Ilaria Fragalà

Filippo Gazzola

Jimmy Lamboley

Michel Pierre

Counterexamples to Symmetry for Partially Overdetermined Elliptic Problems

Keywords: AMS 1991 subject classification: Primary: 35J70; Secondary: 35B50, 49Q10 overdetermined boundary value problems, shape optimization

We exhibit several counterexamples showing that the famous Serrin's symmetry result for semilinear elliptic overdetermined problems may not hold for partially overdetermined problems, that is when both Dirichlet and Neumann boundary conditions are prescribed only on part of the boundary. Our counterexamples enlighten subsequent positive symmetry results obtained by the first two authors for such partially overdetermined systems and justify their assumptions as well.

Introduction

Let Ω be an open bounded connected subset of R n with smooth enough boundary, and let Γ be a nonempty connected (relatively) open subset of ∂Ω. Let also ν denote the unit outer normal to ∂Ω, c be a positive constant and f : R → R be a smooth function. By "overdetermined problem", we mean any boundary value problem of the following kind:

   -∆u = f (u) in Ω u = 0 and u ν = -c on Γ u = 0 on ∂Ω \ Γ , (1)
or

   -∆u = f (u) in Ω u = 0 and u ν = -c on Γ |∇u| = c on ∂Ω \ Γ , (2)
where u ν denotes the normal derivative of u on ∂Ω. Here and in the sequel, by a solution u to problem (1) (resp. ( 2)), we always mean that u ∈ C 0 (Ω) ∩ C 1 (Ω ∪ Γ) ∩ C 2 (Ω) (resp. C 1 (Ω) ∩ C 2 (Ω)). The choice of the word "overdetermined" is justified by the presence of both the Dirichlet and Neumann conditions on a same nonempty part Γ of the boundary in problems ( 1)-( 2): this makes them in general not well-posed. Thus the existence of a solution to (1) or ( 2) is not always guaranteed, and, if existence happens to hold, it is actually supposed to imply some severe geometric constraint on Ω.

This kind of problem was studied by Serrin [START_REF] Serrin | A symmetry problem in potential theory[END_REF]. His celebrated result states that, in the case of totally overdetermined problems, that is when Γ ≡ ∂Ω, then existence of a solution implies that Ω is a ball (and u is radially symmetric).

More recently, the case of partially overdetermined problems, that is when Γ ∂Ω, has been studied by the first two authors in [START_REF] Fragalà | Partially overdetermined problems[END_REF], where they investigate the following natural question: "If Γ ∂Ω, can we still conclude that Ω is a ball whenever (1) or (2) admits a solution?" The answer is trivially no without any extra natural geometric restriction on Ω. Assume, for instance, that Ω is an annulus, that is Ω = {x ∈ R n ; 0 < a < |x| < b}. Then, the solution of -∆u = 1 on Ω, with u = 0 on its boundary, is radially symmetric. Therefore, u ν is equal to a constant on each piece of the boundary, but with different constants for each of them.

On the other hand, if ∂Ω is assumed to be connected, the problem becomes much more significant and delicate. In fact there are many different situations where the answer to the above question is yes, so that Serrin's symmetry result continues to hold. This occurs under suitable additional assumptions, involving both regularity and geometric features, on the source term f and the overdetermined region Γ: for the detailed statements, as well as for a more extensive bibliography about overdetermined problems, we refer to [START_REF] Fragalà | Partially overdetermined problems[END_REF].

The goal of this note is to show that there are nontrivial cases (meaning in particular that ∂Ω is connected) when the requirements of [START_REF] Fragalà | Partially overdetermined problems[END_REF] are not satisfied and problems like (1)-( 2) admit a solution in domains Ω different from a ball.

The counterexamples we construct for problems of type (1) or (2) are of different kind. Problems of type [START_REF] Briançon | Regularity of optimal shapes for the Dirichlet's energy with volume constraint[END_REF] are treated in Section 2 by an approach based on shape optimization and domain derivative. More precisely, we consider the problem of minimizing the Dirichlet energy of domains with prescribed volume and confined in a planar box, that is

|Ω * | = α, Ω * ⊂ D, J(Ω * ) = min |Ω|=α,Ω⊂D J(Ω), (3)
where D = (-1, 1) 2 and

J(Ω) := inf v∈H 1 0 (Ω) Ω 1 2 |∇v| 2 -v dx . (4)
Choosing α in a suitable range and applying the regularity results in [START_REF] Briançon | Regularity of optimal shapes for the Dirichlet's energy with volume constraint[END_REF][START_REF] Briançon | Lipschitz continuity of state functions in some optimal shaping[END_REF], we obtain that (3) admits an optimal open shape Ω * with a nonempty smooth "free boundary" ∂Ω * ∩ D. Then, writing down the optimality conditions by using shape derivatives, we are lead to a problem of type [START_REF] Briançon | Regularity of optimal shapes for the Dirichlet's energy with volume constraint[END_REF] on Ω * , with f ≡ 1 and Γ = ∂Ω * ∩ D. Problems of type ( 2) are treated in Section 3 by a different approach, which works in any dimension n ≥ 2. In this case, the counterexamples are derived through some explicit computations. They are based on the idea of studying the zero level surfaces of radial functions u built so as to satisfy both an elliptic equation of the type -∆u = f (u) on the whole R n and the eikonal equation |∇u| = c on the complement of a ball. Such construction can be adapted to treat also the case of a partially overdetermined problem similar to (2), but stated on an exterior domain (see Section 3.2).

Counterexamples using shape optimization

In this section we use shape optimization in order to prove the following.

Theorem 2.1

There exists an open starshaped planar domain Ω ⊂ (-1, 1) 2 , different from a disk, such that, for a nonempty connected analytic subset Γ of ∂Ω, the problem

   -∆u = 1 in Ω u = 0 on ∂Ω u ν = -c on Γ, (5)
admits a solution. In particular, Proposition 2.3 implies that the analytic piece Γ of the boundary of Ω found in Theorem 2.1 cannot be continued into a globally analytic closed "curve" (namely the boundary of another open set Ω). In the counterexample provided here, ∂Ω is piecewise analytic and globally at most C 1, 1 2 as analyzed in [START_REF] Lamboley | About Hölder regularity of the optimal convex planar shape for λ 2[END_REF].

Proof of Theorem 2.1: Let D = (-1, 1) 2 and α ∈ (π, 4). We will construct Ω as an optimal set for the shape minimization problem (3).

From [4, Theorem 2.4.6] (see also [START_REF] Henrot | Variation et optimisation de forme[END_REF]), we know there exists a quasi-open optimal set Ω * which solves problem [START_REF] Briançon | Regularity of the optimal shape for the first eigenvalue of the Laplacian with volume and inclusion constraint[END_REF]. In view of [2, Corollary 1.2], Ω * is in fact an open set. It is known that, for any open bounded set Ω (and in particular for Ω * ), the functional J defined in (4) satisfies

J(Ω) = Ω 1 2 |∇u Ω | 2 -u Ω dx
where u Ω denotes the unique solution of the homogeneous Dirichlet problem

-∆u = 1 in Ω u = 0 on ∂Ω . (6)
Since α < 4, Ω * cannot be equal to D so that the free boundary Γ := ∂Ω * ∩ D is nonempty. Moreover, by [1, Section 5], we infer that Γ is analytic because f ≡ 1 is positive and analytic. On this "free boundary" Γ, using the notion of shape derivative (see for instance [START_REF] Henrot | Variation et optimisation de forme[END_REF]), we classically obtain the Euler-Lagrange equation of problem (3), namely, [START_REF] Cianchi | Steiner symmetric extremals problems in Pólya-Szegö-type inequalities[END_REF] with

Ω = Ω * , u = u Ω * together with |∇u Ω * | = Λ > 0 on ∂Ω * ∩ D. (7)
Since f (u) = 1 > 0, the positivity of the Lagrange multiplier Λ follows from [1, Proposition 6.1]. By elliptic regularity, we know that there exists a unique solution u Ω * ∈ C ∞ (Ω ∪ Γ) to [START_REF] Silva | A singular energy minimizing free boundary[END_REF]. We now prove the geometric properties of solutions of (3). First, since α > π, Ω * is not a disk. Second, we show that Ω * is starshaped, or at least that it may be replaced by an optimal starshaped set. To this end, we introduce Ω := S X S Y (Ω * ), where S X and S Y denote the Steiner symmetrization about the axes OX and OY respectively, see e.g. [START_REF] Henrot | Variation et optimisation de forme[END_REF], [START_REF] Kawohl | Rearrangements and convexity of level sets in[END_REF]. Because of the symmetry of the square D with respect to these axes, we have Ω ⊂ D. Moreover, | Ω| = |Ω * | = α and, by well-known properties of Steiner symmetrization, J( Ω) ≤ J(Ω * ). Therefore, Ω is also a solution of the shape optimization problem (3) so that, as for any optimal set, Γ = ∂ Ω ∩ D is smooth and u e Ω satisfies [START_REF] Caffarelli | Global energy minimizers for free boundary problems and full regularity in three dimensions[END_REF]. To verify that it is starshaped, we may denote

∀x ∈ [-1, 1], A(x) := {y ∈ [-1, 1]; (x, y) ∈ S Y (Ω * )}.
As a consequence of the definition of the Steiner symmetrization, we have [0

≤ x ≤ x] ⇒ [A(x) ⊂ A(x)].
We may also write

S X S Y (Ω * ) = (x, y); |y| ≤ 1 2 measA(x) . Since x ∈ [0, 1] → measA(x) is nonincreasing, we have |y| ≤ 1 2 measA(x), λ ∈ (0, 1) ⇒ |λy| ≤ 1 2 measA(x) ≤ 1 2 measA(λx) .
This proves that Ω is starshaped. Therefore, Ω = Ω, u = u e Ω , c = Λ, Γ = any connected component of ∂Ω∩D satisfy the statement of Theorem 2.1.

We conclude this section by mentioning some possible extensions of Theorem 2.1.

