Maximization of the second non-trivial Neumann eigenvalue

c

2019 by Institut Mittag-Leffler. All rights reserved

Maximization of the second non-trivial Neumann eigenvalue

by

Dorin Bucur

Universit´e Savoie Mont Blanc Le-Bourget-Du-Lac, France

Antoine Henrot

Universit´e de Lorraine Vandoeuvre-l`es-Nancy, France

1. Introduction and statement of the results

Let N>2 and Ω⊆R^N be a bounded open set such that the Sobolev space H¹(Ω) is compactly embedded inL²(Ω) (for instance Ω Lipschitz). Those sets are calledregular throughout the paper. On such domains, the spectrum of the Laplace operator with Neumann boundary conditions consists only on eigenvalues that we denote (counting their multiplicities)

0 =µ0(Ω)6µ1(Ω)6µ2(Ω)6...!∞.

For everyk>1, we have

µ_k(Ω) = min

S∈S_kmax

u∈S

R

Ω|∇u|²dx R

Ωu²dx , whereSk is the family of all subspaces of dimensionk in

H¹(Ω)_/R:={u∈H¹(Ω) :R

Ωu dx= 0}.

If Ω is connected, thenµ1(Ω)>0.

In 1954 Szeg¨o proved that among simply connected, 2-dimensional, smooth sets Ω⊆R² the ball maximizesµ₁(see [20], [3] and [5])), i.e.(¹)

|Ω|µ₁(Ω)6|B|µ₁(B).

Dorin Bucur is member of the Institut Universitaire de France. His work is part of the ”Geometry and Spectral Optimization” GeoSpec, Persyval Lab research programme.

(¹) Weinberger noted in [21] that the proof of Szeg¨o gives a stronger result, namely that the disc minimizes the sum 1/µ1(Ω)+1/µ2(Ω) among 2-dimensional, smooth, simply connected sets of given area.

Two years later, Weinberger [21] removed the topological constraint and the dimension restriction, and he proved that, for every N>2 and Ω⊆R^N regular, the following in- equality holds:

|Ω|^2/Nµ1(Ω)6µ^∗₁, whereµ^∗₁=|B|^2/Nµ1(B).

Maximizing the Neumann eigenvalues under volume constraint is also related to the celebrated conjecture of P´olya [18] asserting that the principal term of the Weyl law provides in fact a bound for the eigenvalues. This conjecture reads, inN dimensions,

µ_k(Ω)64π² k

ωN|Ω|) 2/N

for allk>1,

where ωN is the volume of the unit ball in R^N. The conjecture was proved to hold only for particular classes of domains, for instance tiling domains inR² (see [16]). For general regular domains, the conjecture holds true in the casek=1, as a consequence of the Szeg¨o–Weinberger inequality, but continues to remain open for arbitrary k. Kr¨oger found in [15] a series of bounds, which are larger than the conjectured ones. For instance, ifk=2 he provedµ2(Ω)616π/|Ω| for 2-dimensional domains. The value 16π/|Ω| is the double of the conjectured one. A natural, related, question is to find the geometry of the domain which maximizes thekth Neumann eigenvalue. This question turns out to be difficult fork=2 and probably impossible to answer analytically fork>3. We refer to [2], [1], [6] for numerical approximations of the (presumably) optimal sets fork610, but there is no proof of the existence of those sets.

We refer the reader to the result of Girouard, Nadirashvili and Polterovich [11], where the authors prove that inR² the union of two equal (and disjoint) disks gives a larger second eigenvalue than any smooth simply connected open set of the same measure.

Moreover, this value is asymptotically attained by two disks with vanishing intersection.

Their proof is based on a combination of topological and analytical arguments and relies on a folding and rearrangement technique introduced by Nadirashvili in [17], taking advantage on the use of conformal mappings. This method cannot be adapted to non- simply-connected sets. The authors left the case of arbitrary (regular) domains ofR²as an open problem.

Several independent numerical computations [2], [1], [6] inR² andR³ brought sup- port in favor of the maximality of the union of the two discs without the simply connect- edness constraint inR² and, moreover, lead to a similar conjecture in three dimensions of the space.

The purpose of this paper is to prove that, in general, the second Neumann eigen- value attains its maximum on a union of two disjoint, equal balls in the class of arbitrary

domains of prescribed measure ofR^N. As a consequence, we prove that the P´olya conjec- ture for Neumann eigenvalues holds fork=2, without any restriction on the dimension, geometry or topology of the domains.

Let us denote the scale invariant quantity µ^∗₂=2^2/N|B|^2/Nµ1(B), where B is any ball. On the union of two disjoint ballsB1 andB2, each of mass ¹₂, we have

µ0(B1∪B2) = 0, µ1(B1∪B2) = 0 and µ2(B1∪B2) =µ^∗₂. The main result of the paper is the following.

Theorem 1. Let Ω⊆R^N be a regular set. Then

|Ω|^2/Nµ2(Ω)6µ^∗₂.

If equality occurs, then Ωcoincides a.e. with the union of two disjoint,equal balls.

As a consequence, we get the following.

Corollary 2. The P´olya conjecture for the Neumann eigenvalues holds for k=2 in any dimension of the space.

In fact, we shall prove a more general result than Theorem 1. Specifically, we shall prove that the result of Theorem1holds true on arbitrary open sets (even non-regular) and, moreover, on L¹∩L^∞-densities in R^N, provided the classical eigenvalues, seen as variational quotients, receive a suitablerelaxed definition (see [10, Chapter 7] and [7]).

