After rearranging terms and partial integration we obtain ̃ 0 ≤ ℰ (ψ) = ∫ |Δψ|̃ 2 − Λ|∇ψ|̃ 2 dx + 2c ∫ Δψ(Δu + Λu) dx Ω

Ω

+ c2 ∫ |Δu|2 − Λ|∇u|2 dx. Ω

234 � 15 Buckling eigenvalue Since u is the first eigenfunction, the last integral vanishes, The overdetermination (15.3.5) then yields 0 ≤ ℰ (ψ) = ∫ |Δψ|̃ 2 − Λ|∇ψ|̃ 2 dx + 2cc0 ∫ Δψ̃ dx. Ω

Ω

Since ψ ∈ 𝒵 and 𝜕ν u = 0 on 𝜕Ω, 0 = ∮ 𝜕ν ψ dS = ∮ 𝜕ν ψ̃ dS = ∫ Δψ̃ dx. 𝜕Ω

Ω

𝜕Ω

This leads to the final form ̃ ψ̃ + Λψ)̃ dx. 0 ≤ ℰ (ψ) = ∫ |Δψ|̃ 2 − Λ|∇ψ|̃ 2 dx = ∫ Δψ(Δ Ω

Ω

Since ψ̃ = (1 − t)u1 + tu2 and u1 and u2 are eigenfunctions, Δψ̃ = −(1 − t)λ1 u1 − tλ2 u2 . Hence, 0 ≤ ℰ (ψ) = (1 − t)2 λ1 (λ1 − Λ) + t 2 λ2 (λ2 − Λ). Since λ1 − Λ < 0 and λ2 − Λ ≤ 0, both summands in ℰ (ψ) have to vanish. Consequently, t = 1 and λ2 (Ω) = Λ(Ω). The case of equality in Payne’s inequality implies that Ω is a ball. This proves the main theorem of the section. Theorem 15.6. Let Ω ⊂ ℝn be a bounded, smooth domain. Assume the boundary 𝜕Ω is connected and Ω minimizes the first buckling eigenvalue Λ(Ω) among all domains of equal volume. Moreover, we assume the simplicity of this eigenvalue. Then Ω is a ball.

15.7 Notes One of the earliest papers on this plate model was published by Weinstock in 1937. In his doctoral thesis [125] he showed that there is a discrete spectrum of positive eigenvalues of finite multiplicity and their only accumulation point is ∞. Pólya and Szegö’s conjecture is not yet proved in its full generality. However, partial results are known. In [116] Szegö proved the conjecture for all smooth plane domains under the additional assumption that u > 0 in Ω. In [127] H. Weinberger and B. Willms proved the following uniqueness result for n = 2. If an optimal simply connected bounded plane domain Ω exists and 𝜕Ω is smooth (at least C 2,α ), then Ω is a disc. The result of H. Weinberger and B. Willms was generalized

15.7 Notes

� 235

to arbitrary dimensions by Stollenwerk and Wagner in [115]. This paper was motivated by the work of E. Mohr [89], and [90] for a related problem. M. S. Ashbaugh and D. Buçur [5] showed that among simply connected plane domains of prescribed volume there exists an optimal domain. In [112–114] K. Stollenwerk applied techniques from the theory of free boundary value problems to prove the existence of an optimal domain with prescribed volume. She also derived geometric properties of the optimal domain and regularity properties of the first eigenfunction.

16 A fourth order Steklov problem In this chapter we investige the domain dependence of the first Steklov eigenvalue of the Bilaplace operator. However several partial results are known. The stability of the ball has not yet been investigated. For volume and area preserving perturbations we compute the first and second variation for the ball. In the plane the first eigenvalue has a local maximum for the disk. In general depending on the perturbation the second variation can be positive or negative.

16.1 The Steklov eigenvalue We consider open bounded domains in ℝn which are at least of class C 4,α for 0 < α ≤ 1. Let ℛ(φ, Ω) :=

∫Ω |Δφ|2 dx

∮𝜕Ω |𝜕ν φ|2 dS

(16.1.1)

on the space W 2,2 ∩ W01,2 (Ω). Set ℛ = ∞ for φ ∈ W02,2 (Ω) and normalize ∮𝜕Ω |𝜕ν φ|2 dS = 1. We denote q1 (Ω) = inf{ℛ(φ, Ω) : φ ∈ W 2,2 ∩ W01,2 (Ω)}. Any minimizer u solves the fourth order Steklov eigenvalue problem Δ2 u = 0

in Ω,

(16.1.2)

u=0

on 𝜕Ω,

(16.1.3)

Δu − q1 (Ω)𝜕ν u = 0

on 𝜕Ω.

(16.1.4)

We list some known properties from the literature. – In [83] (see also [10]) it was shown that there exists a minimizer u ∈ W 2,2 ∩ W01,2 (Ω) which is of constant sign in Ω and which is unique up to a multiplicative constant. Thus, the first eigenvalue q1 is simple. – The eigenvalue problem (16.1.2)–(16.1.4) admits infinitely many eigenvalues. All eigenfunctions change sign except the one corresponding to the first eigenvalue q1 (Ω). – The lowest eigenvalue of the ball is differentiable with respect to perturbations (see [30]). – All eigenfunctions are as smooth as the boundary permits (i. e., u ∈ C 4,α (Ω)) (see [10, Theorem 1] and [50, Theorem 1]). https://doi.org/10.1515/9783111025438-016

16.2 Shape derivatives

�

237

16.2 Shape derivatives For the reader’s convenience we prove the following result. Lemma 16.1. Let u ∈ W 2,2 ∩ W01,2 (Ω) be a minimizer of ℛ(φ, Ω). Then u is unique up to a multiplicative constant. Moreover it is of constant sign and q1 is therefore simple. Proof. 1. Suppose that u ∈ W 2,2 ∩ W01,2 (Ω) is a minimizer which changes sign. Consider the solution w ∈ W 2,2 ∩ W01,2 (Ω) of Δw = −|Δu| in Ω,

w = 0 on 𝜕Ω.

By the strong maximum principle, w > 0 in Ω. Moreover, Δ(w ± u) = −|Δu| ± Δu ≤ 0 in Ω,

w ± u = 0 on 𝜕Ω.

We may assume that Δ(w ± u) does not vanish identically in Ω. Otherwise by the strong maximum principle w ± u would vanish identically in Ω. Since u changes sign and w > 0 this is a contradiction. As a consequence, w ± u > 0 in Ω. Hopf’s lemma and the smoothness of w ± u then imply 𝜕ν (w ± u) ≤ 0 on 𝜕Ω

and

𝜕ν (w ± u) < 0 on Γ ⊂ 𝜕Ω

for some Γ with positive surface measure. This implies |𝜕ν w| ≥ |𝜕ν u| on 𝜕Ω and |𝜕ν w| > |𝜕ν u| on Γ. Consequently q1 >

∫Ω (Δw)2 dx

∮𝜕Ω (𝜕ν w)2 dS

.

