EXISTENCE AND REGULARITY OF SOLUTIONS TO OPTIMAL ...

EXISTENCE AND REGULARITY OF SOLUTIONS TO OPTIMAL PARTITION PROBLEMS INVOLVING LAPLACIAN EIGENVALUES

arXiv:1403.6313v1 [math.AP] 25 Mar 2014

MIGUEL RAMOS, HUGO TAVARES, AND SUSANNA TERRACINI Abstract. Let Ω ⊂ RN be an open bounded domain and m ∈ N. Given k1 , . . . , km ∈ N, we consider a wide class of optimal partition problems involving Dirichlet eigenvalues of elliptic operators, including the following ( ) m X inf Φ(ω1 , . . . , ωm ) := λki (ωi ) : (ω1 , . . . , ωm ) ∈ Pm (Ω) , i=1

where λki (ωi ) denotes the ki –th eigenvalue of (−∆, H01 (ωi )) counting multiplicities, and Pm (Ω) is the set of all open partitions of Ω, namely Pm (Ω) = {(ω1 , . . . , ωm ) : ωi ⊂ Ω open, ωi ∩ ωj = ∅ ∀i 6= j} .

We prove the existence of an open optimal partition (ω1 , . . . , ωm ) ∈ Pm (Ω), proving as well its regularity in the sense that the free boundary ∪m i=1 ∂ωi ∩ Ω is, up to a residual set, locally a C 1,α hypersurface. In order to prove this result, we first treat some general optimal partition problems involving all eigenvalues up to a certain order. This class of problems includes the one with 1/p Pm Pki p , whose solutions approach cost function Φp (ω1 , . . . , ωm ) := i=1 j=1 (λj (ωi )) the solutions of the original problem as p → +∞. The study of this new class of problems is done via a singular perturbation approach with a class of Schr¨ odinger-type systems which models competition between different groups of possibly sign-changing components. An optimal partition appears in relation with the nodal set of the limiting components, as the competition parameter becomes large. The proofs also involve new boundary Harnack principles on NTA and Reifenberg flat domains, as well as an extensive use of Almgren– type monotonicity formulas.

Contents 1. Introduction 2. A general existence result. An approximated problem depending on p 2.1. Preliminaries 2.2. Uniform bounds in H¨ older spaces. Proof of the general existence theorem 2.3. Almgren’s Monotonicity Formula 3. Regularity of the nodal set of the approximated solution 3.1. Compactness of blowup sequences

2 9 9 15 22 30 31

Date: March 26, 2014. Key words and phrases. elliptic competitive systems, optimal partition problems, Laplacian eigenvalues, segregation phenomena, extremality conditions, regularity theory on NTA and Reifenberg flat domains, regularity of free boundary problems, blowup techniques, Almgren’s Monotonicity Formula, Boundary Harnack Principle. Acknowledgments. H. Tavares was supported by Funda¸ca õ para a Ciˆ encia e Tecnologia through the program Investigador FCT. H. Tavares and S. Terracini are partially supported through the project ERC Advanced Grant 2013 n. 339958 “Complex Patterns for Strongly Interacting Dynamical Systems - COMPAT”. 1

2

M. RAMOS, H. TAVARES, AND S. TERRACINI

3.2. Hausdorff dimension estimates for the nodal and singular sets. Definition of R(u,v) , the regular part of the nodal set 3.3. A Clean-Up result. R(u,v) is locally flat 3.4. The set R(u,v) is locally a regular hypersurface 4. The limit of the approximating solutions as p → ∞. Existence and regularity results for the initial problem 4.1. Properties of the approximating solutions up , vp . Proof of Theorems 1.3 and 1.5 4.2. Proof of property (H1): Strong convergence of the last components of up , vp 4.3. Proof of properties (H2) and (H3): Regularity of the last components of u˜, v˜, and of the partition (e ω1 , ω e2 ) Appendix A. Boundary behavior on NTA and Reifenberg flat domains A.1. Results on NTA domains A.2. Results on Reifenberg flat domains References

38 41 47 54 54 57 60 63 63 64 70

1. Introduction Let Ω be an open bounded domain of RN (N > 2) and take m ∈ N and k1 , . . . , km ∈ N. The starting point and first aim of this paper is to prove existence and regularity of solutions to problems of type: (1.1) where

inf

(ω1 ,...,ωm )∈Pm (Ω)

m X

λki (ωi ),

i=1

Pm (Ω) = {(ω1 , . . . , ωm ) ⊆ Ωm : ωi open ∀i, ωi ∩ ωj = ∅ ∀i 6= j}

denotes the class of all partitions of Ω with m open sets, and λk (ω) is the k–th eigenvalue of −∆ in H01 (ω), counting multiplicities. Observe that (1.1) is an example of an optimal partition problem (see for instance [6, Section 5.5] or [7]) with cost function Φ(ω1 , . . . , ωm ) := P m i=1 λki (ωi ), and class of admissible sets given by A(Ω) := {ω ⊆ Ω : ω open set}. Optimal partition problems belong to the wide family of shape optimization problems, and have many applications in physics and engineering. They also play a fundamental role in the study of the nodal sets of eigenfunctions of Schrödinger operators [21], as well as in the proof of monotonicity formulas [1, 3, 4, 20, 22, 23, 30, 33]. In general, these kind of problems may only have solution in a relaxed sense [8, 9], except when one imposes certain geometric constraints on the admissible domains, or some monotonicity properties on the cost function (we refer the reader to the book by Bucur and Buttazzo [6] for a good survey on these issues). Problem (1.1) falls into the second case: as Φ is monotone decreasing with respect to set inclusion, as well as lower semi continuous for the γ–convergence, the general abstract result of [7] implies the existence of solution for (1.1) in the class of quasi-open sets. In general, it is hard to pass from quasi-open to open sets. By using a penalization technique with partition of unity functions, Bourdin, Bucur and Oudet [5] give a different proof for the existence of quasi-open solutions, while proving the existence of open solutions for the bidimensional case ˇ ak [31] which only holds in dimension two). N = 2 (by using a compactness result by Sver´ Much more is known for k1 = . . . = km = 1 or, more generally, when 1 6 ki 6 2 for all i, but we will leave the discussion of these particular cases for later.

OPTIMAL PARTITION PROBLEMS INVOLVING LAPLACIAN EIGENVALUES

3

In this paper we are concerned not only with existence results in the class of open sets, but also with the regularity of the optimal partition, or, more specifically, of the common boundaries of the elements of the partition itself. Let us state rigorously what we intend by regularity. Definition 1.1. An open partition (ω1 , . . . , ωm ) ∈ Pm (Ω) is called regular if: Sm 1. denoting Γ = Ω \ i=1 ωi , there holds1 Hdim (Γ) 6 N − 1; 2. there exists a set R ⊆ Γ, relatively open in Γ, such that - Hdim (Γ \ R) 6 N − 2; - R is a collection of hypersurfaces of class C 1,α (for some 0 < α < 1), each one separating two different elements of the partition. Moreover we ask that, if N = 2, then actually Γ \ R is a locally finite set, and that R consists of a locally finite collection of curves meeting at singular points with equal angles. Remark 1.2. In particular, observe that a regular partition exhausts the whole domain, as it satisfies Ω = ∪m i=1 ωi . A key point in the proof of the regularity will be to derive proper extremality conditions on Γ for the eigenfunctions associated with the optimal partition. Observe that for (1.1) this fact is difficult and not obvious at all, as no domain variation formula is available in the unfortunate (yet possible) case of degenerate eigenvalues (observe that we will take any prescribed k1 , . . . , km ∈ N). This fact, together with the initial lack of regularity of the interfaces, makes Hadamard type conditions unavailable for this problem. Our first main result deals with existence and regularity of solutions to a wide class of optimal partition problems which includes (1.1), providing as well regularity and extremality conditions for the corresponding eigenfunctions. Take a function F : Rm → R such that ∂F (F1) F ∈ C 1 ((R+ )m ), and ∂x > 0 in (R+ )m for every i = 1, . . . , m; i (F2) Given i, and for every fixed x¯j > 0, j 6= i, we have Consider (1.2)

F (¯ x1 , . . . , x ¯i−1 , xi , x ¯i+1 . . . , x ¯m ) → +∞ inf

(ω1 ,...,ωm )∈Pm (Ω)

as xi → +∞.

F (λk1 (ω1 ), . . . , λkm (ωm )).

Theorem 1.3. The optimal partition problem (1.2) admits a regular solution (e ω1 , . . . , ω em ) ∈ Pm (Ω). Moreover, let (˜ uin )n be the sequence of eigenfunction of ω ei , ordered according to the increasing sequence of eigenvalues. Then for each i = 1, . . . , m there exists 1 6 li 6 ki such that are Lipschitz continuous, ˜iki u˜ili , u˜ili +1 , . . . , u and   ki  X   ω ei = int  (˜ uij )2 > 0  .   j=li

Finally, on the regular part of the common boundary, we have the following extremality condition: for each i = 1, . . . , m there exist coefficients a ˜ili , . . . , a ˜iki > 0 such that 1Here, H dim (·) denotes the Hausdorff dimension of a set.

4


given x0 ∈ R, denoting by ω ei and ω ej the two adjacent sets of the partition at x0 , lim

ki X

x→x0 x∈e ωi n=li

kj X

a ˜in |∇˜ uin (x)|2 = lim

x→x0 x∈e ωj n=lj

a ˜jn |∇˜ ujn (x)|2 6= 0.

Remark 1.4. Both in this theorem and in the following main results, the regularity of the free boundary can be extended up to the boundary ∂Ω under appropriate regularity assumptions on the set Ω. We refer the reader to [33, Section 7] for the details. The coefficients ain appear as the result of a double penalization/approximation of the Pm problem. To simplify the presentation, we deal from now on with F (x1 , . . . , xm ) = i=1 xi , hence with (1.1) - the general case follows exactly in the same way, as the proofs only use properties (F1) and (F2); check also Remark 4.2 ahead. In order to motivate our approach, let us review first the cases ki = 1, or ki = 2 for every i. The study of (1.1) in the case k1 = . . . = km = 1 is the simplest, as it can be characterized by an absolute minimization of an energy functional in a singular space, namely (m Z ) Z X 2 2 1 (1.3) inf ui dx = 1, ui · uj ≡ 0 ∀i 6= j . |∇ui | dx : ui ∈ H0 (Ω), i=1

Ω

Ω

This minimization problem was the main object of study in [11, 16], where it is proved the existence of a positive Lipschitz continuous solution (u1 , . . . , um ) to (1.3), and that Pmthe partition ({u1 > 0}, . . . , {um > 0}) is a regular element of Pm (Ω), achieving inf Pm (Ω) i=1 λ1 (ωi ). The existence result was first proved in [16] together with the regularity for N = 2. The regularity result in any space dimension was then stated in [11], see also [33, Section 8] for a detailed proof. We would like to stress that the proofs in these papers cannot be applied to the case of higher eigenvalues. We also refer to the papers [14, 15], where related optimal partition problems involving sums of first nonlinear eigenvalues are considered. Another natural approach to the problem is to consider limiting profiles of solutions to the singularly perturbed problem m X ui u2j , ui ∈ H01 (Ω), i = 1, . . . , m, (1.4) −∆ui = λi ui − β j=1 j6=i

under the constraints

Z

Ω

u2i dx = 1,

i = 1, . . . , m.

Theses systems have been the object of an intensive study in the last decade, in particular in the case of competitive interaction β > 0, mainly because of its interesting mathematical features, as well as for its physical applications (for instance in the study of Bose-Einstein condensation). Their relation with optimal partition problems has also been addressed, for instance in [2, 11, 13, 16, 21, 34]. It has been shown in [29, 34] that, in some situations, phase separation occurs between different components as the competition parameter increases, i.e., β → ∞. In particular it is shown that by taking an L∞ bounded family of solution (uβ )β , and corresponding bounded coefficients (λiβ )β , then there exists a limiting profile ui := limβ→+∞ uiβ such that ({u1 6= 0}, . . . , {um 6= 0}) ∈ Pm (Ω), and −∆˜ ui = λi u˜i

in {ui 6= 0}.


5

This clearly illustrates the relation between optimal partitions involving eigenvalues and the system of Schrödinger equations (1.4). In particular, it is known that (1.3) can be well approximated (as β → ∞) by the ground state (least energy) levels of (1.4), namely:   Z m  Z X X u2i dx = 1 . u2i u2j : ui ∈ H01 (Ω) and inf |∇ui |2 + β   Ω Ω i 0 = int  (uijp )2 > 0  .     

j=1

On the regular part of the common boundary, we have the following extremality condition: for each i = 1, . . . , m there exist coefficients ai1p , . . . , aiki p > 0 such that given x0 ∈ R, denoting by ωip and ωjp the two adjacent sets of the partition at x0 , lim

ki X

x→x0 x∈ωip n=1

ainp |∇uinp (x)|2 = lim

kj X

ajnp |∇ujnp (x)|2 x→x0 x∈ωjp n=1

Finally, as p → +∞,

ainp → 0 ∀n = 1, . . . , li − 1,

6= 0

ainp → a ˜in ∀n = li , . . . , ki ,

and we have strong convergence of the last ki − li + 1 eigenfunctions: uinp → u˜in strongly in H01 (Ω) ∩ C 0,γ (Ω),

∀0 < γ < 1, ∀n = li , . . . , ki

(where a ˜in and u˜in are the coefficients and functions appearing in Theorem 1.3. The existence and regularity results of the previous theorem will come out as a particular case of a more general result, for optimal partition problems that depend on all eigenvalues k up to a certain order. More precisely, take k ∈ N and a function ψ : (R+ 0 ) → R satisfying (F1), (F2), that is ∂ψ (ξ1 , . . . , ξk ) > 0 for every i, whenever ξ1 , . . . , ξk > 0; - ∂ξ i - Given i, and for every fixed ξ¯j > 0, j 6= i, we have

ψ(ξ¯1 , . . . , ξ¯i−1 , ξi , ξ¯i+1 . . . , ξ¯m ) → +∞

as ξi → +∞.