Precisely, let%∈L¹(R^N,[0,1]). For everyk>1, we define

˜

µk(%) := inf

S∈Lk

maxu∈S

R

R^N%|∇u|²dx R

R^N%u²dx , whereL_k is the family of all subspaces of dimension kin

{u·1{%(x)>0}:u∈C_c^∞(R^N) and R

R^N%u dx= 0}. (1) We have the following.

Theorem 3. Let %∈L¹(R^N,[0,1])be non-identically-zero. Then,

• (k=1, extension of the Szeg¨o–Weinberger inequality) Z

R^N

% dx 2/N

˜

µ₁(%)6µ^∗₁, (2)

with equality if and only if %=1B a.e., for some ball B of R^N.

• (k=2)

Z

R^N

% dx 2/N

˜

µ2(%)6µ^∗₂, (3)

with equality if and only if %=1_B]+1_B^∗ a.e., where B^] and B^∗ are two disjoint (open) balls of equal measure.

Fork=1, Theorem3 above is a generalization of the Szeg¨o–Weinberger inequality, and fork=2 is a generalization of Theorem1.

We notice the following.

• If Ω is a bounded, open Lipschitz set, then taking%=1Ω, one gets ˜µk(%)=µk(Ω).

• Let us remove a smooth manifold Γ from Ω, such that H¹(Ω\Γ) is compactly embedded in L²(Ω\Γ). This is for example the case when Ω is Lipschitz and the crack Γ is itself Lipschitz. Then, for %=1Ω, one has ˜µk(%)>µk(Ω\Γ), since C_c^∞(R^N)|_Ω\Γ⊆ H¹(Ω\Γ). From this perspective, Theorem3covers the inequality proved in Theorem1 even in this less regular case.

• If the set Ω={%>0}is smooth and there existsα>0 such that%>α1_{%>0}(i.e.%is not degenerating on its support and preserves ellipticity), then ˜µ_k(%) are the eigenvalues associated with the well-posed problem

−div(%∇u) =µk%u in Ω, ∂u

∂n= 0 on∂Ω.

• If Ω is just a bounded open set, without any smoothness, the spectrum of the Neumann Laplacian on Ω may be continuous. Theorem 3 still applies to %=1Ω, but we do not have any spectral interpretation of ˜µk(1Ω). The same occurs if either % is degenerating loosing ellipticity on its support, and/or if its support is not smooth enough.

Concerning the ideas of the proof, it is worth to recall what happens for the Dirichlet Laplacian. The Faber–Krahn inequality for the first Dirichlet eigenvalue of the Laplacian λ1(Ω) asserts that the minimum of|Ω|^2/Nλ1(Ω) is attained on balls. A simple argument analysing the positive and negative parts of a second eigenfunction, leads to the conclu- sion that the minimum of|Ω|^2/Nλ2(Ω) is achieved on a set consisting on two equal and disjoint balls. We refer to [8] for a detailed description of the history of the result, which is attributed to Krahn [14], Hong [13] and Szeg¨o [19].

A similar argument for the Neumann Laplacian is not valid. The proof of Theorem1 (and of Theorem 3) is based on a suitable construction of a set of N test functions which are simultaneously orthogonal to the constant functionand to the first Neumann eigenfunction on a regular set Ω. The structure of these N functions is inspired by the functions built by Weinberger, that we briefly describe below (see, for instance, [12] and [21] for more details).

We denote throughout the paperR_Ωthe radius of a ball of the same volume as Ω, by r_Ωthe radius of a ball of volume¹₂|Ω|, byB_Ra ball centered at the origin of radiusR, and byB_A,Rthe ball centered at pointAof radiusR. We denote byga non-negative, strictly increasing solution(²)of the following differential equation on (0, R) (see the paper by

(²) The functiongis explicitly given byg(r)=r^1−N/2J_N/2(kr/R), wherek=p

µ1(BR) is the first positive zero of r7!(r^1−N/2J_N/2(r))⁰ sometimes denotedp_N/2,1 as in [4], and J_N/2 is the standard Bessel function.

Weinberger [21], or [12,§7.1.2] for details):

g⁰⁰(r)+N−1 r g⁰(r)+

µ1(BR)−N−1 r²

g(r) = 0, g(0) = 0, g⁰(R) = 0. (4) Given a pointA∈R^N and a valueR>0, Weinberger introduced the function

gA:R^N−!R^N, x7−!GR(dA(x))

dA(x)

−→Ax,

(5)

whereG_R: [0,∞)!Ris defined by

G_R(r) :=g(r)1_[0,R]+g(R)1_[R,∞). (6) BydA(x) we denoted the distance fromxtoA.

Using Brouwer’s fixed point theorem, Weinberger proved forR=RΩthe existence of a pointAsuch that the set of functions

x7−!gA(x)·ei, i= 1, ..., N,

are orthogonal to the constants inL²(Ω). Here, (ei)i are the vectors of an orthonormal basis. As a consequence, those functions can be taken as tests in the Rayleigh quotient forµ1(Ω). By summation, this lead to

µ1(Ω)6 Z

Ω

G⁰²_R

Ω(d_A(x))+(N−1)G²_R

Ω(dA(x)) d²_A(x)

dx Z

Ω

G²_R

Ω(d_A(x))dx

. (7)

The functionr7!GR_Ω(r) is strictly increasing on [0, RΩ] (and then constant), while r7−!G⁰²_R

Ω(r)+(N−1)G²_R

Ω

r²

is decreasing. Consequently, the right-hand side of (7) is not larger than

µ₁(B_A,RΩ) = Z

BA,RΩ

G⁰²_RΩ(dA(x))+(N−1)G²_R

Ω(d_A(x)) d²_A(x)

dx Z

BA,RΩ

G²_RΩ(dA(x))dx

. (8)

In order to observe thatµ₁(Ω)6µ₁(B_A,RΩ), one formally has to move, one to one, the points of Ω\B_A,RΩ toward the points of the B_A,RΩ\Ω, pushing forward the measure

A B M

rΩ

HA HB

HAB

Figure 1. The geometry of the test functionsg^AB.