Since w is admissible, this is contradictory. Hence, the minimizer is of constant sign. 2. Next we show that the minizer is unique up to a multiplicative constant. If there exist two linearly independent minimizers u1 , u2 ∈ W 2,2 ∩ W01,2 (Ω), then any linear combination c1 u1 + c2 u2 =: u is again a minimizer. It is always possible to find c1 and c2 such that u changes sign. This contradicts the previous observation in Step 1. Let {Ωt }|t| 0 sufficiently small, Φ(t, ⋅) : Ω → Ωt is a diffeomorphism. As in Section 2.4.2, δ(⋅, t) is the signed distance function with respect to 𝜕Ωt . Then νt (y) := −∇y δ(y, t) for the outer unit normal vector in y ∈ 𝜕Ωt . We denote by ut (y) := u(y, t) the first Steklov eigenfunction on Ωt , and we write for short q1 (t) := q1 (Ωt ) and q1 := q1 (0) = q1 (Ω). Then ut solves

238 � 16 Steklov eigenvalue Δ2 ut = 0

ut = 0

Δut − q1 (t)𝜕νt ut = 0

in Ωt ,

(16.2.1)

on 𝜕Ωt ,

(16.2.2)

on 𝜕Ωt .

(16.2.3)

The shape derivative u′ (x) = 𝜕t u(Φt (x), t)|t=0 solves the following boundary value problem. Lemma 16.2. The shape derivative u′ satisfies Δ2 u′ = 0

u + υ ⋅ ∇u = 0 ′

2

in Ω,

(16.2.4)

in 𝜕Ω,

(16.2.5)

Δu − q1 𝜕ν u − q1 (ν ⋅ D uυ) = −υ ⋅ ∇Δu + q̇ 1 (0)𝜕ν u ′

′

in 𝜕Ω.

(16.2.6)

Proof. The assertions (16.2.4) and (16.2.5) are straightforward. We therefore prove only (16.2.6). Let y = Φt (x) for x ∈ 𝜕Ω. Then ut (y) = u(Φt (x), t). Thus for all |t| < t0 , (16.2.3) may be rewritten as an equation on 𝜕Ω : Δy u(Φt (x), t) − q1 (t)ν(Φt (x), t) ⋅ ∇y u(Φt (x), t) = 0

∀ x ∈ 𝜕Ω,

where ν(Φt (x), t) = νt (y). We now differentiate this equation with respect to t. Hence for all x ∈ 𝜕Ω and |t| < t0 , 0=

d {Δ u(Φt (x), t) − q1 (t)ν(Φt (x), t) ⋅ ∇y u(Φt (x), t)}. dt y

In order to evaluate the first term we observe that the partial derivatives with respect to t and y commute. Thus, d Δ u(Φt (x), t) = 𝜕t Δy u(Φt (x), t) + 𝜕t Φt ⋅ ∇y Δy u(Φt (x), t) dt y = Δy 𝜕t u(Φt (x), t) + 𝜕t Φt ⋅ ∇y Δy u(Φt (x), t). For t = 0 this reads as 󵄨󵄨 d 󵄨 Δy u(Φt (x), t)󵄨󵄨󵄨 = Δu′ + υ ⋅ ∇Δu. 󵄨󵄨t=0 dt In addition we have 󵄨󵄨 d 󵄨 ν(Φt (x), t)󵄨󵄨󵄨 = ν′ + (υ ⋅ Dν ) 󵄨󵄨t=0 dt and

16.3 First domain variation

� 239

󵄨󵄨 d 󵄨 ∇y u(Φt (x), t)󵄨󵄨󵄨 = ∇u′ + υD2 u. 󵄨󵄨t=0 dt By (2.4.12) we have (ν′ ⋅ ∇u) = 0. Moreover, since u = 0 on 𝜕Ω it follows from (2.4.1) that υ ⋅ Dν ∇u = 𝜕ν u(υ ⋅ Dν ν) = 0. Summation of all the contributions leads to (16.2.6) and establishes the proof of the lemma.

16.3 First domain variation In the sequel we will consider perturbations which are volume or perimeter preserving of second order. We shall apply the moving surface technique to establish the next result. Theorem 16.1. 1. Let Ωt be a family of volume preserving perturbations of Ω such that 𝒱̇ (0) = 0. Then Ω is a critical point of the eigenvalue q1 (Ωt ), i. e., q̇ 1 (0) = 0, if and only if (Δu)2 − (n − 1)q1 (Ω)(𝜕ν u)2 H − 2(𝜕ν u)𝜕ν Δu = const. 2.

on 𝜕Ω.

(16.3.1)

̇ = 0. Then Let Ωt be a family of perimeter preserving perturbations of Ω such that 𝒮 (0) Ω is a critical point of the eigenvalue q1 (Ωt ) if and only if 1 2 (Δu)2 − q1 (Ω)(𝜕ν u)2 − (𝜕 u)𝜕 Δu = c H n−1 n−1 ν ν

on 𝜕Ω.

(16.3.2)

for some constant c. Proof. We write q1 (t) as the Rayleigh quotient q1 (Ωt ) =

∫Ω |Δut |2 dy t

∮𝜕Ω |𝜕νt ut |2 dSt

,

(16.3.3)

t

where ut is the first Steklov eigenfunction of problem (16.2.1)–(16.2.3). The proof will be done in several steps. Step 1 For the differentiation of the numerator we apply Reynolds’ theorem (see (4.2.1)) and find 󵄨󵄨 d 󵄨 ∫ |Δut |2 dy󵄨󵄨󵄨 = 2 ∫ ΔuΔu′ dx + ∮ |Δu|2 (υ ⋅ ν) dS. 󵄨󵄨t=0 dt Ωt

Ω

𝜕Ω

If we integrate the first integral on the right side twice by parts and use the fact that u is biharmonic and u = 0 on 𝜕Ω, we obtain 2 ∫ ΔuΔu′ dx = 2 ∮ 𝜕ν uΔu′ dS. Ω

𝜕Ω

240 � 16 Steklov eigenvalue Hence, 󵄨󵄨 d 󵄨 ∫[Δut ]2 dy󵄨󵄨󵄨 = 2 ∮ 𝜕ν uΔu′ dS + ∮ |Δu|2 (υ ⋅ ν) dS. 󵄨󵄨t=0 dt Ωt

𝜕Ω

𝜕Ω

Step 2 For the differentiation of the denominator we apply (4.2.2) and get d ∮ |𝜕νt ut |2 dSt = ∮ 𝜕t (|𝜕νt ut |2 ) dSt + ∮ 𝜕νt (|𝜕νt ut |2 )(𝜕t Φt ⋅ νt ) dSt dt 𝜕Ωt