Remark 2.4

The construction done in the proof of Theorem 2.1 is valid in any dimension and one finds as well an optimal open set Ω * ⊂ (-1, 1) n (see [START_REF] Briançon | Regularity of optimal shapes for the Dirichlet's energy with volume constraint[END_REF] for a proof), which is different from a ball if α > ω n (the measure of the unit ball). But, the full regularity of the boundary is not proved -and probably does not hold-in any dimension. According to some recent papers ( [START_REF] Caffarelli | Global energy minimizers for free boundary problems and full regularity in three dimensions[END_REF][START_REF] Silva | A singular energy minimizing free boundary[END_REF][START_REF] Weiss | Partial regularity for weak solutions of an elliptic free boundary problem[END_REF][START_REF] Weiss | Partial regularity for a minimum problem with free boundary[END_REF]), it is very likely that full regularity of the boundary may be extended to dimensions greater than 2 (up to 6? but not more?).

However, as proved in [START_REF] Briançon | Regularity of optimal shapes for the Dirichlet's energy with volume constraint[END_REF], the reduced boundary of this Ω * is an analytic hypersurface and this regular part of the boundary is of positive (n -1)-Hausdorff measure if α < 2 n , whereas Ω * is not a ball if α > ω n . Therefore, this also provides a (generalized) counterexample in any dimension by choosing Γ to be this reduced boundary.

Remark 2.5 In view of [START_REF] Briançon | Regularity of the optimal shape for the first eigenvalue of the Laplacian with volume and inclusion constraint[END_REF] (see also [START_REF] Henrot | Extremum problems for eigenvalues of elliptic operators[END_REF]Section 3.4]), it is possible to extend the statement of Theorem 2.1 to the case when J is replaced by the shape functional Ω → λ 1 (Ω), the first eigenvalue of the Laplace operator on Ω with homogeneous Dirichlet boundary conditions. This provides one more example of an optimal domain Ω * where u Ω * , the first normalized eigenfunction, solves [START_REF] Briançon | Regularity of optimal shapes for the Dirichlet's energy with volume constraint[END_REF] with f (u) = λu (here, λ = λ 1 (Ω * )). The proof is similar and we do not reproduce it here. It is possible that one could go further and extend the same construction to more general sources f (u), for instance of power-type such as f (u) = u p . Remark 2.6 The minimal shape Ω * for the second Dirichlet eigenvalue λ 2 (Ω) of the Laplace operator, among all planar convex domains of given area, is also a natural candidate for another nice counterexample. It is proved that Ω * is not a "stadium" (the convex envelope of two identical tangent balls), see [START_REF] Henrot | Minimizing the second eigenvalue of the Laplace operator with Dirichlet boundary conditions[END_REF]. However, it is expected that it looks like a stadium (see [START_REF] Henrot | Minimizing the second eigenvalue of the Laplace operator with Dirichlet boundary conditions[END_REF]). If it is the case, as explained in [START_REF] Fragalà | Partially overdetermined problems[END_REF], then the first order optimality condition would lead to an overdetermined problem in which the expected overdetermined part Γ would be the strictly convex part of ∂Ω * . The exact regularity and shape of Ω * is still to be completely understood: see [START_REF] Henrot | Minimizing the second eigenvalue of the Laplace operator with Dirichlet boundary conditions[END_REF]Theorems 4,[START_REF] Cianchi | Steiner symmetric extremals problems in Pólya-Szegö-type inequalities[END_REF][START_REF] Fragalà | Partially overdetermined problems[END_REF] and [START_REF] Lamboley | About Hölder regularity of the optimal convex planar shape for λ 2[END_REF].

Remark 2.7 In the proof of Theorem 2.1, we started with some optimal shape Ω * and adapted it so that it satisfies the required conditions. We may wonder whether all optimal shapes have the same symmetry properties. This question is related to the nontrivial question of equality case in the Steiner symmetrization, namely: is it true that J(Ω) = J(S X (Ω)) implies that Ω = S X (Ω) up to a translation? We refer to [START_REF] Cianchi | Steiner symmetric extremals problems in Pólya-Szegö-type inequalities[END_REF] for this question.

Counterexamples via explicit construction

In this section we provide an explicit example of a problem of type (2) which admits a solution on a domain different from a ball. We also exhibit a similar example for an analogous exterior problem.

A counterexample in an interior domain

Remark 2 . 2

 22 Note that a nonempty analytic subset Γ of ∂Ω is relatively open in ∂Ω. The interest of this negative result should be considered in the light of the following extension of Serrin's result proved in [8]: Proposition 2.3 Let Ω be open and bounded with ∂Ω connected. Let Γ ⊂ ∂Ω nonempty and (relatively) open. Assume there exists an open set Ω with a connected analytic boundary containing Γ. If there exists a solution u of (1) with f analytic, then Ω = Ω, Ω is a ball, and u is radially symmetric.

Theorem 3 . 1

 31 There exist a Lipschitz continuous and strictly increasing functionf : R → (0, +∞) and u ∈ C 2 (Ω) solution of    -∆u = f (u) in Ω |∇u| =where Ω ⊂ R n is open, bounded, simply connected, different from a ball, with ∂Ω globally C ∞ , and Γ ⊂ ∂Ω is nonempty, connected, relatively open and included in a sphere of R n .Proof: Fix an integer n ≥ 2 and consider the function f : R → (0, +∞) defined byf (s) =        64(n -1) 8s if s ≤ 0 4 (n + 2) √ s + 4 -6 if s ≥ 0 .

Then, f is globally Lipschitz continuous and strictly increasing over R.

Consider also the (radial) function u defined on R n by

Then, u ∈ C 2 (R n ); to see this, it suffices to write u = u(r) as a function of the real variable r = |x| and to note that

are continuous functions in [0, ∞). Moreover, some computations show that u satisfies

where B denotes the unit ball.

Let 1); for a suitable choice of Ω 2 one has ∂Ω ∈ C ∞ . Let Γ = ∂Ω 1 ∩ ∂Ω, then u satisfies (8) but Ω is not a ball. Theorem 3.1 should be compared with the following result obtained in [START_REF] Fragalà | Partially overdetermined problems[END_REF], and similar to Proposition 2.3:

Let Ω be open and bounded with ∂Ω connected. Let Γ ⊂ ∂Ω nonempty and (relatively) open. Assume there exists an open set Ω with a connected analytic boundary containing Γ. If there exists a solution u of (8) with f analytic and nonincreasing, then Ω = Ω, Ω is a ball, and u is radially symmetric.

Note in particular that: the overdetermined part Γ in Theorem 3.1 satisfy the hypothesis in Proposition 3.2 (analytically continuable according to the definition in [8, Section 3.1]), but f is neither analytic, nor nonincreasing.

Similarly, Theorem 3.1 should also be compared with the statements (b) in Theorems 3 and 7 in [START_REF] Fragalà | Partially overdetermined problems[END_REF] which gives more various sufficient conditions to obtain symmetry in overdetermined problems of type [START_REF] Fragalà | Partially overdetermined problems[END_REF]. Again, Theorem 3.1 provides an example where all these hypothesis are satisfied, except the fact that f be nonincreasing.

A counterexample in an exterior domain

where Ω ⊂ R n is open, bounded, simply connected, different from a ball, with ∂Ω globally C ∞ , and Γ ⊂ ∂Ω is nonempty, connected, relatively open and included in a sphere.

Proof: Fix an integer n ≥ 2 and consider the function f : R → R defined by

Then, f is globally Lipschitz continuous over (0, 3 2 ); moreover, if n ≥ 4 then f is positive and strictly increasing.

Consider also the (radial) function u defined on R n \ {0} by

Then, u ∈ C 2 (R n \ {0}); to see this, it suffices to write u = u(r) as a function of the real variable r = |x| and to note that

are continuous functions in (0, ∞). Moreover, some computations show that u satisfies

where B denotes the unit ball. Take any smooth domain Ω B such that 0 ∈ Ω and {x ∈ ∂B; x 1 < 1 2 } ⊂ ∂Ω (see figure 2). Let Γ = ∂Ω ∩ ∂B, then u satisfies (9) but Ω is not a ball.

Remark 3.4 Again, Theorem 3.3 should be compared with the results in [START_REF] Fragalà | Partially overdetermined problems[END_REF], similarly to what we did for Theorem 3.1.


Jimmy Lamboley

Arian Novruzi

Polygons as optimal shapes with convexity constraint

Keywords: Shape optimization, convex constraint, optimality conditions

In this paper, we focus on the following general shape optimization problem:

where S ad is a set of 2-dimensional admissible shapes and J : S ad → R is a shape functional. Using a specific parameterization of the set of convex domains, we derive some extremality conditions (first and second order) for this kind of problem. Moreover, we use these optimality conditions to prove that, for a large class of functionals (satisfying a concavity like property), any solution to this shape optimization problem is a polygon.

Introduction

In this paper, we are mainly interested in questions related to a convexity constraint in shape optimization. We deal with the following general shape optimization problem:

min{J(Ω), Ω convex, Ω ∈ S ad }, (1)
where J is a shape functional defined on a class S ad of subsets of R 2 .

Our goal is, on one hand, to write down explicit first and second order optimality conditions for general 2-dimensional shape optimization problems with convexity constraint and, on the other hand, to use them to exhibit a family of shape functionals for which optimal shapes are polygons.

As it is well-known, dimension 2 allows to write the convexity constraint through the positivity of a linear operator with respect to the shape. More precisely, if one uses polar coordinates representation (r, θ) for the domains, namely

Ω u := (r, θ) ∈ [0, ∞) × R ; r < 1 u(θ) , (2)
where u is a positive and 2π-periodic function, then

Ω u is convex ⇐⇒ u ′′ + u ≥ 0.
As a consequence, we look at shape optimization problems of the form

u 0 ∈ F ad , j(u 0 ) = min{j(u) := J(Ω u ), u > 0, u ′′ + u ≥ 0, u ∈ F ad } (3)
where F ad is a set of convenient 2π-periodic admissible functions.