Now, given m functions ψ1 , . . . , ψm satisfying these assumptions (eventually for different k1 , . . . , km ), we will consider the following class of optimal partition problems (1.6)

inf

(ω1 ,...,ωm )∈Pm (Ω)

m X

ψi (λ1 (ωi ), . . . , , λki (ωi )),

i=1

Pm Qki As examples we have (1.5), the optimization with cost function i=1 j=1 λj (ωi ), or any combination of the two, among many other. We will prove that (1.6) can be reformulated as a constrained minimization problem of an energy functional: (1.7)   ui ∈ H01 (Ω; Rki ), uik · ujl ≡ 0 ∀k, l, i 6= j       Z  m Z   X Z i i i 2 i 2 uk ul dx = δkl ∀k, l ψi |∇u1 | dx, . . . , |∇uki | dx : and, for every i, inf   Ω Z Ω Ω   i=1       ∇uik · ∇uil dx = 0 ∀k 6= l Ω

and that this, in turn, can be approximated by a solution to a competitive system of k1 + . . . + km equations, where competition occurs between groups of components. More precisely,


take the penalized energy functional Eβ : Rk1 × . . . × Rkm defined as Z Z m X ψi |∇ui1 |2 dx, . . . , |∇uiki |2 dx Eβ (u1 , . . . , um ) = Ω

i=1

+

X 2δij β i 0, i = 1, . . . , m, j, l = 1, . . . , ki , it is a solution of the system (1.9)

−ail,β ∆uil =

ki X j=1

m 2q −1 X q2 µijl uil − βuil (ui1 )2 + . . . + (uiki )2 (uj1 )2 + . . . + (ujkj )2 , j6=i j=1

with

Z Z ∂ψi |∇uiki ,β |2 dx > 0. |∇ui1,β |2 dx, . . . , ∂ξl Ω Ω (ii) There exists (¯ u1 , . . . , u ¯m ), with Lipschitz continuous components, such that ail,β =

uiβ → u ¯i

and (1.10)

strongly in H01 ∩ C 0,γ (Ω; Rki ) ∀0 < γ < 1,

−ail ∆uil = µil u ¯il

in

 ki X 

j=1

as β → ∞;

  ki  X   (¯ uij )2 > 0 = int  (¯ uij )2 > 0  .     

j=1

for ail = limβ→∞ ail,β > 0 and µil = limβ→∞ µill,β > 0. (iii) The vector (¯ u1 , . . . , u¯m ) achieves (1.7), and such level is the limit of cβ as β → ∞; (iv) The partition     km k1   X X 2  ∈ Pm (Ω) (¯ um (¯ u1j )2 > 0 , . . . , (¯ ω1 , . . . , ω ¯ m ) :=  j ) >0     j=1

j=1

is regular, and solves the optimal partition problem (1.5). (v) For ail = limβ→∞ ailβ , we have that, whenever x0 belongs to the regular part of the common nodal set of (¯ u1 , . . . , u ¯m ), if ω ¯i, ω ¯ j are the corresponding adjacent sets, then

(1.11)

lim

ki X

x→x0 x∈¯ ωi n=1

ain |∇uin (x)|2 = lim

kj X

ajn |∇ujn (x)|2 x→x0 x∈¯ ωj n=1

6= 0

8


Observe that the existence of minimizersR in (1.7) and (1.8) is not a trivial fact (unlike in (1.3)) due to restriction on the gradients Ω ∇uik · ∇uil dx = 0, k 6= l. We will be able to remove this condition by extending the ψi ’s to symmetri functions defined on the space of self-adjoint matrices. The coefficients ail , which give the extremality conditions, are related to the chosen functions ψi , thus are related to the penalization. We recall that for the choice 

ψi,p (ξ1 , . . . , ξki ) := 

ki ψip : (R+ → R, 0)

ki X j=1

1/p

ξjp 

we are dealing with (1.5) and thus, with an approximation of (1.1). It turns out that the coefficients in the extremality condition of Theorem 1.3 come out from the double approximation as being a limit as β → +∞, followed from a limit as p → +∞, of the coefficients provided in the previous theorem. The asymptotic study of systems of type (1.9), that is, competitive systems which lead to phase segregation between groups of components, was done by Caffarelli Lin [12] in a simpler situation. There, all components are taken to be nonnegative, µijl = 0, and the problem corresponds to the minimization of an associated free energy. The fact that our components may change sign, as well as the fact that at the limit we get an eigenvalue problem (1.10) with (possibly) different parameters, instead of an harmonic equation, makes the study of the free boundary much more delicate. In particular, we will prove an Harnack Boundary Principle for quotients of two components of each vector ui which is not contained in the celebrated paper by Jerison and Kenig [25]. Moreover, our proof of Theorem 1.6 will make it clear that the its conclusions (i)-(ii)-(iii) also hold for any (u1β , . . . , um β )

L∞ –bounded solution,

with 0 < A 6 ail,β 6 B,

|µijl,β | 6 C.

The minimality of the energy is only used explicitly in Proposition 2.13 ahead, which only provides a more precise understanding of the optimal partition, and it is not necessary in order to obtain the convergence and regularity results stated in (i)-(ii)-(iii). Finally, let us now come to other possible applications of our strategy and future directions of research. Recall that in [21] spectral minimal partitions like (1.12)

max {λ1 (ωi )}

inf

(ω1 ,...,ωm )∈Pm (Ω) i=1,...,m

were introduced (see the survey [20] for a description on the subject). Our approach can be applied to the following generalization of (1.12) inf

max {λki (ωi )},

(ω1 ,...,ωm )∈Pm (Ω) i=1,...,m

for k1 , . . . , km ∈ N,

(which is not contained in the assumptions of Theorem 1.3) by taking the following approximating problem (for p and q large):  p/q 1/p  ki m X X   (λj (ωi ))q   inf  .  (ω1 ,...,ωm )∈Pm (Ω)

i=1

j=1


9

This, in turn, can also be well approximated by constrained minimal energy solutions of a competitive Schrödinger system. Another interesting question, that we wish to tackle in the near future, is to check if, in (1.1), every optimal partition has a regular representative (in the sense of a class of equivalence involving the capacity of sets). This was proved in [21] for (1.12), and it is our belief that the strategy followed there can be sucessfully combined with our approach for (1.1). The paper is organized as follows. - We start to deal with the general problem (1.6). In Section 2 we prove existence of an open partition for it, introducing system (1.9) and the minimization problem cβ (Subsections 2.1), proving afterwards uniform H¨ older bounds for the solutions of the system, as well as qualitative properties for the limiting configuration such as phase separation and Lispchitz regularity. The proof that (¯ ω1 , . . . , ω ¯ m ) ∈ Pm (Ω) solves (1.6) is concluded in Subsection 2.2, while in Subsection 2.3 we give another characterization of the optimal sets, and check the validity of an Almgren’s monotonicity formula. - Almgren’s formula will be the key to Section 3, where we end the proof of Theorem 1.6 proving that the optimal partition is regular. Here, although with different proofs, we follow the structure of [12, 33], using the Almgren’s quotient to characterize the regular and singular part of the free boundary (Subsection 3.2) after a careful study of blowup sequences (Subsection 3.1), and then in the last two sections we first shown the extremality condition (1.11) and construct explicitly the normal vector to the surface, which then leads to a non degeneracy property. - As already said, Theorem 1.6 yields in particular existence and regularity of an optimal partition for (1.5), giving also some properties for the associated eigenfunctions: Theorem 1.5. Section 4 is then dedicated to the study of the limit p → ∞, which yields our first main result, Theorem 1.3. - Finally in Appendix A we prove generalizations of some of the results of [25], namely Boundary Harnack Principles for the quotient between two functions satisfying two different problems, with one of the functions being possibly sign-changing. Remark 1.7. To simplify the notation, we will present the proofs of the theorems for the simpler case k1 = . . . , km =: k, and deal with partitions with m = 2 components. It will become clear that the general case follows exactly in the same way, without any extra work. R Notation. We will denote the H01 –norm simply the k · k, namely kuk2 := Ω |∇u|2 dx. 2. A general existence result. An approximated problem depending on p

2.1. Preliminaries. As previously mentioned, to simplify the presentation we will work from now on with m = 2 and k1 = k2 =: k ∈ N. Define Sk (R) and Ok (R) as the spaces of symmetric and orthogonal k × k matrices with real entries, respectively. In order to deal with (1.5) and (1.6), we will work with a special class of functions with domain in Sk (R). Definition 2.1. We say that ϕ ∈ F if ϕ : Sk (R) → R is C 1 in Sk (R) \ {0} and ϕ(M ) = ϕ(P T M P )

for all M ∈ Sk (R) and P ∈ Ok (R).

Moreover, consider the restriction ψ of ϕ to the space of diagonal matrices, that is ψ(a1 , . . . , ak ) := ϕ(diag(a1 , . . . , ak )). Then we ask that ∂ψ > 0 on (R+ )k for every i = 1, . . . , k; - ∂a i

10


- for each i and a ¯1 , . . . , a ¯i−1 , a ¯i+1 , . . . , a ¯k > 0, we have ψ(¯ a1 , . . . , a ¯i−1 , ai , a ¯i+1 , . . . , a ¯k ) → +∞

as ai → +∞.

Remark 2.2. Along this paper, whenever we say that ϕ : Sk (R) → R is a C 1 function, ∂ϕ we are identifying Sk (R) with Rk(k+1)/2 . The partial derivative ∂ξ with i > j denotes the ij derivative of ϕ along the direction of the symmetric matrix A having all entries zero except for aij = aji = 1. Examples. Functions as in Definition 2.1 just depend on the set of the eigenvalues of a 1/p matrix. To fix our mind, let us think of the examples ϕ(M ) = (trace(M p )) for p ∈ N, or P 1/p Qk k p ϕ(M ) = det(M ). In these cases, ψ(a1 , . . . , ak ) = and ψ(a1 , . . . , ak ) = i=1 ai , i=1 ai

respectively. We will use the first of these examples to treat (1.5). The main aim of this section is to prove the following result. Theorem 2.3. Given ϕ ∈ F , the level (2.1)

c∞ : = =

inf

(ω1 ,ω2 )∈P2 (Ω)

inf

(ω1 ,ω2 )∈P2 (Ω)

is achieved.

2 X

i=1 2 X

ϕ (diag(λ1 (ωi ), . . . , λk (ωi ))) ψ(λ1 (ωi ), . . . , λk (ωi ))

i=1

Corollary 2.4. In particular, going through the previous examples, we have that the optimal partition problems 1/p  2 k X X  λj (ωi )p  , p ∈ N; inf (ω1 ,ω2 )∈P2 (Ω)

i=1

j=1

and

inf

(ω1 ,ω2 )∈P2 (Ω)

admit a solution.

k 2 Y X

i=1 j=1

λj (ωi )

We will solve problem (2.1) by approaching it with a penalized problem having an additional parameter (recall system (1.9) in the Introduction). More precisely, the proof of Theorem 2.3 in this section will imply Theorem 1.6, (i)-(ii)-(iii), as well as the existence part of (iv). The proof of the remaining statements will be complete in Section 3. At this point it is useful to recall that, if {vn }n∈N represents the sequence of eigenfunctions of (−∆, H01 (Ω)), then Z Z 2 λk (Ω) = inf ∇v · ∇vi dx = 0, i = 1, . . . , k − 1 . |∇v| : Ω

Given u ∈ (2.2)

H01 (Ω; Rk ),

Ω

define the k × k symmetric matrix Z M (u) = ∇ui · ∇uj dx Ω

i,j=1,...,k

.


11

Fix 2 < 2q < 2∗ , where 2∗ = +∞ if N 6 2, 2∗ = 2N/(N − 2) whenever N > 3. For each β > 0, we take the energy Eβ : H01 (Ω; Rk ) × H01 (Ω; Rk ) → R defined as Z q q 2β (u2 + . . . + u2k ) 2 (v12 + . . . + vk2 ) 2 dx Eβ (u, v) = ϕ(M (u)) + ϕ(M (v)) + q Ω 1 and consider the energy levels cβ := inf Eβ (u, v) : u, v ∈ Σ(L2 ) ,

where

Z o n wi wj dx = δij for every i, j Σ(L2 ) := w = (w1 , . . . , wk ) ∈ H01 (Ω; Rk ) : Ω

(δij denotes the Kronecker symbol). In order to prove that cβ is achieved, we need three auxiliary lemmas. Lemma 2.5. Let ϕ : Sk (R) → R be a map satisfying ϕ(M ) = ϕ(P T M P )


Given u ∈ Σ(L2 ), let M (u) be defined as in (2.2). Then there exists u ˜ ∈ Σ(L2 ) such that M (˜ u) = diag k˜ u1 k2 , . . . , k˜ uk k2 , Pk Pk and ϕ(M (u)) = ϕ(M (˜ u)); moreover, i=1 u2i = i=1 u ˜2i pointwise. In particular, Eβ (˜ u, v˜) = Eβ (u, v). Proof. Let P be an orthogonal matrix such P T M (u)P is diagonal. Take the new elements     u1 u˜1 k X  ..  T  .  pji uj , i = 1, . . . , k. ˜i =  .  = P  ..  ⇔ u j=1

uk

u ˜k

Then, given any bilinear form b : H01 (Ω) × H01 (Ω) → R and its associated matrix B(u) = (b(ui , uj ))ij , we have b(˜ ui , u ˜j ) = b

k X l=1

pli ul , R

k X

m=1

k X pli pmj b(ul , um ) = (P T B(u)P )ij . pmj um = l,m=1

w1 w2 dx, then as u ∈ Σ(L2 ) we have B(u) = Id and Z u ˜i u ˜j dx = (P T P )ij = δij , Ω R hence u ˜ ∈ Σ(L2 ). If one takes b(w1 , w2 ) = Ω ∇w1 · ∇w2 dx instead, then

If one takes b(w1 , w2 ) :=

Ω

(M (˜ u))ij = (P T M (u)P )ij ,

hence M (˜ u) is a diagonal matrix. As for the final statement of the lemma, we have k X

u ˜2i

=

k k X X

pli pmi ul um =

i=1 l,m=1

i=1

=

k X

l,m=1

ul um (P P T )lm =

k X

ul um

l,m=1 k X

l,m=1

k X

pli pmi

i=1

ul um δlm =

k X l=1

u2l .

12


This lemma implies that cβ has the equivalent characterization shown in the Introduction - see (1.8) - namely that it is the infimum of Eβ (u, v) for Z Z u, v ∈ Σ(L2 ) with ∇ui · ∇uj dx = ∇vi · ∇vj dx = 0 ∀i 6= j. Ω

Ω

For such (u, v), observe that

Eβ (u, v) > 2ψ(0, . . . , 0),

and so cβ is finite.

Lemma 2.6. Let ϕ : Sk (R) → R be a C 1 function such that ϕ(M ) = ϕ(P T M P )

Then

∂ϕ ∂ξij


(D) = 0 for every diagonal matrix D, and j > i.

Proof. Let us check it for instance for i = 1 6= 2 = j. such that d1 6= d2 . Consider P (t) ∈ Ok (R) given by  cos t − sin t  sin t cos t   0 P (t) =  0  .. ..  . . 0

0

First take a matrix D = diag(d1 , . . . , dk ) 0 ... 0 0 ... 0 1 ... 0 .. . . . 0 . 0 ... 1

which satisfies P (0) = Id and

   P (0) =    ′

Then



0 1 0 .. . 0

−1 0 0 0 0 0 .. .. . . 0 0

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

0 0 0

      



   .  0  ... 0 

  d  T ′ T T ′ (P (t) DP (t)) = P (0) DP (0) + P (0) DP (0) =   dt 

0 d2 − d1 0 .. .

d2 − d1 0 0 .. .

0 0 0 .. .

0

0

0

and since by assumption the map t 7→ ϕ(P (t)T DP (t)) is constant, ∂ϕ (D)(d2 − d1 ) = 0, ∂ξ12

and

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

0 0 0



     0  ... 0

∂ϕ (D) = 0 since d1 6= d2 . ∂ξ12

As the set of diagonal matrices satisfying d1 6= d2 is obviously dense among the set of all ∂ϕ diagonal matrices (identified with Rk ), since ϕ is C 1 we get that ∂ξ (D) = 0 for every 12 diagonal matrix D. The proof in the other cases is analogous, by suitably choosing the family of matrices P (t). Let us check that Σ(L2 ) is a manifold.