𝜕Ωt

𝜕Ωt

+ (n − 1) ∮ |𝜕νt ut |2 Ht (y)(𝜕t Φt ⋅ νt ) dSt . 𝜕Ωt

Thus, 󵄨󵄨 d 󵄨 ∮ |𝜕νt ut |2 dSt 󵄨󵄨󵄨 󵄨󵄨t=0 dt 𝜕Ωt

= ∮ 2𝜕ν u(𝜕ν u′ + ∇u ⋅ ν′ ) + (υ ⋅ ν)[2𝜕ν u𝜕ν2 u + (n − 1)(𝜕ν u)2 H] dS. 𝜕Ω

By (2.4.13), ∇u ⋅ ν′ = 0; hence, 󵄨󵄨 d 󵄨 ∮ |𝜕νt ut |2 dSt 󵄨󵄨󵄨 = 2 ∮ 𝜕ν u𝜕ν u′ dS + 2 ∮(υ ⋅ ν)𝜕ν u𝜕ν2 u dS 󵄨󵄨t=0 dt 𝜕Ωt

𝜕Ω

𝜕Ω

+ (n − 1) ∮(υ ⋅ ν)(𝜕ν u)2 H dS. 𝜕Ω

Step 3 We now compute the first variation q̇ 1 (0). The normalization ∮ |𝜕νt ut |2 dSt = 1

(16.3.4)

𝜕Ωt

together with (16.3.3) implies q̇ 1 (t) =

d d ∫ |Δut |2 dy − q1 (t) ∮ |𝜕νt ut |2 dSt . dt dt Ωt

𝜕Ωt

For t = 0 the computations in Steps 1 and 2 yield q̇ 1 (0) = 2 ∮ 𝜕ν uΔu′ dS + ∮ |Δu|2 υ ⋅ ν dS − 2q1 ∮ 𝜕ν u𝜕ν u′ dS 𝜕Ω

𝜕Ω

− 2q1 ∮(υ ⋅ 𝜕Ω

ν)𝜕ν u𝜕ν2 u dS

𝜕Ω

− (n − 1)q1 ∮(υ ⋅ ν)(𝜕ν u)2 H dS. 𝜕Ω

(16.3.5)

16.3 First domain variation

� 241

Some of the integrals in this expression can be simplified. This is done in the next two steps. Step 4 First we multiply equation (16.2.6) by 𝜕ν u, integrate over 𝜕Ω, and use the normalization (16.3.4) for t = 0. After rearranging the terms we get 2 ∮ 𝜕ν uΔu′ dS − 2q1 ∮ 𝜕ν u𝜕ν u′ dS = 2q̇ 1 (0) + 2q1 ∮ 𝜕ν u(ν ⋅ D2 uυ) dS 𝜕Ω

𝜕Ω

𝜕Ω

− 2 ∮ 𝜕ν u(υ ⋅ ∇Δu) dS. 𝜕Ω

The two terms on the left side correspond to the first and third integrals in (16.3.5). As a consequence all integrals in the expression for (16.3.5) which depend on u′ can be replaced by integrals depending only on u and υ. Consequently, q̇ 1 (0) = − ∮ |Δu|2 (υ ⋅ ν) dS + (n − 1)q1 ∮(υ ⋅ ν)(𝜕ν u)2 H dS 𝜕Ω

𝜕Ω 2

− 2 ∮ 𝜕ν u[q1 (ν ⋅ D uυ) − υ ⋅ ∇Δu − q1 (υ ⋅ ν)𝜕ν2 u] dS.

(16.3.6)

𝜕Ω

Step 5 In this step we replace the term containing 𝜕ν2 u. Since Δu = q1 𝜕ν u on 𝜕Ω, this equality remains after tangential differentiation: υτ ⋅ ∇Δu = q1 υτ ⋅ ∇𝜕ν u. Replacing υτ on the right side of this equation by υ − (υ ⋅ ν)ν we get υτ ⋅ ∇Δu = q1 (υ ⋅ D2 uν) − q1 (υ ⋅ ν)𝜕ν2 u. Consequently, −υ ⋅ ∇Δu = −(υ ⋅ ν)𝜕ν Δu − (υτ ⋅ ∇Δu)

= −(υ ⋅ ν)𝜕ν Δu − q1 (υ ⋅ D2 uν) + q1 (υ ⋅ ν)𝜕ν2 u.

We insert this into (16.3.6): q̇ 1 (0) = − ∮ |Δu|2 (υ ⋅ ν) dS + (n − 1)q1 ∮(υ ⋅ ν)(𝜕ν u)2 H dS 𝜕Ω

𝜕Ω

+ 2 ∮ 𝜕ν u(υ ⋅ ν)𝜕ν Δu dS. 𝜕Ω

If the vector fields are volume preserving in the sense 𝒱̇ (0) = 0, then q̇ 1 (0) = 0 if and only if (16.3.1) holds. ̇ = 0. This proves (16.3.2). An analogous result holds if 𝒮 (0)

242 � 16 Steklov eigenvalue

16.4 First variation for the ball From now on we consider a family {Ωt }|t| 2 then q̈ 1 (0) ≥ 0 for all perturbations of the form t2 Φt (x) = x + tρ ν + w + o(t 2 ), 2

k ∗∗ (n)

where ρ = ∑ ck Yk (ξ). k=2

Moreover q̈ 1 (0) ≤ 0 for all perturbations for which ρ=

∞

∑

k=k ∗∗ (n)+1

ck Yk (ξ)

The ball is therfore neither stable nor unstable.

16.6 Notes It is known that among all convex domains in ℝn of given volume or perimeter there exists an optimal one which minimizes q1 (see [31, Theorem 4.6] and [3, Theorem 5]). Kuttler [83] showed – in two dimensions – that a square has a first eigenvalue q1 , which is strictly smaller than the one of a disc. Ferrero, Gazzola, and Weth (see [50, formulas (1.14) and (1.15)]) improved this result. More recently Antunes and Gazzola [3] gave numerical evidence that the optimal planar shape is the regular pentagon. Buçur, Ferrero, and Gazzola [30] showed that the ball is a critical domain of q1 among all volume and area preserving smooth perturbations. By means of the second domain variations it was shown in [84] that the ball is unstable.

