Gisella Croce
email: gisella.croce@univ-lehavre.fr

Antoine Henrot
email: antoine.henrot@univ-lorraine.fr

A Poincaré type inequality with three constraints

Keywords: Calculus of variations, Euler equation, Poincaré type inequality

In this paper, we consider a problem in calculus of variations motivated by a quantitative isoperimetric inequality in the plane. More precisely, the aim of this article is the computation of the minimum of the variational problem

where a function u ∈ W is a H 1 (-π, π) periodic function, with zero average on (-π, π) and orthogonal to sine and cosine.

Introduction

In this article we are interested in the following variational problem :

inf u∈W π -π [(u ′ ) 2 -u 2 ]dθ π -π |u|dθ 2
where W denotes the subspace of functions in the Sobolev space H 1 (-π, π) that are 2π-periodic, satisfying the following constraints: Our aim is to compute the value of the minimum and to identify the minimizer. The difficulty comes here from the nonlinear term in the denominator of the functional together with the three constraints (L1), (L2), (L3). We will prove the following result. Let

m = inf u∈W J(v) , J(v) = π -π (|v ′ | 2 -|v| 2 ) π -π |v| 2 .
(

) 1
Theorem 1 Let m be defined by [START_REF] Alvino | A sharp isoperimetric inequality in the plane[END_REF]. Then m = 1 2(4 -π)

and the minimizer u of the functional J is the odd and π periodic function defined on [0, π/2] by u(θ) = cos θ + sin θ -1.

We remark that the minimization problem ( 1) is a variant of the Wirtinger inequality :

inf u∈H 1 per (-π,π): π -π u=0 π -π |u ′ | 2 dθ π -π |u| 2 dθ = 1 .
At first glance, one could think that cos 2θ is a minimizer of the functional in [START_REF] Alvino | A sharp isoperimetric inequality in the plane[END_REF], as for the Wirtinger-type inequality

inf u∈W π -π [(u ′ ) 2 -u 2 ]dθ π -π u 2 dθ
, but this is not true.

In the literature one can find various generalizations of the Wirtinger inequality, without our constraints (L2) and (L3). For example, the problem inf u∈W 1,p per (-π,π):

π -π |u| r-2 u=0 π -π |u ′ | p dθ π -π |u| q dθ
for all values of p, q, r greater than 1, has been studied by several authors in the series of papers [START_REF] Belloni | A symmetry problem related to Wirtinger's and Poincaré's inequality[END_REF], [START_REF] Buslaev | On a family of extremal problems and related properties of an integral[END_REF], [START_REF] Croce | On a generalized Wirtinger inequality[END_REF], [START_REF] Dacorogna | Sur une généralisation de l'inégalité de Wirtinger[END_REF], [START_REF] Egorov | On a Kondratiev problem[END_REF], [START_REF] Gerasimov | Best constant in a three-parameter Poincaré inequality[END_REF], [START_REF] Ghisi | Symmetry-breaking in a generalized Wirtinger inequality[END_REF], [START_REF] Kawohl | Symmetry results for functions yielding best constants in Sobolev-type inequalities[END_REF], [START_REF] Mukoseeva | On the symmetry of extremal in several embedding theorems[END_REF], [START_REF] Nazarov | On exact constant in the generalized Poincaré inequality[END_REF] (where a constraint on the average of u has been considered). We recommend the nice recent paper [START_REF] Nazarov | Steklov-type 1D inequalities[END_REF] that makes a good survey on these problems. We also mention [START_REF] Ferone | A remark on optimal weighted Poincaré inequalities for convex domains[END_REF] in which the authors study, in any dimension N ≥ 1, the inequality inf u∈W 1,p (Ω): Ω |u| p-2 uω=0 Ω |∇u| p ω Ω |u| p ω with a positive log-concave weight ω, on a convex bounded domain Ω ⊂ R N . In our article, the Rayleigh quotient that we minimise is not "too nonlinear" as in these cited papers. The difficulty comes from the orthogonality to sine and cosine.

We will explain the strategy of the proof in the next section. The proof will be developed in Sections 3, 4, 5 and 6. In the last section we will explain the geometrical motivation of this minimization problem and how we used the value of m to study a quantitative isoperimetric inequality.

Strategy of the proof of Theorem 1

The strategy to prove our result is the following. It is immediate to see that the minimization problem (1) has a solution. Thus we write the Euler equation that any minimizer u satisfies:

-u ′′ -u = m • sgn(u) + λ 0 + λ 1 cos θ + λ 2 sin θ .
For that purpose, we introduce the three Lagrange multipliers, related to the three constraints (L1), (L2) and (L3) : they can be written as a function of sgn(u). By homogeneity, we can assume that the L 1 (-π, π) norm of u equals 1 and, by a translation, that λ 2 is zero. Our aim is to prove that

λ 0 = λ 1 = 0
and that u has four nodal intervals, of same length. This allows us to fully determine the minimizer and compute the value of m. Indeed, by using the explicit expression of u on any of the four nodal domain, as a function of the endpoints of the interval, we can easily deduce the explicit expression of the solution u on the whole interval [-π, π] and thus compute the value of m (see Section 6).

For that purpose, the most involved step is to prove that λ 1 is zero (see Proposition 4). We are now going to give an idea of the strategy. Let I k = [a k-1 , a k ], I k+1 = [a k , a k+1 ], I k+2 = [a k+1 , a k+2 ], I k+3 = [a k+2 , a k+3 ] four consecutive intervals of lengths ℓ k , ℓ k+1 , ℓ k+2 , ℓ k+3 , respectively. We assume that u is alternatively positive, negative, positive and negative on these intervals. From the explicit expression of u on any nodal domain, as a function of the endpoints of the interval, it is easy to deduce an equality involving the length of an interval and its consecutive :

λ 0 sin ℓ k + ℓ k+1 2 -m sin ℓ k+1 -ℓ k 2 = λ 1 2 cos ℓ k 2 cos ℓ k+1 2 A(I k , I k+1 ) . λ 0 sin ℓ k+1 + ℓ k+2 2 + m sin ℓ k+2 -ℓ k+1 2 = λ 1 2 cos ℓ k+1 2 cos ℓ k+2 2 A(I k+1 , I k+2 ) λ 0 sin ℓ k+2 + ℓ k+3 2 -m sin ℓ k+3 -ℓ k+2 2 = λ 1 2 cos ℓ k+2 2 cos ℓ k+3 2 A(I k+2 , I k+3 ) ,
where

A(I k , I k+1 ) = ℓ k sin ℓ k sin a k-1 - ℓ k+1 sin ℓ k+1 sin a k+1
(see Section 3.2). Assuming by contradiction that λ 1 = 0 and that ℓ k = ℓ k+2 , ℓ k+1 = ℓ k+3 , this 3 × 3 system in (λ 0 , λ 1 , m) has necessarily a null determinant, that provides, after some manipulations the identity

C sin ℓ k+3 2 cos ℓ k 2 cos ℓ k+1 2 sin ℓ k+3 2 sin ℓ k + ℓ k+2 2 + cos ℓ k 2 sin ℓ k+2 2 sin ℓ k+1 + ℓ k+3 2 = A(I k , I k+1 ) cos ℓ k+1 2 cos ℓ k+2 2 -A(I k+2 , I k+3 ) cos ℓ k+2 2 cos ℓ k+3 2 sin ℓ k+1 2 sin ℓ k+3 2 ,
where C = 2(m + λ 0 ) λ 1 . After proving that the length of each nodal domain is less than π, we will be able to study the sign of each term and arrive to the contradiction that C is both positive and negative. The proof that the length of each nodal domain is strictly less than π is more difficult that we expected and is done in Section 4.

Preliminaries

Existence of a minimizer and Euler equation

The existence of a minimizer of the functional in (1) and the optimality conditions follow easily from the direct methods of the calculus of variations.

Notice that we will assume that

π -π |u| = 1 (2)
in all the paper.

Proposition 1

The minimization problem (1) has a solution u. If K denotes the set of points where u vanishes, then K has zero Lebesgue measure and u satisfies the following Euler equation almost everywhere in (-π, π):

-u ′′ -u = m • sgn(u) + λ 0 + λ 1 cos θ + λ 2 sin θ , (3)
where the Lagrange multipliers are given by

λ 0 = - m 2π π -π sgn(u(θ))dθ λ 1 = - m π π -π sgn(u(θ)) cos θdθ λ 2 = - m π π -π sgn(u(θ)) sin θdθ. ( 4
)
In particular, the function

u is C 1 (-π, π).
Proof The existence of a solution to problem (1) is straightforward using the classical methods of the calculus of variations. We are going to write the optimality condition. For this purpose, we introduce the open set

ω = K c = {x ∈ (-π, π), u(x) = 0}. We fix now a function ϕ ∈ H 1 ((-π, π)) satisfying π -π ϕ(x)dx = π -π ϕ(x) cos xdx = π -π ϕ(x) sin xdx = 0 .
Therefore u + tϕ ∈ W, that is, it can be used as a test function for our

functional J(v) = π -π (|v ′ | 2 -|v| 2 ) π -π |v| 2 . We observe that π -π |u + tϕ| = ω |u + tϕ| + |t| K |ϕ|.
Now, on the set ω where u is not zero, we have the expansion

ω |u + tvp| = ω |u| + t ω sign(u)ϕ + o(t).
Therefore, we get

J(u + tϕ) = J(u) + 2t π -π (u ′ ϕ ′ -uϕ) - ω msign(u)ϕ -2|t|m K |ϕ| + o(t) . (5)
Let us denote by I 0 the term

π -π (u ′ ϕ ′ -uϕ) - ω msign(u)ϕ. If I 0 > 0,
we choose t < 0 small enough and get a contradiction. If I 0 < 0, we choose t > 0 small enough and get the same contradiction. Therefore I 0 = 0 for all admissible ϕ providing on ω the desired Euler equation. At last, coming back to (5) we necessarily get K |ϕ| = 0 proving that K has zero measure and the Euler equation holds almost everywhere. The expression of the Lagrange multipliers is obtained by integrating the Euler equation after multiplication by 1, cos θ, sin θ.

The C 1 regularity of the function u comes from the Euler equation that shows that its second derivative is

L ∞ , implying that u ∈ W 2,∞ (-π, π) ⊂ C 1 (-π, π).
Remark 1 Using (4) we see that

m + λ 0 = m 2π π -π (1 -sgn(u(θ)))dθ > 0 (6)
and

-m + λ 0 = - m 2π π -π (1 + sgn(u(θ)))dθ < 0 . (7)
Up to a translation on θ, we can assume that one Lagrange multiplier is zero. Indeed, by periodicity, replacing u(θ) by u(θ + a) amounts to replace λ 2 by cos aλ 2 -sin aλ 1 . Thus we can choose a such that λ 2 = 0. Therefore in the sequel, we will assume:

the Lagrange multiplier λ 2 is zero.

We introduce the measure of the sets where u is respectively positive and negative:

ℓ + = |{x ∈ (-π, π) : u(x) ≥ 0}|, ℓ -= |{x ∈ (-π, π) : u(x) < 0}| . (9)
With these notations, we can rewrite m + λ 0 and m -λ 0 :

m + λ 0 = mℓ - π , m -λ 0 = mℓ + π = m π (2π -ℓ -). ( 10
)
Note also that if u(x) is a minimizer of our problem, then -u(x) or u(-x) or -u(-x) are also minimizers. Therefore, without loss of generality, we can assume, from now on, that ℓ + ≥ ℓ -.