Lemma 2.7. Let G : H01 (Ω; Rk ) → R

13

k(k+1) 2

defined by Z G(w) = (gij (w))j>i := wi wj dx Ω

,

j>i

so that Σ(L2 ) = {w ∈ H01 (Ω; Rk ) : gij (w) = δij ∀j > i}. Then G′ (u) is onto for each u ∈ Σ(L2 ). In particular, the set Σ(L2 ) is a submanifold of H01 (Ω; Rk ) of codimension k(k + 1)/2. Proof. Given u ∈ Σ(L2 ), we have Z G′ (u)(ϕ1 , . . . , ϕk ) = (ϕm un + um ϕn ) Ω

′ = (gmn (u)(ϕ1 , . . . , ϕk ))n>m .

n>m

In order to prove that G′ (u) is onto, let us prove that every vector of the canonical basis of k(k+1) R 2 is the image through G′ (u) of an element in H01 (Ω; Rk ). Fix i, j ∈ {1, . . . , k} with j > i; then by taking ϕ¯ := (ϕ¯1 , . . . , ϕ¯k ) with ϕ¯j = ui , ϕ¯l = 0 for l 6= i, we obtain Z Z ′ ′ 2u2i dx = 2 (if i = j), u2i dx = 1 (if i 6= j), or gij (u)ϕ¯ = gij (u)ϕ¯ = Ω

Ω

and

′ gmn (u)ϕ¯ = 0 ∀(m, n) 6= (i, j), n > m. ′ Thus G (u) is onto and the result follows.

Let us now fix once and for all some ϕ ∈ F , and denote by ψ its restriction to the space of k × k diagonal matrices. Theorem 2.8. Given β > 0, the infimum cβ is attained at uβ , vβ ∈ Σ(L2 ) such that Z Z ∇ui,β · ∇uj,β dx = ∇vi,β · ∇vj,β dx = 0 whenever i 6= j. Ω

Ω

Moreover, for each i we have  P 2q −1 P q2 Pk  k k 2 2  u µ u − βu −a ∆u = v  i,β i,β i,β j=1 j,β j=1 ij,β j,β j=1 j,β (2.3) q q2 −1 P   k  −bi,β ∆vi,β = Pk νij,β vj,β − βvi,β Pk v 2 2 u2 j,β

j=1

j=1

j=1

j,β

with

ai,β =

∂ϕ ∂ψ (M (uβ )) = (M (uβ )), ∂ξii ∂ai

and (2.4) µij,β = µji,β = δij ai,β νij,β = νji,β = δij bi,β

Z

Z

2

Ω

Ω

bi,β =

|∇ui,β | dx + β 2

|∇vi,β | dx + β

Z

∂ϕ ∂ψ (M (vβ )) = (M (vβ )), ∂ξii ∂ai

ui,β uj,β

Ω

Z

k X

u2l,β

l=1

vi,β vj,β

Ω

k X l=1

k 2q −1 X

2 vl,β

l=1

2 vl,β

k 2q −1 X l=1

u2l,β

q2

q2

dx,

dx.

Remark 2.9. We observe that, although the quantity q/2 − 1 needs not to be positive, there occur no singularities in the expressions above. In fact, for every u ∈ H01 (Ω; Rk ) we have that Z X Z k k k k q2 X 2q 2q q2 −1 X X u2l vl2 dx 6 vl2 dx < ∞, |ui uj | u2l Ω

l=1

l=1

Ω

l=1

l=1

14


as 2q < 2∗ . Proof of Theorem 2.8. Fix β > 0 and let us take a minimizing sequence un = (u1,n , . . . , uk,n ), vn = (v1,n , . . . , vk,n ). By the Ekeland’s variational principle we can suppose that un , vn ∈ Σ(L2 ) is such that Eβ (un , vn ) → cβ

and

Eβ |′Σ(L2 ) (un , vn ) → 0

as n → ∞.

(taking also in consideration Lemma 2.7). Consider the functions u ñ , vñ defined in Lemma 2.5. As Eβ (u, v) = Eβ (P u, P v) for every u, v ∈ H01 (Ω; Rk ), P ∈ Ok (R), then Eβ (˜ un , vñ ) = Eβ (un , vn ) → cβ

(2.5)

un , vñ ) → 0. Eβ |′Σ(L2 ) (˜

and

The first convergence implies that Z v1,n k2 , . . . , k˜ vk,n k2 Eβ (˜ un , vñ ) = ψ k˜ u1,n k2 , . . . , k˜ uk,n k2 + ψ k˜ Ω

Z X k k q2 2q X 2β v˜j,n dx 6 cβ + 1, ∀n large. u ˜2j,n + p Ω j=1 j=1 R R 2 ui,n |2 dx > δ > 0 for all n, and thus since ψ → +∞ ˜i,n dx = 1, we must have Ω |∇˜ Since Ω u R ui,n |2 dx 6 C for all i and n. Analogously as for some i, we deduce that Ω |∇˜ R si → +∞ 2 vi,n | dx 6 C and hence there exists (˜ u, v˜) ∈ Σ(L2 ) such that, up to a subsequence, Ω |∇˜ u˜i,n ⇀ u ˜i ,

v˜i,n ⇀ v˜i

weakly in H01 (Ω), strongly in Lr (Ω) (1 6 r < 2∗ ), a.e x ∈ Ω.

The second condition in (2.5) together with Lemma 2.6 implies the existence of µ ˜ij,n , ν˜ij,n such that k k k q2 −1 X X 2q X ∂ϕ 2 u ˜2j,n µ ˜ij,n u ˜j,n − β u˜i,n (M (˜ un ))∆˜ ui,n = − v˜j,n + on (1) ∂ξii j=1 j=1 j=1 2 2 −1 X X X ∂ϕ 2 u ˜2j,n + on (1) v˜j,n ν˜ij,n v˜j,n − β˜ vi,n (M (˜ vn ))∆˜ vi,n = − ∂ξii j=1 j=1 j=1 k

k

q

k

q

in H −1 (Ω). Observe that Z Z k k q2 2q −1 X X ∂ϕ 2 µ ˜ij,n = δij dx + o(k˜ uj,n k) v˜l,n (M (˜ un )) |∇˜ ui,n |2 dx + β u˜i,n u˜j,n u˜2l,n ∂ξii Ω Ω l=1

l=1

is bounded independently of n. Thus, by multiplying the i–th equation for u ñ by u ˜i,n − u ˜i we obtain Z ∂ϕ ∇˜ ui,n · ∇(˜ ui,n − u˜i ) dx = on (1). (M (˜ un )) ∂ξii Ω R R uk,n |2 dx lies in a compact set not containing the origin and As Ω |∇˜ u1,n |2 dx, . . . , Ω |∇˜

∂ϕ > 0 on (R+ )k , we have that ∂ξ (M (˜ un )) > δ > 0 for all i and n. Hence u ˜i,n → u ˜i ii 1 strongly in H0 (Ω) and, analogously, v˜i,n → v˜i . In conclusion, for any i 6= j we have Z Z Z Z ∇˜ ui ·∇˜ uj dx = lim ∇˜ ui,n ·∇˜ uj,n dx = 0, ∇˜ vi ·∇˜ vj dx = lim ∇˜ vi,n ·∇˜ vj,n dx = 0, ∂ψ ∂ai

Ω

n

Ω

Ω

n

Ω

and Eβ (˜ ui , v˜i ) = cβ . The last paragraph of the statement is a consequence of the Lagrange multiplier rule, by recalling once again Lemma 2.7.


15

Observe that (2.3) is a type of competitive system where competition occurs between groups of components. In fact, if one thinks of each ui and vi as being a population density, then no competition occurs between different ui ’s, or between different vi ’s, but each element of {u1 , . . . , uk } competes with each element of {v1 , . . . , vk }. A Lotka-Volterra (non variational) type competition of this kind was considered in the paper by Caffarelli and F.-H. Lin [12]. 2.2. Uniform bounds in H¨ older spaces. Proof of the general existence theorem. Our next goal is to pass to the limit β → ∞ in (2.3), relating the limiting profiles with problem c∞ . Having this in mind, first we establish some uniform bounds (in β) in the L∞ –norm, and then in H¨ older spaces. Throughout this subsection, we take (uβ , vβ ) as in Theorem 2.8. Lemma 2.10. We have cβ 6 c∞ for every β > 0. In particular this implies that k(uβ , vβ )kH01 (Ω) and k(uβ , vβ )kL∞ (Ω) are bounded independently of β. Proof. 1. Given (ω1 , ω2 ) ∈ P2 (Ω), consider the first k eigenfunctions u ¯ = (¯ u1 , . . . , u ¯k ) of −∆ in H01 (ω1 ), orthonormal in L2 , and the first k L2 -orthonormal eigenfunctions v¯ = (¯ v1 , . . . , v¯k ) of −∆ in H01 (ω2 ), so that u¯, v¯ ∈ Σ(L2 ). Then, since ω1 ∩ ω2 = ∅, ψ(λ1 (ω1 ), . . . , λk (ω1 )) + ψ(λ1 (ω2 ), . . . , λk (ω2 )) = ϕ(M (¯ u)) + ϕ(M (¯ v )) = Eβ (¯ u, v¯) > cβ

for all β > 0, hence cβ 6 c∞ . Z k k X q q X 2 2 2 vj,β ) 2 dx > 0 and ψ|(R+ )k > ψ(0, . . . , 0), we obtain that uj,β ) ( β( 2. As Ω

j=1

j=1

ψ ku1,β k2 , . . . , kuk,β k2 ,

and also

Z

β

Ω

k X

u2j,β

j=1

ψ kv1,β k2 , . . . , kvk,β k2

k 2q X j=1

2 vj,β

q2

dx

are bounded from above, independently of β. Since uβ , vβ ∈ Σ(L2 ), we have kui,β k2 , kvi,β k2 > δ > 0, and by the assumptions made for ψ we obtain uniform H01 bounds for uβ , vβ . 3. To prove the L∞ bounds, observe first of all that the conclusions of the previous point imply the existence of µ ¯ such that |µij,β | 6 µ ¯ for every i, j and β, where µij,β are the Lagrange multipliers defined in (2.4). Moreover, there exists a > 0 such that ai,β > a. If we combine equation (2.3) with the Kato inequality 2 we deduce that −∆|ui,β | 6 −sign(ui,β )∆ui,β =

k X µij,β j=1

ai,β

k

sign(ui,β )uj,β −

µ ¯X |uj,β |. 6 a j=1

2 −1 X X 2 β 2 sign(ui,β )ui,β u2j,β vj,β ai,β j=1 j=1 k

q

k

q

Summing up in i yields

k k kµ X ¯X |u | 6 |ui,β |. −∆ i,β a i=1 i=1 2The Kato inequality (see for instance [35, Th. 21.19]) states that given u ∈ L1 with ∆u ∈ L1 , then loc loc sign(u)∆u 6 ∆|u|.

16


Pk Since a similar statement holds for i=1 |vi,β |, we can conclude by using a standard BrezisKato type argument together with the H01 bounds. Theorem 2.11. There exists (u, v) ∈ H01 (Ω; Rk ) ∩ C 0,α (Ω; Rk ) (∀0 < α < 1) such that, up to a subsequence, as β → +∞, (i) uβ → u, vβ → v in C 0,α (Ω) ∩ H01 (Ω); 2

(ii) ui · vj ≡ 0 in Ω ∀i, j; u, v ∈ Σ(L ), and (iii) Moreover,

Z

Ω

k k X q X q 2 u2j,β ) 2 ( vj,β ) 2 dx → 0. β( j=1

j=1

in ωu := {x ∈ Ω : u21 + . . . + u2k > 0},

− ai ∆ui = µi ui

in ωv := {x ∈ Ω : v12 + . . . + vk2 > 0}

− bi ∆vi = νi vi

for ai = limβ ai,β , bi = limβ bi,β , µi = limβ µii,β , νi = limβ νii,β . Finally, for every i = 1, . . . , k, x0 ∈ Ω, r ∈ (0, dist(x0 , ∂Ω)), Z Z Z 2 ai |∇ui | dx = ai ui ∂n ui dσ + µi u2i dx Br (x0 )

and bi

Z

Br (x0 )

∂Br (x0 )

|∇vi |2 dx = bi

Z

Br (x0 )

vi ∂n vi dσ + νi

∂Br (x0 )

Z

Br (x0 )

vi2 dx.

Proof. (i) The proof of this point is made by contradiction. Namely, we assume that |uj,β (x) − uj,β (y)| |vi,β (x) − vi,β (y)| Lβ := max , max max → +∞ i,j=1,...,k x,y∈Ω |x − y|α |x − y|α x,y∈Ω and will reach a contradiction. The argument is very close to the one in [29, Section 3] (there the authors deal with positive solutions and moreover q = 2, k = 1 and ai,β = bi,β = 1) and therefore we merely focus on the technical differences with respect to [29]. Step A. We may assume that Lβ is achieved by the component u1,β , say at the pair (xβ , yβ ) with |xβ − yβ | → 0. Accordingly, for a given sequence rβ → 0 to be chosen later on, we define the rescaled functions 1 1 ui,β (xβ + rβ x), v¯i,β (x) = vi,β (xβ + rβ x), u¯i,β (x) = Lβ rβα Lβ rβα Ω−x

for x ∈ Ωβ := rβ β . It follows that  P q2 −1 P q2 Pk k k  2  ¯2j,β ¯j,β ui,β = j=1 rβ2 µij,β u ¯j,β − βMβ u ¯i,β  −ai,β ∆¯ j=1 u j=1 v (2.6) P 2q −1 P q2  Pk  k k 2 2  −bi,β ∆¯ v¯ vi,β = r νij,β v¯j,β − βMβ v¯i,β u ¯2 j=1 β

j=1

j,β

j=1

j,β

with

2α(q−1)+2

Mβ = L2q−2 rβ β

We observe that

,

δ 6 ai,β , bi,β 6 C,

|µij,β |, |νij,β | 6 C.

rβ2 k¯ uj,β kL∞ (Ωβ ) + rβ2 k¯ vj,β kL∞ (Ωβ ) → 0


and also that ( max

i,j=1,...,k

|¯ uj,β (x) − u¯j,β (y)| |¯ vi,β (x) − v¯i,β (y)| , max max |x − y|α |x − y|α x,y∈Ωβ x,y∈Ωβ =

17

)

|¯ u1,β (0) − u¯1,β ( |

yβ −xβ rβ )|

yβ −xβ α rβ |

= 1.

We next prove two claims which correspond to Lemmas 3.4 and 3.5 in [29]. Claim 1. Whenever rβ is such that (2.7)

lim inf βMβ > 0

and

β→∞

lim sup β→∞

|xβ − yβ | |xβ − yβ |/rβ . For sufficiently large β, one has B2R (0) ⊆ Ωβ . Take η a smooth cutoff function such that η = 1 in BR (0) and η = 0 in RN \ B2R (0). By testing the first equation in (2.6) with u ¯i,β η 2 and the second one with v¯i,β η 2 , we find out that !q/2 !q/2 Z Z k k k X X X 2 2 u¯2i,β dx dx 6 C v¯i,β u ¯i,β βMβ BR (0)

and

(2.8)

βMβ

Z

BR (0)

k X

B2R (0) i=1

i=1

i=1

u ¯2i,β

i=1

!q/2

k X i=1

2 v¯i,β

!q/2

dx 6 C

Z

k X

B2R (0) i=1

2 v¯i,β dx.