A General remarks A.1 Jacobi’s formula for determinants Let A(t) be an n × n-matrix depending on the parameter t. Then by Jacobi’s formula d d det A(t) = det A(t) ⋅ T(A−1 (t) A(t)), dt dt where T(C) denotes the trace of the matrix C. Set A(t) = I + tB + A−1 (t) = I − tB −

(A.1.1) t2 C. 2

Then for small |t|,

t2 C + t 2 B2 + o(t 2 ), 2

d A(t) = B + tC, dt d 3CB A−1 (t) A(t) = B + t(C − B2 ) + t 2 (B3 − ) + o(t 2 ). dt 2 We apply these computations to the Taylor expansion of det(I + tB +

t2 t2 C) = 1 + a0 t + a1 + O(|t|3 ), 2 2

|t| ≪ 1.

Then 󵄨 a0 = det(A(t))󵄨󵄨󵄨t=0 = T(B), a1 =

󵄨󵄨 d d d 󵄨 󵄨 (det A(t)T(A−1 (t) A(t)))󵄨󵄨󵄨 = T 2 (B) + T(B + t(C − B2 ) + o(t))󵄨󵄨󵄨t=0 󵄨󵄨t=0 dt dt dt

= T 2 (B) − T(B2 ) + T(C).

A.2 Hölder continuity Let f (x) be a function defined in Ω ⊂ ℝn and let βi , i = 1, 2, . . . , n, be nonnegative integers. Set Dβ f = – – –

β

𝜕f |β| β

n

β

𝜕x1 1 𝜕x2 2 . . . 𝜕xn n

,

where βi ≥ 0, β = (β1 , β2 , . . . , βn ), and |β| = ∑ βi . i=1

The function f is in the class C k if ‖f ‖C k := max|β|≤k supx∈Ω |Dβ f | < ∞. The function f is in the class C 0,α , 0 < α < 1, if the Hölder coefficient |f |C 0,α = (y)| is bounded. supx,y∈Ω |f (x)−f |x−y|α The function f is in the Hölder space C k,α if ‖f ‖C k,α = ‖f ‖C k + max|β|=k |Dβ f |C 0,α < ∞.

https://doi.org/10.1515/9783111025438-017

254 � A General remarks The following definition is taken from [65]. Definition A.1. A bounded domain Ω or equivalently 𝜕Ω is in C k or C m,α if at any point x0 ∈ 𝜕Ω there is a ball Br0 centered at x0 and a one-to-one map Ψ : Br0 → D ∈ ℝn such that: 1. Ψ(x) > 0 if x ∈ Br0 ∩ Ω, 2. Ψ(x) = 0 if x ∈ Br0 ∩ 𝜕Ω, 3. Ψ and Ψ−1 in C k or C m,α . Denote by − inf{|x − z| : z ∈ 𝜕Ω, x ∈ ℝn \ Ω},

δ(x, 𝜕Ω) := {

inf{|x − z| : z ∈ 𝜕Ω, x ∈ Ω}

the signed distance function. In [65, Lemma 14.16] it is shown that if 𝜕Ω ∈ C k or C k,α , k ≥ 2, and α ≥ 0, then there is a neighborhood of 𝜕Ω such that δ(⋅, 𝜕Ω) is in C k or in C k,α , respectively.

A.2.1 Whitney’s extension theorem The deformation of domains can be described by the perturbation of the boundary. Especially for the change of variables method we need their extensions to the whole domain. This is possible by Whitney’s theorem. ̃ Theorem A.1. Let 𝜕Ω ⊂ ℝn be locally parametrized by x(ξ), where ξ ∈ ℝn−1 are local ̃ coordinates. On 𝜕Ω consider a vector field υ(ξ) ∈ C m . Then by Whitney’s theorem [126], υ̃ can be extended into Ω such that its extension υ : Ω → ℝn with υ = υ̃ on 𝜕Ω and υ ∈ C m (Ω). A particular extension of a Hadamard diffeomorphism Φt : 𝜕Ω → 𝜕Ωt is constructed in the next theorem. ̃ Theorem A.2. Let 𝜕Ω ⊂ ℝn be locally parametrized by x(ξ), where ξ ∈ ℝn−1 are local coordinates. Let Φt : 𝜕Ω → 𝜕Ωt be a diffeomorphism such that t2 ̃ ν(ξ) ̃ + w(ξ). ̃ ̃ + t ρ(ξ) Φ̃ t (ξ) = x(ξ) 2 Then there exists a diffeomorphism Φt : Ω → Ωt such that Φt |𝜕Ω = Φ̃ t |𝜕Ω .

A.2 Hölder continuity

� 255

Proof. We follow the arguments in [117, Section 3.4]. We set Ω+η := {x ∈ ℝn \ Ω : dist(x, 𝜕Ω) < η},

Ω−η := {x ∈ Ω : dist(x, 𝜕Ω) < η}.

It is well known (cf. the remark in the previous section) that if 𝜕Ω ∈ C 2,α , there exists a sufficiently small η0 > 0 such that 𝜕Ω−η0 ∈ C 2,α as well. As a consequence, there exists a ̃ ̃ − ην(ξ) local coordinate system such that any x ∈ Ω−η0 can be written in the form x = x(ξ) where 0 < η < η0 . Next we assume that t0 is chosen so small that for |t| < t0 , 𝜕Ωt ⊂ Ω+η0 ∪ Ω−η0 . This η implies that for 0 < η < 20 , ̃ + (Φ̃ t (ξ) − x(ξ)) ̃ − ην(ξ) ̃ x(ξ) ∈ Ω+η0 ∪ Ω−η0 . Finally, we choose a cutoff function ϑ ∈ C ∞ (Ω) with 0 ≤ ϑ ≤ 1 such that: – 0 < ϑ < 1 in Ω−η0 \ Ω−η0 – –

ϑ = 0 in Ω \ Ω−η0 , ϑ = 1 in Ω−η0 .

2

2

We then define the following extension: ̃ Φt (x) := x + ϑ(x)(Φ̃ t (ξ) − x(ξ)).

B Geometry B.1 Some concepts from differential geometry Denote by {xi }ni=1 the Cartesian coordinates and let (a ⋅ b) be the Euclidean scalar product of the vectors a, b ∈ ℝn . Let 𝜕Ω ⊂ ℝn be an (n − 1)-dimensional surface given in local ̃ coordinates by x(ξ), ξ ∈ ℝn−1 . The vectors x̃ξi , i = 1, . . . , n − 1, form a basis in the tangent space. Let ν be the outer unit normal of 𝜕Ω. Two tensors play an important role in geometry, namely: – the metric tensor gij = (x̃ξi ⋅ x̃ξj ), g ij is the inverse of gij , –

the second fundamental form Lij = −(x̃ξi ξj ⋅ ν) = 21 ((xξ ̃ ⋅ νξj ) + (x̃ξj ⋅ νξi )). i

The surface element of 𝜕Ω is given by dS = √det gij dξ,

dξ := dξ1 dξ2 ⋅ ⋅ ⋅ dξn−1 .