Expression of the solution

As usual, we call nodal domain, each interval on which u has a constant sign. We observe that u can be zero in some points in the interior of a nodal domain. By periodicity, there is an even number of nodal domains. A straight consequence of the Sturm-Hurwitz theorem (see [START_REF] Katriel | From Rolle's theorem to the Sturm-Hurwitz theorem[END_REF] and [START_REF] Hurwitz | Uber die Fourierschen konstanten integrierbarer funktionen[END_REF]) applied to any minimizer (satisfying (L1), (L2), (L3)) is that Proposition 2 A minimizer u has at least four nodal domains.

The main difficulty will be to prove that there are exactly four nodal domains with same length. It will be a consequence of the fact that the Lagrange multipliers are all zero and will be done in Section 6.

On each nodal domain, we can integrate the Euler equation and get an explicit expression of the solution. We are going to write explicitly u on two consecutive intervals

u(x) = A 0 cos x + B 0 sin x -(m + λ 0 ) - λ 1 2 x sin x, x ∈ [a, b]
where

A 0 = (m + λ 0 ) cos( a+b 2 ) cos( b-a 2 ) - λ 1 2 (b -a) sin a sin b sin(b -a) , B 0 = (m + λ 0 ) sin( a+b 2 ) cos( b-a 2 ) + λ 1 2 b sin b cos a -a sin a cos b sin(b -a) . (11)
Assume that u ≤ 0 on an interval [b, c], with u(b) = u(c) = 0. We find

u(x) = A 1 cos x + B 1 sin x -(-m + λ 0 ) - λ 1 2 x sin x, x ∈ [b, c]
where

A 1 = (-m + λ 0 ) cos( c+b 2 ) cos( c-b 2 ) - λ 1 2 (c -b) sin c sin b sin(c -b) , B 1 = (-m + λ 0 ) sin( b+c 2 ) cos( c-b 2 ) + λ 1 2 c sin c cos b -b sin b cos c sin(c -b) . (12)
Now we can obtain another expression of the solution on [b, c] using the C 1 regularity of u in b. This gives

A 1 cos b + B 1 sin b = λ 0 -m + λ 1 2 b sin b and (A 0 -A 1 ) sin b = (B 0 -B 1 ) cos b .
We then get a different expression for A 1 and B 1 :

A 1 = (λ 0 -m) cos b + λ 1 2 b sin b cos b -B 0 sin b cos b + A 0 sin 2 b B 1 = (λ 0 -m) sin b + λ 1 2 b sin 2 b + B 0 cos 2 b -A 0 sin b cos b .
Replacing A 0 , B 0 of formulas [START_REF] Ghisi | Symmetry-breaking in a generalized Wirtinger inequality[END_REF] in the above expressions of A 1 and B 1 , one gets:

A 1 = λ 0 cos( a+b 2 ) cos( b-a 2 ) -m cos( 3b-a 2 ) cos( b-a 2 ) - λ 1 2 (b -a) sin a sin b sin(b -a) B 1 = λ 0 sin( a+b 2 ) cos( b-a 2 ) -m sin( 3b-a 2 ) cos( b-a 2 ) + λ 1 2 b sin b cos a -a sin a cos b sin(b -a) . (13)
Expressions ( 12) and ( 13) of A 1 give

λ 0 sin b sin( c-a 2 ) cos( b-a 2 ) cos( c-b 2 ) -m sin b sin( a+c-2b 2 ) cos( b-a 2 ) cos( c-b 2 ) = λ 1 2 sin b (b -a) sin a sin(b -a) - (c -b) sin c sin(c -b) .
Let us assume now that b = 0. If ℓ 1 = b -a and ℓ 2 = c -b, this equality can be written as

λ 0 sin ℓ 1 + ℓ 2 2 -m sin ℓ 2 -ℓ 1 2 = λ 1 2 cos ℓ 1 2 cos ℓ 2 2 ℓ 1 sin ℓ 1 sin a - ℓ 2 sin ℓ 2 sin c . (14)
Let us assume now that b = 0. Expressions ( 12) and ( 13) of B 1 give

(-m + λ 0 ) tan c 2 + λ 1 2 c = (λ 0 + m) tan a 2 + λ 1 2 a that is, λ 0 sin c -a 2 -m sin a + c 2 + λ 1 2 (c -a) cos c 2 cos a 2 = 0
This is exactly equation [START_REF] Kawohl | Symmetry results for functions yielding best constants in Sobolev-type inequalities[END_REF] written in the case b = 0.

Here we have assumed the lengths of the intervals not equal to π. The case of an interval of length π will be considered in Section 4.

4 The length of the nodal intervals cannot be greater than π

In this section we prove that the length of any nodal interval of the solution u is strictly less than π. We argue by contradiction, mainly by considering the integral of u on a nodal domain. We assume that there exists a nodal interval (a, b) of length ℓ greater than π. Without loss of generality, we can assume that

-u ≥ 0 on (a, b); -a ∈ [-π, 0] and b ∈ (0, π] (since the function x → u(x + π) is also a minimizer satisfying λ 2 = 0); - a + b 2 ≤ 0 (since u(-x) is also a minimizer); this implies a ≤ - π 2 .
In the sequel we will call negative interval (resp. positive interval) any interval where u is negative (resp. positive). On a negative interval (a j , b j ) of length ℓ j = π, we have bj aj

u(x)dx = (-m+λ 0 ) 2 tan ℓ j 2 -ℓ j + λ 1 2 2 sin ℓ j 2 cos a j + b j 2 1 + ℓ j sin ℓ j , (15) while
, on a positive interval (a k , b k ) of length ℓ k = π, we have b k a k u(x)dx = (m+λ 0 ) 2 tan ℓ k 2 -ℓ k + λ 1 2 2 sin ℓ k 2 cos a k + b k 2 1 + ℓ k sin ℓ k . (16)
In the case of a nodal domain of length π, let (a, a + π) be such an interval where we suppose u ≥ 0. Now the Euler equation

-u ′′ -u = m + λ 0 + λ 1 cos x on (a, a + π) u(a) = 0, u(a + π) = 0
has not a unique solution, since 1 is an eigenvalue on the interval. Moreover, by the Fredholm alternative, the right-hand side of the equation must be orthogonal to the eigenfunction sin(x -a), providing the relation

m + λ 0 = λ 1 4 π sin a. ( 17
)
Lemma 1 The Lagrange multiplier λ 1 is negative.

Proof We first study the case where (a, b) has length ℓ > π. Let us analyse equation ( 16). We recall that m + λ 0 > 0 by [START_REF] Dacorogna | Sur une généralisation de l'inégalité de Wirtinger[END_REF]; for ℓ > π, both terms 2 tan(ℓ/2) -ℓ and 1 + ℓ/ sin ℓ are negative. If λ 1 ≥ 0, the integral is negative: this is a contradiction with the sign of u on (a, b).

In the case of a nodal domain of length π, one has λ 1 < 0 by equation ( 17), since m + λ 0 > 0 and sin a < 0.

Lemma 2

The Lagrange multiplier λ 1 satisfies

λ 1 2 ≤ 2 sin ℓ - 2 m π = 2 ℓ - sin ℓ - 2 (m + λ 0 ) < m + λ 0 , (18)
where ℓ -is the measure of {x : u(x) < 0}.

Proof Let us introduce the two numbers:

ℓ b -= |{t > b, u(t) < 0}|, ℓ a -= |{t < a, u(t) < 0}| . Obviously ℓ a -+ ℓ b -= ℓ -. By (6), m + λ 0 = mℓ - π . Now, since λ 1 < 0 by the previous lemma, λ 1 2 = m 2π π -π sign(u) cos tdt and π -π sign(u) cos tdt = a -π sign(u) cos tdt + b a cos tdt + π b sign(u) cos tdt.
By the bathtub principle (see [START_REF] Lieb | Analysis[END_REF]), the value of a -π sign(u) cos tdt is maximum when we choose sign(u) = -1 on the left, namely on (-π, -π + ℓ b -] (because cos is increasing on [-π, a]) and similarly for the last integral. Therefore, we get

πλ 1 m ≤ - -π+ℓ b - -π cos tdt+ π+ℓ a - -π+ℓ b - cos tdt- π π-ℓ a - cos tdt = 2(sin ℓ a -+sin ℓ b -). (19) Since sin ℓ a -+ sin ℓ b -= 2 sin ℓ - 2 cos ℓ a --ℓ b - 2 ≤ 2 sin ℓ - 2 ≤ ℓ -,
we finally get estimate (18), using ( 6) and ( 19).

Let us introduce the following positive quantity :

A = |λ 1 /2| m + λ 0 . ( 20
)
Our strategy to get a contradiction is based on the following

Proposition 3 If A < 2 π or A cos( a+b 2 ) < 2 π , then we cannot have a nodal interval of length ℓ ≥ π.
Proof Let us start with the case b -a = ℓ = π. In that case we have A = 2 π| sin a| by [START_REF] Nazarov | On exact constant in the generalized Poincaré inequality[END_REF]. Therefore the assumption A < 2 π provides immediately a contradiction. In the same way, if A cos( a+b 2 ) < 2 π we deduce | sin a| > cos( a+b 2 ) = cos( a+a+π 2 ) = -sin a that is also a contradiction. Now, let us assume that b -a = ℓ > π. We use the following claim: the function g :

ℓ → ℓ -2 tan(ℓ/2) + 4 π sin(ℓ/2)[1 + ℓ/ sin ℓ] is positive on (π, 2π). Indeed g is positive if and only if k(t) = t cos t -sin t + 1 π [2t + sin(2t)] is negative on π 2 , π . Observe that k( π 2 ) = 0 and k(π) < 0. Now, the derivative of k, k ′ (t) = -t sin(t) + 2 π [1 + cos(2t)],
is the difference between two functions which intersect in only one point t 0 . Since k ′ is negative near π 2 and positive near π, k is minimal at t 0 and therefore k < 0 on π 2 , π . We are able to get a contradiction by using the expression [START_REF] Mukoseeva | On the symmetry of extremal in several embedding theorems[END_REF] of the integral of u on the interval (a, b), that can be written as

b a u(x)dx = (m + λ 0 ) 2 tan ℓ 2 -ℓ -2A cos a + b 2 sin ℓ 2 1 + ℓ sin ℓ . Therefore, if A < 2 π or A cos( a+b 2 ) < 2 π , we obtain b a u(x)dx ≤ -g(ℓ)
≤ 0 (note that 1 + ℓ/ sin ℓ < 0 for ℓ > π). Thus we have the desired contradiction.

We are now going to find some estimates on A, in order to apply Proposition 3. This is quite technical and for that reason, we postpone all these computations to the Appendix. After proving an estimate on ℓ -(see Proposition 6), we distinguish the cases where u has at least 6 nodal domains (see Propositions 7 and 8) and u has exactly 4 nodal domains (see Proposition 9).

5 The Lagrange multipliers are zero and the nodal domains have same length

Now we enter into the heart of the paper. We are going to prove that the Lagrange multipliers λ 0 and λ 1 are zero (we already know that λ 2 = 0) and that all the nodal domains have the same length. For that purpose, we will use the relation ( 14) on different intervals.