1_Ω\BA,R

Ωdx to 1B_A,R

Ω\Ωdx. Throughout the paper, we call this procedure amass dis- placement argument.

In order to prove Theorem 1, we shall use a somehow similar strategy, searching a set ofN suitable test functions. The new difficulty is that the set of functions that we have to build should be orthogonal to both the constant function and to a first Neumann eigenfunction on Ω (which is unknown). In the same time, the associated Rayleigh quotient should not exceedµ^∗₂.

Given two different pointsA, B∈R^N, we introduce the linear part of the symmetry operator with respect to the mediator hyperplaneHAB of the segmentAB:

TAB:R^N−!R^N, v7−!v−2(−→

ab·v)−→ ab, where −→

ab=−−→

AB/kABk. Denoting H_A and H_B the half-spaces determined by HAB and containingA andB, respectively, we build the function

g^AB:R^N−!R^N,

x7−!1H_A(x)gA(x)+1H_B(x)TAB(gB(x)).

(9) The functionsgA and gB are the functions of Weinberger introduced in (5), associated with GrΩ. Roughly speaking, g^AB is a suitable gluing along HAB of two Weinberger functions corresponding to different points.

Our main purpose will be to justify the existence of two pointsA andB such that the set ofN scalar functions

x7−!g^AB(x)·e_i, i= 1, ..., N,

are simultaneously orthogonal inL²(Ω) to the constant function and to a first eigenfunc- tionu1 of the Neumann Laplacian on Ω, i.e.

Z

Ω

g^AB·eidx= Z

Ω

g^AB·eiu1dx= 0 for alli= 1, ..., N. (10) The proof of existence of A and B with such properties relies on a topological degree argumentand requires the most of the attention (it is worth mentioning that every result on maximization for Neumann eigenvalues in the literature relies on a strong topological argument).

Once the pointsAandBare found, the proof of Theorem1follows in its main lines the one of Weinberger, being based on the mass displacement argument. In fact, on each of the half-spacesHAandHB, the restriction ofg^AB acts like a Weinberger function (5) associated with a ball of half measure.

Structure of the paper.

• In the next section we prove Theorem 1 for regular sets. We give the detailed construction of the function g^AB, prove the existence of a couple of points A and B makingg^AB·eisuitable as test functions forµ2(Ω), and prove the inequality. The equality case will be a direct consequence.

• In§3, we give the proof of Theorem3. We start by proving that the classical Szeg¨o–

Weinberger inequality holds true as well for densities and arbitrary domains. Concerning the second eigenvalue, we rely on the main ideas introduced in§2 and focus on the new difficulties raised by the possible absence of a first eigenfunction and by the possible unboundedness of the support.

Although it would have been more natural to prove first the general case and de- duce the inequality for regular domains as a consequence, we have chosen to start by giving a detailed proof of Theorem1 in the classical framework, as most of readers are presumably interested in this case. The new difficulties raised by the proof of Theo- rem3 are exclusively related to the ill-posedness of the eigenvalue problem in the non- smooth/degenerate/unbounded setting. The fact that we deal with a density instead of a geometric domain does not raise any supplementary difficulty, being handled by mass displacement.

2. Proof of Theorem 1

In this section we prove Theorem1. Let Ω⊆R^N be regular. We split the proof in several parts.

The validity of the test functions. Recall thatrΩis the radius of the ball of volume

1

2|Ω|. A set of eigenfunctions associated with the first non-zero eigenvalue µ1(Br_Ω) on the ballBr_Ω are{(g(r)/r)xi:i=1, ..., N}, whereg solves the differential equation (4) for R=rΩ.

LetAand B be two distinct points inR^N. We recall the function g^AB introduced in (9):

g^AB(x) = 1_HA(x)g_A(x)+1_HB(x)g_B(x)−2·1_HB(x) g_B(x)·−→ ab−→

ab.

The function g^AB is well defined, and continuous across HAB. Indeed, it is enough to observe that, forx0∈∂HA∩∂HB=HAB, we have

gA(x0) =TAB(gB(x0)), which comes from direct computation.

We notice that, for allx∈R^N,

kg^AB(x)k6g(rΩ) and

k∇g^AB(x)k62N²

sup

r>0

G_rΩ(r) r +sup

r>0

G⁰_r

Ω(r)

<∞. (11)

As a conclusion, we get g^AB∈W^1,∞(R^N,R^N), with a uniform bound on their norm, independent onAandB. The functionsg^AB·eiare therefore admissible as test functions on Ω.

The use of the test functions. Assume for a moment that we have found two different points A, B∈R^N such that the orthogonality relations (10) hold. Let us show that we can prove Theorem1.

For somei∈{1, ..., N}, let us take in the definition ofµ2(Ω) the subspace S= span{u1,g^AB·ei}.

We can write

µ2(Ω)6 R

Ω|∇(g^AB·ei)|²dx R

Ω|g^AB·ei|²dx for alli= 1, ..., N.

As a consequence,

µ₂(Ω)6 PN

i=1

R

Ω|∇(g^AB·e_i)|²dx PN

i=1

R

Ω|g^AB·ei|²dx .