1

A first contribution is to write down explicitly the first and second order optimality conditions on u 0 for some specific choices of F ad . Then, we use these conditions to address the following question: let us consider the functional

J(Ω u ) = j(u) = 2π 0 G (θ, u(θ), u ′ (θ)) dθ
where G : R × (0, +∞) × R → R is C 2 , 2π-periodic in the first variable, and j is defined on some set of functions F ad as above. Then, is it possible to write down sufficient conditions on G so that any optimal shape of (3) be a polygon?

This question is motivated by two preliminary results in this direction arising from two quite different fields:

• first a result by M. Crouzeix in [START_REF] Crouzeix | Une famille d'inégalités pour les ensembles convexes du plan[END_REF], motivated by abstract operator theory: a problem of the form ( 3) is considered where G(θ, u, u ′ ) = h(u ′ /u) with h strictly concave and even, and F ad = {u regular enough ; 0 < a ≤ u ≤ b}. All optimal shapes are shown to be polygons,

• then a result by T. Lachand-Robert and M.A. Peletier for a shape optimization arising in the modeling of the movement of a body inside a fluid (Newton's problem, see [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF] and references therein). Here G(θ, u, u ′ ) = h 1 (u)u ′2 h 2 (u) and F ad = {u regular enough ; 0 < a ≤ u ≤ b}.

Again, with convenient assumptions on h 1 and h 2 , they prove that optimal shapes are polygons.

We also refer to T. Bayen [START_REF] Bayen | Optimisation de forme dans la classe des corps de largeur constante et des rotors[END_REF] for results about minimizing functionals of type j with similar constraints, seen as controls.

Our goal here is to generalize these two results and to find rather general sufficient conditions on G which will imply that optimal shapes are necessarily polygons. We state three results in this direction in the next section. It turns out that a main step in the proof is based on the use of the second order optimality conditions with convexity constraint. This is the main reason why we write down explicitly these conditions, which are actually interesting for themselves and which may also be useful in some other problems (see [START_REF] Henrot | Minimizing the second eigenvalue of the Laplace operator with Dirichlet boundary conditions[END_REF] for the use of the first order optimality condition on a particular problem of optimal eigenvalue with convexity constraint). They imply that optimal shapes are necessarily polygons inside the constraints (see Theorem 2.1). Next, to deal with the solution on the constraint, additional assumptions are needed on the boundary of the constraints (see Theorems 2.2 and 2.3). The sufficient conditions that we obtain on G, are rather sharp as shown through several examples and counterexamples.

We state sufficient conditions on G for solutions to be polygons in the following section. Then, Section 3 is devoted to the "abstract" first and second order optimality conditions for convexity constraint. Proofs of the results in Section 2 are given in Section 4. Finally, we give examples and counterexamples in Section 5 which show how sharp our Section 2 results are.

Main results

Notation: T := [0, 2π). Throughout the paper, any function defined on T is considered as the restriction to T of a 2π-periodic function on R, with the same regularity.

Let W 1,∞ (T) := {u ∈ W 1,∞ loc (R) / u is 2π-periodic}. If u ∈ W 1,∞ (T), we say that u ′′ + u ≥ 0 if ∀ v ∈ W 1,∞ (T) with v ≥ 0, T (uv -u ′ v ′ ) dθ ≥ 0. (4)
In this case, u ′′ + u is a nonnegative 2π-periodic measure on R; we then denote S u = Supp(u ′′ + u) the support of this measure.

As explained in the introduction, using the parameterization (2), we consider all open bounded shapes (Ω u ) u>0 . A simple calculus of the curvature gives:

κ(Ω u ) = u ′′ + u (1 + u ′2 ) 3/2 ,
which leads to the characterization of the convexity of Ω u by the linear condition u ′′ + u ≥ 0. Moreover, straight lines in ∂Ω u are parameterized by the set {u ′′ + u = 0}, and corners in the boundary are seen as Dirac masses in the measure u ′′ + u.

We consider, as in [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF][START_REF] Crouzeix | Une famille d'inégalités pour les ensembles convexes du plan[END_REF], the geometric constraint ∂Ω u ⊂ A(a, b) where A(a, b) := {(r, θ) / 1/b ≤ r ≤ 1/a} is a closed annulus. So we consider the problem min j(u)

:= J(Ω u ), u ∈ W 1,∞ (T), u ′′ + u ≥ 0, a ≤ u ≤ b , (5)
where j : W 1,∞ (T) → R, 0 < a < b are given. We are interested in sufficient conditions on j (less restrictive as possible) such that the problem (5) has for solution a polygon. We also look at the same question for the following problem with the volume constraint |Ω u | = m 0 where m 0 is given, namely

min j(u), u ∈ W 1,∞ (T), u ′′ + u ≥ 0, m(u) := 1 2 T dθ u 2 = m 0 , (6)
with m 0 > 0. Note that m(u) is the measure of the domain inside the curve {(1/u(θ), θ), θ ∈ T}.

Theorem 2.1 Let G : (θ, u, p) ∈ T×R×R → G(θ, u, p) ∈ R be of class C 2 and set j(u) = T G(θ, u, u ′ ).
Let u 0 be a solution of [START_REF] Henrot | Variation et optimisation de formes : une analyse géométrique[END_REF] or [START_REF] Ioffe | Theory of extremal problems[END_REF] and assume that G is strongly concave in the third variable at u 0 , that is to say

G pp (θ, u 0 , u ′ 0 ) < 0, ∀θ ∈ T. (7)
• If u 0 is a solution of (5), then S u0 ∩ I is finite, for any I = (γ 1 , γ 2 ) ⊂ {θ ∈ R, a < u 0 (θ) < b}, and in particular Ω u0 is locally polygonal inside the annulus A(a, b),

• If u 0 > 0 is a solution of (6), then S u0 ∩ T is finite, and so Ω u0 is a polygon.

Here S u0 denotes the support of the measure u ′′ 0 + u 0 .

See sections 4.1 and 4.2 for a proof.

Remark 2.1

We choose to analyze a volume constraint in [START_REF] Ioffe | Theory of extremal problems[END_REF] because this one is classical, and also to show that our approach can be adapted to nonlinear constraints. With a few adjustments, this approach can be adapted to some other constraints, regular enough in terms of u, see Proposition 3.3 and Section 4.2.

Remark 2.2

The result is still true if u 0 is only a local minimum of (5) or [START_REF] Ioffe | Theory of extremal problems[END_REF], since the proof only use the optimality conditions stated in Section 3.

Remark 2.3 With the only assumptions of Theorem 2.1, it is not true that Ω u0 is a polygon if u 0 is solution of [START_REF] Henrot | Variation et optimisation de formes : une analyse géométrique[END_REF]. Indeed, a solution can saturate the constraint u ≥ a or u ≤ b, and in these cases, ∂Ω u0 contains an arc of circle. In some particular cases, a solution can also have an infinite number of corners. We refer to Section 5 for explicit examples.

In the following results, we want to go deeper in the analysis, in order to find conditions on G for the solution of (5) to be a polygon. As mentioned in Remark 2.3, we need to avoid that ∂Ω u0 touches the boundary of A(a, b) in an arc of circle, and also an accumulation of corners of ∂Ω u0 in a neighborhood of ∂A(a, b). We treat two kinds of technical assumptions:

Theorem 2.2 Let j(u) = T G(u, u ′ ) with G : (0, ∞) × R → R,
and let u 0 be a solution of [START_REF] Henrot | Variation et optimisation de formes : une analyse géométrique[END_REF]. Assume that (i) G is a C 2 function and G pp < 0 on {(u 0 (θ), u ′ 0 (θ)), θ ∈ T}, (ii) The function p → G(a, p) is even and one of the followings holds (ii.1) G u (a, 0) < 0 or

(ii.2) G u (a, 0) = 0 and G u (u 0 , u ′ 0 )u 0 + G p (u 0 , u ′ 0 )u ′ 0 ≤ 0, (iii) The function p → G(•, p) is even and G u ≥ 0 near (b, 0). Then S u0 is finite, i.e. Ω u0 is a polygon.
The proof of this theorem follows from Theorem 2.1 and Proposition 4.1.

Example 2.1 We can give the following geometric example :

J(Ω) = λ|Ω| -P (Ω),
where | • | denotes the area, P (•) denotes the perimeter, and λ ∈ [0, +∞]. The minimization of J within convex sets whose boundary is inside the annulus A(a, b) is in general non trivial.

When λ = 0, the solution is the disk of radius 1/a (see [START_REF] Buttazzo | Shape optimization problems over classes of convex domains[END_REF] for a monotony property of perimeter with convex sets). When λ = +∞, the solution is the disk of radius 1/b.

We can easily check (see section 5 for more detailed examples) that j(u) = J(Ω u ) satisfies hypothesis of Theorem 2.1, so any solution is locally polygonal inside A(a, b). And from Theorem 2.2, if λ ∈ (a, b) (in order to get conditions (ii) and (iii)), any solution is a polygon.

We can prove the same result as in Theorem 2.2 with a weaker condition than the uniform condition given in (i), namely when G pp (a, p) = 0, like in [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF]. Lemma 4.1) and let u 0 be a solution of [START_REF] Henrot | Variation et optimisation de formes : une analyse géométrique[END_REF]. We assume that •,p) is even and G u ≥ 0 near (b, 0). Then S u0 is finite, i.e. Ω u0 represents a polygon.

Theorem 2.3 Let j(u) = T G(u, u ′ ) with G : (0, ∞) × R → R, C(b) = 2πb (see
(i) G is a C 3 function, G pp = 0 in {a} × [-C(b), C(b)], and G pp < 0 in (a, b] × [-C(b), C(b)], (ii) p → G(a, p) is even, G u (a, p) < 0 for all p ∈ [-C(b), C(b)] and pG up (a, p) = z(p)G upp (a, p) for p ∈ (0, C(b)], with a certain function z ≥ 0, (iii) p → G(
The proof of this theorem follows from Propositions 4.2 and 4.3.