The Laplace–Beltrami operator on 𝜕Ω is expressed in terms of the metric tensor as follows: Δ∗ =

1 𝜕 𝜕 (√gg ij ), 𝜕ξj √g 𝜕ξi

g := det gij .

The formulas of Gauss and Weingarten are x̃ξi ξj = Γsij x̃ξs − Lij ν, ν̃ξj = g ki Lji x̃ξk .

𝜕giq 𝜕gij 𝜕gjq 1 where Γsij = g sq ( − + ), 2 𝜕ξj 𝜕ξq 𝜕ξi

(B.1.1)

The operator g ik Lkj is called Weingarten operator . The eigenvalues κi of g ik Lkj are called principal curvatures and the eigenvectors are tangent to the lines of principal curvatures. An important notion in our study is the mean curvature H. It is defined as H(ξ) =

1 ik g Lik (ξ). n−1

The mean curvature is invariant with respect to changes of the coordinates. Therefore, H=

1 n−1 ∑κ. n − 1 i=1 i

If the local coordinates are chosen such that xξs is an eigenvector, then g ℓi Lik

𝜕x̃k (ξ) 𝜕x̃ (ξ) = κs δℓk k 𝜕ξs 𝜕ξs

https://doi.org/10.1515/9783111025438-018

󳨐⇒

Lik

𝜕x̃k (ξ) 𝜕x̃ (ξ) = κs gik k . 𝜕ξs 𝜕ξs

258 � B Geometry Moreover, for s ≠ r we have gik 𝜕ξk

𝜕x 𝜕xi s 𝜕ξr

= 0.

Example B.1. Let 𝒮 be a surface represented by a graph (x ′ , f (x ′ )), where x ′ ∈ ℝn−1 . In this case gij = δij + 𝜕i f 𝜕j f , Lij = −

𝜕i 𝜕j f √1 + |∇f |2

ν̃ =

(−∇f , 1) √1 + |∇f |2

,

.

Here ν̃ is the outer normal of the domain {x : x ∈ ℝn , xn < f (x ′ )}. The principal curvatures κi are the eigenvalues of −

𝜕i 𝜕j f √1 + |∇f |2

.

We can always choose a coordinate system such that the origin lies on P ∈ 𝜕Ω, ℝn−1 coincides with the tangent space at P, and the en -axis is the outer normal. In this case ∇f (0) = 0, gij (0) = δij , and Lij (0) = −𝜕i 𝜕j f (0).

B.1.1 Curvilinear coordinates in ℝn Spherical coordinates The spherical coordinates (r, θ1 , θ2 , . . . , θn−1 ), r > 0, θj ∈ (0, π), if j = 2, . . . , n − 1 and θ1 ∈ (0, 2π), are defined as follows (see Figure B.1): n−1

x1 = r(∏ sin θj ) cos θ1 , j=2

n−1

x2 = r(∏ sin θj ) sin θ1 , j=2

n−1

x3 = r(∏ sin θj ) cos θ2 , .. .

j=3

xn−1 =r sin θn−1 cos θn−2 , xn = r cos θn−1 .

B.1 Some concepts from differential geometry

� 259

Figure B.1: Spherical coordinates.

The volume element is dx = r n−1 sin(θ1 )n−2 sin(θ2 )n−3 . . . sin(θn−2 ) dr dθ1 dθ2 . . . dθn−1 . Set θ := (θ1 , θ2 , . . . , θn−1 ). A generic point in ℝn is given by x(r, θ) = rξ(θ), where ξ(θ) is a point on S n−1 := {x ∈ ℝn : |x| = 1}. Clearly, (ξ(θ) ⋅ ξθj (θ)) = 0. that

Hence, the square of the line element is |dx|2 = dr 2 + r 2 (ξθi ⋅ ξθj ) dθi dθj . This implies gij = (

1 0

0 ) r 2 G(S n−1 )

and

g ij = (

1 0

0 ), r −2 G−1 (S n−1 )

where G(S n−1 ) is the metric tensor on S n−1 . In this case g = r 2(n−1) det G(S n−1 ) and thus Δx =

𝜕2 n−1 𝜕 1 + + Δ∗ , r 𝜕r r 2 𝜕r 2

where Δ∗ = Laplace–Beltrami operator on S n−1 . (B.1.2)

B.1.2 Normal coordinates ̃ Let 𝜕Ω ∈ C k for k > 2. In a neighborhood of x(ξ) ∈ 𝜕Ω we introduce the coordinates (ξ, r) as follows: ̃ ̃ + r ν(ξ), y(ξ, r) = x(ξ)

̃ where x(ξ) ∈ 𝜕Ω, ξ = (ξ1 , ξ2 , . . . , ξn−1 ).

The square of the line element is 2 󵄨 ̃ dr 󵄨󵄨󵄨󵄨 |dy|2 = 󵄨󵄨󵄨(x̃ξi + r ν̃ξi ) dξi + ν(ξ)

= [x̃ξi ⋅ x̃ξj + r (⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ν̃ξi ⋅ x̃ξj + ν̃ξj ⋅ x̃ξi ) +r 2 ν̃ξi ⋅ ν̃ξj ] dξi dξj + dr 2 . 2Lij

260 � B Geometry The volume element is then given by n−1

dy = √det(gij + 2rLij + r 2 νξi ⋅ νξj ) ∏ dξi dr. i=1

From Jacobi’s formula (cf. Section A.1) and the definition of the mean curvature (2.2.3) we have for small |r| det[gij + 2rLij + o(r)] = det[gij (I + 2rg ik Lik ) + o(r)]

= det[gij ] det[I + 2rg ik Ljk + o(r)] = det[gij ][1 + 2r(n − 1)H + o(r)].

Since √1 + 2r(n − 1)H + o(r)] = (1 + (n − 1)rH + o(r)), the volume element assumes now the form dy = (1 + (n − 1)rH + o(r)) dr dS𝜕Ω .

(B.1.3)

Suppose that Ω is represented locally as a graph as in Example B.1. We also assume that ℝn−1 is in the tangent space at P ∈ 𝜕Ω and that the coordinate axes {ej }n−1 j=1 point in the direction of the principal curvatures. Then, evaluating |dy|2 at P we obtain n−1

|dy|2 = [δij + 2rκj δij + r 2 κi κj δij ] dxi dxj + dr 2 = ∑ (1 + rκi )2 dxi2 + dr 2 . i=1

In order to compute the Laplacian, we need √g = ∏n−1 i=1 (1 + κi r) and (1 + rκ1 )−2 0 .. ( ( . g ij = ( ( 0 (

0

0 (1 + rκ2 )−2 .. . .. . 0

... 0 .. . .. . .. .

(1 + rκn−1 )−2

0 0 ) ) ). ) 0

0

1)

... ... .. .