Since u ¯i,β and v¯i,β are H¨ older continuous and (2.7) holds, there exist C1 , C2 > 0 (depending only on R > 0) such that ! !q/2 !q/2 k k k X X X 2 2 2 u ¯i,β (x) + C2 v¯i,β (x) dx 6 C1 u ¯i,β (x) i=1

i=1

i=1

and

k X

!q/2

u ¯2i,β (x)

i=1

k X i=1

!q/2

2 v¯i,β (x)

dx 6 C1

k X i=1

for every x ∈ BR (0). In particular,

!

2 v¯i,β (x)

+ C2 .

(dβ eβ )q 6 (C1 dβ + C2 )(C1 eβ + C3 ).

Since q > 1, we cannot have that both sequences are unbounded. Assume first of all that dβ → +∞ (up to a subsequence), and eβ is bounded, so that in particular k k X X 2 v¯i,β is bounded, u ¯2i,β → +∞, and sup inf B2R (0)

i=1

B2R (0) i=1

18


and actually sup

k X

B2R (0) i=1

Combining this with (2.8) gives βMβ

Z

BR (0)

k X i=1

2 v¯i,β →0

u ¯2i,β

as β → ∞.

!q/2

k X i=1

2 v¯i,β

!q/2

dx → 0.

Going back to the system, we conclude that k∆¯ ui,β kL1 (BR (0)) → 0 for every i = 1, . . . , k and every large R > 0. In particular, the sequence u ˜β (x) := u ¯1,β (x) − u¯1,β (0) converges uniformly in compact sets to a non constant, harmonic, H¨ older continuous function in Ω∞ , where Ω∞ is either RN or a half-space of RN . Arguing as in [29, p. 282], this contradicts the Liouville type theorem stated in [29] In conclusion, we have shown that (dβ )β is bounded. The case when (eβ )β can be handled in an analogous way, and thus the claim is established. Claim 2. We have that βL2q−2 |xβ − yβ |2α(q−1)+2 → ∞. β Indeed, assuming the contrary, by letting 2α(q−1)+2 rβ−1 = βL2q−2 →∞ β

we find ourselves in the situation displayed in (2.7), with βMβ = 1. According to Claim 1 and up to subsequences, we conclude that u¯i,β → ui,∞ , v¯i,β → vi,∞ uniformly in compact sets, and  2q 2q −1 P P k k  2 2  −ai ∆ui,∞ = −ui,∞ v u  j=1 j,∞ j=1 j,∞ q2 2q −1 P   k 2  −bi ∆vi,∞ = −vi,∞ Pk v 2 u . j=1 j,∞ j=1 j,∞

Moreover, all these functions are H¨ older continuous and u1,∞ is non constant. As explained in [29, p. 283], the case where the system holds in a half-space is easier to handle with, and N so wePassume henceforth P∞that this limit problem holds in the whole space R . Let us denote ∞ u = i=1 |ui,∞ |, v = i=1 |vi,∞ |. Since ai , bi > δ > 0 ∀i and since   q2 −1   2q k k X X 1 |u|  u2j,∞  > |u|q−1 , u2j,∞  > c|u|q , δ j=1 j=1 for some c > 0 (and similarly for v), we find that (2.9)

−∆|u| 6 −c|u|q−1 |v|q ,

−∆|v| 6 −c|v|q−1 |u|q .

We may assume that c = 1 by possibly replacing u and v with c1/(2q−2) u and c1/(2q−2) v respectively. Suppose first that |u| |v| ≡ 0, so that −∆|u| 6 0 and −∆|v| 6 0. It follows then from [29, Proposition 2.2] that v = 0 and so u1,∞ is a non constant, harmonic, H¨ older continuous function in RN , in contradiction with [29, Corollary 2.3]. Assume now that |u(x0 )| |v(x0 )| 6= 0 for some x0 ∈ RN ; since the problem is invariant by translation, we may assume that x0 = 0. Then we argue precisely as in the proof of [29, Proposition 2.6] to conclude that again v = 0, thus finishing the proof of Claim 2.


19

For the sake of clarity, we explicitly mention the only part of the argument where an extra care with respect to [29] is in order, due to the fact that here possibly q 6= 2. The main step in the proof of [29, Proposition 2.6] consists in showing (cf. [29, Lemma 2.5]) that, given ε > 0, the function Z Z 1 2 q q f (|x|) |∇u| + |u| |v| dx f (|x|) |∇v|2 + |u|q |v|q dx J(r) := 4−ε r Br (0) Br (0)

associated to (2.9) is increasing for r sufficiently large, where f is a suitable cut-off function. This property, in turn, relies on the study of the quantities R 2 2 q q dσ ∂B1 (0) |∇θ u(r) | + r |u(r) | |v(r) | R , Λ1 (r) := 2 ∂B1 (0) u(r) dσ R 2 2 q q dσ ∂B1 (0) |∇θ v(r) | + r |u(r) | |v(r) | R , Λ2 (r) := 2 ∂B1 (0) v(r) dσ

where r(r) (θ) := u(rθ), v(r) (θ) := v(rθ). Arguing as in [29, p. 275], assume that Λ1 (rn ) and Λ2 (rn ) are bounded sequences along a sequence rn → ∞. In particular, Z Z Z 2 2 2 q q dσ. v(r u(rn ) dσ and C |u(rn ) | |v(rn ) | dσ is less than or equal to C rn n) ∂B1 (0)

∂B1 (0)

∂B1 (0)

By multiplying these inequalities we find that Z 2 rn |u(rn ) |q |v(rn ) |q dσ 6 C 2 ku(rn ) kL2 (∂B1 (0)) kv(rn ) kL2 (∂B1 (0)) . ∂B1 (0)

In particular, since p > 1 and since both sequences ||u(rn ) ||L2 (∂B1 (0)) and ||v(rn ) ||L2 (∂B1 (0)) are bounded away from zero (this a consequence of the fact that both |u| and |v| are subharmonic functions such that u(0) 6= 0, v(0) 6= 0), we deduce from the previous inequality that Z rn2 |u(rn ) |q |v(rn ) |q dσ 6 C ′ ku(rn) kqL2 (∂B1 (0)) kv(rn ) kqL2 (∂B1 (0)) ∂B1 (0)

′

for some C > 0. By using this estimate we can apply the remaining argument in [29, p. 275] to conclude the proof of Claim 2. Step B. Once Claims 1 and 2 are established, one proceeds as follows. Let rβ = |xβ − yβ |. With this choice of rβ we have that βMβ → ∞ (cf. Claim 2) and so, according to Claim 1 and up to subsequences, we have that u ¯i,β → ui,∞ and v¯i,β → vi,∞ uniformly in compact subsets of RN , and u1,∞ is non constant. Moreover, by proceeding exactly 1 as in Lemmas 3.6 and 3.7 in [29], we deduce that these convergences hold in Hloc (RN ), P q/2 P P q/2 P k k k k 2 2 2 ≡ 0, ¯2i,β ¯i,β → 0 in L1loc (RN ), βMβ i=1 u i=1 v i=1 vi,∞ i=1 ui,∞ Pk P k 2 2 −ai ∆ui,∞ = 0 in {x : i=1 ui,∞ (x) > 0}, and −bi ∆vi,∞ = 0 in {x : i=1 vi,∞ (x) > 0}. A contradiction is then reached as in [29, p. 293]. We outline the argument. The main tool in reaching such a contradiction is a monotonicity formula applied to the Almgren quotients N (x0 , r) =

E(x0 , r) H(x0 , r)

(x0 ∈ RN , r > 0),

20


where, in the present setting, E(x0 , r) :=

1 rN −2

Z

Br (x0 )

and H(x0 , r) :=

1 rN −1

k X i=1

Z

∂Br (x0 )

ai |∇ui,∞ |2 + k X i=1

ai u2i,∞ +

k X i=1

k X i=1

bi |∇vi,∞ |2

2 bi vi,∞

!

!

dx

dσ.

Namely, one can prove (cf. [29, Proposition 3.9]) that N (x0 , r) is a non decreasing function and d 2N (x0 , r) log(H(x0 , r)) = ∀x0 ∈ RN , r > 0. dr r ∂ϕ ∂ϕ Due to the presence of the terms ai = ∂ξ (M (u∞ )) and bi = ∂ξ (M (v∞ )) (which are absent ii ii in [29]), this monotonicity formula will be proved in full detail in the next subsection, in a slightly different context which will also be needed later on. Then we conclude as follows. We Pk Pk 2 denote U = i=1 u2i,∞ , V = i=1 vi,∞ , so that U V ≡ 0, −∆U 6 0 and −∆V 6 0 in RN . Since U 6= 0, V must vanish somewhere in RN . Also, U must vanish at some point, otherwise u1,∞ would a non constant, harmonic, H¨ older continuous function in RN , in contradiction in [29, Corollary 2.3]. Thus, by continuity, U and V must have a common zero point. Since moreover they are both non negative, sub-harmonic and H¨ older continuous in RN , we deduce from [29, Proposition 2.2] that V =0

and

{x : U (x) > 0} is a proper connected subset of RN .

Now, let Γ := {x : U (x) = 0}. Since U is α-Hölder continuous, by using the above mentioned Almgren monotonicity formula one finds out that N (x0 , r) = α

∀x0 ∈ Γ, r > 0,

which turns out to be equivalent to the statement that ui,∞ is given in polar coordinates around x0 by ui,∞ (x) = rα gi (θ) for some function gi (i = 1, . . . , k, r = |x−x0 |, θ = (x−x0 )/r). Since this procedure can be applied to any given x0 ∈ Γ, we conclude that Γ is a cone with respect to each of its points, and so Γ is a (proper) subspace of RN . Moreover, RN \ Γ is connected and so Γ has dimension at most N − 2; in particular, it has zero capacity. As a consequence, u1,∞ is a non constant, harmonic, H¨ older continuous function in the whole of RN , in contradiction in [29, Corollary 2.3]. This final contradiction establishes the property (i) in Theorem 2.11. (ii) The proof of this fact is much simpler and follows precisely as in [29, p. 294]. (iii) We multiply the i–th equation for u in (2.3) by ui , integrate by parts and pass to the limit, by observing that, thanks to (2.4), µij,β → 0 for i 6= j as β → ∞. A full proof of a stronger result will be presented ahead in Proposition 2.25. The following corollary implies in particular Theorem 2.3, as well as Theorem 1.6 (i)-(ii)(iii) except the statement of Lipschitz continuity and the equality between sets in (ii). Corollary 2.12. We have the following alternative characterization for c∞ : c∞ = inf ϕ(M (w)) + ϕ(M (z)) : w, z ∈ Σ(L2 ), wi · zj = 0 ∀i, j .


21

Moreover, for ωu := {u21 + . . . + u2k > 0}

and

ωv := {v12 + . . . + vk2 > 0},

where u and v are given by Theorem 2.11, we have that (ωu , ωv ) ∈ P2 (Ω) achieves c∞ . In particular, Theorem 2.3 is proved. Proof. As a result R of the strong Rconvergences proved in the previous theorem, we have that ui vj = 0 ∀i, j, Ω ∇ui · ∇uj = Ω ∇vi · ∇vj = 0 ∀i 6= j and u, v ∈ Σ(L2 ). Thus, by the monotonicity property of ψ, c∞ > lim cβ β→∞

Z p p 2β 2 2 (u2 + . . . + u2k,β ) 2 (v1,β + . . . + vk,β ) 2 dx = lim ϕ(M (uβ )) + ϕ(M (vβ )) + β→∞ p Ω 1,β = ϕ(M (u)) + ϕ(M (v)) = ψ ku1 k2 , . . . , kuk k2 + ψ kv1 k2 , . . . , kvk k2

> ψ(λ1 (ωu ), . . . , λk (ωu )) + ψ(λ1 (ωv ), . . . , λk (ωv )) > c∞ .

We end this subsection by showing that, as a consequence of the minimization, the equations in Theorem 2.11-(iii) for ui and vi hold in the (apparently) larger sets ω eu := Ω \ ωv , ω ev := Ω \ ωu . It is easy to check that ωu ⊆ ω e u , ωv ⊆ ω ev ; later on we will check that these sets coincide (cf. Corollary 2.24). Proposition 2.13. The following holds for every i:

− ai ∆ui = µi ui in the open set ω eu ,

− bi ∆vi = νi vi in the open set ω ev .

Proof. 1. As we will see, this is an immediate consequence of the Lagrange multiplier rule, and it is related to the fact that (u, v) solves a minimization problem. To fix ideas, let us just prove that −a1 ∆u1 = µ1 u1 in Ω \ ωv .

Consider the functional J : H01 (w˜u ) → R defined by J(w) = ϕ(M (w, u2 , . . . , uk )) + ϕ(M (v)) and the set Z Z Z 1 2 wuk dx = (1, 0, . . . , 0) . wu2 dx, . . . , M = w ∈ H0 (˜ ωu ) : G(w) := w dx, ω ˜u

ω ˜u

ω ˜u

We observe that M is a manifold in a neighborhood of u, since G′ (u) is onto. Indeed, G′ (u)ui is equal to 2ei if i = 1, and G′ (u)ui = ei if i > 2, where ei denotes the i–th vector of the canonical basis of Rk . 2. The infimum inf{J(w) : w ∈ M}

is attained at w = u1 . In fact, suppose in view of a contradiction that there exists w ¯∈M satisfying J(w) ¯ < J(u1 ) = ϕ(M (u)) + ϕ(M (v)). Denote by w ˜ the extension of w ¯ by 0 to Ω. Then (w, ˜ u2 , . . . , uk ) ∈ Σ(L2 ) and w ˜ · vj ≡ 0 for every j, as w ˜ = 0 on {v12 + . . . + vk2 > 0}. Thus we get a contradiction with the first statement of Corollary 2.12.

22


3. By the Lagrange multipliers rule, there exists µ ˜1 such that −a1 ∆u1 = µ ˜ 1 u1

in ω ˜u.

LetR us now check that µ ˜1 = µ1 . Since by passing to the limit as β → +∞ in (2.4) yields a1 Ω |∇u1 |2 dx = µ1 , the result follows.

2.3. Almgren’s Monotonicity Formula. Let (u, v) be the limiting profile provided by Theorem 2.11, achieving c∞ . Our aim is to continue to characterize this limit. We will prove an Almgren Monotonicity Formula (Theorem 2.21), which will imply the Lipschitz continuity of u and v, and will have a central role in the proof of the regularity of the free boundary in the next section. As a first step towards the monotonicity formula, we need to establish some general integral identities, which are the subject of the following lemmas. ∗

Lemma 2.14. Let H : R2k → R be a C 1 function such that H|(s, t)| 6 C(1 + |s| + |t|)2 for every s, t ∈ Rk , and take li , mi ∈ R, i = 1, . . . , k. Let w = (w1 , . . . , wk ), z = (z1 , . . . , zk ) be a smooth solution of −li ∆ui = Hwi (w, z),

(2.10)

−mi ∆vi = Hzi (w, z).