Theorem 2

The Lagrange multipliers λ 0 , λ 1 are equal to zero and all the nodal intervals have the same length.

The proof will be done in two main steps. First, we prove that λ 1 = 0 in Proposition 4. Then, we prove that λ 0 = 0 and the nodal domains have same length in Proposition 5. Let us first introduce some notations and give a preliminary lemma. Let I k = [a k-1 , a k ] and I k+1 = [a k , a k+1 ] be two consecutive intervals of length respectively ℓ k , ℓ k+1 . We introduce:

A(I k , I k+1 ) = ℓ k sin ℓ k sin a k-1 - ℓ k+1 sin ℓ k+1 sin a k+1 .
Note that, using a k-1 = a k -ℓ k and a k+1 = a k + ℓ k+1 we can also write

A(I k , I k+1 ) = ℓ k tan ℓ k - ℓ k+1 tan ℓ k+1 sin a k -(ℓ k + ℓ k+1 ) cos a k . ( 21
)
Lemma 3 There exist three consecutive intervals, say I j , I j+1 , I j+2 , such that A(I j , I j+1 ) ≥ 0, A(I j+1 , I j+2 ) ≥ 0 and there exist three consecutive intervals, say

I i , I i+1 , I i+2 , such that A(I i , I i+1 ) < 0, A(I i+1 , I i+2 ) < 0. Proof Let us consider I i = (a i-1 , a i ), I i+1 = (a i , a i+1 ), I i+2 = (a i+1 , a i+2
), with a i < 0 < a i+1 . Without loss of generality we can assume that

I i ∪ I i+1 ∪ I i+2 ⊂ [-π, π] (up to consider u(-x) instead of u(x))
. Since sin a i-1 < 0 and sin a i+1 > 0

A(I i , I i+1 ) = ℓ i sin ℓ i sin a i-1 - ℓ i+1 sin ℓ i+1 sin a i+1 < 0 .
Since sin a i < 0 and sin a i+2 > 0

A(I i+1 , I i+2 ) = ℓ i+1 sin ℓ i+1 sin a i - ℓ i+2 sin ℓ i+2 sin a i+2 < 0 .
Let us consider I j = (a j-1 , a j ), I j+1 = (a j , a j+1 ), I j+2 = (a j+1 , a j+2 ), with a j < -π < a j+1 . Assume that I j ∪ I j+1 ∪ I j+2 ⊂ [-2π, 0]. Since sin a j-1 > 0 and sin a j+1 < 0

A(I j , I j+1 ) = ℓ j sin ℓ j sin a j-1 - ℓ j+1 sin ℓ j+1 sin a j+1 > 0 .
Since sin a j > 0 and sin a j+2 < 0

A(I j+1 , I j+2 ) = ℓ j+1 sin ℓ j+1 sin a j - ℓ j+2 sin ℓ j+2 sin a j+2 > 0 .
Proposition 4 The Lagrange multiplier λ 1 is zero.

Proof Let I k = [a k-1 , a k ], I k+1 = [a k , a k+1 ], I k+2 = [a k+1 , a k+2 ], I k+3 = [a k+2 , a k+3
] four consecutive intervals of lengths ℓ k , ℓ k+1 , ℓ k+2 , ℓ k+3 , respectively. We assume that u is alternatively positive, negative, positive and negative. In Section 3 we have seen that (see ( 14))

λ 0 sin ℓ k + ℓ k+1 2 -m sin ℓ k+1 -ℓ k 2 = λ 1 2 cos ℓ k 2 cos ℓ k+1 2 A(I k , I k+1 ) . ( 22
)
We can reproduce this identity for the other intervals:

λ 0 sin ℓ k+1 + ℓ k+2 2 + m sin ℓ k+2 -ℓ k+1 2 = λ 1 2 cos ℓ k+1 2 cos ℓ k+2 2 A(I k+1 , I k+2 ) (23)
λ 0 sin ℓ k+2 + ℓ k+3 2 -m sin ℓ k+3 -ℓ k+2 2 = λ 1 2 cos ℓ k+2 2 cos ℓ k+3 2 A(I k+2 , I k+3 ) .
(24) Assume by contradiction that λ 1 = 0. We divide the proof into three cases, according to the lengths of the nodal intervals.

1. Let us assume that ℓ k = ℓ k+2 and ℓ k+1 = ℓ k+3 . Equations ( 22), (23) can be seen as a system in λ 0 and m from which we get

λ 0 = λ 1 2 cos ℓ k+1 2 cos ℓ k 2 sin ℓ k+2 -ℓ k+1 2 A(I k , I k+1 ) + cos ℓ k+2 2 sin ℓ k+1 -ℓ k 2 A(I k+1 , I k+2 ) sin ℓ k+1 sin ℓ k+2 -ℓ k 2 m = λ 1 2 cos ℓ k+1 2 -cos ℓ k 2 sin ℓ k+2 +ℓ k+1 2 A(I k , I k+1 ) + cos ℓ k+2 2 sin ℓ k+1 +ℓ k 2 A(I k+1 , I k+2 ) sin ℓ k+1 sin ℓ k+2 -ℓ k 2 .
We observe that

λ 0 + m = λ 1 2 cos ℓ k 2 cos ℓ k+2 2 sin ℓ k+2 -ℓ k 2 [A(I k+1 , I k+2 ) -A(I k , I k+1 )] . (25)
Similarly, if one chooses equations ( 23), (24) to solve with respect to λ 0 , m, he gets

λ 0 + m = λ 1 2 cos ℓ k+2 2 sin ℓ k+2 2 sin ℓ k+3 -ℓ k+1 2 × × cos ℓ k+1 2 sin ℓ k+3 2 A(I k+1 , I k+2 ) -cos ℓ k+3 2 sin ℓ 2 2 A(I k+2 , I k+3 ) . (26)
We now set C = 2(m + λ 0 ) λ 1 . We use (25) to get A(I k+1 , I k+2 ) in terms of A(I k , I k+1 ):

A(I k+1 , I k+2 ) = A(I k , I k+1 ) + C sin ℓ k+2 -ℓ k 2 cos ℓ k 2 cos ℓ k+2 2 . ( 27
)
We also use (26) to get A(I k+1 , I k+2 ) in terms of A(I k+2 , I k+3 ):

A(I k+1 , I k+2 ) = tan ℓ k+1 2 tan ℓ k+3 2 A(I k+2 , I k+3 ) + C tan ℓ k+2 2 sin ℓ k+3 -ℓ k+1 2 sin ℓ k+3 2 cos ℓ k+1 2 . ( 28
)
Since m is non-zero, the 3×3 determinant of the system in (m, λ 0 , λ 1 ) given by equations ( 22), ( 23), (24) has to be equal to zero. Now, the computation of this determinant with respect to its third column gives the following equality after some simplification:

A(I k , I k+1 ) cos ℓ k 2 cos ℓ k+1 2 sin ℓ k+2 sin ℓ k+1 -ℓ k+3 2 +A(I k+2 , I k+3 ) cos ℓ k+2 2 cos ℓ k+3 2 sin ℓ k+1 sin ℓ k+2 -ℓ k 2 -A(I k+1 , I k+2 ) cos ℓ k+1 2 cos ℓ k+2 2 × × sin ℓ k+1 -ℓ k+3 2 sin ℓ k +ℓ k+2 2 + sin ℓ k+2 -ℓ k 2 sin ℓ k+1 +ℓ k+3 2 = 0. ( 29
)
Now we replace A(I k+1 , I k+2 ) in (29) by using both (27) (for the first term), (28) (for the second) and we get, after use of trigonometric formulae

0 = A(I k , I k+1 ) cos ℓ k+1 2 cos ℓ k+2 2 sin ℓ k+1 -ℓ k+3 2 sin ℓ k+2 -ℓ k 2 + +A(I k+2 , I k+3 ) cos ℓ k+2 2 cos ℓ k+3 2 sin ℓ k+3 -ℓ k+1 2 sin ℓ k+2 -ℓ k 2 sin ℓ k+1 2 sin ℓ k+3 2 -C sin ℓ k+1 -ℓ k+3 2 sin ℓ k+2 -ℓ k 2 sin ℓ k+3 2 cos ℓ k 2 cos ℓ k+1 2 sin ℓ k+3 2 sin ℓ k +ℓ k+2 2 + cos ℓ k 2 sin ℓ k+2 2 sin ℓ k+1 +ℓ k+3 2
.

(30) Simplifying by sin ℓ k+1 -ℓ k+3 2 sin ℓ k+2 -ℓ k 2 (this is possible since we are assuming ℓ k+2 = ℓ k and ℓ k+3 = ℓ k+1 ) we finally get

C sin ℓ k+3 2 cos ℓ k 2 × (31) × cos ℓ k+1 2 sin ℓ k+3 2 sin ℓ k + ℓ k+2 2 + cos ℓ k 2 sin ℓ k+2 2 sin ℓ k+1 + ℓ k+3 2 = A(I k , I k+1 ) cos ℓ k+1 2 cos ℓ k+2 2 -A(I k+2 , I k+3 ) cos ℓ k+2 2 cos ℓ k+3 2 sin ℓ k+1 2 sin ℓ k+3 2
.

Note that C has the same sign as λ 1 , since by definition of λ 0 (see section 3), λ 0 +m > 0. Moreover, in equation ( 31), the coefficients of C, A(I k , I k+1 ) and -A(I k+2 , I k+3 ) are all positive, since the length of each nodal domain is less than π, as we have seen in Section 4. Now we claim that we can choose four consecutive intervals such that -A(I k , I k+1 ) is positive and A(I k+2 , I k+3 ) is negative (with u positive on I k ). and we can choose four intervals such that -A(I j , I j+1 ) is negative and A(I j+2 , I j+3 ) is positive (with u positive on I j ). If this claim is true, we get a contradiction since (31) would show that C (and then λ 1 ) is both positive and negative. To prove our claim, we set In conclusion, we get a contradiction. 2. Assume ℓ k = ℓ k+2 . Since the left-hand sides of equations ( 22) and ( 23) coincide, the right-hand sides are equal that implies necessarily (since we are assuming λ 1 = 0)

I -= {(I k , I k+1 ) : A(I k , I k+1 ) < 0}, I + = {(I k , I k+1 ) : A(I k , I k+1 ) ≥ 0} .
ℓ k sin ℓ k (sin a k-1 + sin a k+2 ) = ℓ k+1 sin ℓ k+1
(sin a k + sin a k+1 ). Now, we observe that since ℓ k = ℓ k+2 , one has (a k-1 + a k+2 )/2 = (a k + a k+1 )/2. Replacing sin a k-1 + sin a k+2 by 2 sin(a k-1 + a k+2 )/2 cos(a k+2a k-1 )/2 and sin a k + sin a k+1 by 2 sin(a k + a k+1 )/2 cos(a k+1 -a k )/2 the above equality gives

ℓ k sin ℓ k cos(ℓ k + ℓ k+1 2 ) = ℓ k+1 sin ℓ k+1 cos ℓ k+1 2 = ℓ k+1 2 sin ℓ k+1 2 . ( 32
)
Assuming ℓ k fixed, we can study the function

g : x → ℓ k sin x cos(ℓ k + x) -x sin ℓ k .
Since g ′ (x) is negative and g(0) = 0, it is not possible to find ℓ k+1 > 0 such that (32) holds. Therefore we have a contradiction. 3. Assuming ℓ k+1 = ℓ k+3 we get a contradiction in the same way as in the previous case (ℓ k = ℓ k+2 ).