Consequently, Δ=

1 𝜕2 (n − 1)H 𝜕 √g 𝜕i { 𝜕 } + + . i 2 2 √g √g 𝜕r (1 + rκi ) 𝜕r

(B.1.4)

B.2 Implicit function theorem

� 261

For small r and for κi ∈ C 1 it follows that Δ = Δ∗ +

𝜕2 𝜕 + (n − 1)H + O(r), 𝜕r 𝜕r 2

Δ∗ = Laplace–Beltrami operator on 𝜕Ω.

(B.1.5)

B.2 Implicit function theorem The fact that geometric quantities in perturbed domains {Ωt }t are differentiable with respect to the parameter t relies on the following version of the implicit function theorem (see, e. g., [23, Sections 3.1.10–3.1.13]). For two Banach spaces X and Z the space of bounded linear mappings from X to Z will be denoted by ℒ(X, Z). Theorem B.1. Let X, Y , and Z be Banach spaces and let U and V be open subsets of X and Y , respectively. Let F ∈ C r (U × V , Z), r ≥ 1, and fix (x0 , y0 ) ∈ U × V . Suppose 𝜕x F(x0 , y0 ) ∈ ℒ(X, Z) is an isomorphism. Then there exists an open neighborhood U1 × V1 ⊂ U × V of (x0 , y0 ) such that for each y ∈ V1 there exists a unique point (ξ(y), y) ∈ U1 × V1 satisfying F(ξ(y), y) = F(x0 , y0 ). Moreover, the map ξ is in C r (V1 , Z) and satisfies ∇ξ(y) = −(𝜕x F(ξ(y), y))

−1

∘ 𝜕y F(ξ(y), y).

C Sobolev spaces and inequalities m,p

Let Ω ⊂ ℝn be a bounded Lipschitz domain. The Sobolev spaces W m,p (Ω) and W0 (Ω), where 1 < p < ∞ and p is a nonnegative integer, are given by W m,p (Ω) := {u ∈ Lp (Ω) : Dα u ∈ Lp (Ω), 0 ≤ |α| ≤ n},

Dα u weak derivative,

m,p

W0 (Ω) := closure of C0∞ (Ω) in W p,m (Ω).

For u ∈ W m,p (Ω) we set 1/p

󵄨 󵄨p ‖u‖W m,p (Ω) := { ∑ 󵄨󵄨󵄨Dα u󵄨󵄨󵄨Lp (Ω) } 0≤|α|≤m

.

Equipped with this norm, W m,p (Ω) is a reflexive Banach space. In particular, the space W 1,2 (Ω) is equipped with the norm 2

1 2

2

‖u‖W 1,2 (Ω) := (∫ |∇u| + u dx) . Ω

The expression 1

2 󵄩󵄩 󵄩 2 󵄩󵄩|∇u|󵄩󵄩󵄩L2 (Ω) = (∫ |∇u| dx)

Ω

is a half-norm on W 1,2 (Ω) which vanishes on the space 𝒫 consisting of the constants. If |u|p is any norm on 𝒫 , then ‖u‖ =

(|u|2P

2

+ ∫ |∇u| dx)

1 2

Ω

is an equivalent norm on W 1,2 (Ω). Poincaré’s inequality states that there exists a positive constant cp such that ∫ u2 dx ≤ cp ∫ |∇u|2 dx Ω

Ω

∀ u ∈ W01,2 (Ω).

(C.0.1)

Hence, ‖|∇u|‖L2 (Ω) is a norm in W01,2 (Ω). A function u ∈ W 1,2 (Ω) possesses a trace satisfying the trace inequality ∮ u2 dS ≤ cT ‖u‖2W 1,2 (Ω) , 𝜕Ω https://doi.org/10.1515/9783111025438-019

(C.0.2)

264 � C Sobolev spaces and inequalities for a suitable positive constant cT which depends only on Ω. A closer look at the proof (see, e. g., [47]) shows that Young’s inequality gives the modified trace inequality: For any positive ϵ there exists a number cϵ > 0 such that ∮ u2 dS ≤ ϵ ∫ |∇u|2 dx + cϵ ∫ u2 dx Ω

𝜕Ω

∀ u ∈ W 1,2 (Ω).

(C.0.3)

Ω

Friedrich’s inequality states that there exists a cF > 0 such that for all u ∈ W 1,2 (Ω), ∫ u2 dx ≤ cF (∫ |∇u|2 dx + ∮ u2 dS). Ω

Ω

(C.0.4)

𝜕Ω

In our studies we shall often need the modified Friedrich inequality: For any ϵ > 0 there exists cϵ such that ∫ u2 dx ≤ ϵ ∫ |∇u|2 dx + cϵ ∮ u2 dS Ω

Ω

∀ u ∈ W 1,2 (Ω).

(C.0.5)

𝜕Ω

In Lipschitz domains the following embedding theorems hold: W 1,2 (Ω) 󳨅→ L2 (Ω), compact, 1,2

(C.0.6)

2

W (Ω) 󳨅→ L (𝜕Ω), compact. For the discussion of fourth order problems we need the spaces W 2,2 (Ω) and W02,2 (Ω), equipped with the norms 2

2

2

‖u‖W 2,2 (Ω) = (∫[(Δu) + |∇u| + u ] dx) Ω

1 2

2

1 2

and ‖u‖W 2,2 (Ω) = (∫(Δu) dx) . 0

Ω

If Ω ⊂ ℝn is bounded, then W02,2 (Ω) 󳨅→ W01,2 (Ω), compact see, e. g., [2, Theorem A8.1].

(C.0.7)

D Bilinear forms D.1 Abstract setting Let ℋ and ℒ be two real Hilbert spaces such that the embedding ℋ ⊂ ℒ is compact. We consider a symmetric bilinear form a : ℋ × ℋ → ℝ,

(u, υ) → a(u, υ)

which satisfies a(u, υ) ≤ c‖u‖ℋ ‖υ‖ℋ for some positive constant c. The inner product on ℋ is denoted by (u, υ)ℋ and induces the norm ‖u‖ℋ . Similarly, we denote by (u, υ)ℒ the inner product on ℒ and the induced norm by ‖u‖ℒ . Then the variational formulation of the eigenvalue problem is as follows. Find a pair (u, λ) ∈ ℋ × ℝ such that a(u, υ) = λ(u, υ)ℒ

∀ υ ∈ ℋ.