Remark 2.4

The hypotheses in Theorem 2.2 and 2.3 are quite general. In Section 5 we give certain examples showing that if one of these hypotheses is not satisfied, then the solutions of ( 5), in general, are not polygons.

Remark 2.5 The condition (ii.2) in Theorem 2.2 (less natural than (ii.1)) has been motivated by the problem in [START_REF] Crouzeix | Une famille d'inégalités pour les ensembles convexes du plan[END_REF], where G(u, p) = h(p/u) with h(•) a C 2 , strictly concave, and even function. Such a G(u, p) satisfies the hypothesis of Theorem 2.2. Indeed, (a)

G pp (u, p) = h ′′ (p/u)u -2 , so G pp (u, p) < 0 and (i) is satisfied, (b) G u (u, p)u + G p (u, p)p = 0 and G u (•, 0) = 0, so (ii.2) is satisfied, (c) G u (u, p) = -h ′ (p/u) p u 2 ≥ 0 so (iii) is satisfied.
Therefore the solution is a polygon. In [START_REF] Crouzeix | Une famille d'inégalités pour les ensembles convexes du plan[END_REF], several more precise statements about the geometric nature of solutions are proven (in this particular case).

Remark 2.6 Similarly, Theorem 2.3 gives a generalization of the problem studied in [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF]. Indeed, in this problem, they have

G(u, p) = h 1 (u) -p 2 h 2 (u) with h 1 , h 2 two C 2 functions satisfying h ′ 1 (a) < 0, h ′ 1 (b) > 0, h 2 (a) = 0, and ∀ t > a, h 2 (t) > 0 (G is not C 3 in
this case, but in fact we only need the existence of G upp , which is clear here). The function G(u, p) satisfies the hypothesis of Theorem 2.3 as p → G(u, p) is even and (a)

G pp (u, p) = -2h 2 (u), so (i) is satisfied. (b) G u (a, p) = h ′ 1 (a) < 0 and G up (a, p) = -2ph ′ 2 (a), G upp (a, p) = -2h ′ 2 (u), so G up (u, p) = pG upp (u, p), and therefore (ii) is satisfied. (c) G u (u, p) = h ′ 1 (u) -p 2 h ′ 2 (u) so G u (b, 0) = h ′ 1 (b) > 0.
This last assumption is not specified in [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF], but according to us, we need this one, see Section 5.2. In fact, it seems that the case of an accumulation of corners in the interior boundary {u 0 = b} is not considered in [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF] (see Proposition 4.1, case (b)). So the solution is a polygon. In [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF], it is also proven that this polygon is regular in this particular case.

Remark 2.7

Les us make some comments on the question of existence. For the problem (5), there always exists a solution, if for example j is continuous in H 1 (T) (see below for a definition). Indeed, the minimization set

{u ∈ W 1,∞ (T) / u ′′ + u ≥ 0, a ≤ u ≤ b} is strongly compact in H 1 (T).
About the problem ( 6) with a measure constraint, the question is more specific. For example, if one looks at the problem of maximization of the perimeter (for which the concavity assumptions is satisfied), with convexity and measure constraints, we are in a case of non-existence (the sequence of rectangles Ω n = (-n/2, n/2) × (-m 0 /2n, m 0 /2n) satisfies the constraints, whereas the perimeter is going to +∞). However, existence may be proved for many further functionals. In Theorem 2.1, we avoid this issue by asking the solution to be positive (and so to represent a convex bounded set of dimension 2).

First and second order optimality conditions

As we noticed in Remark 2.7, the minimization set is compact. So there are very few directions to write optimality. However, we are able in this section to write general optimality conditions for our problem.

Let us first introduce an abstract setting (see [START_REF] Ioffe | Theory of extremal problems[END_REF], [START_REF] Maurer | First and second order necessary and sufficient optimality conditions for infinite-dimensional programming problems Mathematical Programming[END_REF]). Let U, Y be two real Banach spaces, let K be a nonempty closed convex cone in Y and let f :

U → R, g : U → Y . We consider the minimization problem min{f (u), u ∈ U, g(u) ∈ K}. (8)
We denote by U ′ (resp. Y ′ ) the Banach space of continuous linear maps from U (resp. Y ) into R (dual spaces of U, Y ), and we introduce

Y ′ + = {l ∈ Y ′ ; ∀ k ∈ K, l(k) ≥ 0 }.
The following result is a particular case of Theorem 3.2 and 3.3 stated in [START_REF] Maurer | First and second order necessary and sufficient optimality conditions for infinite-dimensional programming problems Mathematical Programming[END_REF] which will be sufficient for our purpose.

Proposition 3.1 Let u 0 ∈ U be a solution of the minimization problem [START_REF] Maurer | First and second order necessary and sufficient optimality conditions for infinite-dimensional programming problems Mathematical Programming[END_REF]. Assume f and g are twice (Fréchet-)differentiable at u 0 and that g

′ (u 0 )(U ) = Y . Then, (i) there exists l ∈ Y ′ + such that f ′ (u 0 ) = l • g ′ (u 0 ) and l(g(u 0 )) = 0, (ii) if F (u) := f (u) -l(g(u)), then F ′′ (u 0 )(v, v) ≥ 0 for all v ∈ T u0
where

T u0 = v ∈ U ; f ′ (u 0 )(v) = 0, g ′ (u 0 )(v) ∈ K g(u0) = {K + λg(u 0 ); λ ∈ R} .
Remark 3.1 When applying the second order optimality condition (ii), we have to check whether well-chosen v ∈ U are in T u0 . This may be done by using (i) and the information on the linear map l.

We may use instead the following: assume g(u 0 + tv) ∈ K for t > 0 small, or, more generally that

u 0 + tv = v t + tε(t) with lim t→0,t>0 ε(t) = 0 and g(v t ) ∈ K; (9) then f ′ (u 0 )(v) ≥ 0 and g ′ (u 0 )(v) ∈ K g(u0) . (10)
To see this, we write the two following lines:

0 ≤ t -1 [f (v t ) -f (u 0 )] = f ′ (u 0 )(v) + ε 1 (t) where lim t→0,t>0 ε 1 (t) = 0, g ′ (u 0 )(v) = t -1 [g(v t ) -g(u 0 )] + ε 2 (t) where lim t→0,t>0 ε 2 (t) = 0,
and we let t tend to zero. If now, (9) is valid for all t small (t > 0 and t < 0), then v ∈ T u0 .

For our purpose, we choose U = H 1 (T) the Hilbert space of functions from R into R which are in H 1 loc (R) and 2π-periodic, equipped with the scalar product

∀u, v ∈ U, (u, v) U×U = T u v + u ′ v ′ .
Let g 0 : U → U ′ be defined by

∀u, v ∈ U, g 0 (u)(v) = T u v -u ′ v ′ .
For l ∈ U ′ we say l ≥ 0 in U ′ if l(v) ≥ 0 for all v ∈ U . Note that, if g 0 (u) ≥ 0 in U ′ then u + u ′′ , computed in the sense of distributions in R , is a 2π-periodic nonnegative measure on R, and we have

g 0 (u)(v) = T u v -u ′ v ′ = T v d(u + u ′′ ). ( 11
)
Note also, for further purposes, that g 0 (U ) is a closed subspace of U ′ which may be described as the "orthogonal" of the kernel of g 0 (because R(g 0 ) = N (g * 0 ) ⊥ , with g * 0 the adjoint of g 0 ), namely

g 0 (U ) = {z ∈ U ′ ; ∀v ∈ Ker g 0 , z(v) = 0} = {z ∈ U ′ ; z(cos) = z(sin) = 0},
(and cos, sin denote the usual cosine and sine functions on R).

Finally, if l is a continuous linear map from g 0 (U ) into R (that is l ∈ g 0 (U ) ′ ), then, thanks to the Hilbert space structure, there exists ζ ∈ U such that

∀z ∈ g 0 (U ), l(z) = z, ζ U ′ ×U , (ζ, cos) U×U = (ζ, sin) U×U = 0. ( 12
)
First problem:

Let j : U → R be C 2 .
We set Y := g 0 (U ) × U × U equipped with its canonical Hilbert space structure whose scalar product writes:

∀y = (z, u 1 , u 2 ), y = ( z, u 1 , u 2 ) ∈ Y , y, y Y ×Y := z, z U ′ ×U ′ + (u 1 , u 1 ) U×U + (u 2 , u 2 ) U×U .
And we define g : U → Y and K ⊂ Y by

g(u) = (g 0 (u), u -a, b -u), K = {(z, u 1 , u 2 ) ∈ Y ; z ≥ 0 in U ′ , u 1 , u 2 ≥ 0 in U }.
We look at the minimization problem (see Lemma 4.1 and Remarks 3.2 and 4.1 for details about the choice of the two functional spaces H 1 (T) and W 1,∞ (T)): and

min{j(u), u ∈ U, g(u) ∈ K}. (13)
∀ v ∈ H 1 (T), j ′ (u 0 )v = ζ 0 + ζ ′′ 0 , v U ′ ×U + T vdµ a - T vdµ b . ( 15
)
Moreover, ∀ v ∈ H 1 (T) such that ∃λ ∈ R with    v ′′ + v ≥ λ(u ′′ 0 + u 0 ) v ≥ λ(u 0 -a), v ≤ λ(u 0 -b), ζ 0 + ζ ′′ 0 , v U ′ ×U + T vd(µ a -µ b ) = 0 we have j ′′ (u 0 )(v, v) ≥ 0. ( 16
)
Remark 3.2 We choose here to work in the space H 1 (T), whereas the problem is more naturally settled in W 1,∞ (T). This choice is motivated by the fact that H 1 (T) is reflexive. If we had worked in W 1,∞ (T), we would have obtained a Lagrange multiplier in the bidual W 1,∞ (T) ′′ , which is not so easy to make explicit. Nevertheless, this choice of H 1 (T) leads to this new difficulty: for G regular, the functional j(u) = T G(θ, u, u ′ ) is generally not well defined on H 1 (T), and so we cannot directly apply Proposition 3.2. We explain in Section 4.1 the adjustments that are needed to apply this one.