Therefore, we conclude that necessarily λ 1 = 0.

To finish the proof of Theorem 2, we need the following proposition.

Proposition 5 The Lagrange multiplier λ 0 is zero and the nodal domains have same length.

Proof Since λ 1 = 0 by the previous proposition, from section 3 we deduce that

λ 0 sin ℓ k + ℓ k+1 2 -m sin ℓ k+1 -ℓ k 2 = 0 , (33)
λ 0 sin ℓ k+1 + ℓ k+2 2 + m sin ℓ k+2 -ℓ k+1 2 = 0 .
The determinant of this homogeneous system is zero, as m = 0. This means

sin ℓ k+1 -ℓ k 2 sin ℓ k+1 + ℓ k+2 2 = sin ℓ k+1 -ℓ k+2 2 sin ℓ k + ℓ k+1 2 that is, cos 2ℓ k+1 -ℓ k + ℓ k+2 2 = cos 2ℓ k+1 + ℓ k -ℓ k+2 2 which implies ℓ k -ℓ k+2 = -ℓ k + ℓ k+2 , that is, ℓ k = ℓ k+2 .
With the same argument on ℓ k+1 , ℓ k+2 , ℓ k+3 we find ℓ k+1 = ℓ k+3 . Therefore all the intervals where u is positive have the same length, say ℓ 1 ; all the intervals where u is negative have the same length, say ℓ 2 . The sum of these lengths give n(ℓ 1 + ℓ 2 ) = 2π.

On the other hand,

λ 0 = - m 2π 2π 0 sign(u) = - m 2π n(ℓ 1 -ℓ 2 ) which gives us ℓ 2 -ℓ 1 = 2πλ 0 mn . ( 34
)
If we replace this equality in (33), we have

λ 0 sin π n = m sin πλ 0 mn .
We now study the function

f (x) = m sin πx mn -x sin π n , for x ∈ [0, m) (recall that λ 0 < m). Since f (0) = 0 = f (m), f ′ (0) > 0, f ′ (
m) < 0 and f ′′ (x) < 0, we deduce that the only zero of f is zero. This means that f (λ 0 ) = 0 if and only if λ 0 = 0. This argument proves that λ 0 = 0. We deduce from (34) that ℓ 1 = ℓ 2 .

Conclusion

We are now in position to prove Theorem 1. We have seen in the previous section that all the nodal intervals have same length, say ℓ < π, and the Lagrange multipliers λ 0 , λ 1 , λ 2 are all zero. We recall that if (a, b) is an interval where u ≥ 0, one has ,c) is an interval where u ≤ 0, one has

u(x) = A 0 cos x + B 0 sin x -m, A 0 = m cos( a+b 2 ) cos ℓ 2 , B 0 = m sin( a+b 2 ) cos ℓ 2 ; if (b
u(x) = A 1 cos x + B 1 sin x + m, A 1 = -m cos( b+c 2 ) cos ℓ 2 , B 1 = -m sin( b+c 2 ) cos ℓ 2
(see subsection 3.2). The solution of the system

u(x) = A 0 cos x + B 0 sin x -m = 0 u(x) = A 1 cos x + B 1 sin x + m = 0 is cos(x -ϕ 0 ) = cos ℓ 2 , tan(ϕ 0 ) = tan a+b 2 cos(x -ϕ 1 ) = cos ℓ 2 , tan(ϕ 1 ) = tan b+c 2 with x -ϕ 0 = ±(π + x -ϕ 1 ) + 2kπ. Since x = a + ℓ, the only possible solution is ℓ = π 2 .
Therefore u(x) is symmetric with respect to a, and

u(x) = m cos x + m sin x -m, x ∈ a, a + π 2 -m cos x -m sin x + m, x ∈ a + π 2 , a + π .
We now compute m defined in [START_REF] Alvino | A sharp isoperimetric inequality in the plane[END_REF]. Recalling that Therefore m = 1 2(4 -π) .

Motivations and final remarks

The minimization of the functional in ( 1) is motivated by a shape optimization problem and more precisely from a quantitative isoperimetric inequality. Indeed, for any open bounded set of R n , let us introduce the isoperimetric deficit:

δ(Ω) = P (Ω) -P (B) P (B) , (35)
where |B| = |Ω|. Let the barycentric asymmetry be defined by:

λ 0 (Ω) = |Ω∆B x G | |Ω|
where B x G is the ball centered at the barycentre

x G = 1 |Ω| Ω x dx of Ω
and such that |Ω| = |B x G |. Fuglede proved in [START_REF] Fuglede | Lower estimate of the isoperimetric deficit of convex domains in R n in terms of asymmetry[END_REF] that there exists a positive constant (depending only on the dimension n) such that

δ(Ω) ≥ C(n) λ 2 0 (Ω), for any convex subsets Ω of R n . (36)
Now, the constant C(n) is unknown (as it is the case in most quantitative inequalities like (36)) and it would be interesting to find the best constant. This leads to consider the minimization of the ratio

G 0 (Ω) = δ(Ω) λ 2 0 (Ω)
among convex compact sets in the plane, in particular. In the study of this minimization problem, one is led to exclude sequences converging to the ball in the Hausdorff metric. The strategy is to prove that on these sequences G 0 is greater than 0.406 which is the value of G 0 (S) where S is a precise set with the shape of a stadium, as computed in [START_REF] Alvino | A sharp isoperimetric inequality in the plane[END_REF]. If a convex planar set E has barycenter in 0, it can be parametrized in polar coordinates with respect to 0, as

E = {y ∈ R 2 : y = tx(1 + u(x)), x ∈ S 1 , t ∈ [0, 1]} , (37)
where u is a Lipschitz periodic function. Then the shape functional G 0 (E) can be written as a functional H of the function u describing E, as follows :

G 0 (E) = H(u) = π 2 π -π (1 + u) 2 + u ′ (θ) 2 -1 dθ 1 2 π -π |(1 + u) 2 -1|dθ 2 . ( 38
)
The constraints of area (fixed equal to π without loss of generality) and barycentre in 0 read in terms of a periodic u ∈ H 1 (-π, π) as:

(NL1) 1 2π π -π (1 + u) 2 dθ = 1; (NL2) π -π cos(θ)[1 + u(θ)] 3 dθ = 0; (NL3) π -π sin(θ)[1 + u(θ)] 3 dθ = 0.
The computation of the minimum of H, under the constraints (NL1), (NL2) and (NL3), seems very difficult. However, for sequences of sets converging to the ball in the Hausdorff metric, the limit of

m ε := inf{H(u), u L ∞ = ε, u ∈ H 1 (-π, π) periodic, satisfying (NL1), (NL2), (NL3)}
as ε → 0, equals the limit of the shape functional G 0 for these sequences. Thus, a possible strategy consists in estimating from below the minimum of H by a simpler functional, namely its linearization. Define

m = inf u∈W π -π [(u ′ ) 2 -u 2 ]dθ π -π |u|dθ 2
where W is the space of periodic H 1 (0, 2π) functions satisfying the constraints:

(L1) π -π u dθ = 0 (L2) π -π u cos(θ) dθ = 0 (L3) π -π u sin(θ) dθ = 0.
In [START_REF] Bianchini | On the quantitative isoperimetric inequality in the plane with the barycentric distance[END_REF] we proved that lim inf

ε→0 m ε ≥ π 4 m. (39)
The value of m found in Theorem 1 allows us to conclude that lim inf

ε→0 m ε ≥ π 4 m > 0.406 .
Remark 2 We observe that one can easily get an estimate from below of m by using the Cauchy-Schwarz inequality

π -π |u|dθ 2 ≤ 2π π -π u 2 dθ.
Then, a Wirtinger-type inequality (or Parseval formula) shows that

m ≥ inf u∈W π -π [(u ′ ) 2 -u 2 ]dθ 2π π -π u 2 dθ ≥ 3 2π .
Unfortunately this estimate on m is not sufficient to prove the desired inequality lim inf ε→0 m ε > 0.406 .

Remark 3 One could be tempted by looking for an approximation of the value of m, considering the subset of W composed by piecewise affine functions, which are 0 on the same set of zeros as a minimizer u. Unfortunately this strategy would give an estimate from above of m. Instead, we need an estimate from below for our quantitative isoperimetric inequality.

Appendix

In this section we prove the most technical results of Section 4, under the assumptions given at the beginning of that section. We recall that our aim is to prove that A < 2 π and thus to get a contradiction, as explained in Proposition 3. Let us denote by m j the midpoint of a nodal interval (a j , b j ).

Lemma 4 Let (a j , b j ) be a negative interval. If m j / ∈ - π 2 , π 2 , then A ≤ l 2 j (2π -ℓ -) 12ℓ -| cos(m j )| . ( 40
)
For a positive interval (a k , b k ), whose midpoint m k ∈ - π 2 , π 2
, one has

A ≤ l 2 k 12 cos(m k ) . ( 41
)
Proof For a negative interval (a j , b j ), as soon as

a j + b j 2 / ∈ - π 2 , π 2 the fact that bj aj u < 0 implies |λ 1 /2| m -λ 0 ≤ h(ℓ j ) | cos( aj +bj 2 )| , where h(x) = 2 tan( x 2 ) -x 2 sin( x 2 ) 1 + x sin x
. We claim that for x ∈ [0, π], one has h(x) ≤

x 2

12 . By using ( 6), ( 7), [START_REF] Gerasimov | Best constant in a three-parameter Poincaré inequality[END_REF] and this bound on h we have the following estimate on A:

A ≤ l 2 j (2π -ℓ -) 12ℓ -| cos( aj +bj 2 )| .
We now prove our claim, that is, the bound on h. The statement is equivalent to the positivity of f (x) = x 2 12 (x + sin x) -2 sin(x/2) + x cos(x/2) in [0, π]. Observe that f (0) = 0. The result will follow if we prove that the derivative of f is positive, that is, k(x) = 3x + 2 sin(x) + x cos(x) -6 sin(x/2) is positive. We remark that k(0) = 0, k( π 2 ) > 0, k(π) > 0. We will split the analysis into two cases.

1. If x ∈ [0, π 2 ] it is easy to see that k ′′ < 0 and therefore k(x) ≥ 0 in [0, π 2 ]. 2. In the case where x ∈ [ π 2 , π], k is decreasing. Indeed it is easy to see that k ′ is negative, since k ′ is convex, k ′ (π) = 0, k ′ ( π
2 ) < 0. In the same way, using [START_REF] Mukoseeva | On the symmetry of extremal in several embedding theorems[END_REF], we can prove that, on a positive interval (a k , b k ) (m k being its midpoint):

A ≤ l 2 k 12 cos(m k ) , when m k ∈ (-π 2 , π 2 ).
Estimate of A involves the midpoint of nodal domains, as we have just seen. The next lemma gives us important information about nodal domains whose midpoint is between -π/2 and π/2 or outside this interval.