(D.1.1)

The number λ is an eigenvalue of a if there exists an element u ∈ ℋ, u ≠ 0, such that (D.1.1) holds. The element u is called eigenvector corresponding to λ. The number of independent eigenvectors corresponding to λ is called the multiplicity of λ. If the multiplicity is one, then the eigenvalue is simple. The smallest (first) eigenvalue λ1 is characterized by the Rayleigh principle λ1 = inf{

a(u, u) : u ∈ ℋ, u ≠ 0}. (u, u)ℒ

The kernel of the bilinear form a is the set of all elements u ∈ ℋ such that a(u, υ) = 0 for all υ ∈ ℋ. A special situation occurs if a is coercive, that is, if there exists a positive number γ such that a(u, u) ≥ γ‖u‖2ℋ . In that case a(⋅, ⋅) defines an inner product on ℋ which is equivalent to (⋅, ⋅)ℋ . As a consequence, (D.1.1) shows that any eigenvalue λ is positive. The following theorem can be found in many textbooks. Theorem D.1. Let a(⋅, ⋅) be a coercive symmetric bounded bilinear form on ℋ × ℋ and let ℋ ⊂ ℒ be compactly embedded. Then the following holds true. (i) There are at most countably many eigenvalues (λk )k with finite multiplicity each. There is no finite accumulation point. (ii) The eigenvalues are ordered λ1 ≤ λ2 ≤ ⋅ ⋅ ⋅ ≤ λk ≤ ⋅ ⋅ ⋅ . https://doi.org/10.1515/9783111025438-020

266 � D Bilinear forms (iii) There are countably infinitely many eigenvectors corresponding to nonzero eigenvalues λ1 ≤ λ2 ≤ ⋅ ⋅ ⋅ ≤ λk ≤ ⋅ ⋅ ⋅ . They are orthonormal with respect to the inner product (⋅, ⋅)ℒ . (iv) Let Eλk be the eigenspace corresponding to the eigenvalue λk . Then dim Eλk < ∞ for all ∞ k ∈ ℕ and ℋ = ⨁∞ k=1 Eλk . Moreover, the eigenfunctions {uj }j=1 form an orthonormal basis for ℋ. (v) The higher eigenvalues are characterized by Poincaré’s principle λi = min {

i−1 a(υ, υ) : υ ⊥ ⨁ Eλk } . (υ, υ)ℒ k=1

An equivalent version of this principle is as follows. Let Lk ⊂ ℋ be a k-dimensional linear subspace. Then λi = min (max

Li−1 ⊂ℋ υ∈Li−1

a(υ, υ) ). (υ, υ)ℒ

D.2 Examples D.2.1 Eigenvalues on the sphere Consider the eigenvalue problem Δ∗ ϕ + Λϕ = 0

on 𝜕B1 ⊂ ℝn .

In this case a(u, υ) = ∮ (∇τ u ⋅ ∇τ υ) dS

and (u, υ) = ∮ uυ dS.

𝜕B1

𝜕B1

,k = Here ℋ = ℒ = W 1,2 (𝜕B1 ) and ℒ = L2 (𝜕B1 ). The eigenvalues are Λk = k(n+k−2) R2 x 0, 1, . . . , and the corresponding eigenfunctions are the spherical harmonics Yk ( |x| ). By Rayleigh’s principle, Λ0 = 0 = inf υ

Λ1 = Λ2 =

∮𝜕B |∇τ υ|2 dS R

∮𝜕B υ2 dS

n−1 = inf υ R2

,

υ ∈ W 1,2 (𝜕BR ),

R

∮𝜕B |∇τ υ|2 dS R

∮𝜕B υ2 dS

,

υ ∈ W 1,2 (𝜕BR ),

∮ υ dS = 0, BR

R

τ

2

∮𝜕B |∇ υ| dS 2n R = inf , 2 υ R ∮ υ2 dS 𝜕BR

υ ∈ W 1,2 (𝜕BR ),

∮ υ dS = 0 and ∮ xυ dS = 0. BR

𝜕BR

D.2 Examples

� 267

D.2.2 Robin eigenvalues Let Ω be a bounded domain. For ℋ = W 1,2 (Ω) and ℒ = L2 (Ω) we consider the following bilinear form: a(u, υ) := ∫(∇u ⋅ ∇υ) dx + α ∮ uυ dS, Ω

α ∈ ℝ.

𝜕Ω

By Friedrich’s inequality (C.0.4), we have for α > 0 a(u, u) ≥ cF−1 min{α, 1} ∫ u2 dx

for all u ∈ W 1,2 (Ω).

Ω

If α < 0, the modified trace inequality (C.0.3) gives a(u, u) ≥ ∫ |∇u|2 dx − |α|[ϵ ∫ |∇u|2 dx + cϵ ∫ u2 dx] ≥ −|α|cϵ ∫ u2 dx. Ω

Ω

Ω

Ω

Hence, for α > 0 the bilinear form a(u, υ) is coercive. If α < 0, there exists a first negative eigenvalue. Theorem D.1 applies in both cases. The eigenspace of λ1 is one-dimensional. In fact, if u is an eigenfunction of λ1 , then a(u,u) as well. Both functions solve the Euler–Lagrange clearly |u| is a minimizer of ‖u‖ 2 L2 (Ω)

equation Δw + λ1 w = 0 in Ω. Classical regularity theory then implies the analyticity of any solution. Hence, u cannot change sign. Any other eigenfunction is orthogonal to u by Theorem D.1(v). Thus, it must change sign. This gives the uniqueness of the first eigenfunction up to a multiplicative constant. Hence, λ1 is simple. D.2.3 Buckling eigenvalue Let Ω be a bounded domain. For ℋ = W02,2 (Ω) and ℒ = W01,2 (Ω) we consider the following bilinear form: a(u, υ) := ∫ ΔuΔυ dx, Ω

(u, υ)ℒ := ∫ (∇u ⋅ ∇υ) dx. Ω

Poincaré’s inequality yields a(u, u) ≥ γ‖u‖2W 2,2 (Ω) . Due to the compact embedding (C.0.7), the statements of Theorem D.1 apply.

268 � D Bilinear forms D.2.4 Steklov eigenvalue of fourth order Let Ω be a bounded Lipschitz domain. We equip the space ℋ = W 2,2 ∩ W01,2 (Ω) with the scalar product (u, υ)ℋ = ∫Ω ΔuΔυ dx (see, e. g., [49]). Let ℒ = L2 (𝜕Ω). The bilinear form is defined as a(u, υ) := ∫ ΔuΔυ dx, Ω

(u, υ)ℒ := ∫ uυ dS. 𝜕Ω

The embedding from W 2,2 ∩ W01,2 (Ω) to L2 (𝜕Ω) is compact due to some results in [92, Theorem 6.2 in Chapter 2].