Then for every vector field Y ∈ Cc1 (Ω; RN ) we have that Z k X 1 2 dx hdY ∇wi , ∇wi i − divY |∇wi | li 2 Ω i=1 Z k X 1 2 mi hdY ∇zi , ∇zi i − divY |∇zi | + dx 2 Ω i=1 Z H(w, z) divY dx = 0. + Ω

Proof. Consider a field Y ∈ Cc1 (Ω; RN ), that is such that Y = 0 ouside of a compact subset of Ω. Multiply the equation for wi in (2.10) by ∇wi · Y and the one for zi by ∇zi · Y . Summing up, one obtains Z Z k k X X mi h∇zi , ∇(∇zi · Y )i dx li h∇wi , ∇(∇wi · Y )i dx + Ω

i=1

− Now

Z

Ω

i=1

k Z X i=1

Ω

Ω

(Hwi (w, z)(∇wi · Y ) + Hzi (w, z)(∇zi · Y )) dx = 0.

h∇wi , ∇(∇wi · Y )i dx = =

Z

ZΩ

h∇wi , (Hwi Y + dY T ∇wi )i dx

(hHwi ∇wi , Y i + hdY ∇wi , ∇wi i) dx ZΩ 1 2 = ∇(|∇wi | ) · Y + hdY ∇wi , ∇wi i dx 2 Ω Z 1 2 hdY ∇wi , ∇wi i − |∇wi | divY dx, = 2 Ω

=


23

where Hwi denotes the Hessian of wi . Analogously, Z 1 2 hdY ∇vi , ∇vi i − |∇vi | divY dx. h∇vi , ∇(∇vi · Y )i dx = 2 Ω Ω

Z Finally,

=

Z

Ω

k Z X i=1

Ω

(Hwi (w, z)(∇wi · Y ) + Hzi (w, z)(∇zi · Y )) dx

∇x (H(w(x), z(x))) · Y (x) dx = −

Z

H(w, z)divY dx.

Ω

Now let us apply the previous result to (u, v), the limit functions obtained in Theorem 2.11. Recall the definitions ai =

∂ϕ ∂ψ (M (uβ )) = (M (uβ )), ∂ξii ∂ai

bi =

∂ϕ ∂ψ (M (vβ )) = (M (vβ )), ∂ξii ∂ai

and µi = lim µii,β , β→∞

νi = lim νii,β . β→∞

Corollary 2.15. Let (u, v) be the limit functions given by Theorem 2.11. Then Z 1 2 dx ai hdY ∇ui , ∇ui i − divY |∇ui | 2 Ω i=1 Z k X 1 hdY ∇vi , ∇vi i − divY |∇vi |2 dx bi + 2 Ω i=1 k Z X 1 + (µi u2i + νi vi2 ) divY dx = 0. 2 i=1 Ω k X

Proof. Since the functions (uβ , vβ ) solve the system (2.3), we can apply the previous result with:  q2   2q  k k k X X X β 1 vj2  . (µij,β ui uj + νij,β vi vj ) +  u2   Hβ (u, v) = 2 i,j=1 q j=1 j j=1

24


By Lemma 2.6, for each β > 0 we have Z k X 1 ai,β (hdY ∇ui,β , ∇ui,β i − divY |∇ui,β |2 ) dx 2 Ω i=1 Z k X 1 bi,β (hdY ∇vi,β , ∇vi,β i − divY |∇vi,β |2 ) dx + 2 Ω i=1 Z (2.11) k 1 X + (µij,β ui,β uj,β + νij,β vi,β vj,β )divY dx 2 i,j=1 Ω  q2  2q   Z k k X X β 2   vj,β divY dx = 0. u2   − q Ω j=1 j,β j=1

By using the convergences stated in Theorem 2.11 we can conclude by passing to the limit in (2.11).

Corollary 2.16 (Local Pohozaev–type identities). Given x0 ∈ Ω and r ∈ (0, dist(x0 , ∂Ω)), we have that k Z X (2 − N ) (ai |∇ui |2 + bi |∇vi |2 ) dx i=1

=

k Z X i=1

∂Br (x0 )

ai r(2(∂n ui )2 − |∇ui |2 ) dσ + +

k Z X i=1

∂Br (x0 )

k Z X i=1

Br (x0 )

∂Br (x0 )

r(µi u2i + νi vi2 ) dσ −

k Z X i=1

bi r(2(∂n vi )2 − |∇vi |2 ) dσ

Br (x0 )

N (µi u2i + νi vi2 ) dx.

Proof. Given x0 ∈ Ω and r > 0 we choose Y = Yδ (x) = (x − x0 )ηδ (x) where ηδ is a cut-off function such that ηδ = 1 on Br (x0 ), ηδ (x) = 0 in RN \ Br+δ (x0 ). Observe that hdYδ ∇ui , ∇ui i = hd(ηδ (x)(x − x0 ))∇ui , ∇ui i = ηδ |∇ui |2 + (∇ui · ∇ηδ )h∇ui , x − x0 i, div(Yδ ) = N ηδ + h∇ηδ , x − x0 i.

By using Corollary 2.15, we have that k Z k Z X X 2−N N 2 2 ηδ (ai |∇ui | + bi |∇vi | ) dx + (µi u2i + νi vi2 ) ηδ dx 2 2 Ω Ω i=1 i=1 Z k X 1 2 |∇ui | h∇ηδ , x − x0 i − (∇ui · ∇ηδ )h∇ui , x − x0 i dx ai = 2 Ω i=1 Z k X 1 bi + |∇vi |2 h∇ηδ , x − x0 i − (∇vi · ∇ηδ )h∇vi , x − x0 i dx 2 Ω i=1 k Z X 1 − (2.12) (µi u2i + νi vi2 )h∇ηδ , x − x0 i dx. 2 Ω i=1


25

We can rewrite the right-hand-side as Z k X ∇ηδ ∇ηδ 1 h∇ui , x − x0 i dx |∇ui |2 , x − x0 − ∇ui · ai |∇ηδ | 2 |∇ηδ | |∇ηδ | Ω i=1 Z k X ∇ηδ ∇ηδ 1 2 h∇vi , x − x0 i dx |∇vi | , x − x0 − ∇vi · bi |∇ηδ | + 2 |∇ηδ | |∇ηδ | Ω i=1 k Z X 1 ∇ηδ 2 2 2 − |∇ηδ |(µi ui + νi vi ) , x − x0 dx 2 |∇ηδ | i=1 Ω which, by the co-area formula, is equal to Z 1Z k X 1 ∇ηδ ∇ηδ 2 ai h∇ui , x − x0 i dσdt |∇ui | , x − x0 − ∇ui · 2 |∇ηδ | |∇ηδ | 0 {ηδ =t} i=1 Z 1Z k X 1 ∇ηδ ∇ηδ 2 bi + h∇vi , x − x0 i dσdt |∇vi | , x − x0 − ∇vi · 2 |∇ηδ | |∇ηδ | 0 {ηδ =t} i=1 k Z 1Z X ∇ηδ 1 (µi u2i + νi vi2 ) , x − x0 dσdt. − |∇ηδ | {ηδ =t} 2 i=1 0 Thus, as δ → 0, (2.12) becomes k Z X

k Z X 2−N N 2 2 (ai |∇ui | + bi |∇vi | ) dx + (µi u2i + νi vi2 ) dx 2 2 Br (x0 ) B (x ) r 0 i=1 Z k X 1 ai (∇ui · ν)h∇ui , x − x0 i − |∇ui |2 hν, x − x0 i dσ = 2 ∂Br (x0 ) i=1 Z k X 1 (∇vi · ν)h∇vi , x − x0 i − |∇vi |2 hν, x − x0 i dσ bi + 2 ∂Br (x0 ) i=1 k Z X 1 + (µi u2i + νi vi2 )hν, x − x0 i dσ. 2 ∂B (x ) r 0 i=1

i=1

As ν =

x−x0 r

k X i=1

ai

on ∂Br (x0 ), the right-hand-side of the previous inequality is equal to

Z

∂Br (x0 )

Z k X r r 2 r(∂n ui ) − |∇ui | dσ + bi r(∂n vi )2 − |∇vi |2 dσ 2 2 ∂Br (x0 ) i=1 Z k X r (µi u2i + νi vi2 ) dσ + 2 ∂B (x ) r 0 i=1

and the conclusion follows.

2

Ahead we will only need the local Pohozaev identities of Corollary 2.16. We have stated as well the more general ones of Corollary 2.15, as we think they are of independent interest (see for instance [17] for an application).

26


Under the previous notations, now define the quantities k Z 1 X ai |∇ui |2 + bi |∇vi |2 − µi u2i − νi vi2 dx, E(x0 , (u, v), r) = N −2 r i=1 Br (x0 ) H(x0 , (u, v), r) = and the Almgren’s quotient by

1 rN −1

k Z X i=1

N (x0 , (u, v), r) =

∂Br (x0 )

(ai u2i + bi vi2 ) dσ

E(x0 , (u, v), r) , H(x0 , (u, v), r)

whenever H(x0 , (u, v), r) 6= 0. Lemma 2.17. Given x0 ∈ Ω and r ∈ (0, dist(x0 , ∂Ω)) we have that k Z 2 X d ai (∂n ui )2 + bi (∂n vi )2 dσ E(x0 , (u, v), r) = N −2 dr r i=1 ∂Br (x0 ) k Z 2 X − N −1 (µi u2i + νi vi2 ) dx. r B (x ) r 0 i=1 Proof. The derivative of r 7→ E(x0 , (u, v), r) is given by k Z k Z 2−N X 2−N X 2 2 (a |∇u | + b |∇v | ) dx − (µi u2i + νi vi2 ) dx i i i i rN −1 i=1 Br (x0 ) rN −1 i=1 Br (x0 ) k Z k Z 1 X 1 X 2 2 (ai |∇ui | + bi |∇vi | ) dσ − N −2 (µi u2i + νi vi2 ) dσ. + N −2 r r ∂B (x ) ∂B (x ) r 0 r 0 i=1 i=1

By using Corollary 2.16, this quantity is equal to k Z k Z 1 X 1 X 2 2 ai (2(∂n ui ) − |∇ui | ) dσ + N −2 bi (2(∂n vi )2 − |∇vi |2 ) dσ rN −2 i=1 ∂Br (x0 ) r i=1 ∂Br (x0 ) k Z k Z 1 X N X + N −2 (µi u2i + νi vi2 ) dσ − N −1 (µi u2i + νi vi2 ) dx r r i=1 ∂Br (x0 ) i=1 Br (x0 ) Z Z k k 2−N X 1 X − N −1 (µi u2i + νi vi2 ) dx + N −2 (ai |∇ui |2 + bi |∇vi |2 ) dσ r r B (x ) ∂B (x ) r 0 r 0 i=1 i=1 k Z X 1 (µi vi2 + νi vi2 ) dσ − N −2 r ∂B (x ) r 0 i=1 or, equivalently, to k Z k Z 2 X 2 X 2 2 (a (∂ u ) + b (∂ v ) ) dσ − (µi u2i + νi vi2 ) dx. i n i i n i rN −2 i=1 ∂Br (x0 ) rN −1 i=1 Br (x0 )


27

Lemma 2.18. Given x0 ∈ Ω and r ∈ (0, dist(x0 , ∂Ω)) we have k Z 2 X d H(x0 , (u, v), r) = N −1 (ai ui (∂n ui ) + bi vi (∂n vi )) dσ. dr r i=1 ∂Br (x0 ) Proof. After a change of variables, H(x0 , (u, v), r) =

k Z X i=1

whence

(ai u2i (x0 + rx) + bi vi2 (x0 + rx)) dσ,

∂B1 (0)

k Z X d H(x0 , (u, v), r) =2 ai ui (x0 + rx)h∇ui (x0 + rx), xi dσ dr i=1 ∂B1 (0) k Z X +2 bi vi (x0 + rx)h∇vi (x0 + rx), xi dσ i=1

=

2 rN −1

∂B1 (0)

k Z X

(ai ui (∂n ui ) + bi vi (∂n vi )) dσ.

∂Br (x0 )

i=1

Now we prove an auxiliary technical result. Lemma 2.19. There exists C > 0 and r¯ > 0, both depending only on N and kui k, kvi k, i = 1, . . . , k, such that k Z 1 X (µi u2i + νi vi2 ) dx 6 Cr(E(x0 , (u, v), r) + H(x0 , (u, v), r)) rN −1 i=1 Br (x0 )

for every x0 ∈ Ω, r ∈ (0, r¯).

Proof. By making β → +∞ in (2.4) we obtain Z µi = a i |∇ui |2 dx 6 κ1 ai , Ω

with κ1 depending only on ku1 k, . . . , kuk k. Thus we have k Z k Z κ1 X 1 X 2 2 (µi ui + νi vi ) dx 6 N −1 (ai u2i + bi vi2 ) dx rN −1 i=1 Br (x0 ) r B (x ) r 0 i=1 ! k Z k Z X X κ1 1 1 2 2 2 2 6 r (ai |∇ui | + bi |∇vi | ) dx + N −1 (ai ui + bi vi ) dσ N −1 rN −2 i=1 Br (x0 ) r i=1 ∂Br (x0 )

by the Poincaré inequality. Next we observe that k Z k Z 1 X 1 X 2 2 = E(x , (u, v), r) + a |∇u | + b |∇v | (µi u2i + νi vi2 ) dx. 0 i i i i rN −2 i=1 Br (x0 ) rN −2 i=1 Br (x0 )

Thus for r small enough such that

κ1 2 N −1 r

6 1/2 the result follows, with C =

2κ1 N −1 .

28


Remark 2.20. For later reference, we mention that in the proof of Lemma 2.19 we have established the following result: There exists r¯ > 0 depending only on N and kui k, kvi k, i = 1, . . . , k, such that k Z k Z 1 X 1 X 2 2 (a |∇u | + b |∇v | ) dx + (ai u2i + bi vi2 ) dσ i i i i rN −2 i=1 Br (x0 ) rN −1 i=1 ∂Br (x0 ) (2.13)

6 2(E(x0 , (u, v), r) + H(x0 , (u, v), r)).

for every x0 ∈ Ω, r ∈ (0, r¯). We now have all the ingredients to state and prove the monotonicity result. Theorem 2.21 (Almgren’s Monotonicity Formula). Let C = C(N, ||ui ||, ||vi ||) be given by ˜ ⋐ Ω, there exists r˜ > 0 such that for every x0 ∈ Ω ˜ and r ∈ (0, r˜] Lemma 2.19. Then, given Ω we have H(x0 , (u, v), r) 6= 0, N (x0 , (u, v), ·) is absolutely continuous function, and d N (x0 , (u, v), r) > −2Cr(N (x0 , (u, v), r) + 1). dr

2

In particular, eCr (N (x0 , (u, v), r) + 1) is a non decreasing function for r ∈ (0, r˜] and the limit N (x0 , (u, v), 0+ ) := limr→0+ N (x0 , (u, v), r) exists and is finite. Also, d 2 log(H(x0 , (u, v), r)) = N (x0 , (u, v), r) dr r Moreover, the common nodal set

∀r ∈ (0, r˜).

Γ(u,v) := {x ∈ Ω : ui (x) = vi (x) = 0 for every i = 1, . . . , k}

has no interior points.