Lemma 5 There are no negative intervals on the "left" of I 0 = (a, b) whose midpoint is between -π/2 and π/2. There is at most one negative interval on the "right" of I 0 whose midpoint lies between -π/2 and π/2.

Proof Since a ≤ -π 2 , there are no negative intervals on the "left" of I 0 whose midpoint is between -π 2 and π 2 . We prove the second statement by contradiction. If there were two negative intervals, say I 1 and I 3 with a positive interval

I 2 = (a 2 , b 2 ) between them, its midpoint would satisfy m 2 ≤ π -ℓ 2 2 . Moreover, since b 2 ≤ π 2 , and 0 ≤ b < a 2 = b 2 -ℓ 2 ≤ π 2 -ℓ 2 , we infer ℓ 2 ≤ π 2
. Now using (41), we see that

A ≤ ℓ 2 2 12 cos(m 2 ) ≤ ℓ 2 2 12 sin(ℓ 2 /2) .
Now, it is immediate to check that x → x 2 / sin(x/2) is increasing. Therefore the previous inequality would imply for

ℓ 2 ≤ π 2 : A ≤ π 2 /(24 √ 2) < 2 π implying b2 a2
udt < 0, that is, a contradiction.

We are now going to find a lower bound for ℓ -, the measure of {u < 0}. We know that

{u>0} u(x)dx + {u<0} u(x)dx = 0 , while π -π |u(x)|dx = {u>0} u(x)dx - {u<0} u(x)dx = 1 . Therefore {u>0} u(x)dx = - {u<0} u(x)dx = 1 2 . ( 42
)
The following proposition gives a lower bound for ℓ -.

Proposition 6 ℓ -≥ 1.55.

Proof We start with a simple estimate on m that will be useful in the proof. Using the explicit function given in Theorem 1 as a test function in the functional defined by (1), we get

m ≤ 1 2(4 -π) . (43)
We now prove the estimate of the statement. We can assume that the length of all negative intervals is less than 1.6 (otherwise, if there is a negative interval of length ℓ j ≥ 1.6, the estimate ℓ -≥ 1.55 is straightforward). By Lemma 5, we can split the analysis into two cases:

1. no negative interval has its midpoint in [-π 2 , π 2 ]; 2. the midpoint of only one negative interval, say I 1 , belongs to [-π 2 , π 2 ]; that we are now going to develop.

1. Assume that no negative interval has its midpoint in -π 2 , π 2 . We are going to estimate the sum of Ij (-u(x))dx on all negative intervals I j . We observe that on all (but possibly one) negative intervals I j whose midpoint

m j / ∈ - π 2 , π 2
, we have

Ij -u(x)dx ≤ (m -λ 0 )[2 tan(ℓ j /2) -ℓ j ] (44)
by [START_REF] Lieb | Analysis[END_REF], since the second term of the right hand side is negative. Now, it is easy to prove that

2 tan(x/2) -x ≤ 0.45x 3 4 , 0 ≤ x ≤ 0.8 . (45)
Therefore inequalities (44) and (42) imply

1 2 = j Ij (-u(x))dx ≤ 0.45(m -λ 0 ) 4 j ℓ 3 j .
The maximization of the convex function t ∈ R + → t 3 and the fact that the sum of the lengths ℓ j of all negative intervals is ℓ -give

j ℓ 3 j ≤ ℓ 3 -.
By [START_REF] Gerasimov | Best constant in a three-parameter Poincaré inequality[END_REF] and (43), we end up with

1 2 ≤ 0.45(2π -ℓ -) 8π(4 -π) ℓ 3 -.
This polynomial inequality provides finally the inequality ℓ -≥ 1.74. 2. Assume that the midpoint of one interval, say I 1 , belongs to -π 2 , π 2 . Notice that by Lemma 5, such a negative interval is unique. We are going to estimate the sum of Ij (-u(x))dx on all negative intervals I j . For j = 1, the midpoint of

I j does not belong to - π 2 , π 2
. Therefore the integral Ij (-u(x))dx, j = 1 can be estimated as in the previous case, that is, using ( 15) and (45). Instead, I1 (-u(x))dx can be estimated by

(m -λ 0 )[2 tan(ℓ 1 /2) -ℓ 1 ] + P, P = - λ 1 2 2 sin ℓ 1 2 cos m 1 1 + ℓ 1 sin ℓ 1 (46)
by formula [START_REF] Lieb | Analysis[END_REF]. We thus obtain, for the sum of Ij (-u(x))dx on all negative intervals I j

1 2 = j Ij (-u(x))dx ≤ 0.45(m -λ 0 ) 4 j ℓ 3 j + P , using (42)
for the first equality. By the same argument as in the previous case, we get

1 2 ≤ 0.45(2π -ℓ -) 8π(4 -π) ℓ 3 -+ P . (47)
We are going to distinguish two cases, according to the values of ℓ 1 : (a) Assume ℓ 1 ≥ 0.228. In this case, we look at the first negative interval

I -1 on the left of I 0 . Since b ≤ π -ℓ 1 2 ≤ π 2 -0.114 and ℓ ≥ π, we have m -1 = a - ℓ -1 2 = b -ℓ - ℓ -1 2 ≤ -0.114 - π + ℓ -1 2 . If ℓ -1 ≥ π 2 , then ℓ -≥ π 2 + 0.228 > 1.75. If, on the contrary, ℓ -1 < π 2 , then m -1 ≥ - 5π 4 necessarily (otherwise a = m -1 + ℓ -1 2 < -π which is impossible). Therefore, | cos(m -1 )| ≥ min √ 2 2 , cos 0.114 + π + ℓ -1 2 := C.
Using this estimate and ℓ -≥ 0.228 + ℓ -1 in (40) we get

A ≤ ℓ 2 -1 (2π -0.228 -ℓ -1 ) 12(0.228 + ℓ -1 )C .
As a function of ℓ -1 , the right-hand side is increasing and for ℓ -1 ≤

1.322 we get the inequality A ≤ 0.636 < 2 π , that is, a contradiction. Therefore, in this case, we deduce ℓ -1 ≥ 1.322 and ℓ -≥ ℓ 1 +ℓ -1 ≥ 1.55. (b) Assume ℓ 1 ≤ 0.228. Lemma 2 and inequalities [START_REF] Gerasimov | Best constant in a three-parameter Poincaré inequality[END_REF] and (43) give

- λ 1 2 ≤ m + λ 0 = m ℓ - π ≤ ℓ - 2π(4 -π)
.

We can use these inequalities to estimate the positive term P defined by (46) :

P ≤ ℓ - 2π(4 -π)
2 sin(0.114) 1 + 0.228 sin(0.228) .

Therefore we get from (47)

1 2 ≤ 0.45(2π -ℓ -) 8π(4 -π) ℓ 3 -+ ℓ - 2π(4 -π)
2 sin(0.114) 1 + 0.228 sin(0.228) .

This polynomial inequality implies ℓ -≥ 1.55.

In the next two propositions we will analyse the case where there would be at least 6 nodal domains.

Proposition 7 Assume that u has at least 6 nodal domains and b ≥ π 2 . Then A < 2/π and the length of no nodal interval is greater than π.

Proof We will consider the two negative intervals next to I 0 = (a, b), namely I -1 on the left of I 0 and I 1 on the right, and prove that both have lengths ℓ -1 and ℓ 1 respectively greater than 1.17.

-Assume by contradiction that ℓ -1 ≤ 1.17. On one hand, the midpoint m -1

of

I -1 satisfies m -1 = a - ℓ -1 2 ≤ - π 2 - ℓ -1 2 .
On the other hand,

m -1 ≥ - 5π 4 (otherwise a = m -1 + ℓ -1 2 < -π which is impossible). Therefore we have | cos(m -1 )| ≥ min 1 √ 2 , cos π + ℓ -1 2 = min 1 √ 2 , sin ℓ -1 2 .
Plugging this estimate in (40) and using Proposition 6 yields

A ≤ ℓ 2 -1 (2π -1.55) 12 • 1.55 • min 1 √ 2 , sin(ℓ -1 /2) . (48)
Now the right-hand side is increasing in ℓ -1 and its value for ℓ -1 = 1.17 is less than 0.631 < 2 π that is absurd, proving the claim.

-Assume by contradiction that ℓ 1 ≤ 1.17. In this case we are going to use that b ≥ π 2 to estimate m 1 :

π 2 + ℓ 1 2 ≤ b + ℓ 1 2 = m 1 = b + ℓ 1 2 < π + 1.17 2 < 5π 4 .
One can now repeat the above argument to get a contradiction.

By the previous estimates on ℓ -1 , ℓ The proof of this proposition follows analogous arguments to the previous one, but uses three intervals on the right of (a, b), since we do not know the position of m 1 .

Proof One can prove that ℓ -1 ≥ 1.17, following exactly the same argument as in the proof of Proposition 7. We will work with the intervals

I 1 = [a 1 , b 1 ], I 2 = [a 2 , b 2 ], I 3 = [a 3 , b 3 ], with b 1 = a 2 , b 2 = a 3 , on the right of (a, b), with b = a 1 . 1. Assume that b 2 ≤ π 2 . Observe that 0 < m 2 ≤ π 2 - ℓ 2 2
where we have used that b > 0. Using (41) we get

A ≤ ℓ 2 2 12 sin(ℓ 2 /2) . Since b 2 ≤ π 2 and 0 ≤ b < a 2 = b 2 -ℓ 2 ≤ π 2 -ℓ 2 , one has ℓ 2 ≤ π 2 .
Therefore the right hand side of the above estimate of A is less than 2 π , that is a contradiction. 2. Assume that b 2 = a 3 ≥ π 2 . We start by proving that ℓ -≥ 1.93. Assume by contradiction that ℓ -≤ 1.93. On one hand we have

m 3 = a 3 + ℓ 3 2 ≥ π 2 + ℓ 3 2 .
On the other hand we claim that m 3 < 5π 4 . We have

m 3 = b + ℓ 1 + ℓ 2 + ℓ 3 2 . Observe that -b < π 2 by hypothesis, -ℓ 2 ≤ π -1.55 since ℓ 2 + ℓ ≤ ℓ + = 2π -ℓ -≤ 2π -1.55, -ℓ 1 + ℓ 3 2 ≤ 0.76 since ℓ -1 + ℓ 1 + ℓ 3 2 ≤ ℓ -≤ 1.
93 and ℓ -1 ≥ 1.17.

Thus m 3 < 3π 2 -1.55 + 0.76 < 5π 4 . We deduce from the previous bounds on m 3 that:

| cos m 3 | ≥ min 1 √ 2 , cos π + ℓ 3 2
which provides, as in case (a)(i), ℓ 3 ≥ 1.17. We deduce that ℓ -≥ ℓ -1 + ℓ 3 > 1.93. We are going to prove that ℓ -1 ≥ 1.532. Assume that this is not the case. Using the same argument as in step (a)(i), one has a contradiction from the following estimate :

A ≤ 1.532 2 (2π -1.93) 12 • 1.93 • min 1 √ 2 , sin(1.532/2) < 2 π . (49)
It is easy to prove that ℓ -> 2.67. Indeed, if this is not the case, replacing 1.93 by 2.67 in (49), we get a contradiction since A is still less than 2 π .