Decomposing the sums over Ω∩HA and Ω∩HB, we get using (9) (the computation is similar with the one in Weinberger’s proof, see [12]))

µ₂(Ω)6 Z

HA∩Ω

G⁰²_r

Ω(dA(x))+(N−1)G²_rΩ(dA(x)) d²_A(x)

dx+

Z

HB∩Ω

G⁰²_r

Ω(dB(x))+(N−1)G²_rΩ(dB(x)) d²_B(x)

dx Z

HA∩Ω

G²_r

Ω(dA(x))dx+

Z

HB∩Ω

G²_r

Ω(dB(x))dx

.

We displace the mass as follows: we split Ω\(BA,rΩ∪BB,rΩ) in two sets Ω_Aand Ω_B, such that|ΩA|+|Ω∩BA,rΩ|=¹₂|Ω|. By monotonicity ofr7!G_rΩ(r) and of

r7−!G⁰²_r

Ω(r)+(N−1)G²_r

Ω(r) r² ,

for any couple of pointsx∈Ω_A andy∈B_A,rΩ\Ω the following inequalities hold:

G⁰²_rΩ(dA(x))+(N−1)G²_rΩ(dA(x)) d²_A(x) d²_A(x) <

G⁰²_rΩ(dA(y))+(N−1)G²_rΩ(dA(y)) d²_A(y) d²_A(y) , G²_r

Ω(d_A(x))> G²_r

Ω(d_A(y)).

We then formally displace the mass from Ω_A to B_A,rΩ\Ω, and increase the Rayleigh quotient. We use the same argument for Ω_B andB_B,rΩ\Ω, finally getting that

Z

HA∩Ω

G⁰²_rΩ(dA(x))+(N−1)G²_rΩ(dA(x)) d²_A(x)

dx+

Z

HB∩Ω

G⁰²_rΩ(dB(x))+(N−1)G²_rΩ(dB(x)) d²_B(x)

dx Z

HA∩Ω

G²_rΩ(dA(x))dx+

Z

HB∩Ω

G²_rΩ(dB(x))dx

6

2

Z

BrΩ

G⁰²_r

Ω(r)+(N−1)G²_r

Ω

r²

dx

2

Z

BrΩ

G²_r

Ω(r)dr

=µ1(BrΩ).

Sinceµ₁(B_rΩ) is the second eigenvalue of the union of two disjoint balls of mass ¹₂|Ω|, the inequality in Theorem1 follows.

If equality occurs, then H_A∩Ω andH_B∩Ω have to be balls of mass ¹₂|Ω|, up to a set of zero Lebesgue measure. Indeed, if there is mass displacement on a set of positive measure, the inequality has to be strict. So, if equality occurs, Ω is a.e. identical to the union of two disjoint balls of mass ¹₂|Ω|.

Remark 4. If, for instance, Ω is Lipschitz and equality occurs, then Ω has to coincide with the union of the two balls. If we work only with regular sets without further geometric assumption, it might be possible that one removes from one ball a set of capacity zero and, from the other ball, a (small) set of positive capacity but of zero measure (say a piece of a smooth manifold of dimensionN−1). The removed set should be small enough such that the second non-trivial eigenvalue of the slitted ball is not smaller than the first eigenvalue of the genuine ball. InR², this situation could occur if one removes a small segment from a diameter.

Existence of the family of test functions. In order to complete the proof, it remains to justify the existence of two pointsAandB such that the orthogonality relations (10) hold true. We shall do this below, but we point out from the beginning that the proof works in an identical way, provided 1Ωis replaced by a measurable function%:R^N![0,1]

with bounded support, and the first eigenfunction u1 is replaced by any measurable functionusuch thatu1_{%=0}=0,R

R^N%u²dx<∞andR

R^N%u dx=0.

We give the following.

Lemma5. Let A6=B be two points of R^N. Then, for all x∈R^N, gA(x)·−→

ab >gB(x)·−→ ab.

Proof. The proof is immediate, by direct comparison.

Lemma6. Assume that A and B are two points of R^N such that Z

Ω

g_A(x)·eidx= Z

Ω

g_B(x)·eidx for all i= 1, ..., N. Then, for all v∈R^N,we have

Z

Ω

g_A(x)·vdx= Z

Ω

g_B(x)·vdx andA=B.

Proof. The first assertion is trivial and the second is a consequence of Lemma5for v=−→

ab.

In the sequel, we shall use a deformation argument in the framework of the topolog- ical degree theory (see for instance [9, Theorem 1]), in order to prove the following.

Proposition 7. There exist two different points A and B such that the orthogo- nality relations (10)hold true.

Proof. By rescaling, we may assume that Ω⊆B₁, the ball centered at the origin of radius equal to 1. Let M>20 be fixed (the value 20 is chosen to be large enough with respect to the radius ofB1). Denote

D={(X, Y) :X, Y ∈R^N andX=Y} ⊆R^2N. We introduce the function

F: [−M, M]^2N\D −!R^2N,

by

F(A, B) :=

Z

Ω

g^AB·eidx, Z

Ω

g^AB·eiu1dx

.

We want to prove that there exists a couple of points (A, B)∈[−M, M]^2N\D which makeF vanish. So, we assume for contradiction thatF does not vanish on its definition domain. We first observe that there existsδ >0 such that, if (A, B)∈[−M, M]^2N\D and

d_R2N((A, B),D)6δ,

thenF cannot vanish at (A, B). Indeed, assume for contradiction that (A_n, B_n)∈[−M, M]^2N\D

is such that

Z

Ω

g^AnBn·eidx= 0 for alli= 1, ..., N

andd_R2N((A_n, B_n),D)!0. Extracting a subsequence, we may assume that An!A, Bn!A and

−−−→AnBn

kAnBnk!v∈S^N−1.