Proof. We apply Proposition 3.1 with the notations just introduced above. The main assumption g ′ (u 0 )(U ) = Y is satisfied since g ′ (u 0 ) = (g 0 , I, -I) where I denotes the identity. By the statement (i) there exists l = (l 0 , l a , l b ) ∈ Y ′ + and thanks to the remarks (11), (12), there exists

(ζ 0 , µ a , µ b ) ∈ U × U ′ × U ′ such that • ∀v ∈ U, j ′ (u 0 )(v) = ζ 0 + ζ ′′ 0 , v U ′ ×U + T vdµ a -T vdµ b , • µ a is a nonnegative measure and T (u -a)dµ a = 0 or Supp(µ a ) ⊂ {u 0 = a}, • µ b is a nonnegative measure and T (b -u)dµ b = 0 or Supp(µ b ) ⊂ {u 0 = b}, • (ζ 0 , cos) U×U = (ζ 0 , sin) U×U = 0, T ζ 0 d(u 0 + u ′′ 0 ) = 0 and ∀v ∈ U with g 0 (v) ≥ 0, T ζ 0 v -ζ ′ 0 v ′ ≥ 0. ( 17
)
Let now v ∈ U with v+v ′′ = ψ(θ)(u 0 +u ′′ 0 ) with ψ Borel measurable and bounded. Then, g 0 ( ψ

L ∞ u 0 ± v) ≥ 0 so that ζ 0 , g 0 ( ψ L ∞ u 0 ± v) U×U ′ ≥ 0. It follows that | ζ 0 , g 0 (v) U×U ′ | ≤ ψ L ∞ ζ 0 , g 0 (u 0 ) U×U ′ = 0. ( 18
)
But this information on ζ 0 is not sufficient to obtain the first property of ( 14), namely ζ 0 (u ′′ 0 + u 0 ) ≡ 0. For this, we now show that it is possible to change ζ 0 into Z 0 = ζ 0 + a cos +b sin so that all same properties remain valid, but also Z 0 (u ′′ 0 + u 0 ) ≡ 0. Since T (v+v ′′ ) cos = 0 = T (v+v ′′ ) sin, we also have (15) for Z 0 in place of ζ 0 . Moreover, (18) is also true for Z 0 , that is to say: for every ψ Borel measurable and bounded such that v + v ′′ = ψ(θ)(u 0 + u ′′ 0 ) for some v ∈ U , T ψZ 0 dν = 0, where we denote ν = u 0 + u ′′ 0 . Let us show that we can find a, b ∈ R so that Z 0 ν ≡ 0 and Z 0 ≥ 0, and the proof of Proposition 3.2 will be complete. Let us choose a, b so that

T cos θZ 0 (θ)dν(θ) = 0 = T sin θZ 0 (θ)dν(θ), (19)
which writes for some λ ∈ R and it is not the case since ν has at least 3 distinct points in its support. Let now ϕ be a Borel measurable bounded function on T. Let ψ := ϕ + α cos +β sin where α, β ∈ R are chosen so that ψν = v + v ′′ for some v ∈ U , or equivalently

T cos θζ 0 (θ)dν(θ) + a T cos 2 θ dν(θ) + b T cos θ sin θdν(θ) = 0, T sin θζ 0 (θ)dν(θ) + a T sin θ cos θ dν(θ) + b T sin 2 θdν(θ) = 0. (20)
T cos θ[ϕ + α cos +β sin](θ)dν(θ) = 0, T sin θ[ϕ + α cos +β sin](θ)dν(θ) = 0. (21)
Again, this is possible since CS -B 2 = 0. Next, we deduce from (19), then from (18) that

T ϕZ 0 dν = T ψZ 0 dν = 0.
By arbitrarity of ϕ, this implies Z 0 ν ≡ 0 as expected. This gives ( 14) and ( 15) with Z 0 in place of ζ 0 . We now prove that Z 0 is nonnegative :

Supp(u ′′ 0 + u 0 ) c = n ω n where ω n are open intervals. Then, if ψ ≥ 0 is regular with a compact support in ω n , we can introduce v ∈ H 1 0 (ω n ) satisfying v ′′ + v = ψ in ω n (possible since diam(ω n ) < π).
We define v by 0 outside ω n . Thus v ′′ + v has Dirac mass at ∂ω n , but since Z 0 vanishes at ∂ω n , we finally get, using (17):

ωn Z 0 ψdθ = T Z 0 d(v ′′ + v) = 0.
Since ψ is arbitrary, we get Z 0 ≥ 0 in ω n , and then Z 0 ≥ 0 in T.

By the statement (ii) of Proposition 3.1, for each v ∈ U satisfying

f ′ (u 0 )(v) = 0, ∃λ ∈ R, g 0 (v) ≥ λg 0 (u 0 ), v ≥ λ(u 0 -a), v ≤ λ(u 0 -b), (22)
we have f ′′ (u 0 )(v, v) ≥ 0 (the constraint g is linear, so g ′′ = 0). Whence Proposition 3.2, with Z 0 in place of ζ 0 .

Remark 3.3 In general, the positivity of ζ 0 on the orthogonal of {cos, sin} does not imply that it is pointwise positive (one can write explicit examples).

Remark 3.4 In the following section, the main difficulty will be to analyze the situation where the convexity constraint is almost everywhere saturated. It would be easy to prove the non-existence of an nonempty interval I ⊂ S u0 ∩ {a < u 0 < b}. However, this is not sufficient to conclude that u ′′ 0 + u 0 is a sum of Dirac masses (we can look at the Lebesgue decomposition of measures to see this). That is why we have to analyze the case of infinitely many corners, or even of a diffuse singular measure (see the proof of Theorem 2.1). Another way to avoid these difficulties has been chosen by M. Crouzeix in [START_REF] Crouzeix | Une famille d'inégalités pour les ensembles convexes du plan[END_REF] for his particular problem (see Remark 2.5): he considers the minimization problem restricted to convex polygons having at most n edges, and proves that with n large enough, any solution in this restricted class has only n 0 edges where n 0 is only determined by a and b. Therefore, using the density of convex polygons in convex sets, the solution for this particular problem ( 5) is still a polygon. Remark 3.5 Our analysis in Section 4 could easily show on some simple examples that the first order equation is not sufficient to get the results of Theorems 2.1, 2.2 or 2.3. It turns out that the second order condition is very helpful.

Second problem:

Similarly, we can give the optimality conditions in the case of the measure constraint:

min j(u), u ∈ H 1 (T), u ′′ + u ≥ 0, m(u) := 1 2 T dθ u 2 = m 0 , (23)
Proposition 3.3 If u 0 solves (23) where j :

H 1 (T) → R is C 2 , then there exist ζ 0 ∈ H 1 (T) nonnegative, µ ∈ R such that ζ 0 = 0 on S u0 , ( 24
)
and

∀ v ∈ H 1 (T), j ′ (u 0 )v = ζ 0 + ζ ′′ 0 , v U ′ ×U -µm ′ (u 0 )(v). ( 25
)
Moreover, for all v ∈ H 1 (T), such that ∃λ ∈ R satisfying

v ′′ + v ≥ λ(u ′′ 0 + u 0 ) ζ 0 + ζ ′′ 0 , v U ′ ×U -µm ′ (u 0 )(v) = 0 , we have j ′′ (u 0 )(v, v) + µm ′′ (u 0 )(v, v) ≥ 0. ( 26
)
Proof. We make the same choices except for

Y = g 0 (U ) × R, g : U → Y, ∀u ∈ U, g(u) = (g 0 (u), m(u) -m 0 ),
and

K = {z ∈ g 0 (U ), z ≥ 0 in U ′ } × {0} ⊂ Y .
Here, using min(u 0 ) > 0, we have First of all, we have to prove that u ′ is bounded by a constant C(b), for all u admissible.

∀v ∈ U, g ′ (u 0 )(v) = (g 0 (v), m ′ (u 0 )(v)) = g 0 (v), - T v dθ u 3 0 and g ′ (u 0 )(U ) = g 0 (U ) × R = Y .
Lemma 4.1 ∀u ∈ H 1 (T), 0 ≤ u ≤ b, u ′′ + u ≥ 0 =⇒ u ′ L ∞ ≤ 2πb =: C(b)
Proof of lemma 4.1 Since u is periodic, there exists x 0 ∈ T such that u ′ (x 0 ) ≥ 0. With x ∈ [x 0 , x 0 +2π] and integrating the inequality u ′′ + u ≥ 0, we get u ′ (x)-u ′ (x 0 )+ x x0 u ≥ 0 which leads to u ′ (x) ≥ -2πb, true for all x ∈ R by periodicity. Similarly with x 1 such that u ′ (x 1 ) ≤ 0 and x ∈ [x 1 -2π, x 1 ], we get u ′ (x) ≤ 2πb which leads to the result with C(b) = 2πb.

(u) = T G(θ, u, u ′ ) is a priori not defined on H 1 (T): if η(u, p) is a C ∞ cut-off function, with 0 ≤ η ≤ 1 and such that η = 1, (u, p) ∈ [a/2, 2b] × [-2C(b), 2C(b)], 0, otherwise,
where C(b) is introduced in Lemma 4.1, then we can set j(u) := T G(θ, u, u ′ )dθ, with G(θ, u, p) := η(u, p)G(θ, u, p). Easily, the new functional j is

C k in H 1 (T) if G is C k in T × R × R.
Moreover, by the choice of η, any solution of the problem ( 5) is still solution for j instead of j, and we can write first and second order necessary conditions for the function j, in terms of G.

We easily check that G still satisfy the hypothesis in Theorem 2.

1, since η = 1 in a neighborhood of [a, b] × [-C(b), C(b)
] (this will also be true for Theorems 2.2 and 2.3). We drop the notation • in all what follows.