We are going to prove that ℓ -1 > π 2 . Assume that this is not the case. By the same argument as in step (a)(i), one has a contradiction from estimate

A ≤ (π/2) 2 (2π -2.67) 12 • 2.67 • √ 2/2 < 2 π .
We are going to prove that ℓ 3 ≥ π 2 . Assume that this is not true. Recall that ℓ -< π. As above, on one hand we have m

3 = a 3 + ℓ 3 2 ≥ π 2 + ℓ 3 2 . On the other hand m 3 < 5π 4 . Indeed m 3 = b + ℓ 1 + ℓ 2 + ℓ 3 2 . Observe that -b < π 2 by hypothesis, -ℓ 2 ≤ π -2.67 -ℓ 1 + ℓ 3 2 ≤ π - π 2 Thus m 3 < 2π -2.67 < 5π 4
. We deduce from the previous bounds on m 3 that:

| cos m 3 | ≥ 1 √ 2
. This gives a contradiction since

A ≤ (π/2) 2 (2π -2.67) 12 • 2.67 • √ 2/2 < 2 π . Therefore ℓ 3 > π 2 . This last estimate implies π ≥ ℓ -> ℓ -1 + ℓ 3 > π, that is, a contradiction.
We consider now the case of four nodal domains.

Proposition 9 Assume that u has 4 nodal domains. Then A < 2/π and the length of no nodal interval is greater than π.

Proof Assume now that u has four nodal domains: I -1 = I 3 , I 1 (two negative ones) and I 0 , I 2 (two positive ones), that we write as I 0 = (a 0 , a 1 ), I 1 = (a 1 , a 2 ), I 2 = (a 2 , a 3 ), I -1 = I 3 = (a 3 -2π, a 0 ). We assume that m 0 , m 1 , m 2 , m 3 are the midpoints of each interval and ℓ 0 , ℓ 1 , ℓ 2 , ℓ 3 are the lengths.

We are going to work with a more explicit expression of λ 1 , λ 2 . Without loss of generality, we can assume that the lengths satisfy 0 < ℓ 1 ≤ ℓ 3 < π (up to replacing u(x) by u(-x)).

As in the proof of Proposition 7, we can get the lower bound ℓ 3 ≥ 1.17. We can assume -π ≤ m 3 ≤ -π 2 by Lemma 5. From λ 2 = 0 we have

0 = π -π sign(u(x)) sin xdx = 2 (cos a 0 -cos a 1 + cos a 2 -cos a 3 ) . (50)
Gathering -cos a 1 + cos a 2 on the one hand and cos a 0 -cos a 3 on the other hand, the right hand side of (50) can be rewritten

sin ℓ 3 2 sin m 3 + sin ℓ 1 2 sin m 1 = 0. (51)
In the same way, coming back to the definition of λ 1 (see ( 4)), we have

λ 1 2 = 2m π sin ℓ 3 2 cos m 3 + sin ℓ 1 2 cos m 1 . (52)
Observing that cos m 3 < 0, by (51) cos m 3 can be rewritten as

cos m 3 = -1 - sin 2 ℓ1 2 sin 2 m 1 sin 2 ℓ3 2 .
(53)

By using (52), ( 53) and ( 6), the quantity A defined in (20) can be rewritten as a function of ℓ 1 , ℓ 3 , m 1 as

A = A(ℓ 1 , ℓ 3 , m 1 ) = 2 sin 2 ℓ3 2 -sin 2 ℓ1 2 sin 2 m 1 -sin ℓ1 2 cos m 1 ℓ 1 + ℓ 3 . ( 54
)
We use now the integral of u on the interval I 3 . We recall that

I3 (-u(x))dx = mℓ + π 2 tan ℓ 3 2 -ℓ 3 - λ 1 2 2 sin ℓ 3 2 cos m 3 1 + ℓ 3 sin ℓ 3 . Moreover 0 < I3 (-u(x))dx ≤ 1 2
by (42). We now use the expression of λ 1 /2

given in (52) and replace sin ℓ1

2 cos m 1 by ± sin 2 ℓ1 2 -sin 2 ℓ3 2 sin 2 m 3 (with a + if m 1 ≤ π 2 and a -if π 2 ≤ m 1 ≤ 1.9
), thanks to (51). This provides an expression of the integral as a function of the three variables ℓ 1 , ℓ 3 , m 3 :

0 < I3 (-u(x))dx = m π I ± (ℓ 1 , ℓ 3 , m 3 ) .
More precisely, we have

I ± (ℓ 1 , ℓ 3 , m 3 ) = (2π -ℓ 1 -ℓ 3 ) 2 tan ℓ 3 2 -ℓ 3 -. . . 4 sin ℓ 3 2 cos m 3 ± sin 2 ℓ 1 2 -sin 2 ℓ 3 2 sin 2 m 3 sin ℓ 3 2 cos m 3 1 + ℓ 3 sin ℓ 3 (with a + if m 1 ≤ π 2 and a -if π 2 ≤ m 1 ≤ 1.9)
. By Lemma 6 below, the function I ± is negative, and thus we get a contradiction.

We recall that we assume that u has four nodal domains: I -1 = I 3 , I 1 (two negative ones) and I 0 , I 2 (two positive ones). We assume that m 0 , m 1 , m 2 , m 3 are the midpoints of each interval and ℓ 0 , ℓ 1 , ℓ 2 , ℓ 3 are the lengths. We recall that

A = A(ℓ 1 , ℓ 3 , m 1 ) = 2 sin 2 ℓ3 2 -sin 2 ℓ1 2 sin 2 m 1 -sin ℓ1 2 cos m 1 ℓ 1 + ℓ 3 .
(see 54) and

cos m 3 = -1 - sin 2 ℓ1 2 sin 2 m 1 sin 2 ℓ3 2 .
(55) Lemma 6 Assume ℓ 3 ≥ 1.17 and 0 < ℓ 1 ≤ ℓ 3 < π. Let m * 3 be the solution to

| sin m 3 | sin ℓ 3 2 = sin ℓ 1 2 . Assume that -π ≤ m 3 ≤ m * 3 . 1. Let I + (ℓ 1 , ℓ 3 , m 3 ) = (2π -ℓ 1 -ℓ 3 ) 2 tan ℓ 3 2 -ℓ 3 -. . . 4 sin ℓ 3 2 cos m 3 + sin 2 ℓ 1 2 -sin 2 ℓ 3 2 sin 2 m 3 sin ℓ 3 2 cos m 3 1 + ℓ 3 sin ℓ 3 . Assume ℓ 1 ≤ π 6 . Then I + (ℓ 1 , ℓ 3 , m 3 ) < 0. 2. Let I -(ℓ 1 , ℓ 3 , m 3 ) = (2π -ℓ 1 -ℓ 3 ) 2 tan ℓ 3 2 -ℓ 3 -. . . 4 sin ℓ 3 2 cos m 3 -sin 2 ℓ 1 2 -sin 2 ℓ 3 2 sin 2 m 3 sin ℓ 3 2 cos m 3 1 + ℓ 3 sin ℓ 3 .
Assume ℓ 1 ≤ 0.62. Then I -(ℓ 1 , ℓ 3 , m 3 ) < 0.

The proof of Lemma 6 is based on the following estimates on ℓ 1 . The above quantity is positive, since 0 ≤ m 1 ≤ π 2 and ℓ 3 ≥ ℓ 1 . Therefore, if

m 1 ≤ π 2 A(ℓ 1 , ℓ 3 , m 1 ) ≤ A(ℓ 1 , ℓ 3 , π 2 ) = 2 sin 2 ℓ3 2 -sin 2 ℓ1 2 ℓ 1 + ℓ 3 . ( 56
)
Assume by contradiction that ℓ 1 > π 6 . We are going to prove that A < 2 π , thus reaching a contradiction. We set This implies, by (53)

x = ℓ 3 -
| cos m 3 | = 1 -sin 2 ℓ1 2 sin 2 m 1 sin 2 ℓ3 2 ≥ 1 -sin 2 (1.9 -ℓ- 2 ) sin 2 1.9 .

Therefore, using (40) we can estimate A from above by

A ≤ ℓ 2 3 (2π -ℓ -)
12ℓ -1 -sin 2 (1.9 -ℓ- 2 ) sin 2 1.9 .

As a function of ℓ -the right-hand side is clearly decreasing. Now, if ℓ 1 ≥ 0.62 (and then ℓ -≥ 1.17 + 0.62), we can see that A < 2 π for any ℓ 3 ∈ [1.17, π -ℓ 1 ] giving the desired contradiction.

We are now able to prove Lemma 6. The derivative with respect to ℓ 1 is ℓ 3 -2 tan ℓ 3 2 +2 cos ℓ 1 2 sin ℓ 3 2 1 + ℓ 3 sin ℓ 3 that is decreasing in ℓ 1 , thus greater than its value for ℓ 1 = π 6 :

Proof (of

∂I + (ℓ 1 , ℓ 3 , -π) ∂ℓ 1 ≥ ℓ 3 -2 tan ℓ 3 2 + 2 cos π 12 sin ℓ 3 2 1 + ℓ 3 sin ℓ 3 > 0.
This shows that I + (ℓ 1 , ℓ 3 , -π) is increasing in ℓ 1 . We deduce that -for 1.17 ≤ ℓ 3 ≤ 5π 6 , I + (ℓ 1 , ℓ 3 , -π) ≤ I + π 6 , ℓ 3 , -π ;

for ℓ 3 ∈ 5π 6 , π , I + (ℓ 1 , ℓ 3 , -π) ≤ I + (π -ℓ 3 , ℓ 3 , -π).

  ) sin(θ) dθ = 0.

  (θ)) sin θdθ.

  [a, b], [b, c], where u(a) = u(b) = u(c) = 0 and u ≥ 0 in [a, b], u ≤ 0 in [b, c] . Assume that u ≥ 0 on (a, b), u(a) = u(b) = 0. By integrating the Euler equation on [a, b] and using that u(a) = u(b) = 0, we find

  We have seen in Lemma 3 that both I -and I + contain pairs of consecutive intervals (or triplet of intervals). Let us now consider the last triplet of intervals for which A < 0. the second one. Let I k-1 , I k , I k+1 be these three intervals. Therefore A(I k+1 , I k+2 ) ≥ 0. If u > 0 on I k-1 we are done, because we can consider (I k-1 , I k ), (I k+1 , I k+2 ). If u is negative on I k-1 we have u negative on I k-1 , I k+1 , I k+3 . . . and positive on I k , I k+2 , I k+4 . . .. We can consider (I k , I k+1 ) for which A < 0. If A(I k+2 , I k+3 ) ≥ 0 we are done. If A(I k+2 , I k+3 ) ≤ 0, then A(I k+3,I k+4 ) ≥ 0 (otherwise the last triplet in I -would be I k+2 , I k+3 ; I k+4 ). If A(I k+4 , I k+5 ) ≥ 0 we are done. If A(I k+4 , I k+5 ) ≤ 0, then A(I k+5,I k+6 ) ≥ 0....after some steps we will get necessarily four consecutive intervals I m-2 , I m-1 , I m , I m+1 (with u positive on I m-2 ) such that A(I m-2 , I m-1 ) < 0, A(I m , I m+1 ) ≥ 0 (because we have to stop before the first triplet of I + ). Therefore, we have proved the first part of our claim. The second part is proved exactly in the same way, starting from the last triplet in I + .