Proof. We use the same argument as in [29, Proposition 4.3]. As a starting point, we observe that Theorem 2.11 (iii) implies that k Z 1 X (2.14) E(x0 , (u, v), r) = N −2 (ai ui (∂n ui ) + bi vi (∂n vi )) dσ. r i=1 ∂Br (x0 ) Suppose first that H(x0 , (u, v), r) > 0 in some interval. For simplicity of notations, we write H(r) := H(x0 , (u, v), r) and similarly for E(r) and N (r). By combining Lemma 2.18 and (2.14) we find that

2 d log H(r) = N (r). dr r Also, by using Lemmas 2.17 and 2.18 together with (2.14), we deduce that 2 d N (r) = 2N−3 2 dr r H (r)

−

Z

k X

∂Br (x0 ) i=1

"Z

Z k X (ai (∂n ui )2 + bi (∂n vi )2 )dσ

∂Br (x0 ) i=1

!2  (ai ui (∂n ui ) + bi vi (∂n vi )) dσ  −

k X

(ai u2i + bi vi2 )dσ

∂Br (x0 ) i=1

2 H(r)r N−1

Z

k X

(µi u2i + νi vi2 ) dx.

Br (x0 ) i=1

As a consequence, thanks to Lemma 2.19, Z k X d E(r) + H(r) 2 (µi u2i + νi vi2 ) dx > −2Cr N (r) > − = −2Cr (N (r) + 1). N −1 dr H(r)r H(r) Br (x0 ) i=1


29

Thus, in order to conclude the proof of Theorem 2.21 we must show that H(x0 , (u, v), r) is ˜ indeed positive for small values of r and x0 ∈ Ω. We first prove that Γ(u,v) has no interior points. Assume the contrary, and let x1 ∈ Γ(u,v) be such that d1 := dist(x1 , ∂Γ(u,v) ) > 0. Thus H(x1 , (u, v), r) > 0 for r ∈ (d1 , d1 + ε) for some small ε. By what we have done so far, the function H(r) := H(x1 , (u, v), r) satisfies the initial value problem H ′ (r) = a(r)H(r) for r ∈ (d1 , d1 + ε) and H(d1 ) = 0, with a(r) = 2N (r)/r, 2 which is continuous also at the point d1 by the monotonicity of the map eCr (N (r)+ 1). Then H(r) ≡ 0 for r ∈ (d1 , d1 + ε), a contradiction with the definition of d1 . This proves our claim that Γ(u,v) has empty interior. ˜ This means that ui = vj = 0 Now, suppose H(x0 , (u, v), r) = 0 for some r > 0 and x0 ∈ Ω. on ∂Br (x0 ) for every i, j. By extending a non zero component of (u, v), say u1 |Br (x0 ) by zero to the whole Ω (such a non zero component exists, as Γ(u,v) has empty interior), call u ˜1 such u1 | in Ω, we find out that an extension, since −∆|˜ u1 | 6 µa11 |˜ Z Z Z µ1 2 2 u21 dx. |∇u1 | dx > λ1 (Br (x0 )) u dx > a1 Br (x0 ) 1 Br (x0 ) Br (x0 ) ˜ and we are done. This cannot hold if r = r(µ1 , a1 ) is small enough, uniformly in x0 ∈ Ω, As a consequence of the Almgren’s Monotonicity Formula, we have the following. Corollary 2.22. Let (u, v) be as before. Then ˜ ⋐ Ω there exist C˜ = C( ˜ Ω, ˜ N, kui k, kvi k) and r˜ > 0 such that, for every x ∈ Ω ˜ 1. given Ω and 0 < r1 < r2 6 r˜, 2C˜ r2 ; H(x, (u, v), r2 ) 6 H(x, (u, v), r1 ) r1 2. the map Ω 7→ R, x 7→ N (x, (u, v), 0+ ) is upper semi-continuous; 3. for every x0 ∈ Γ(u,v) we have N (x0 , (u, v), 0+ ) > 1.

Proof. The proof of this result follows word by word the ones of Corollaries 2.6, 2.7 and 2.8 in [33]. The Almgren’s monotonicity formula allows us to improve the regularity result for (u, v) and to give and alternative characterization of the nodal set. Corollary 2.23. The functions u and v are Lipschitz continuous. Proof. Once Theorem 2.21 and Corollary 2.22 are settled, we can reason exactly as in [29, Proposition 4.1] (see also [32, Subsection 2.4]). Corollary 2.24. If A is a connected component of ω eu (respectively ω ev ), then there exists a component ui such that A ⊆ {u2i > 0} (respectively A ⊆ {vi2 > 0}). As a consequence, we have ωu = ω e u , ωv = ω ev , and Γ(u,v) := {x ∈ Ω : ui (x) = vi (x) = 0 for every i = 1, . . . , k} = ∂ωu ∩ Ω = ∂ωv ∩ Ω

Proof. Take A a connected component of ω eu , so that in particular vj ≡ 0 in A for every j. Since all ui ’s are eigenfunctions on the set A, and by the minimality provided by Corollary 2.12, then either uj ≡ 0 in A for every j, or A ⊆ {u2i > 0} for some i. The first case cannot occur since Γ(u,v) has an empty interior, and thus the first statement of the corollary is proved. All the other statements now follow in a straightforward way.

30


We end this subsection with a more detailed description of the equation satisfied by the functions ui , vi in the whole Ω. Proposition 2.25. Let (u, v) be the limiting configuration obtained before. Then for each i ∈ {1, . . . , k} there exist measures Mi and Ni , both supported on Γ(u,v) 3, such that (2.15) −ai ∆ui = µi ui − Mi ,

−bi ∆vi = νi vi − Ni

in D ′ (Ω) = (Cc∞ (Ω))′ , i = 1, . . . , k.

Proof. By using Kato’s inequality in (2.3), we deduce that −aij,β ∆|ui,β | 6

k X j=1

|µij,β ||uj,β | − β|ui,β |

Integrating over Ω, we deduce that Z Z k k p2 −1 X X p2 2 u2j,β β|ui,β | vj,β dx 6 Ω

j=1

∂Ω

j=1

k X j=1

u2j,β

k p2 −1 X j=1

ai,β ∂n |ui,β | dσ +

k Z X j=1

Ω

2 vj,β

p2

.

|µij,β ||uj,β | dx 6 C,

where we have used the fact that ∂n |ui,β | 6 0, and C depends only on the uniform bounds p2 P p2 −1 P k k 2 2 v is bounded in L1 (Ω) u of kui,β k, kvi,β k, i = 1, . . . , k. Thus βui,β j=1 j,β j=1 j,β independently of β, whence there exists Mi ∈ M(Ω) = (C0 (Ω))′ , a Radon measure, such that βui,β

k X j=1

u2j,β

k p2 −1 X j=1

2 vj,β

p2

⇀ Mi

weak – ∗ in M(Ω).

Reasoning in the same manner with the equation for vi,β , we obtain the existence of Ni such that k k p2 −1 X X p2 2 vj,β βvi,β u2j,β ⇀ Ni weak – ∗ in M(Ω). j=1

j=1

Thus by Theorem 2.11 we conclude that (2.15) holds; we recall that, thanks to (2.4), we have that µij,β → 0, νij,β → 0 for i 6= j. Let x0 ∈ Ω \ Γ(u,v) . Then there exists δ > 0 such that either Bδ (x0 ) ⊆ ωu or Bδ (x0 ) ⊆ ωv and the hence the measures are concentrated on Γ(u,v) . 3. Regularity of the nodal set of the approximated solution Along this section, (u, v) will always denote the limiting function obtained in Theorem 2.11. Our aim is to prove a regularity result for the common nodal set Γ(u,v) , completing the proof of Theorem 1.6. Although with different proofs, initially we follow the structure of [12, 33]: first we make a detailed study of blowup sequences, and afterwards we use the Almgren’s quotient to split the nodal set Γ(u,v) and give a first characterization of its regular part. On this subset, locally we prove that locally Ω is a Reifenberg flat and NTA domain, and that Γ separates exactly two different connected components of the partition. Finally through an iterative procedure we construct the normal vector to Γ(u,v) , proving an extremality condition and a non degeneracy property. 3That is, M (A) = M (A ∩ Γ i i (u,v) ) and Ni (A) = Ni (A ∩ Γ(u,v) ) for each measurable set A.


31

3.1. Compactness of blowup sequences. In order to understand the local behavior of the functions (u, v) around any given point, let us study the behavior of blowup sequences. To ˜ ⋐ Ω and take some sequences xn ∈ Ω, ˜ tn → 0+ . We define a blowup that purpose, let Ω sequence by ui,n (x) :=

ui (xn + tn x) , ρn

vi,n (x) =

vi (xn + tn x) ρn

∀ x ∈ Ωn :=

Ω − xn , tn

where we have normalized using the quantity ρ2n

= H(xn , (u, v), tn ) =

1 −1 tN n

k Z X i=1

∂Btn (xn )

(ai u2i + bi vi2 ) dσ

k X ai kui (xn + tn ·)k2L2 (∂B1 (0)) + bi kvi (xn + tn ·)k2L2 (∂B1 (0)) . = i=1

We observe that

H(0, (un , vn ), R) = H(xn , (u, v), tn R)ρ−2 n

(3.1)

∀R > 0,

k X ai kui,n k2L2 (∂B1 (0)) + bi kvi,n k2L2 (∂B1 (0)) = 1

(3.2)

i=1

N

and, for every x0 ∈ R , r > 0 (cf. Theorem 2.11 (iii)) Z Z Z (3.3) ai |∇ui,n |2 dx = ai ui,n ∂n ui,n dσ + t2n µi Br (x0 )

(3.4)

bi

Z

Br (x0 )

∂Br (x0 )

|∇vi,n |2 dx = bi

Z

∂Br (x0 )

Br (x0 )

vi,n ∂n vi,n dσ + t2n νi

Z

Br (x0 )

u2i,n dx,

2 vi,n dx.

It is clear that −ai ∆ui,n = t2n µi ui,n − Mi,n ,

−bi ∆vi,n = t2n νi vi,n − Ni,n

in D ′ (Ωn ),

where Mi,n and Ni,n are measures defined by Mi,n (A) =

1 Mi (xn + tn A), −2 ρ n tN n

Ni,n (A) =

1 Ni (xn + tn A) −2 ρ n tN n

for every measurable set A ⊆ Ωn . Theorem 3.1. Under the previous notations there exists (¯ u, v¯) 6≡ (0, 0) such that u ¯i · v¯j ≡ 0 in RN ∀i, j and, up to a subsequence, (un , vn ) → (¯ u, v¯)

0,α in Cloc (RN )

1 for all 0 < α < 1, strongly in Hloc (RN ).

Moreover, there exist M¯i , N¯i ∈ Mloc (RN ), both concentrated on the set

Γ(¯u,¯v) := {x ∈ RN : u¯i (x) = v¯i (x) = 0, i = 1, . . . , k},

such that Mi,n ⇀ M¯i ,

Ni,n ⇀ N¯i

and −ai ∆¯ ui = −M¯i ,

weakly in Mloc (RN ),

−bi ∆¯ vi = −N¯i

in D ′ (RN ).

32


In particular, Z Z |∇¯ ui |2 dx = (3.5)

u¯i ∂n u ¯i dσ

and

Br (x0 )

∂Br (x0 )

Br (x0 )

Z

|∇¯ vi |2 dx =

Finally, for each x0 ∈ RN and r > 0, k Z k Z X X (ai |∇¯ ui |2 + bi |∇¯ vi |2 ) dx = (2 − N ) (3.6)

i=1

Br (x0 )

i=1

+

∂Br (x0 )

k Z X i=1

Z

v¯i ∂n v¯i dσ.

∂Br (x0 )

ai r(2(∂n u¯i )2 − |∇¯ ui |2 ) dσ

∂Br (x0 )

bi r(2(∂n v¯i )2 − |∇¯ vi |2 ) dσ.

In order to prove Theorem 3.1 we will need to prove first a technical result. Define Γ(un ,vn ) := {x ∈ Ωn : ui,n (x) = 0 = vi,n (x)

∀i = 1, . . . , k}.

Lemma 3.2. Let r˜ be given by Theorem 2.21. Given R > 0 then, for n sufficiently large, H(x, (un , vn ), r) 6 Cr2 ˜ N, kui k, kvi k). with C = C(Ω,

∀0 < r < r˜/2, x ∈ B2R (0) ∩ Γ(un ,vn ) ,

˜ we have that Proof. As long as we have r < r˜, tn < 1 and xn + tn x ∈ Ω, ρ2n H(x, (un , vn ), r) = H(xn + tn x, (u, v), rtn )

and so (cf. Theorem 2.21) d d ρ2 H(x, (un , vn ), r) 2 = tn [log H(xn + tn x, (u, v), ·)] |r=tn x − log n dr r2 dr r 2 = (N (xn + tn x, (u, v), rtn ) − 1), r implying that (3.7)

2 2 r d ρ2 H(x, (un , vn ), r) = (N (xn + tn x, (u, v), rtn ) + 1)eC(rtn ) e−C(rtn ) − 2. log n 2 dr r2 Now, applying Theorem 2.21, 2

(N (xn + tn x, (u, v), rtn ) + 1)eC(tn r) > N (xn + tn x, (u, v), 0+ ) + 1. But since, by assumption, xn + tn x ∈ Γ(u,v) , it follows from Corollary 2.22 that 2

(N (xn + tn x, (u, v), rtn ) + 1)eC(rtn) > 2. Going back to (3.7), we conclude that 2 ′ 2 4 ρ2 H(x, (un , vn ), r) 4 d > (e−C(rtn ) − 1) > (e−C r − 1) log n ∀0 < r < r˜. 2 dr r r r This implies, after an integration, that, for every 0 < r < r¯ < r˜, Z r¯ ′ 2 H(x, (un , vn ), r¯) 4 H(x, (un , vn ), r) 6 exp( (1 − e−C s ) ds) 6 C(¯ r ) < ∞, r2 r¯2 s r

thus proving the lemma.

Proof of Theorem 3.1. We follow the argument in the proof of [33, Theorem 3.3].