Notation Φt : Ω → Ωt

diffeomorphism

Ωt = Φt (Ω) = {x + tυ(x) + ̃ : ξ ∈ ℝn−1 } {x(ξ)

t2 w(x) 2

ρ = (υ ⋅ ν)

+ o(t ) : x ∈ Ω} domain perturbation

n

if x ∈ ℝ \ Ω,

−dist(x, 𝜕Ω)

δ(x) := {

2

dist(x, 𝜕Ω)

if x ∈ Ω

̃ := f (x(ξ)) f (ξ) ∇τ f ̃ := gij f ̃ x ̃

normal component of a vector field on 𝜕Ω signed distance function function defined on 𝜕Ω tangential gradient

ξj ξi

G = gij := (xξi ⋅ xξj ), G

:= g

−1 2

t 2

̃ + t υ(ξ) ̃ + 𝜕Ωt = {x(ξ)

ij

̃ w(ξ) + o(t 2 ) : ξ ∈ ℝn−1 }

metric tensor on 𝜕Ω parametric representation of Ωt derivative

𝜕 𝜕xi

𝜕i :=

parametric representation of 𝜕Ω

(Dυ )ij := 𝜕i υj

Jacobi matrix

(DΦt )ij = 𝜕i (xj + tυj (x) + 𝜕i∗ υ̃ := gij υ̃ξj

t2 w (x)) 2 j

div𝜕Ω υ̃ := (𝜕i∗ υ̃ ⋅ υ̃ξi )

tangential divergence

Δ∗

Laplace–Beltrami operator on 𝜕Ω

T (A) := aii

trace of the matrix (A)ij = aij

Lij

second fundamental form ik

ℒ = g Lkj

Weingarten’s operator

(At )ij := aji

transpose of A

Dυ : Dυ := 𝜕i υj 𝜕j υi = D2 u(y) :=

𝜕2 u 𝜕yi 𝜕yj

T (D2υ )

̃ := u (x + tυ(x) + u(t) bt :=

1 |Ωt

trace of the square of D2υ Hessian matrix

2

t 2

w(x) + o(t 2 ))

∫Ω y dy t

∮𝜕Ω (υ ⋅ ν)x dS = 0

Qg (u ) ′

pullback of u(y) to Ω barycenter of Ωt Φt satisfies the barycenter condition quadratic form

f ± (x) = max{±f (x), 0}

𝕊n and ℍn

spherical and hyperbolic space

̇ ̈ 𝒱(0) and 𝒱(0)

first and second variation of the volume

̇ and 𝒮(0) ̈ 𝒮(0)

first and second variation of the surface area

https://doi.org/10.1515/9783111025438-021

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Index Alexandrov’s uniqueness theorem 25 annular domain 45 annulus 214 barycenter 29 barycenter condition 30, 142, 168, 249 bilinbear form 265 Bossel–Daners inequality 192 buckling eigenvalue 4, 217 buckling plate 49 capacity 209 Cartesian coordinates 7, 17 change of variables 52 change of variables method 52 compact self-adjoint elliptic operators 38 compatibility condition 195 convolutions 81 cost function 92 critical point 167, 221 curvilinear coordinates 258 Dirichlet eigenvalue 191 Dirichlet energy 114, 121, 125, 144, 201 Dirichlet integral 57 distance 32 domain variations 7 dynamical boundary conditions 188 Einstein convention 7 energy 100 Euler–Lagrange equation 1, 99 extension Theorem 14, 16 exterior domains 179 exterior eigenvalue problem 178 Fichera’s principle 248 first and second order perturbations 19 first fundamental form 9 first variation 18 first variation of the surface area 24 first variation of 𝒱(t) 18 formulas of Gauss and Weingarten 257 Fourier series 248 fourth order Steklov eigenvalue problem 50, 236 fourth order Steklov problem 236 Fraenkel asymmetry function 168 https://doi.org/10.1515/9783111025438-023

Frenet’s formula 27 Friedrich inequality 133 Friedrich’s inequality 264 Funk–Hesse formula 42, 87 Gamov energy 91 Gauss’ theorem 224 Gauss theorem on surfaces 13 Gelfand problem 159 generalized Steklov eigenvalue problem 131 graph 258 Green’s function 4 Hadamard perturbations 7, 14, 20, 121, 214 Hadamard’s formula 1 harmonic homogeneous polynomial 38 Helmholtz equation 48 Hesse matrix 54 Hölder continuity 253 homogeneous polynomial of degree k 38 hyperbolic space 79 implicit function theorem 158, 261 isoperimetric inequality 78, 80, 191 Jacobi matrix 8 Jacobi’s formula 23, 54, 253, 260 kernel 29, 126, 127, 141, 249, 265 Laplace–Beltrami operator 13, 39, 257 local maximizer 178 local orthonormal frame 10 material derivative 34 maximum principle 237 mean curvature 10, 257 membrane eigenvalue problems 48 metric 23 metric tensor 257 modified Friedrich inequality 132 moment of inertia 72 moving surface method 52, 59 nearly spherical domain 135, 174, 197, 204, 212 Neumann eigenvalues 222 Neumann energy 128

278 � Index

normal coordinates 259 parallel set 32 Payne’s inequality 4, 217, 232 perimeter 23 Pohozaev’s identity 120, 130 Poincaré’s inequality 263 Poincaré’s principle 173, 266 Pólya and Szegö’s conjecture 4, 217 principal curvatures 10, 257 projections 12 quantitative inequality 167 Rayleigh principle 2, 132, 265 Rayleigh quotient 4, 38, 132, 163, 239 Rayleigh–Faber–Krahn inequality 3, 166 reaction–diffusion problem 188 Reynolds 54 Reynolds formula 56 Reynolds theorem 239 Reynolds’ transport theorem 220 Robin eigenvalue 2, 173, 191 Robin eigenvalue problem 171, 191 Robin energy 116, 126, 135, 170, 201 rotation 21, 119 Schwarz symmetrization 203 second fundamental form 10, 19, 257 second variation of the surface area (perimeter) 25 second variation of 𝒱(t) 18 shape derivative 34, 220, 238, 242 signed distance 32, 254 Simpson–Spector inequality 177 Sobolev spaces 263 spherical coordinates 258 spherical exterior domain 178

spherical harmonic of degree k 39 spherical harmonics Yk,i (ξ) 40 stability 167 stability criterion 169 Steklov eigenvalues 248 Steklov problem 43, 175 Steklov type eigenvalue problem 131 Steklov–Dirichlet eigenvalue problem 46 stereographic projection 75, 79 stream function 209 surface area 23 surface element 23, 257 Szegö–Weinberger inequality 193 tangential divergence 12 tangential gradient 11, 12 tangential perturbations 20, 123 theorem of Stone–Weierstrass 40 torsion energy 200 torsion problem 1, 93, 143, 195, 202, 213 trace 10, 23 trace inequality 216, 263 translation 119 Turan inequality 183 turning point 157 volume preserving perturbations 19, 125, 197, 221, 239 weighted isoperimetric inequalities 74 weighted surface area 74 weighted volume 74 Weingarten operator 10, 257 Weingarten’s equations 10 Weingarten’s formula 30 Whitney’s theorem 254

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