Lemma 7 2 . Then ℓ 1 ≤ π 6 . 1   1 - cos m 1 sin ℓ1 2 sin 2 ℓ3 2 -sin 2 ℓ1 2 sin 2 m 1   , that is, the same sign as sin m 1 sin 2 ℓ 3 2 -

 7216112212 Assume that u has 4 nodal domains. Assume that m 1 ≤ π Proof Let us first look at the dependence of A with respect to m 1 . The derivative of A(ℓ 1 , ℓ 3 , m 1 ) with respect to m 1 has the same sign as sin m sin 2 ℓ 1 2 .

Lemma 6) 1 .m 3 sin 2 ℓ 1 2 -sin 2 ℓ 3 2 sin 2 m 3 1/ 4 + sin ℓ3 2 cos m 3 sin 2 ℓ1 2 -sin 2 ℓ3 2 sin 2 m 3 1/4 2 . 2 , sin m 3 2 -ℓ 3

 12342322323 We look first at the dependence with respect to m 3 . The derivative of I + (ℓ 1 , ℓ 3 , m 3 ) with respect to m 3 has the sign of sin Since -π ≤ m 3 ≤ -π is negative. Therefore m 3 → I(ℓ 1 , ℓ 3 , m 3 ) is decreasing, which impliesI + (ℓ 1 , ℓ 3 , m 3 ) ≤ I + (ℓ 1 , ℓ 3 , -π). Now, I + (ℓ 1 , ℓ 3 , -π) = (2π-ℓ 1 -ℓ 3 ) 2 tan ℓ 3

  1 , we get ℓ -≥ ℓ -1 + ℓ 1 ≥ 2.34.With the same arguments of the two steps above, we can prove that ℓ -1 ≥

			π
	2 and 1.55 by 2.34 in the , that is a contradiction. We deduce . Indeed it is sufficient to replace 1.17 by π 2 estimate of A. This gives A ≤ 0.49 < and ℓ 1 ≥ π 2 2 π that ℓ -1 ≥ π 2 and ℓ 1 ≥ π 2 that implies ℓ -≥ π: this is a contradiction, since ℓ + > π.
	Proposition 8 Assume that u has at least 6 nodal domains and b < A < 2/π and the length of no nodal interval is greater than π.	π 2	. Then

  The map x → G(x, y) is increasing while y → G(x, y) is decreasing on this set. We remark that the assumption ℓ 1 > Indeed the function t → G(t, t+ π 6 ) is first increasing then decreasing and satisfies the above inequality for its maximum that is approximately at 0.6627206.From the properties of G, we infer that G(x, y) < Proof Let us suppose first that m 1 ≥ 1.9, therefore | cos m 1 | ≥ | cos 1.9| and, following (40) we inferA(ℓ 1 , ℓ 3 , m 1 ) ≤ l 2 1 (2π -ℓ 1 -ℓ 3 ) 12(ℓ 1 + ℓ 3 )| cos 1.9| .Now, this expression is decreasing in ℓ 3 and increasing in ℓ 1 , thus it is always less than its value for ℓ 1 = ℓ 3 < π 2 :A(ℓ 1 , ℓ 3 , m 1 ) ≤ A(ℓ 1 , ℓ 1 , 1.9) = l 1 (π -ℓ 1 ) ≤ 1.9. Expressing m 3 in terms of m 1 , we have m 3 = m 1 -ℓ - 2 -ℓ 0 ≤ 1.9 -ℓ - 2-π, thus | sin m 3 | ≤ sin(1.9 -ℓ- 2 ) (recall that -π ≤ m 3 ≤ -

	ℓ 1 We can therefore assume that 2 ≤ m 1 π and y = π 2 ). By identity (51), we have ℓ 3 + ℓ 1 2 2 and observe that 0 ≤ x < π 2 , 1.17 2 ≤ y < π 2 . The quantity A(ℓ 1 , ℓ 3 , rewritten as a function of x, y: | sin 1.9| sin ℓ 1 2 ≤ sin m 1 sin ℓ 1 2 = | sin m 3 | sin ℓ 3 2 ≤ sin ℓ 3 2 sin 1.9 -π 2	) can be ℓ -2 .
	G(x, y) =	√ sin x sin y y	, (x, y) ∈ 0,	π 2	×	1.17 2	,	π 2	.
	the rectangle 0,	π 2	×	1.17 2	,	π 2	π 6 this implies that x ∈ 0, is equivalent to y > x + π 3 . Now, it can be π . In 6
	checked that
				∀t ∈ 0,	π 3	, G(t, t +	π 6	) ≤	2 π	.
	y > x + A < 2 π as soon as ℓ 1 > π 6 , y ∈ [ 1.17 2 , π 2 π 6 .	2 π . We have thus proved that for all (x, y) such that , that is a contradiction. Therefore ℓ 1 is less or equal ], x ∈ [0, π 3 ], that is, ℓ 1 > π 6 π 6
	Lemma 8 Assume that u has 4 nodal domains. Assume that m 1 > ℓ 1 ≤ 0.62.	π 2	. Then
									12| cos 1.9|	≤	π 2 48| cos 1.9|	<	2 π
	and the contradiction is obtained in this case.

) are decreasing and negative.

Acknowledgements This work was partially supported by the project ANR-18-CE40-0013 SHAPO financed by the French Agence Nationale de la Recherche (ANR). We kindly thank the anonimouos referee for his precious remarks and suggestions.

Conflict of interest

The authors declare that they have no conflict of interest.


Andrea Guarriello
email: andrea.guarriello@insa-rennes.fr

Daniele Bresciani

Hervé Legay

George Goussetis

Renaud Loison

Design of Circularly Polarized and Highly Depointing Reflectarrays with high polarization purity

Keywords: reflectarray, optimization, circular polarization, tilted beam

This work aims to investigate the limitations and possible solutions in designing highly tilted radiation pattern reflectarrays working in double circular polarization and fed by an onset single primary source. To this extent, three single-layer monolithic reflectarrays, producing a pencil beam at increasing radiation angles with high X-polarization discrimination (XPD) have been designed and analyzed. A direct optimization technique shows that an XPD=20 dB can be attained for radiation patterns pointing at sixty degrees, at the cost of co-polarised directivity degradation. An analysis of the target aperture fields, obtained through backscattering techniques, shows that an amplitude modulation on the aperture field components is necessary to obtain highly tilted radiation patterns with high XPD. Since reflectarray cells are not able to produce such amplitude components unbalance, an elliptically polarized source is considered, producing an incident field components amplitude unbalance that results in a radical reduction of the X-polarization levels.

I. INTRODUCTION

High-gain antennas, which produce narrow beams with elevated gain values able to support high data rate capabilities demanded in space telecommunications, are essential in many applications, spanning telecom space missions, scientific space missions, radars, and long-distance communications. Highgain antennas traditionally employed in the space sector rely on metallic parabolic or shaped reflectors and antenna arrays. Beyond several advantages, both systems suffer from some limitations related to the antenna volume, cost, mass, and efficiency, which are of capital importance for antenna integration in the spacecraft system.

Reflectarray antennas (RAs) are an alternative technology able to alleviate most of the limitations described above while benefiting from the advantages of the conventional reflectors and array antennas for high gain applications [START_REF] Huang | Reflectarray Antenna[END_REF].

For space application, dual circularly polarized RAs are of high interest due to their several advantages. They allow an increment of the communication link capability by using the two different right hand (RH) and left hand (LH) circular polarization (CP), enabling two different communication channels. Moreover, CP is less sensitive to interference and multipath losses than linear polarization, and it is less affected by atmospheric and ionospheric losses. The use of two orthogo-nal CP components demands high polarization discrimination (XPD) between the RHCP and LHCP.

Tilted RAs at high radiation angles can find applications in missions requiring object tracking through radar systems or missions requiring to direct the beam toward off-nadir ground stations. It is frequent, especially in low Earth orbit applications, where the satellites are often in ground station visibility at low elevation angles.

In this paper, the issue of designing highly depointing dual-CP RA is addressed. The main objective consists in analyzing the reasons for the intrinsic difficulties in obtaining depointing radiation patterns with high cross-polarization discrimination. The problem analysis can be exploited to find efficient solutions for the design of tilted beam or broad beam RAs.

Highly tilted RA has been addressed in several works, but few references highlight the dual-polarization capabilities. In [START_REF] Nayeri | Bifocal design and aperture phase optimizations of reflectarray antennas for wideangle beam scanning performance[END_REF] and [START_REF] Wu | Wide-angle beamscanning reflectarray with mechanical steering[END_REF], the feed displacement technique is adopted in order to achieve respectively ±30 • and ±45 • scanning ranges, without addressing the issue of double circular polarization use. In [START_REF] Wu | Lowcost 1-d beam-steering reflectarray with ±70°scan coverage[END_REF] a scanning angular range of ±70 • is successfully achieved through combined feed and RA rotation, at the cost of 3.7 dB scan losses. Once again, the dual-polarization issue is not addressed in this work. Another work [START_REF] Yang | A broadband high-efficiency reconfigurable reflectarray using mechanically rotational elements[END_REF] deals with the scanning capabilities of mechanically reconfigurable elements RA antennas, showing a visible degradation of the polarization purity as the scanning angle increases. The scanning losses are evaluated at 3.7 dB for a scanning range of ±60 • , while the Xpolarization increases drastically from 20 • and above. In [START_REF] Kamoda | 60-ghz electronically reconfigurable large reflectarray using single-bit phase shifters[END_REF] a 60 GHz electronically reconfigurable RA for imaging systems is presented. With scanning capabilities over a ±25 • range, it presents XPD levels that are excellent for beams scanned at the azimuth angle of 0 • , while those at the azimuth angle of 20 • are degraded by up to 22 dB.

This paper shows and discusses the issues of X-polarization discrimination for highly depointing passive RAs. In the first instance, three different onset single-feed RAs, with the same geometrical features and in the same configuration, are designed and optimized for beam pointing angles of θ 1 = 0 • , θ 2 = 30 • and θ 3 = 60 • . The design and optimization methodology are briefly presented.

Subsequently, the target aperture field analysis is performed through advanced analysis methodologies, notably backscatter-ing techniques. This second step allows to identify the requirements needed to design low X-polarization highly depointing RA and helps to highlight the reasons for the difficulties when dealing with flat apertures passive antennas.

II. ANALYSIS AND DESIGN OF TILTED PENCIL BEAM RA

A. RA features 1) RA geometry: The RA (Fig. 1) comprises a single layer monolithic panel fed by a conventional onset horn working in the frequency band f = [25.5; 27.00] GHz with a center frequency of 26.25 GHz. The RA is circular with a diameter fixed to 300 mm, corresponding to roughly 26.3λ (λ the freespace wavelength at the center frequency). In this work, the bandwidth issue is not addressed, so only the central frequency is considered in the following.

The feeder illuminates the RA panel with an onset-centered configuration with f/D = 0.52. The feed illuminates in RHCP and LHCP with a taper of 11.5 dB at θ inc = 45 • . The RA parameters are summarized in Table I referring to the coordinate system and definitions shown in Figure 1.