33

Step 1. We first prove that given R > 1 then, for n sufficiently large, k(un , vn )kH01 (BR (0)) 6 C. Indeed, if n is so large that tn R 6 r˜ (cf. Theorem 2.21) we have k Z X

∂BR (0)

i=1

2 (ai u2i,n + bi vi,n ) dσ =

k Z X 1 (ai u2i + bi vi2 ) dσ −1 ρ2n tN n ∂B (x ) tn R n i=1

= RN −1

H(xn , (u, v), tn R) ˜ 6 RN −1 R2C H(xn , (u, v), tn )

(cf. Corollary 2.22), yielding that H(0, (un , vn ), R) 6 C(R). On the other hand, 1 RN −2

k Z X i=1

BR (0)

(ai |∇ui,n |2 + bi |∇vi,n |2 ) dx

k Z C(R) 1 X (ai |∇ui,n |2 + bi |∇vi,n |2 ) dx H(0, (un , vn ), R) RN −2 i=1 BR (0) k Z 1 X C(R) (ai |∇ui,n |2 + bi |∇vi,n |2 ) dx = H(0, (un , vn ), R) RN −2 i=1 BR (0) ! k Z 1 X 2 + N −1 (ai u2i,n + bi vi,n ) dσ − C(R) R i=1 ∂BR (0) k Z X 1 C(R)ρ−2 n (ai |∇ui |2 + bi |∇vi |2 ) dx = H(0, (un , vn ), R) (tn R)N −2 i=1 Btn R (xn ) ! Z k X 1 2 2 + (ai ui + bi vi ) dσ − C(R) (tn R)N −1 i=1 ∂Btn R (xn ) 2C(R)ρ−2 n E(xn , (u, v), tn R) + H(xn , (u, v), tn R) − C(R) 6 H(0, (un , vn ), R)

6

where we have used (2.13) in the last inequality. By taking (3.1) into account, we conclude that k Z 1 X (ai |∇ui,n |2 + bi |∇vi,n |2 ) dx RN −2 i=1 BR (0) 2C(R) E(xn , (u, v), tn R) + H(xn , (u, v), tn R) − C(R) 6 H(xn , (u, v), tn R) = 2C(R)(N (xn , (u, v), tn R) + 1) − C(R) ˜

2

6 2C(R)(N (xn , (u, v), tn R) + 1)eC(tn R) − C(R) ˜

2

6 2C(R)(N (xn , (u, v), r˜) + 1)eC r˜ − C(R) 6 C ′ (R),

34


where we have used Theorem 2.21 in the last inequality. The new constant C ′ (R) depends on supy∈Ω˜ |N (y, (u, v), r˜)| < ∞. Step 2. We next show that given R > 0 then, for n sufficiently large, k(un , vn )kL∞ (BR (0)) 6 C ′ . Pk Indeed, it follows from (3.8) that, by letting Un = i=1 |ui,n |, then −∆Un 6 CUn . This, together with the estimate in Step 1 and a standard Brezis-Kato bootstrap argument, proves our claim for kun kL∞ (BR (0)) and for kvn kL∞ (BR (0)) as well. Step 3. We next prove that given R > 0 the measures Mi,n and Ni,n are bounded in fi,n and Nf M(BR (0)). Take positive measures M i,n satisfying

fi,n , (3.8) −ai ∆|ui,n | 6 t2n |µi | |ui,n | − M and

fi,n , |Mi,n | 6 M

(3.9)

−bi ∆|vi,n | 6 t2n |νi | |vi,n | − Nf i,n

in D ′ (Ωn )

|Ni,n | 6 Nf i,n .

Multiply the first equation in (3.8) by a smooth cut-off function ϕ such that 0 6 ϕ 6 1, ϕ = 1 in BR (0) and ϕ = 0 in RN \ B2R (0). It holds Z fi,n f f ϕ dM kMi,n kM(BR (0)) = Mi,n (BR (0)) 6 B2R (0) Z Z 6 −ai h∇|ui,n |, ∇ϕi dx + t2n |µi | |ui,n | ϕdx. B2R (0)

B2R (0)

fi,n kM(B (0)) )n is a bounded sequence. Thanks to the conclusion in Step 1, we deduce that (kM R It follows then from (3.9) that also (kMi,n kM(BR (0)) )n is bounded. A similar conclusion holds for (kNf i,n kM(BR (0)) )n and (kNi,n kM(BR (0)) )n .

Step 4. We now turn to the proof of a Lipschitz estimate. Namely, we prove that, given R > 0 then, for n sufficiently large, k(un , vn )kC 0,1 (BR (0)) 6 C ′′ .

To this aim, suppose without loss of generality, that      |vi,n (x) − vi,n (y)|  |ui,n (x) − ui,n (y)| , sup sup [(un , vn )]C 0,1 (BR (0)) = max  i=1,...,k  |x − y| |x − y| x,y∈BR (0)  x,y∈BR (0) x6=y

=

x6=y

|u1,n (yn ) − u1,n (zn )| , |yn − zn |

with |yn | 6 R, |zn | 6 R. Let rn := |yn − zn |,

2Rn := max{dist(yn , Γun ), dist(zn , Γun )} = dist(zn , Γun ),

where we have denoted Γun := {x ∈ RN : ui,n (x) = 0

∀i = 1, . . . , k}.

In view of proving our claim, we may already assume that Rn > 0 and, thanks to Step 2, that rn → 0. Moreover, since −a1 ∆u1,n = t2n µ1 u1,n in Ωn \ Γun , the desired conclusion follows


35

easily from standard elliptic estimates in case lim inf n Rn > 0. In conclusion, we may already assume that rn → 0 and 0 < Rn → 0.

A crucial remark in our subsequent argument is the observation that d(zn , Γun ) = d(zn , Γ(un ,vn ) ),

(3.10) where we recall the definition

Γ(un ,vn ) := {x ∈ RN : ui,n (x) = 0 = vi,n (x) ∀i = 1, . . . , k}. P P k k 2 2 This, of course, is a consequence of the fact that v u i=1 i,n ≡ 0 and Rn > 0. i=1 i,n Pk Case 4a. Suppose first that (Rn /rn )n is a bounded sequence. By letting Un := i=1 u2i,n we have −∆Un 6 2Ct2n Un in Ωn . By applying [33, Lemma 3.9] to this inequality we get that Z 1 Ct2n Rn2 Un (zn ) 6 kUn kL∞ (BRn (zn )) Un (x) dx + |BRn (zn )| BRn (zn ) 2(N + 2) Z 1 (3.11) Un (x) dx + C ′ Rn2 . 6 |BRn (zn )| BRn (zn ) Now, thanks to (3.10) we can choose wn ∈ Γ(un ,vn ) ∩ B2R (0) in such a way that d(zn , Γ(un ,vn ) ) = |zn − wn | = d(zn , Γn ) = 2Rn .

It follows then from Lemma 3.2 that, for large n and small r, H(wn , (un , vn ), r) 6 Cr2 . In particular, we have 1 rN −1 thus also

1 rN

Z

Un (σ) dσ 6 C ′ r2 ,

∂Br (wn )

Z

Un (x) dx 6 C ′′ r2 .

Br (wn )

Since Rn → 0, we get from (3.11) that Un (zn ) 6 C ′′′ Rn2 . In particular, since by assumption the sequence (Rn /rn )n is bounded, u21,n (zn ) 6 Crn2 , with C independent of n. This proves our desired estimate (3.12)

|u1,n (zn ) − u1,n (yn )|2 6 C|yn − zn |2 = Crn2

at least in the case where u1,n (yn ) = 0. But in case u1,n (yn ) 6= 0 we have 0 < d(yn , Γn ) 6 d(zn , Γn ) = 2Rn → 0,

and we get as above that

u21,n (yn )

6 Cd(yn , Γn )2 6 C ′ rn2 , leading still to (3.12).

Case 4b. Suppose secondly that rn /Rn → 0. By applying elliptic estimates to the equation −a1 ∆u1,n = t2n µ1 u1,n together with the bound in Step 2 we get that (3.13)

ku1,n kC 0,1 (BRn /2 (zn )) 6 C(Rn−1 ku1,n kL∞ (BRn (zn )) + 1).

Arguing as in Case 1 above we prove the existence of C > 0 such that for large n and every x ∈ BRn (zn ) it holds u21,n (x) 6 Cd(x, Γn )2 6 C ′ Rn2 . By (3.13) we deduce that the

36


sequence ku1,n kC 0,1 (BRn /2 (zn )) is bounded. Since by assumption yn ∈ BRn /2 (zn ) for large n, this completes the argument if Case 2 holds and finishes the proof of the estimates in Step 4. Step 5. By what we have proved so far, by taking subsequences we can find (un , vn ) → (¯ u, v¯)

0,α in Cloc (RN )

1 for all 0 < α < 1, weakly in Hloc (RN ),

as well as measures M¯i , N¯i , with the properties stated in Theorem 3.1. The normalization property (3.2) implies that (¯ u, v¯) 6≡ (0, 0), while the fact that M¯i and N¯i are concentrated on N ˜ Γ(¯u,¯v) follows from the convergence (un , vn ) → (¯ u, v¯) in L∞ loc (R ). Next we show that (un , vn ) → (¯ u, v¯)

1 in strongly in Hloc (RN ).

To show this, we multiply the equation −ai ∆(ui,n − u ¯i ) = t2n µ1 ui,n + M¯i −Mi,n by (ui,n − u ¯i )ϕ, where a smooth cut-off function ϕ such that 0 6 ϕ 6 1, ϕ = 1 in BR (0) and ϕ = 0 in RN \ B2R (0). Since Z Z |ui,n − u ¯i |ϕ d|Mi,n | 6 kui,n − u ¯i kL∞ (B2R (0)) |Mi,n |(B2R (0)) (ui,n − u ¯i )ϕ dMi,n 6 RN

RN

R ¯i )|2 dx → 0, and the and similarly for | RN (ui,n − u¯i )ϕ dMi |, we get that ai BR (0) |∇(ui,n − u conclusion follows. In particular, we obtain (3.5) by passing to the limit in (3.3) and in (3.4). R

Step 6. Let us now check (3.6). From Corollary 2.16 we deduce, by taking suitable rescailings, that k Z X (ai |∇ui,n |2 + bi |∇vi,n |2 ) dx (2 − N ) i=1

=

k Z X i=1

∂Br (x0 )

+

k Z X

ai r(2(∂n ui,n )2 − |∇ui,n |2 ) dσ + k Z X i=1

∂Br (x0 )

i=1

∂Br (x0 )

2 rt2n (µi u2i,n + νi vi,n ) dσ −

Then (3.6) follows by letting n → ∞ in this identity.

Br (x0 )

k Z X i=1

bi r(2(∂n vi,n )2 − |∇vi,n |2 ) dσ

Br (x0 )

2 N t2n (µi u2i,n + νi vi,n ) dx.

Having established a strong convergence result for any blowup sequence, based on the properties of (u, v) we can characterize in a better way the possible blowup limits. First we observe that the Almgren’s monotonicity formula valid from (u, v) is transported to any limit (¯ u, v¯). Corollary 3.3 (Almgren’s formula for (¯ u, v¯)). For any x0 ∈ RN , r > 0, let k Z 1 X E(x0 , (¯ u, v¯), r) = N −2 (ai |∇¯ ui |2 + bi |∇¯ vi |2 ) dx, r i=1 Br (x0 ) H(x0 , (¯ u, v¯), r) = and define the Almgren’s quotient by

1 rN −1

k Z X i=1

N (x0 , (¯ u, v¯), r) =

∂Br (x0 )

(ai u ¯2i + bi v¯i2 ) dσ,

E(x0 , (¯ u, v¯), r) . H(x0 , (¯ u, v¯), r)


37

Then H(x0 , (¯ u, v¯), r) 6= 0 for every r > 0, r 7→ N (x0 , (¯ u, v¯), r) is a nondecreasing map, and 2 d log(H(x0 , (¯ u, v¯), r)) = N (x0 , (¯ u, v¯), r). dr r

(3.14) In particular, the set

Γ(¯u,¯v) = {x ∈ RN : u ¯i (x) = v¯i (x) = 0 for every i = 1, . . . , k} has no interior points. Proof. The proof is similar to the one in Theorem 2.21. We provide the full details. We simplify the notations by writing E(r) = E(x0 , (¯ u, v¯), r) and similarly for H(r) and N (r). At first, thanks to (3.5) we see that k Z 1 X (3.15) E(r) = N −2 (ai u ¯i ∂n u ¯i + bi v¯i ∂n v¯i ) dσ r i=1 ∂Br (x0 ) On the other hand, by a simple computation similar to the the one in Lemma 2.18, 2

′

H (r) =

rN −1

k Z X i=1

(ai u ¯i ∂n u¯i + bi v¯i ∂n v¯i ) dσ ∂Br (x0 )

and, as in Lemma 2.17, the derivative E ′ (r) =

d u, v¯), r) dr E(x0 , (¯

is given by

k Z k Z 1 X 2−N X 2 2 (a |∇¯ u | + b |∇¯ v | ) dx + (ai |∇¯ ui |2 + bi |∇¯ vi |2 ) dσ i i i i rN −2 i=1 Br (x0 ) rN −2 i=1 ∂Br (x0 )

that is, thanks to (3.6)

E ′ (r) =

(3.16)

2 rN −2

k Z X i=1

∂Br (x0 )

ai (∂n u ¯i )2 + bi (∂n v¯i )2 dσ.

By putting together (3.15) to (3.16) we get that, as long as H(r) > 0, (log H)′ (r) =

(3.17)

2 N (r) r

and k

(3.18)

X 1 2 H (r)r2N −3 N ′ (r) = 2 i=1 ×

k Z X i=1

∂Br (x0 )

(ai u ¯2i

+

Z

(ai (∂n u ¯i )2 + bi (∂n v¯i )2 ) dσ

∂Br (x0 )

bi v¯i2 ) dσ

−

k Z X i=1

∂Br (x0 )

(ai u¯i ∂n u ¯i + bi v¯i ∂n v¯i ) dσ

!2

> 0,

by the Cauchy-Schwarz inequality. Now, arguing precisely as in Theorem 2.21, (3.17) together with the monotonicity of the map N (r) implies that Γ(¯u,¯v) has empty interior. In particular, thanks also to (3.5), we must have that H(r) > 0 for every r > 0. We now introduce a very important definition, which will help us in the future to lighten the notation.

38


Definition 3.4. Given y ∈ Ω, we define the set of all possible blowup limits at y by  ∃ xn → x0 , tn → 0 such that, for every i,     vi (xn + tn ·) ui (xn + tn ·) →u ¯ and vi,n := p → v¯ u, v¯) : ui,n := p BUy = (¯  H(xn , (u, v), tn ) H(xn , (u, v), tn )    0,α 1 strongly in Hloc (RN ) ∩ Cloc (RN ), ∀0 < α < 1

        

From now on we will talk about the associated sequences xn , tn defined for each (¯ u, v¯) ∈ BUy , with obvious meaning. Observe that at this point we cannot say that the blowup limit is unique, and it could a priori depende on the chosen sequences. Particular choices of the sequence (xn )n provide additional information on the limit (¯ u, v¯), as a consequence of the Almgren’s Formula. Corollary 3.5. Let x0 ∈ Ω and (¯ u, v¯) ∈ BUx0 with associated sequences xn , tn . Assume that one of the following conditions hold: 1. (xn )n is a constant sequence, i.e. xn = x0 for every n; 2. (xn )n ⊆ Γ(u,v) , and x0 ∈ Γ(u,v) is such that N (x0 , (u, v), 0+ ) = 1.

Then N (0, (¯ u, v¯), r) ≡ N (x0 , (u, v), 0+ ) =: α for every r > 0, and there exist gi , hi : S N −1 → R such that u ¯i = rα gi (θ), v¯i = rα hi (θ), where r = |x|, θ = x/|x|. Proof. The proof that N (r) := N (0, (¯ u, v¯), r) has constant value equal to α = N (x0 , (u, v), 0+ ) follows exactly as in [33, Corollary 3.12], on the ground of Theorem 2.21, Corollary 2.22, and the fact that N (0, (un , vn ), r) = N (xn , (u, v), tn r). In this situation we have N ′ (r) ≡ 0 and from (3.18) we know that, for a.e. r > 0 there exists C(r) > 0 such that (∂n u ¯ 1 , . . . , ∂n u ¯k , ∂n v¯1 , . . . , ∂n v¯k ) = C(r)(¯ u1 , . . . , u ¯k , v¯1 , . . . , v¯k )

on ∂Br (0).