Proof of Theorem 2.1, case of inclusion in A(a, b):

Assume by contradiction that u 0 does not satisfy the conclusion. Therefore there exists an interval I ⊂ {a < u 0 < b} and θ 0 an accumulation point of S u0 ∩ I.

(a) Case a < u 0 (θ 0 ) < b.

Without loss of generality we can assume θ 0 = 0 and also that there exists a decreasing sequence (ε n ) tending to 0 such that S u0 ∩ (0, ε n ) = ∅. Then we follow an idea of T. Lachand-Robert and M.A. Peletier (see [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF]). We can always find 0 < ε i n < ε n , i = 1, . . . , 4, increasing with respect to i, such that

S u0 ∩ (ε i n , ε i+1 n ) = ∅, i = 1, 3. We consider v n,i solving v ′′ n,i + v n,i = χ (ε i n ,ε i+1 n ) (u ′′ 0 + u 0 ), v n,i = 0 in (0, ε n ) c , i = 1, 3.
Such v n,i exist since we avoid the spectrum of the Laplace operator with Dirichlet boundary conditions. Next, we look for

λ n,i , i = 1, 3 such that v n = i=1,3 λ n,i v n,i satisfy v ′ n (0 + ) = v ′ n (ε - n ) = 0. 1 a 1 b ε n Ω u0 ε i n θ 0 A(a, b) Figure 1: Case (a)
The above derivatives exist since v n,i are regular near 0 and ε n in (0, ε n ). We can always find such λ n,i as they satisfy two linear equations. It implies that v ′′ n does not have any Dirac mass at 0 and ε n . Since S u0 ∩ (ε i n , ε i+1 n ) = ∅, we have v n = 0. From (14) and Supp(v n ) ⊂ {a < u 0 < b} it follows that for such v n we have

T v n (ζ 0 + ζ ′′ 0 ) = T v n dµ a = T v n dµ b = 0.
Using the first order Euler-Lagrange equation (15), we get j ′ (u 0 )(v n ) = 0. Consequently, v n is eligible for the second order necessary condition (it is easy to check the other conditions required in Proposition 3.2). So, using (16), we get

0 ≤ j ′′ (u 0 )(v n , v n ) = T G uu (θ, u 0 , u ′ 0 )v 2 n + 2G up (θ, u 0 , u ′ 0 )v n v ′ n + G pp (θ, u 0 , u ′ 0 )v ′ n 2 .
Using the concavity assumptions [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF] on G, it follows that

0 ≤ j ′′ (u 0 )(v n , v n ) ≤ T K uu v 2 n + 2K up |v n ||v ′ n | -K pp |v ′ n | 2 ≤ ε n π 2 K uu + 2 ε n π K up -K pp v ′ n 2 L 2 , (27)
where, if we set

R := T × [a, b] × [-C(b), C(b)],
we have

K uu = sup R |G uu |, K up = sup R |G up |, K pp = inf T |G pp (θ, u 0 (θ), u ′ 0 (θ))| > 0. (28)
In order to get (27), we have used Poincaré's inequality

∀ v ∈ H 1 0 (0, ε), ε 0 u 2 ≤ ε π 2 ε 0 u ′2 , with ε = ε n .
As ε n tends to 0, the inequality (27) becomes impossible and proves that S u0 has not interior accumulation points. It follows that u ′′ 0 + u 0 is a sum of positive Dirac masses, u ′′ 0 + u 0 = n∈N α n δ θn in {a < u 0 < b}.

(b) Case u 0 (θ 0 ) = a. From (a), it follows that near θ 0 and at least from one side of it we have u ′′ 0 + u 0 = n∈N * α n δ θn where {θ n } is a sequence such that θ n → θ 0 , θ n ∈ S u0 ∩ I and α n > 0. Without restriction, we may take θ 0 = 0 and assume that θ n > 0 is decreasing. For every n we consider

v n ∈ H 1 0 (θ n+1 , θ n-1 ) satisfying v ′′ n + v n = δ θn in (θ n+1 , θ n-1
). In T, the measure v ′′ n + v n is supported in {θ n+1 , θ n , θ n-1 }, and since these points are in S u0 , and since u 0 does not touch a in a neighborhood of [θ n+1 , θ n-1 ], we can choose λ ≪ 0 (depending on n) such that Moreover, since v n is supported in {a < u 0 < b}, we finally get, using (14), T vd(ζ 0 + ζ ′′ 0 + µ aµ b ) = 0, and so the function v n is admissible for the second order necessary condition. Proceeding as in (a) above, we find a contradiction which proves that this case is impossible.

v ′′ n + v n ≥ λ(u ′′ 0 + u 0 ) v n ≥ λ(u 0 -a), v n ≤ λ(u 0 -b). 1 a 1 b θ n θ 0 ∂Ω u0
(c) Case u 0 (θ 0 ) = b. This case is treated similarly to the case (b).

Corollary 4.1 We have u ′ 0 L ∞ ≤ 2b(b -a). More generally, if u ∈ H 1 (T), 0 < α ≤ u ≤ β < ∞ and u ′′ + u ≥ 0 with |{α < u < β} ∩ Supp(u ′′ + u)| = 0 then u ′ L ∞ ≤ 2β(β -α).
Proof. We have

T = ∪ n ω n ∪ ({α < u < β} ∩ Supp(u ′′ + u)) ∪ F α ∪ F β , where F α := {u = α}, F β := {u = β} and ω n ⊂ {α < u < β} open interval with u ′′ + u = 0 in ω n . As u ′ = 0 a.e. in F α ∪ F β and |{α < u < β} ∩ Supp(u ′′ + u)| = 0, it's enough to estimate u ′ only in ω n . From u ′′ + u = 0 in ω n we get |u ′ | 2 + u 2 = γ 2 with α 2 ≤ γ 2 ≤ β 2 . Therefore |u ′ | 2 = γ 2 -|u| 2 ≤ 2β(β -α),
= #{θ n ∈ ω} ≥ 3, with n → θ n increasing. Consider v ∈ H 1 0 (θ 1 , θ 3 ) satisfying v ′′ + v = δ θ2 .
The function v is admissible for the second order necessary condition. Similarly to the case (a) we find the following estimation:

θ 3 -θ 1 π ≥ K pp K up + K 2 up + K uu K pp =: C(G,
|g uu (u 0 , u ′ 0 )(v, v)| ≤ K uu v 2 L ∞ , |g up (u 0 , u ′ 0 )(v, v ′ )| ≤ K up v L ∞ v ′ L 2 , g pp (u 0 , u ′ 0 )(v ′ , v ′ ) ≤ -K pp v ′ 2 L 2
for some K uu , K up , K pp > 0, the main argument (27) still works (with a more precise Poincaré inequality, valid in dimension 1, namely

u L ∞ (0,ε) ≤ √ ε u ′ L 2 (0,ε) , ∀u ∈ H 1 0 (0, ε)).

Proof of Theorem 2.1, case of volume constraint

First, we point out that as 0 < u 0 ∈ H 1 (T), we may assume that there exist 0 < a < b such that a < u 0 < b. Therefore, similarly to the case of inclusion in the annulus (see Remark 4.1), we introduce a cut-off function to get a new G and a new functional j, which is equal to j on {u ∈ H 1 (T) ; a < u < b and |u ′ | ≤ C(b)} and therefore, any solution of the problem ( 6) is still solution of

min j(u), u ∈ H 1 (T), a < u < b, u ′′ + u ≥ 0, m(u) = m 0 . (29)
We can apply Proposition 3.3 and write first and second order necessary conditions for the function j, in terms of G (the constraint a < u < b does not appear in the optimality condition, because these constrains are not saturated). It is easy to check that G still satisfies the hypothesis in Theorem 2.1.

In the following, we denote by j, resp. G, the function j, resp. G. Now, we assume by contradiction that u 0 does not satisfy the theorem. Therefore there exits at least one accumulation point θ 0 of S u0 . Without loss of generality we can assume θ 0 = 0, and that there exists a decreasing sequence {ε n > 0} tending to 0 such that S u0 ∩ (0, ε n ) = ∅. Then we can always find 0 < ε i n < ε n , i = 1, . . . , 5, decreasing with respect to i, such that S u0 ∩ (ε i+1 n , ε i n ) = ∅, i = 1, 4. We consider v n,i solving

v ′′ n,i + v n,i = χ (ε i+1 n ,ε i n ) (u ′′ 0 + u 0 ), v n,i = 0 in (0, ε n ) c , i = 1, 4.
Next, we extend the same idea of [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF] that we used in the first part of the proof (section 4.1) as follows: we look for λ n,i , i = 1, 4 such that

v n = i=1,4 λ n,i v n,i satisfies v ′ n (0 + ) = v ′ n (ε - n ) = m ′ (u 0 )(v n ) = 0.
Note that the derivatives at 0 + and ε - n are well defined as v n,i are regular nearby 0 and ε n in the interval (0, ε n ). Such a choice of λ n,i is always possible as λ n,i satisfy three linear equations. Moreover,

v n is not zero since S u0 ∩ (ε i n , ε i+1 n ) = ∅. Using (24), we get T v n (ζ 0 + ζ ′′ 0 ) = 0, which implies 0 = j ′ (u 0 )(v n ) = T v n (ζ 0 + ζ ′′ 0 ) = m ′ (u 0 )(v n ).
As v ′′ n + v n ≥ λ(u ′′ 0 + u 0 ) for λ ≪ 0, it follows that v n is eligible for the second order necessary condition. Then, using [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF],

0 ≤ j ′′ (u 0 )(v n , v n ) = T G uu (θ, u 0 , u ′ 0 ) + 3µ u 4 0 v 2 n + 2G up (θ, u 0 , u ′ 0 )v n v ′ n + G pp (θ, u 0 , u ′ 0 )v ′ n 2 ≤ T K uu + 3|µ| a 4 v 2 n + 2K up |v n ||v ′ n | -K pp |v ′ n | 2 ≤ (o(1) -K pp ) v ′ n 2
L 2 , with o(1) → 0 as n → ∞, where we have used Poincaré's inequality in H 1 0 (0, ε n ) (see (28) for the notation K uu , K up and K pp ). As n tends to ∞, the inequality 0 ≤ j ′′ (u 0 )(v n , v n ) becomes impossible and this proves the theorem.