Then, the a.e.-limit of the sequence of functions (g^AnBn·v_n)_n, denoted for convenience g^AA·v, has a constant sign, vanishing only on a zero-measure set. This contradicts

Z

Ω

g^AnBn·vndx= 0.

So, let us denote

V={(A, B)∈R^N×R^N:d_R2N((A, B),D)6δ}, and restrict the functionF to [−M, M]^2N\V.

LetB_X^∗_,Rbe a ball of some radius 0<R61 (the choice is free, but we should have in mindr_Ω), with a center X^∗ carefully chosen, that will be specified in the proof. For simplicity of the notation, we denote this ballB^∗.

First deformation. We introduce fort∈[0,1] the following family of functions Ft: [−M, M]^2N\V−!R^2N,

by

Ft(A, B) = Z

Ω

g^AB·eidx,(1−t) Z

Ω

g^AB·eiu1dx+t Z

B^∗

g^AB·eidx

.

Clearly, this family is continuous in t, and F0≡F. We shall prove that, for a specific position of the centerX^∗of the ballB^∗, for everyt∈[0,1] the functionFtcannot vanish on∂([−M, M]^2N\V).

Assume that, for some (A, B)∈∂([−M, M]^2N\V) and somet∈(0,1], we have Ft(A, B) = 0.

We shall first focus only on the firstN coordinates ofF_t(A, B), which depend neither on tnor onB^∗. This will give important information on the possible positions of the points (A, B).

Indeed, we start observing that (A, B)∈∂V/ . This is a consequence of the choice ofδ above. It remains that (A, B)∈∂([−M, M]^2N). In other words, at least one of the points AorB is at distance at leastM from the origin (hence at leastM−1 from Ω).

Case 1. Assume the point B is at distance at least M−1 from Ω. Let C be the cone with vertexB, tangent to the ballB1. If the point Adoes not belong to the cone C, then denotingO⁰ the projection ofO on the lineAB, we cannot have

Z

Ω

g^AB·−−→

O⁰O dx= 0, since the functiong^AB·−−→

O⁰O has constant sign on Ω. Therefore,A has to belong to the cone. Moreover, in this situation, the point A has to belong as well to the annulus BB,M+1\BB,M−1. Indeed, ifAdoes not belong to this annulus, we cannot have

Z

Ω

g^AB·−−→ BO dx= 0, since, this time, the functiong^AB·−−→

BO has constant sign on Ω (positive if A is between the ball B₁ and B, negative if the ball B₁ is between A and B). This means that A∈C ∩(B_B,M+1\B_B,M−1), which by simple computation leads toA∈B^√₂.

The main consequence is that the distance fromAtoO is not larger than √ 2, and hence, by the construction of the function g^AB, its action on the domain Ω is entirely given by A, since Ω is covered by H_A only. In other words, the point B does not influence the integrals in (10), and g^AB=g_A on Ω. Moreover, from Lemma 6, we get that the position of A satisfying Ft(A, B)=0 is uniquely determined, for every B far away from Ω.

ForB far away andAfixed as above, let us look now to the linear form v7−^L!Z

Ω

g^AB·vu1dx.

This form is not identically vanishing, otherwise for the couple (A, B) the function F vanishes. Consequently, the kernel of this form is an hyperplane, denotedK. Letξ∈S^N−1 be orthogonal toK such that R

ΩgA·ξu1dx>0. We choose the center of the ballX^∗ to be given by A+3√

2ξ. With this choice, the ballB^∗ does not intersectB1 and is fully covered byHA (recall thatM>20). Consequently,

v7−!Z

B^∗

GR(dA(x)) d_A(x)

−→Ax·vdx

has the same kernelK and has the same sign asL. In other words, for everyt∈[0,1], v7−!(1−t)

Z

Ω

g^AB·vu₁dx+t Z

B^∗

GR(dA(x)) d_A(x)

−→Ax·vdx

vanishes only forv∈K.

At least one of the vectors e1, ...,eN does not belong to K. Consequently, among theN terms

(1−t) Z

Ω

g^AB·eiu1dx+t Z

B^∗

GR(dA(x)) dA(x)

−→Ax·eidx, i= 1, ..., N,

at least one is not vanishing. The conclusion is that, for everyt∈[0,1], the function F_t cannot vanish on∂([−M, M]^2N\V).

Case 2. Assume the point A is at distance at least M−1 from Ω. In this case, only the pointB acts on Ω, as in the previous case, andB∈B^√₂. The only thing which differs, is the expression of the functionF_t, which becomes

Ft(A, B) = Z

Ω

gB·ei−2 gB·−→ ab −→

ab·ei

dx,

(1−t) Z

Ω

gB·eiu1−2 gB·−→ ab −→

ab·ei

u1dx+t Z

B^∗

gB·ei−2 gB·−→ ab −→

ab·ei)

. In other words, we have

Ft(A, B) = Z

Ω

gB·TAB(ei)dx,(1−t) Z

Ω

gB·TAB(ei)u1dx+t Z

B^∗

gB·TAB(ei)

. Again, as in case 1, ifF_t(A, B)=0, thenB has to coincide with the same pointA, in the preceding case.

For the pointX^∗introduced in case 1, we cannot have, for alle_i, (1−t)

Z

Ω

gB·TAB(ei)u1dx+t Z

B^∗

gB·TAB(ei) = 0, since the range ofT_AB is of dimensionN.