2) Array constitutive elements and 1D look-up table handling:

As array elements, the first and second-order Phoenix cells [START_REF] Moustafa | The phoenix cell: A new reflectarray cell large bandwidth and rebirth capabilities[END_REF], [START_REF] Guarriello | A general equivalent circuit model for phoenix cells[END_REF] are selected. The square patch/slot loop periodic cycle on the bottom of figure 2 is considered. This cycle avoids sharp geometrical transitions for adjacent cells, and it is composed of first and second-order inductive Phoenix cells. Figure 2 shows the behavior of the cells as a function of the auxiliary periodic variable ξ defined as: The variation of the phase curves shows that an almost complete 360 • phase cycle is achieved, with a phase gap of ≈ 15 • , considered as acceptable.

ξ(d 1 , d 2 ) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ d 1 -d 1max d 1max -d 1min π ,if d 2 = 0 d 2 -d 2min d 2max -d 2min π ,if d 1 = d 1max (

B. Design through aperture field phase and far-field amplitude optimization 1) Co-polarization optimization:

The preliminary layout design is performed with a classical phase only (PO) synthesis. The layout obtained is used as first guess solution for the next step that consists in minimizing the quadratic error on the aperture field components phase through a multiobjective optimization of the RA aperture field phase distribution, namely the aperture field optimization (AFO). It is done by describing the RA preliminary geometry distribution (obtained with the PO synthesis) through the auxiliary periodic parameter ξ and representing the unwrapped ξ distribution through a continuous splines surface. This representation allows avoiding sharp geometrical transition among contiguous RA elements, thus respecting the quasi-periodic assumption made in the lookup table cells characterization. The results of the phaseonly optimization are presented in figure 3. The radiation pattern is computed through the aforementioned Floquet's mode matching software and a Physical Optics software. We can notice from figure 3 that the three RAs radiation pattern show a notable increment of the X-polarization levels as the beam tilting angle increases. To this extent, the RA radiating at broadside (specular direction of the incident field) shows high level of X-polarization discrimination of 38.65 dB. As the beam tilting angle increases, the X-polarization discriminantion deteriorate, by reaching 21.73 dB for the RA pointing at θ = 30 • and a very low XPD for the RA in the tilting angle configuration θ = 60 • , with an XPD of 8.85 dB.

2) X-polarization optimization: The elementary cell considered for the single circular polarization design is described through one degree of freedom. By considering rectangular elements, it is possible to control the impinging wave delay and depolarize it slightly [START_REF] Zhao | A subwavelength element for broadband circularly polarized reflectarrays[END_REF].

A direct optimization [START_REF] Zhou | The generalized direct optimization technique for printed reflectarrays[END_REF] with respect to a mask for the RHCP and LHCP is performed, namely the far-field amplitude optimization (FFO). This time, the objective is to minimize the maximum error between the radiation pattern and the prescribed mask by distorting the square cells geometry into rectangles (d x = d y ) at global level, i.e. by guaranteeing the local periodicity assumption through splines projection of the parametrized layout and by keeping the original cells square lattice. More in detail, for each of the three cases, the RA layout is described through a continuous and derivable spline projection of the parameter ξ, identifying the cell index in the lookup table according to equations 1, and a cells distortion index h. The optimization variables are therefore the spline coefficients of the two surfaces Ξ and H describing the parametrized cells distribution. A minimax algorithm allows to minimize the maximum of the vectorial cost function described as the error on the radiation pattern amplitude with respect to the mask and the maximization of the XPD levels.

The initial condition is the single polarization design of section II-B1. The optimization is performed at the center frequency f 0 = 26.25 GHz for each of the three RA.

By looking at the plots of figure 4, we can notice that the XPD levels are kept well above 20 dB for the three RAs. The levels of X-polarization for the case θ 1 = 0 • and θ 2 = 30 • are reduced below -30 dB by preserving low levels of side lobes. The case θ 3 = 60 • shows difficulties to increase the XPD levels above 20 dB. Moreover, we can see the apparition of an important radiation in the angular range θ ∈ [-30 • ; 20 • ]. These results show that the reduction of the X-polarization levels obtained through the far-field direct optimization is generally accompanied by a degradation of the maximum levels of directivity. The co-polarization levels degradation is more accentuated as the beam tilting angle increases.

III. APERTURE FIELD RECONSTRUCTION THROUGH X-POLARIZATION CANCELLATION

Backscattering techniques [START_REF] Legay | A 1.3 m facetted reflectarray in ku band[END_REF] are adopted to obtain a RA ideal aperture field that allows obtaining a far-field with low X-polarization levels by keeping adequate co-polarization levels. It consists of enforcing the amplitude of the far-field X-polarization to be canceled for both RHCP and LHCP polarisation, and then derive the aperture field and the 2 × 2 reflection matrices of the RA elements that are the ideal reflection matrices to be realized.

The aperture field reconstruction is derived from the θ 3 = 60 • radiation pattern of figure 3 with the X-polarization component forced to be null everywhere. Figures 5a and5b map the amplitude of the incident field Cartesian components while figures 5c and 5d show the amplitude of the aperture field components obtained through backscattering technique.

As we may notice, the amplitude of the target aperture field is not realizable through a phase-only synthesis. The targeted RA aperture field has an amplitude higher or lower compared to the incident field. We can expect that RA cells cannot realize such an aperture field since the reflection matrix direct coefficients cannot have an amplitude greater than 0 dB. For a more exhaustive analysis, the comparison of the incident field components and the targeted aperture field components for the three reflectarray is shown in figure 6. We see that x and y components umbalance to achieve low X-polarization radiation pattern become more and more accentuated as the beam tilting angle increases. This explains why as the beam tilting angle increase, the realizable XPD degrades.

Examples of aperture field amplitude control can be found in literature in the context of broad-beam antennas, i.e., radiating at a highly tilted direction. In [START_REF] Fouany | New concept of telemetry x-band circularly polarized antenna payload for cubesat[END_REF] the antenna aperture amplitude is locally controlled through parasitic elements to mitigate the X-polarization levels at high radiation angles. In [START_REF] Minatti | Synthesis of modulated-metasurface antennas with amplitude, phase, and polarization control[END_REF] the amplitude of the aperture field is managed by controlling the leakage parameter of the metasurface elements. It allows a modulation of the aperture field amplitude to compensate for the polarization losses at high pointing radiation angles θ. From the other references in the literature (for instance [START_REF] Ravanelli | Multiobjective optimization of an isoflux antenna for leo satellite downhandling link[END_REF], [START_REF] Minatti | A circularly-polarized isoflux antenna based on anisotropic metasurface[END_REF], [START_REF] Minatti | Modulated metasurface antennas for space: Synthesis, analysis and realizations[END_REF]), we can find consistency with the latter explanation since the amplitude of the aperture field is not controlled, the broad-beam antennas suffer from high X-pol levels for radiation angles above 40 • .

As show in the previous section, even if a passive RA element cannot efficiently control the amplitude levels of the aperture field, a direct optimization in which the RA square patches and slots are deformed into rectangles can allow phase or amplitude compensation that at a global level can enhance the XPD levels at high radiation angles, at cost of a copolarization deterioration.

IV. REFLECTARRAY FED BY AN ELLIPTICALLY POLARIZED

SOURCE

The analysis of the target aperture field derived through backscattering techniques revealed that an amplitude unbalance between the components of the aperture field is necessary to obtain RA radiating low X-polarization at highly tilted angles. In this section, we explore an alternative feed design able to produce an incident field amplitude on the RA close to the one of figures 5c and 5d. The case RA radiating at 60 • is considered, for the other cases the process is equivalent. The incident field amplitude unbalance means that the feed should radiate with highly elliptical polarization, with field components in phase quadrature and the amplitude of the components unbalanced according to the amplitude of the target aperture field components. This is obtained by exciting each of the two feed ports with a different normalized weight w x and w y .

A RA fed by the new primary source is designed, the layout is derived through AFO. Figure 7 shows the comparison of the three different designs of a RA radiating at θ = 60 • . In the present case the excitation weights are given by the amplitude relative unbalance of the target aperture field x and y components, w x ≈ 0.89 and w y ≈ 0.45. We can notice how the XPD levels have been enhanced drastically in the new design, reaching values superior to 40 dB. The beam tilting operated by the flat surface RA allows circularizing the elliptical polarization components coming from the feed at the radiation angle direction 60 • , greatly enhancing the circular polarization purity of the radiation pattern. V. CONCLUSION Pencil beam RAs with various beam tilting angles have been analyzed and designed. The reduction of the X-polarization levels was confirmed to be challenging for RA radiating at highly tilted angles. Substantial improvements on the XPD levels can be achieved for each case surveyed through a direct optimization technique, at the cost of an increasing degradation on the co-polarization levels as the beam tilting angle increases. This study confirmed the impact of the limitations of passive RA cells to modulate the aperture field amplitude, which is necessary to achieve high X-polarization discrimination at highly tilted radiating angles. A solution to overcome the limitations on amplitude modulation imposed on passive reflectarray cells is to shape the incident field components coming from the primary source by balancing the source components excitation. To obtain a highly tilted beam radiation pattern with high circular polarization purity, a highly elliptically polarized feed should be designed. A radical improvement of the XPD> 40 dB has been observed in the RA fed by an elliptically polarized source and radiating at θ = 60 • . The recommended ellipticity rate of the onset source polarization is more accentuated as the target beam tilting angle increases.

Fig. 1 .

 1 Fig. 1. RA antenna geometry

1 ) where d 1 ,

 11 d 2 are shown in the inset of figure 2. The geometrical parameters are set according to standard fabrication tolerances, toll = 0.1 mm. Therefore, the geometrical parameters are set to d 1min = 0.1 mm, d 1max = 3.7 mm, d 2min = 0.1 mm and d 2max = 3.5 mm. In figure2, the reflection phase is calculated under normal plane wave incidence and displayed as a function of ξ. The lookup table is computed for all the incidences relative to the RA geometry as seen from the feed. The cells characterization is performed though periodic surface analysis program based on the Floquet expansion of the unit cell scattered field, derived through the Method of Moments. The local periodicity of the RA layout must be therefore respected.

Fig. 2 .

 2 Fig. 2. Phase of the scattering coefficient of a square loop/slot combination element in a periodic environment as function of the periodic parameter ξ describing the geometry. In the inset the relevant parameters defining the cell's geometry.

Fig. 3 .

 3 Fig. 3. Radiation pattern of the RAs obtained through aperture field optimization (AFO).

Fig. 4 .

 4 Fig. 4. Radiation pattern of the RAs obtained through far-field optimization (FFO).

Fig. 5 .

 5 Fig. 5. Field amplitude distribution on the RA surface.

Fig. 6 .

 6 Fig. 6. Comparison of the incident field components and the targeted aperture field components E targ x,y for the three reflectarray generating tilted beams with low X-polarization. Cut along y = 0.

Fig. 7 .

 7 Fig. 7. Comparison of the radiation patterns for RA depointing at 60 • obtained through aperture field phase optimization (blue), far-field direct optimization (red), aperture field phase optimization and RA illuminated by an elliptically polarized source (yellow).

TABLE I RA FEATURES

 IFEATURES

ACKNOWLEDGMENT

This work was supported by the European Commission under H2020 project REVOLVE (MSCA-ITN-2016-722840).