Proof of Theorem 2.2

If j satisfies the hypotheses of Theorem 2.2, we can apply Theorem 2.1 (see also Remark 4.1). Therefore, it remains to prove the following result: Proof. Assume by contradiction there exists θ 0 an accumulation point of {(u 0a)(u 0b) = 0}. (a) First case : u 0 (θ 0 ) = a. Without loss of generality we can assume that θ 0 = 0 and that there exists a sequence {ε n > 0} of S u0 tending to 0, with u 0 (ε n ) = a and S u0 ∩ (0, ε n ) = ∅. (a.1) First subcase: assume by contradiction that there exists a sequence

θ n ∈ S u0 ∩ (0, ε n ) such that θ n → θ 0 and a < u 0 (θ n ) < b. As {θ, a < u 0 (θ) < b} is open, there exists an open connected set ω n , θ n ∈ ω n ⊂ {a < u 0 < b}, diam(ω n ) → 0, u 0 (∂ω n ) = a. Consider the function v n given by v n ∈ H 1 (T), v ′′ n + v n = u ′′ 0 + u 0 = Ni i=1 α i δ θ i n in ω n (where N i is finite), v n = 0 in ω c n (from Theorem 2.1, u ′′ 0 + u 0 is a finite sum of Dirac masses in ω n ).
It follows that for n large v n is admissible (again using (14), and also that u 0 = a on ∂ω) for Proposition 3.2, since u ′′ 0 + u 0 has some Dirac masses in ∂ω n . Then we can apply the second order necessary condition, as in (b), Section 4.1, which leads to a contradiction, since diam(ω n ) is going to 0.

θ 1 n θ 1 n-1 θ 2 n-1 θ 0 ε n 1 a ω n-1 ω n Figure 3: Case (a.1) θ n θ 0 ε n 1 a F a F a ω i-1 ω i Figure 4: Case (a.2) (a.2) Second subcase: (0, ε n ) = F a ∪ i ω i with F a = {u 0 = a} ∩ (0, ε n ) relatively closed and ω i ⊂ (0, ε n ) open intervals with u 0 (∂ω i ) = a and u ′′ 0 + u 0 = 0 in ω i . Let v n given by v ′′ n + v n = -(u ′′ 0 + u 0 ) in (0, ε n ), v n = 0 in (0, ε n ) c . We have v n > 0 on (0, ε n ): indeed, as (u 0 + v n ) ′′ + (u 0 + v n ) = 0 in (0, ε n ) (so u 0 + v n represents a line), u 0 + v n = u 0 in ∂(0, ε n ) and u 0 represents a convex curve, it follows that u 0 < u 0 + v n on (0, ε n ) (v n ≡ 0 because S u0 ∩ (0, ε n ) = ∅).
Then for n large and t ≥ 0 small the function u n = u 0 + tv n satisfies a ≤ u n ≤ b, u ′′ n + u n ≥ 0 (we use that u ′′ 0 + u 0 has positive Dirac masses at 0 and ε n ). Therefore, we can use the first order inequality (see Remark 3.1) j ′ (u 0 )(v n ) ≥ 0, which gives

0 ≤ j ′ (u 0 )(v n ) = T G u (u 0 , u ′ 0 )v n + G p (u 0 , u ′ 0 )v ′ n .
If (ii) holds we have Fa G u (u 0 , u ′ 0 )v n + G p (u 0 , u ′ 0 )v ′ n ≤ 0 because u 0 = a and u ′ 0 = 0 a.e. in F a , G u (a, 0) ≤ 0 and G p (a, 0) = 0 (as p → G p (a, p) is odd). So, if one of (ii) conditions holds, we have

0 ≤ j ′ (u 0 )(v n ) ≤ i ωi G u (u 0 , u ′ 0 )v n + G p (u 0 , u ′ 0 )v ′ n .
Note that we have ωi G u (u 0 , u ′ 0 )u ′ 0 + G p (u 0 , u ′ 0 )u ′′ 0 = [G(u 0 , u ′ 0 )] ∂ωi = 0, since u ′ 0 (∂ + ω i ) = -u ′ 0 (∂ -ω i ) (where ω i = (∂ -ω i , ∂ + ω i )) and G(a, •) is even. Therefore, from

v n = α n,i u 0 + β n,i u ′ 0 in ω i , α n,i = ωi u 0 v n ωi u 2 0 > 0, β n,i = ωi u ′ 0 v n ωi |u ′ 0 | 2 ,
we get that if (ii) holds then

0 ≤ j ′ (u 0 )(v n ) ≤ i α n,i ωi G u (u 0 , u ′ 0 )u 0 + G p (u 0 , u ′ 0 )u ′ 0 . (30)
We now prove that

v n → 0, u n → a in W 1,∞ (T) as n → ∞, (31)
where u n = u 0 in (0, ε n ) and u n = a in (0, ε n ) c . Indeed, the statement for u n follows from Corollary 4.1 because we have u na L ∞ → 0 as n → ∞ (from |u 0 (θ) -a| ≤ √ ε n u ′ 0 L 2 for θ ∈ (0, ε n ) and u ′′ n + u n ≥ 0). Next, from (u n + v n ) ′′ + (u n + v n ) = 0 in (0, ε n ) and u n + v n = a in (0, ε n ) c , using again Corollary 4.1, we find out that (u n + v n )a W 1,∞ (T) → 0, which proves the statement for v n .

Assume (ii.1) holds. We have u n → a in W 1,∞ (T) as n → ∞, so G p (u 0 , u ′ 0 ) = o(1) as n → ∞, and then

0 ≤ j ′ (u 0 )(v n ) ≤ i α n,i ωi (G u (a, 0)u 0 + o(1)) ,
which is impossible as n → ∞ because G u (a, 0) < 0 and α n,i > 0. Now assume (ii.2) holds. In this case, we need a second order information: for n large we have

0 ≤ j(u 0 + v n ) -j(u 0 ) = j ′ (u 0 )(v n ) + 1 2 j ′′ ( u n )(v n , v n ) = εn 0 G u (u 0 , u ′ 0 )v n + G p (u 0 , u ′ 0 )v ′ n + 1 2 εn 0 G uu ( u n , u ′ n )v 2 n + 2G up ( u n , u ′ n )v n v ′ n + G pp ( u n , u ′ n )|v ′ n | 2 ≤ i α n,i ωi G u (u 0 , u ′ 0 )u 0 + G p (u 0 , u ′ 0 )u ′ 0 + 1 2 εn 0 (o(1) -K pp )|v ′ n | 2 .
Here u 0 = u 0 + σ n v n , u ′ n = u ′ 0 + σ n v ′ n with a certain σ n ∈ (0, 1), and we used the estimation (30) for j ′ (u 0 )(v n ), which holds as it uses only the fact G u (a, 0) ≤ 0, and G pp ( u n , u ′ n ) ≤ -K pp < 0. The existence of K pp > 0 follows from hypothesis (i), continuity of G pp at (a, 0) and the W 1,∞ (T) convergence in (31). From (ii.2) we have ωi G u (u 0 , u ′ 0 )u 0 + G p (u 0 , u ′ 0 )u ′ 0 ≤ 0 and therefore we get

0 ≤ j(u 0 + v n ) -j(u 0 ) ≤ 1 2 εn 0 (o(1) -K pp )|v ′ n | 2 ,
which is impossible for n large and proves that this case is cannot occur.

(b) Second case : u 0 (θ 0 ) = b. Without loss of generality we may assume θ 0 = 0 and that there exists a sequence ε n > 0 decreasing and tending to 0 such that u 0 (2ε n ) = b. From Theorem 2.1, it follows that (0, 2ε n ) = ∪ i∈Nn ω n,i ∪ {θ i n , i ∈ N n } ∪ F b with F b = {u = b} ∩ (0, 2ε n ) relatively closed,

Proposition 3 . 2

 32 If u 0 is a solution of (13) where j :H 1 (T) → R is C 2 , then there exist ζ 0 ∈ H 1 (T) nonnegative, µ a , µ b ∈ M + (T) (spaceof nonnegative Radon measure on T) such that ζ 0 = 0 on S u0 , Supp(µ a ) ⊂ {u 0 = a}, Supp(µ b ) ⊂ {u 0 = b} (14)

  This is possible since CS -B 2 = 0 where C = T cos 2 θ dν(θ), S = T sin 2 θ dν(θ), B = T cos θ sin θ dν(θ). Indeed, by Schwarz' inequality, we have B 2 ≤ CS and equality would hold only if we had cos θ = λ sin θ νa.e θ,

4 Proofs 4 . 1

 441 Therefore, we may apply Proposition 3.1, and similarly to the proof of Proposition 3.2, we get the result. Proof of Theorem 2.1, case of inclusion in A(a, b)

Remark 4 . 1

 41 With the help of this lemma, let us explain how we can use Proposition 3.2, whereas j

Figure 2 :

 2 Figure 2: Case (b)

Remark 4 . 4

 44 a, b), Therefore, we get #{θ n ∈ ω} ≤ 2 2π C(G, a, b) + 1, where [•] denotes the floor function. Theorem 2.1 and its proof are valid for non integral operators: if j(u) = g(u, u ′ ) with g : (u, p) ∈ W 1,∞ (T) × L ∞ (T) → g(u, p) ∈ R, of class C 2 and satisfying

Proposition 4 . 1

 41 Under the assumptions of Theorem 2.2, the sets {u 0 = a} and {u 0 = b} are finite.

  In Theorem 2.1 we have to work in an open interval I of {a < u 0 < b} as, at this stage, it is not true in general that S u0 ∩ {a ≤ u 0 ≤ b} is finite (see Section 5). This property will be proved later with extra assumptions on G at the boundary (see the proofs of Theorems 2.2 and 2.3). Assume that ω ⊂ ω ⊂ {a < u 0 < b}, with ω an open connected set, and that n ω

	statement. Remark 4.2 Remark 4.3	which proves the

Acknowledgments :

The two authors would like to thank professor Michel Pierre for introducing them in this interesting subject and for some very helpful comments and discussions about this paper.