At that stage, we have proved that the topological degrees of F₀ and F₁ coincide:

d(F₀,[−M, M]^2N\V,0)=d(F₁,[−M, M]^2N\V,0). We are now going to consider a second continuous deformation which will further simplify the functional.

Second deformation. Let B^] be the ball obtained by symmetry ofB^∗ with respect to the origin. We define the continuous deformationG: [0,1]×R^2N!R^2N by

Gt(A, B) =

(1−t) Z

Ω

g^AB·eidx+t Z

B^]

g^AB·eidx, Z

B^∗

g^AB·eidx

.

Similarly to case 1, we do not vary the lastN coordinates ofGt, which are of the same nature as the firstN coordinates ofFt. Consequently, ifGt(A, B)=0 for sometand one of the point (A, B)∈∂([−M, M]^2N\V), then the other one has to be the center of B^∗, which is the only point which satisfiesR

B^∗g^AB·vdx=0 for allv. Note that we possibly decrease the valueδin the computation ofV, such thatV is also suitable forB^∗. Taking a unit vectorvparallel with the lineX^∗O, one can notice that

(1−t) Z

Ω

g_X^∗·vdx+t Z

B^]

g_X^∗·vdx

cannot vanish, since both integrals have the same sign.

AsG₀=F₁, we can glue the two continuous deformations and notice that they do not vanish on∂([−M, M]^2N\V), so in view of [9, Theorem 1] they have the same topological degree:

d(F,[−M, M]^2N\V,0) =d(F₁,[−M, M]^2N\V,0) =d(G₁,[−M, M]^2N\V,0).

We shall compute in the sequel the topological degree ofG1and we shall prove that it equals 2. As a consequence,F has at least one zero, so we conclude the proof.

Computation of the topological degree ofG₁ at zero. There are two steps.

Step 1. The zeros of the function G₁. We may assume, without loosing generality, that the center ofB^∗ isX^∗=(3√

2,0, ...,0) and the center ofB^] isX^]=(−3√

2,0, ...,0).

We claim that the only zeros of the functionG₁are the couples (X^∗, X^]) and (X], X^∗).

Assume thatAandB are such thatG₁(A, B)=0. Then, Z

B^]

g^AB·vdx= Z

B^∗

g^AB·vdx= 0 for allv∈R^N.

Assume for contradiction that X^∗∈AB. Denoting by/ X⁰ the projection of X^∗ on the lineAB, and taking v=−−−→

X⁰X^∗, one has Z

B^∗

g^AB·vdx6= 0,

as a consequence of the structure of the function g^AB and the symmetry of the ball.

Indeed, the functiong^AB·vis odd with respect to the hyperplane containing the lineAB and orthogonal tov, and has constant sign on each half-space defined by this hyperplane.

As this hyperplane does not cut the ball into two half-balls, the integralR

B^∗g^AB·vdx cannot vanish. Consequently,X^∗∈AB, and similarly X^]∈AB.

LetxA andxB be the abscissas of the pointsAand B, and letxM=¹₂(xA+xB) be the abscissa of the midpointM. We discuss with respect to the possible values ofxM.

• If xM∈[−3√

2+R,3√

2−R], then each ball is completely contained in one of the two half-spaces, consequently, using the uniqueness given by Lemma6, we get that the two points have to coincide with the two centres of the balls.

• Let us prove that it is not possible that x_M∈(−inf,−3√

2+R)∪(3√

2−R,∞).

Indeed, in that case, the two balls would be in the same half-space, and the uniqueness result of Lemma6would imply that the same pointAorB should be the center of each ball.

• At last if x_M∈[−3√

2−R,−3√

2+R]∪[3√

2−R,3√

2+R], one of the two balls is completely contained in the half-spaceH_AorH_Bwhich fixes its center atAorB. Taking now v=−−→

AB, and since the other ball is between A and B, we see that the function g^AB·−−→

ABhas a constant sign on this ball which prevents the integral to be zero and make this case impossible.

Thus, we are always in the first case: this gives the conclusion.

Step 2. Computation of the sign of the Jacobian of G1 at its zeros. The partial derivatives of the functionG1, as function ofAandB, can be computed explicitly.

We have the following general formula. Let h: [0,∞)!R^∗+ be of classC¹ andBO,R

be the ball centered atO of radiusR. For everyi=1, ..., N, we denote f_i^O,R(A) =

Z

B_O,R

h(dA(Y))(yi−xi)dy,

whereA=(xi)i andY=(yi)i. Then, we compute the derivatives at the center of the ball A=O:

∂f_i^O,R

∂xj

_A=O

= Z

B_O,R

h⁰(dO(Y))(xj−yj)(yi−xi)

dO(Y) +h(dO(Y))(−δij)

dy.

Fori6=j, we get

∂f_i^O,R

∂xj

(O) = 0.

Fori=j, we get

∂f_i^O,R

∂x_{i A=O}

=− Z

BO,R

h⁰(dO(Y))(yi−xi)²

d_O(Y) +h(dO(Y))dy

=− Z

BO,R

∂

∂y_i[h(d_O(Y))(y_i−x_i)]dy

=− Z

∂BR

h(dO(Y))(yi−xi)nidσY

=−Rh(R) Z

∂B_O,R

n²_idσ_Y

=−Rh(R)P(B(0, R)) N

=−ωNR^Nh(R).

In a similar manner, for a fixed point A=(x^A_i), and for variable points B=(xi)i, we consider the generic function

f_i^A,O,R(B) = Z

B_O,R

h(d_B(Y))(x_i−x^A_i) −−→ AB·−−→

BY kABk² dy, We assume thatAand O are both on the first axis, so that−→

AO=βe1, for some β∈R^∗. DenotingOε=(x^O₁+ε, x^O₂, ..., x^O_N), we have

∂f₁^A,O,R

∂x1

_B=O

= d dε

Z

BO,R

h(dO_ε(Y))(β+ε)((β+ε)e1·−−→

OεY)

(β+ε)² dy

_ε=0

=−R N

Z

∂B_O,R

h(dO(Y))dσY. Fori>2, we have−→

AO_ε=−→

AO+εe_i and, plugging in the definition off₁, we get

∂f₁^A,O,R

∂xi

_B=O

= Z

B_O,R

h⁰(d_O(Y))(−y_i)

dO(Y)(y₁−x^O₁)dy+

Z

B_O,R

h(d_O(Y))y_i β dy= 0.

In order to compute (∂f₂^A,O,R/∂x1)|B=O, we consider the perturbation Oε= (x^O₁+ε, x^O₂, ..., x^O_N),

and notice that

f₂^A,O,R(Oε) = 0.

In order to compute (∂f₂^A,O,R/∂x₂)|_B=O, we consider the perturbation O_ε= (x^O₁, x^O₂+ε, ..., x^O_N),

and notice that

f₂^A,O,R(Oε) = Z

BO,R

h(dO_ε(Y))ε(β(y1−x^O₁)+ε(y2−ε)) β²+ε² , and

∂f₂^A,O,R

∂x_{2 B=O}

= Z

BO,R

h⁰(dO(Y)) −y2

d_O(Y)·0dy+ Z

BO,R

h(dO(Y))y1−x^O₁ β dy= 0.

In order to compute (∂f₂^A,O,R/∂x_i)|B=O,i>3, we consider the perturbation

−−→AO_ε=βe₁+εe_i,

and notice that

f₂^A,O,R(Oε) = 0.

For the computation of the Jacobian ofG₁at the points (X^], X^∗) and (X^∗, X^]), we recall that

G1(A, B) = Z

B^]

g^AB·eidx, Z

B^∗

g^AB·eidx

, i= 1, ..., N.

Around the zero (X^], X^∗), the expression ofG₁is Z

B^]

g_A·e_idx, Z

B^∗

g_B·e_idx−2g_B·−−→ AB

−−→ AB·e_i kABk²

, i= 1, ..., N,

or, in terms of our notation,

G₁(A, B) = (f_i^X],R(A), f_i^A,X∗,R(B)), i= 1, ..., N,

withh(r)=GR(r)/r. Then, the Jacobian matrix at (X^], X^∗) is diagonal, with all elements on the diagonal equal to−ωNR^Nh(R), except the one on position (N+1, N+1), which equalsωNR^Nh(R). Its determinant equals

−(ωNR^Nh(R))^2N, which is a negative number.

The same value is obtained at the point (X^∗, X^]), as the sign ofβ does not influence the value of the derivatives.

As conclusion, the topological degree of F at zero is equal to 2, which leads to the existence of (at least) two solutions ofF(A, B)=0 in [−M, M]^2N\V.

3. Proof of Theorem 3

In this section we shall prove Theorem3. We start with the following observation. In (1), one can replace C_c^∞(R^N) with W^1,∞(R^N). Indeed, for a function u∈W^1,∞(R^N) such thatR

R^N%u dx=0, both termsR

R^N%u²dxandR

R^N%|∇u|²dxare well defined. Moreover, there exists a sequence of functionsϕn∈C_c^∞(R^N) such that

Z

R^N

%ϕ_ndx= 0 and lim

n!∞

Z

R^N

%(|∇ϕ_n−∇u|²+|ϕ_n−u|²)dx= 0.

The construction is standard by cut-off and convolution, one has to be careful only to the orthogonality on %. Let ϕ∈C_c^∞(R^N,[0,1]) be such that ϕ=1 on B1 and ϕ=0 on R^N\B2. We introduce, for everyδ >0,ϕδ(x):=ϕ(δx) and the constantcδ such that

Z

R^N

%ϕδ(u−cδ)dx= 0.

We observe that

cδ

Z

R^N

%ϕδdx= Z

R^N

%ϕδu dx, and forδ!0 we getcδ!0. This is a consequence of

lim

δ!0

Z

R^N

%(|∇(ϕδu)−∇u|²+|ϕδu−u|²)dx= 0.

Now, for fixedδ >0, we consider a convolution kernel (ξε)εand a constant cδ,ε such that Z

R^N

%ξ_ε∗(ϕδ(u−cδ,ε))dx= 0.

On the one hand

cδ,ε

Z

R^N

%ξε∗ϕδdx= Z

R^N

%ξε∗(ϕδu)dx, and hence, forε!0, we getcδ,ε!cδ. On the other hand,

limε!0

Z

R^N

%(|∇(ξε∗(ϕδu))−∇(ϕδu)|²+|ξε∗(ϕδu)−ϕδu|²)dx= 0, which concludes the proof, by a diagonal argument.

The case k=1 (extension of the Szeg¨o–Weinberger result). Since inequality (2) that we want to prove is scale invariant, we may assume thatR

R^N% dx=1. Letr₁be the radius of the ball of volume equal to 1.

If %has bounded support, the proof follows step by step the geometric case. The existence of a pointAsuch that

Z

R^N

%gA(x)·eidx= 0 for alli= 1, ..., N (12)

is done using the same fixed point argument used by Weinberger (see [12, Lemma 6.2.2]).

The functionG=Gr₁, which enters in the definition ofgAabove, is associated tor1. Then, the proof follows step by step, the final argument being the displacement of the mass of%towards 1B_r1.