N n ⊂ N ∪ {∞}, and u ′′ 0 + u 0 = 0 in the open intervals ω n,i (see Figure 5).

Consider the function u n ∈ H 1 (T) given by

From the assumption of accumulation point, we must have σ n < ε n < τ n .

The first inequality is clear. For the other inequality we point out that 0 = u ′ 0 < |u ′ n | a.e. in F b , and

We also note that as in the case (a.2), u n → b in W 1,∞ (T). As u n satisfies a ≤ u n ≤ b, u ′′ n + u n ≥ 0, and p → G(u, p) is even near (b, 0) we get

with 0 < t, s < 1. But from the parity of p → G(•, p) and G pp < 0 near (b, 0), it follows that G p (•, p) < 0 for p > 0 near (b, 0). Then from the assumption G u ≥ 0 near (b, 0) the last inequality leads to a contradiction, so this case is impossible.

Remark 4.5 Theorem 2.2 can be extended to more general integral operators. More precisely, let

where u 0 is a solution of problem [START_REF] Henrot | Variation et optimisation de formes : une analyse géométrique[END_REF]. Then S u0 is finite, i.e. Ω u0 is a polygon.

The proof of this results is very similar to the proof of Theorem 2.2, except for the analysis on the boundary {u 0 = a}, which requires certain particular estimations.

Proof of Theorem 2.3

Conditions of Theorem 2.1 are satisfied, so it's enough to prove: Proposition 4.2 Assume the conditions (i), (ii) of Theorem 2.3 hold. Then, for any solution u 0 of (5), and for I = (γ 1 , γ 2 ) ⊂ {a < u 0 < b}, there exists n 0 ∈ N such that

Proof. The proof follows closely the one of Theorem 2.1. In fact the proof of steps (a) and (c) are identical, since we have

Let us deal with the step (b), which needs a new proof. (b) Assume by contradiction that there exists θ 0 an accumulation point of S u0 ∩ I with u 0 (θ 0 ) = a (see Figure 2). Without restriction we may take θ 0 = 0 and assume there exists a decreasing sequence {θ n > 0} tending to 0 such that u ′′ 0 + u 0 = n∈N α n δ θn and u 0 > a in {0 < θ ≪ 1} and α n > 0. Like in [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF], we consider v n ∈ H 1 0 (T)) given by

Since u ′′ 0 + u 0 has some Dirac mass at {θ n+1 , θ n , θ n-1 }, and u 0 > a in {0 < θ ≪ 1}, the function v n is admissible for the first and second order necessary conditions of Proposition 3.2. From the first order condition we get

θ denotes the jump at θ). We now prove the following consequence:

We will prove (33) using the technique used in [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF] for a particular functional G(u, p). First we point out that

where we have used that fact that p → G p (u, p) is decreasing (consequence of G pp ≤ 0), G pp (a, p) = 0 and [u ′ 0 ] θn > 0. If by absurd (33) does not hold, there exists a constant c > 0 such that

) and the

, where τ k = θ kθ k+1 , (from an elementary lemma on series, see [START_REF] Lachand-Robert | Newton's Problem of the Body of Minimal Resistance in the Class of Convex Developable Functions, Modeling, Analysis and Simulation[END_REF]).

Therefore, from (34) we obtain

As

to n, with a summation, we get:

The contradiction proves (33). The important corollary of (33) is

Indeed, from (33) and (ii) it follows that 0 = G up (a, u ′ 0 (0 + ))u ′ 0 (0 + ) < 0. As u 0 (0) ≤ u 0 (θ) implies u + 0 (0) ≥ 0, it follows that u ′ 0 (0 + ) > 0 and G up (a, u ′ 0 (0 + )) < 0. Using once more (ii) gives 0 > G u (a, u ′ 0 (0

Using v n in the second order condition of Proposition 3.2 gives

Since G pp (a, 0) = 0, we need further developments allowing to use (35). Namely

). From (35), the last inequality contradicts the second order condition (36) and proves that this case is impossible.

Proposition 4.3 Under the assumptions of Theorem 2.3 the sets {u 0 = a} and {u 0 = b} are finite.

Proof. The proof of proposition follows closely the proof of Proposition 4.1, except for the case (a.1) which needs another proof as G pp (u, p) is not strictly negative near u = a. Note that the case (a.2) of Proposition 4.1 when using only condition (ii.1) (which is the case in this proposition) does not require G pp < 0 (but only G u (a, 0) < 0 and the parity of p → G(a, p)). Furthermore, the case (b) of Proposition 4.1 requires only the (even) parity of p → G(u, p), G pp (u, p) < 0 and G u ≤ 0 near (b, 0). (a.1) We assume by contradiction that 0 is an accumulation point of S u0 ∩ {u 0 = a}, and that there exists a sequence {ε n > 0} tending to 0, with u 0 (ε n ) = a and S u0 ∩(0, ε n )∩{a < u 0 < b} = ∅ (see Figure 3). Then, there exists an open interval ω n ⊂ (0, ε n ) ∩ {a < u 0 < b}, with S u0 ∩ ω n = ∅ and u 0 (∂ω n ) = a. From Theorem 2.1 it follows that S u0 ∩ ω n is finite. Therefore, we can denote ω n = (θ n+1 , θ n-1 ) and find θ n ∈ (θ n+1 , θ n-1 ) ∩ S u0 . We then consider

16

The function v n is admissible for the first order condition, since u ′′ 0 + u 0 has some positive Dirac mass on ∂ω n . We can proceed exactly as in step (b) of Proposition 4.2 and we prove that (35) holds, so u ′ 0 (0 + ) > 0. However, from the fact that θ 0 = 0 is an accumulation point from the right, it's easy to show that u ′ 0 (0 + ) = 0. The contradiction proves the claim.

Sharpness of conditions

The conditions of Theorem 2.2, 2.3 are optimal in the sense that there exist counterexamples with G(u, u ′ ) not satisfying one of (i)-(iii) and such that the corresponding solution of ( 5) is not a polygon. We will provide some counterexamples for Theorems 2.2, 2.3.

Counterexamples for Theorem 2.2

It is obvious that the corresponding solution of ( 5) is not a polygon, but rather the circle {u 0 = c}.

, so G(u, p) satisfies the conditions (i) and (iii), but it does not satisfy (ii.1), neither (ii.2). The solution of ( 5) corresponding to this G(u, p) is the circle u 0 = a. Indeed, for admissible u we have

which proves that u 0 ≡ a is the minimizer of j(u).

Another counterexample is using the perimeter.

u 2 then j(u) := T G(u, u ′ )dθ = -P (u), where P (u) is the perimeter of the domain inside the curve {(1/u(θ), θ), θ ∈ T}. Therefore, solution of ( 5) is u 0 ≡ a, which corresponds to the circle {r = 1/a}. On the other side, G(u, p) satisfies the conditions (i) and (iii) but none of conditions (ii). Indeed,

Condition (iii)

Set G(u, p) = -1 2 (u 2 + p 2 ). Since G u = -u and G pp = -2, G(u, p) satisfies (i), (ii.1), but it does not satisfy (iii). A solution of the corresponding minimization problem is u 0 ≡ b. In fact, any u 0 representing a convex polygon with edges tangent to the circle {u 0 = b} is a solution! We can also add some piece of circle in the boundary. Indeed, first let v be a function such that 1/v represents a straight line with v ≤ b. For such v, we have

For θ 0 such that v ′ (θ 0 ) = 0 the value of 1/v(θ 0 ) gives the distance of the origin from the line v, so we must have 1/v(θ 0 ) ≥ 1/b, which proves the claim. Now, every admissible u can be approached for the H 1 (T) norm by a sequence of convex polygons u n satisfying a ≤ u n ≤ b. Then

which proves that u 0 ≡ b is a minimizer. This example provides some optimal shapes having an infinite number of corners inside {a < u < b} (because we can have an infinite number of edges, tangent to the circle of radius 1/b).

Counterexamples for Theorem 2.3

With minor modifications, the counterexamples given in (i), (ii) and (iii) above can easily be updated for Theorem 2.3.

The condition (i) is not satisfied as G pp = 2(ua) 2 (note that G pp (a, p) = 0). For u admissible we have j(u) ≥ 0 = j(c), so u 0 ≡ c minimizes j(u).

Condition (ii) Let G(u, p) and j(u) be as in the first example of Condition (ii) of Section 5.1. We consider

and let j(u) = T G(u, u ′ ). The function G satisfies the (i), (iii) of Theorem 2.3, but not (ii). For u admissible we have j(u) = T G(u, u ′ ) ≥ T G(u, u ′ ) = j(u) ≥ j(a) = j(a), so u 0 ≡ a minimizes j(u).

Condition (iii) Again, let G(u, p) and j(u) be as in the Condition (iii) of Section 5.1. We consider G(u, p) = -1 2 (u 2 + ϕ(u)p 2 ) and j(u) = T G(u, u ′ ). The function G satisfies the (i), (ii) of Theorem 2.3, but not (iii). Similarly as above, for u admissible we have j(u) T G(u, u ′ ) ≥ T G(u, u ′ ) = j(u) ≥ j(b) = j(b), so u 0 ≡ b minimizes j(u). Same remarks as in the previous subsection can be done. We can construct some optimal shapes locally polygonal inside {a < u < b} (necessary because of Proposition 4.2), but having an infinite number of corners in {a < u < b} (the only condition to be a minimizer is that every edges of these shapes are tangent to the circle of radius 1/b, and inside the domain {ϕ = 1}).