If %has unbounded support, the existence of a point Asatisfying (12) can be done by approximation. Note that, for everyi, the functiongA(x)·ei belongs toW^1,∞(R^N).

LetRn!∞ and consider a pointAn satisfying the orthogonality relations (12) for the density%1B_Rn (which has bounded support) and for theg-functions defined withr1. If, for a subsequence, (An)n remains bounded, then by compactness we find a limitAsuch thatAn!A. It can be easily observed that the orthogonality relations (12) pass to the limit, sincekgAn·eik∞6G(r₁). Hence,A satisfies (12) for%.

Assume for contradiction thatd_O(A_n)!∞. We fix a radius Rsuch that Z

B_R

% dx=2 3. Fornlarge enough such thatr_n>R, we denote

vn= 1

−−→AnO

−−→AnO.

By the choice ofAn, we have Z

B_Rn

gA_n·vndx= 0, which gets in contradiction with

nlim!∞

Z

B_R

g_An·v_ndx=2

3G(r₁) and Z

B_Rn\B_R

|g_An·v_n|dx61 3G(r₁).

Hence, (An)n remains bounded and we can build the functions of (12). The proof ends using a mass displacement argument, pushing forward the measure% dxon 1B_r1dx.

The case k=2. Assume thatR

R^N% dx=1 and that ˜µ₂(%)>µ^∗₂. Let us denote byr_1/2 the radius of the ball of volume ¹₂.

There are two difficulties. Along with the fact that the support of % may be un- bounded, there is a new difficulty: there is not necessarily existence of aneigenfunction associated with ˜µ₁(%), by eigenfunction understanding a function for which the infi- mum is attained in the definition of ˜µ₁(%). The orthogonality on the first eigenfunction, both in L² and H¹, was an important point of the proof in the geometric case. In- deed, in the Rayleigh quotient estimating the second eigenfunction, the scalar product R

R^N%∇u₁∇g_idxwas not present, being equal to zero.

Let us fixε>0 and consider u1∈W^1,∞(R^N) such thatR

R^N%u1dx=0,R

R^N%u²₁dx=1 and

˜ µ1(%)6

Z

R^N

%|∇u1|²dx <µ˜1(%)+ε. (13) Let us prove the existence of two pointsA6=B (one of them being possibly at infinite distance from the origin) such that

Z

R^N

%g^AB·eidx= Z

R^N

%g^AB·eiu1dx= 0 for alli= 1, ..., N. (14) Above, we abuse the notation g^AB even if one of the points A and B is formally at infinite distance from the origin. The exact meaning is given below.

LetR_n!∞. We apply step by step the method of§2to the functions 1B_Rn% and 1B_Rn%u1,

and find a couple of points (An, Bn) such that Z

B_Rn

%g^AnBn·eidx= Z

B_Rn

%g^AnBn·eiu₁dx= 0 for alli= 1, ..., N. (15) If both sequences (An)n and (Bn)n stay bounded, we may assume (up to extracting a subsequence) that An!Aand Bn!B. If A6=B, then all equalities in (15) pass to the limit to (14). IfA=B, then, taking a further subsequence such that

−−−→AnBn

kAnB_nk!v∈S^N−1, we would get in the limit that

Z

R^N

%g^AA·vdx= 0,

whereg^AAis the pointwise limit of the sequenceg^AnBn. This is not possible, sinceg^AA·v is a negative function.

If (An)n stays bounded, and dO(Bn)!∞, we may assume that An!Aand obtain that the limit ofg^AnBn equalsgA:=g^A∞. Then, the functions (gA·ei)i satisfy (14). A similar assertion holds ifBn!B, dO(An)!∞and

−−−→AnBn

kA_nB_nk!v∈S^N−1, in which case the limit is described by

g_B·e_i−2g_B·v(v·e_i) :=g^∞B·e_i, i= 1, ..., N

satisfy the orthogonality relations (14).

We now prove that both sequences (An)nand (Bn)n cannot go unbounded simulta- neously, since the orthogonality on constants (in relations (15)) would be contradicted.

Indeed, denote by On the projection of O on the line AnBn. FixR large enough such

that Z

B_R

% dx=3 4.

We may assume (possibly exchanging the notation and extracting further subsequences) thatkA_nO_nk6kB_nO_nk.

If, for an infinite number of indices, we have O\_nOA_n6π

4, then, takingvn=−−→

OnO/kOnOk, we get

lim inf

n!∞

Z

B_R

%g^AnBn·vndx> 1

√2 3

4G(r_1/2)>1

2G(r_1/2), contradicting the hypotheses (15), as it is not possible that R

B_Rn%g^AnBn·v_ndx=0.

If for an infinite number of indices we have O\_nOA_n>π

4 ⇐⇒ OA\_nO_n6π 4, we takevn=−−−→

AnBn/kAnBnkand arrive to the same conclusion.

We finally conclude with the validity of (14). By abuse of notation, we continue to denote the set of such functions (g^AB·ei)i, even though one of the points is at ∞ (i.e.

the functions g^A∞·ei=gA·ei and g^∞B·ei=gB·ei−2gB·v(v·ei)). Let us denote by G the set of (gi)^N_i=1such that there existAandB so that

gi= g^AB·ei

(R

R^N%(g^AB·e_i)²dx)^1/2, and

Z

R^N

%g^AB·e_idx= 0 for alli= 1, ..., N.

The familyG is not empty, and moreover there exist at least one package ofN functions (gi)i such thatR

R^N%giu1=0, as we have proved above. We have, in particular, Z

R^N

%g²_idx= 1 for alli= 1, ..., N.