By inserting this information in (3.14) we get that Pk R ¯i (∂n u¯i ) + bi v¯i (∂n v¯i )) dσ 2 i=1 ∂Br (0) (ai u H ′ (r) 2 α= = = 2C(r) Pk R r H(r) (ai u ¯2i + bi v¯i2 ) dσ i=1 ∂Br (0)

and C(r) = α/r. As on ∂Br (0) the normal derivative corresponds to the derivative with respect to the radius, the conclusion follows. 3.2. Hausdorff dimension estimates for the nodal and singular sets. Definition of R(u,v) , the regular part of the nodal set. In this subsection we provide upper estimates for the Hausdorff dimension of the nodal set as well as for some of its subsets. Recall from Corollary 2.22 that N (x, (u, v), 0+ ) > 1 whenever x ∈ Γ(u,v) . Definition 3.6. Let (u, v) be the limit given by Theorem 2.11. We split the nodal set Γ(u,v) into the following two sets: R(u,v) = {x ∈ Γ(u,v) : N (x, (u, v), 0+ ) = 1} and S(u,v) = Γ(u,v) \ R(u,v) = {x ∈ Γ(u,v) : N (x, (u, v), 0+ ) > 1}.


39

We will prove in the subsequent subsections that R(u,v) turns out to be the regular part of the nodal set, being of class C 1,α for some 0 < α < 1, whereas the singular part S(u,v) has small Hausdorff measure: more precisely at most N −2. In the spirit of [12, Lemma 4.1] or [33, Lemma 6.1], we prove a jump condition for the value of the map x ∈ Γ(u,v) 7→ N (x, (u, v), 0+ ), which will yield in particular that S(u,v) is relatively close in Γ(u,v) . Here, the fact that we have groups of components, which eventually change sign, makes the proof more delicate. Proposition 3.7. Let (u, v) be the limit given by Theorem 2.11. Then there exists δN > 0 (depending only on the dimension) such that, for every x0 ∈ Γ(u,v) , either N (x0 , (u, v), 0+ ) = 1

or

N (x0 , (u, v), 0+ ) > 1 + δN

Proof. Consider a blowup sequence at a fixed center x0 ∈ Γ(u,v) : ui,n (x) =

ui (x0 + tn x) , ρn

vi,n (x) =

vi (x0 + tn x) , ρn

where ρ2n = H(x0 , (u, v), tn ) and tn ↓ 0. By Theorem 3.1 and Corollary 3.5 (case 1) we have convergence (up to a subsequence) to a limiting configuration (¯ u, v¯), where u ¯i = rα gi (θ),

v¯i = rα hi (θ),

with α = N (x0 , (u, v), 0+ ).

Moreover, these functions are harmonic in RN \ Γ(¯u,¯v) , whence −∆S N −1 gi = λgi

and

− ∆S N −1 hi = λhi

in ∂B1 (0) \ Γ(¯u,¯v) ,

λ = α(α + N − 2). In particular, if A is a connected component of either ∂B1 (0) ∩ {ui 6= 0} or ∂B1 (0) ∩ {vi 6= 0} for some i, then we have λ = λ1 (A), the first eigenvalue of −∆S N −1 in H01 (A). We now split the proof in two cases: Case 1: We have u ¯i 6≡ 0 and v¯j 6≡ 0 for some i, j. Without loss of generality, suppose that for 1 6 l, m 6 k we have u ¯1 , . . . , u ¯l 6≡ 0,

and that all the remaining components are 0.

v¯1 , . . . , v¯m 6≡ 0,

Subcase 1.1. Suppose that for one component, say u1 , the set {u1 6= 0} has at least two connected components: A1 and A2 . Let B be a connected component of {v1 6= 0}. As ui · vj ≡ 0, the sets A1 , A2 and B are disjoint, and thus there exists C ∈ {A1 ∩ ∂B1 (0), A2 ∩ ∂B1 (0), B ∩ ∂B1 (0)} such that H N −1 (C) 6 H N −1 (∂B1 (0))/3.

Whence λ = λ1 (C) > λ1 (E(π/3)) = λ1 ({x ∈ ∂B1 (0) : arccos(hx, (0, . . . , 0, 1)i < π/3)}).

Since E(π/2) (the half-sphere) has first eigenvalue equal to N − 1, we obtain the existence of γ > 0 such that λ1 (E(π/3)) = N − 1 + γ, and s 2 N −2 N −2 +λ− > 1 + δN for some δN > 0. α= 2 2 Subcase 1.2. Suppose that the sets {ui 6= 0}, {vi 6= 0} have at most one connected component. Then each nontrivial component is signed, and to fix ideas we can suppose that the nontrivial components are u ¯1 , . . . , u¯l > 0, and v¯1 , . . . , v¯m > 0.

40


If there exists i, j such that {ui > 0} ∩ {uj > 0} = ∅ (or the same for v), then we can argue exactly as in Subcase 1.1 and get α > 1 + δN . Thus, in the remaining case, {ui > 0} ∩ {uj > 0} 6= ∅ for every i, j = 1, . . . , l. Since each component ui is harmonic in ∪j {uj > 0}, by the maximum principle we must have {uj > 0} = {u1 > 0}. Analogously, {vi > 0} = {v1 > 0} ˜ ˜ for every i = q 1, . . . , m. In thisq case uj = lj u1 , vq j = lj v1 (with l1 = l1 = 1) and the pPm Pk P P l m 2 2 ˜ = 2 ˜2 functions u ˜ = i=1 ai li u1 = i=1 ai ui , v i=1 bi li v1 = i=1 bi vi belong to N the class Gloc (R ) introduced in [33, Definition 3.2]. Thus by [33, Lemma 6.1] we have that N (0, (¯ u, v¯), 0+ ) = N (0, (˜ u, v˜), 0+ ) = α is equal to 1 and moreover that Γ(¯u,¯v) is a hyperplane. Case 2: Either u¯1 ≡ . . . ≡ u¯k ≡ 0 or v¯1 ≡ . . . ≡ v¯k ≡ 0. Suppose that u ¯1 , . . . , u ¯l 6≡ 0, l 6 k, and that all the other functions are zero. If Γ(¯u,¯v) ∩ S N −1 = ∅, then all nontrivial components are harmonic, and thus α ∈ N. Moreover, if α = 1, then u ¯i = x · νi for some νi ∈ S N −1 and in particular Γu¯i is a hyperplane and Γ(¯u,¯v) is a vector space of dimension at most N − 1. Let us now suppose that Γ(¯u,¯v) ∩ S N −1 6= ∅. We will prove this case via an induction argument which follows very closely [33, Section 6]. Because of that, here we will be rather sketchy, providing however all the precise references for the details. We will show that either α > 1 + δN , or α = 1 and Γ(¯u,¯v) is a hyperplane. Observe that, in dimension 2, each connected component of S N −1 \ {ui 6= 0} is an arc, and that all these arcs must have the same length. Moreover, Γu¯i ∩ S N −1 cannot consist of a single point, as in that case λ = λ1 (E(π)) =

1 < 1, 4

a contradiction.

Take x ¯ ∈ Γ(¯u,¯v) ∩ S N −1 , and let A be a connected component of {¯ u1 6= 0} ∩ S N −1 (we take the first component, without loss of generality) so that x ¯ is one end of the arc A. Let y¯ denote the other end of A, so that u ¯1 (¯ y ) = 0. Clearly x ¯ 6= y¯, and thus there exists another arc B disjoint from A, having y¯ as one end point. Let z¯ be the other end of B. If z¯ 6= x ¯, we get the existence of at least three disjoint arcs and, as in Subcase 1.1, α > 1 + δ2 for some i. If z¯ = x ¯ then we conclude that each arc must be a half circle, and in this case we have α = 1 and Γ(¯u,¯v) is a hyperplane. Suppose now that we can prove that the same situation occurs for N − 1, Case 2. Take (¯ u, v¯) = rα (G(θ), H(θ)) in dimension N , belonging to BUx0 . First of all, we can suppose once again that S N −1 \ Γ(¯u,¯v) has at least one, and at most two connected components. As in [33, p. 306], for each y0 ∈ Γ(¯u,¯v) ∩ S N −1 we can take a blowup sequence centered at y0 : u¯i (y0 + tn x) , u˜i,n (x) = p H(y0 , (¯ u, v¯), tn )

which converges to

u ˜i = rγy0 g˜i (θ),

v¯i (y0 + tn x) v˜i,n (x) = p H(y0 , (¯ u, v¯), tn ) ˜ i (θ) v˜i = rγy0 h

u, v˜) only depends on for some γy0 > 0. By the homogeneity of u¯, v¯, as in [33, Lemma 6.3], (˜ N − 1 variables, and by the induction hypothesis either γy0 = 1 (and Γu˜,˜v is a hyperplane) or γy0 > 1 + δN −1 . Now either γy0 > 1 + δN −1 for some y0 – which gives α > 1 + δN −1 by the upper semicontinuity of x 7→ N (x, (¯ u, v¯), 0+ ) as in [33, Lemma 6.5] – or γy0 = 1 for every y0 . In the latter case, we can reason exactly as in [33, p. 307] to conclude that S N −1 ∩ Γ(¯u,¯v) is a hyperplane, and α = 1.


41

Remark 3.8. From the previous proof we learn that, given (¯ u, v¯) ∈ BUx0 with x0 ∈ R(u,v) , we have that Γ(¯u,¯v) is a vector space having dimension at most N − 1. Moreover, it is actually a hyperplane except if possibly either u ¯ ≡ 0 or v¯ ≡ 0. Via a “Clean-Up” result, in the next subsection we will check that neither u ¯ nor v¯ can vanish when we blowup at a point of R(u,v) . This will yield a more complete characterization of (¯ u, v¯), cf. Theorem 3.16 ahead. Corollary 3.9. The set S(u,v) is closed in Ω for every N > 2. Proof. This is a direct consequence of Proposition 3.7 combined with the upper semicontinuity of the map x 7→ N (x, (u, v), 0+ ). Theorem 3.10. For any N > 2 we have that: 1. Hdim (Γ(u,v) ) 6 N − 1; ˜ ⋐ Ω we 2. Hdim (S(u,v) ) 6 N − 2. If N = 2 then moreover for any given compact set Ω ˜ have that S(u,v) ∩ Ω is a finite set. Proof. Once one knows that both the sets Γ(u,v) and S(u,v) are relatively close in each open subset of Ω, and taking in considerations the results on the previous and current subsection, the proof follows almost word by word the one presented in [33, Theorem 4.5] for a similar situation (see also Remark 4.7 therein). The proof there, in turn, is based on the Hausdorff Reduction Principle. 3.3. A Clean-Up result. R(u,v) is locally flat. Having established that S(u,v) is negligible, we will now focus on the remaining part of nodal set, namely R(u,v) . We prove in this and in the following subsection that R(u,v) is a regular hypersurface. Here we will prove a Clean-Up result, which will imply that both u ¯, v¯ 6≡ 0, whenever (¯ u, v¯) ∈ BUx0 with x0 ∈ R(u,v) . This will allow us to prove that, after a blowup at any point, R(u,v) is a hyperplane, and it is locally flat. Given x ∈ Ω, ν ∈ S N −1 and ε > 0, we define the strip S(x, ν, ε) := {y ∈ RN : |(y − x) · ν| 6 ε}.

We saw in the previous subsection (cf. Remark 3.8) that, whenever we take a blowup sequence centered at a point of R(u,v) , the limiting nodal set is a vector space, being at most a hyperplane. This implies that R(u,v) is locally contained in a small strip. Lemma 3.11. Given x0 ∈ R(u,v) , let R0 > 0 be such that B2R0 (x0 ) ∩ R(u,v) = B2R0 (x0 ) ∩ Γ(u,v) (recall that R(u,v) is an open set of Γ(u,v) ). Then

∀ε > 0, ∃0 < r0 < R0 : ∀0 < r < r0 , ∀x ∈ Γ(u,v) ∩ BR0 (x0 ), ∃ν = ν(x, r) ∈ S N −1 such that Γ(u,v) ∩ Br (x) ⊂ S(x, ν(x, r), εr).

Proof. Supposing that the conclusion does not hold, then there exists ε > 0, rn → 0, xn ∈ R(u,v) ∩ BR0 (x0 ) such that Γ(u,v) ∩ Brn (xn ) 6⊆ S(xn , ν, εrn ),

∀ν ∈ S N −1 .

Consider the blowup sequences ui,n (x) =

ui (xn + rn x) , ρn

vi,n (x) =

vi (xn + rn x) . ρn

42


Then we have ∀ν ∈ S N −1 .

Γ(un ,vn ) ∩ B1 (0) 6⊆ S(0, ν, ε)

(3.19)

Suppose that xn → x¯ ∈ R(u,v) ∩ BR0 (x0 ). We have (up to a subsequence) ui,n → u¯i ,

vi,n → v¯i

strongly in C 0,α (B 1 (0)) ∩ H 1 (B1 (0))

and, as N (¯ x, (u, v), 0+ ) = 1, we know (Remark 3.8) that Γ(¯u,¯v) is a vector space of dimension at most N − 1. Thus there exists ν¯ ∈ S N −1 and γ > 0 such that k X (¯ u2i + v¯i2 ) > 2γ,

in B1 (0) \ S(0, ν¯, ε) = {y ∈ B1 (0) : |y · ν¯| > ε}

i=1

and hence, for large n,

k X

2 (¯ u2i,n + v¯i,n ) > γ,

i=1

in B1 (0) \ S(0, ν¯, ε),

which contradicts (3.19).

Next we state an independent result, based on a similar one for harmonic functions present in [10, Lemma 12]. Lemma 3.12. Consider the ball B1 (0) ⊆ RN . Then for every a > 0 there exist C, ε¯ > 0 such that, for every 0 < ε < ε¯, and every w ∈ C(B1 (0)) nonnegative, satisfying −∆w 6 aw in B1 (0) and w ≡ 0 in B1 (0) \ S(0, ν, ε) for some ν ∈ S N −1 , we have sup w 6 e−C/ε sup w. B1/2 (0)

B1 (0)

Proof. Take κ > 0 large so that

1 , and take ε¯ = min{ 2κ and

q

|Bκ (0) ∩ S(0, ν, 1)| 6 N +2 2κ2 a }.

1 |Bκ (0)| 4

Given 0 < ε < ε¯, and x ∈ B1/2 (0), we have Bκε (x) ⊂ B1 (0), 1 |Bκε (x) ∩ S(0, ν, ε))| 6 . |Bκε (x)| 4

Then, given x ∈ B1/2 (0), as −∆w 6 aw in B1 (0),4 Z 1 a(κε)2 w(x) 6 sup w w dx + |Bκε (x)| Bκε (x) 2(N + 2) Bκε (x) 6 6

1 |Bκε (x) ∩ S(0, ν, ε)| sup w + sup w |Bκε (x)| 4 Bκε (x) Bκε (x) 1 sup w, 2 Bκε (x)

4The first inequality is an easy adaptation of the standard proof of the mean value theorem for subharmonic functions, see for instance [19, Theorem 2.1].


We can now iterate this procedure int

1−|x| κε

times (where int( · ) denotes the integer part

of a number), and hence 1 2κε 1−|x| 1 1 κε sup w 6 sup w w(x) 6 2 2 B1 (0) B1 (0) Thus the conclusion holds for C :=

log 2 2k

43

∀x ∈ B1/2 (0),

> 0.

Given x0 ∈ R(u,v) and 0 < ε < ε¯, let R0 , r0 be as in Lemma 3.11. For x ∈ Γ(u,v) ∩ BR0 (x0 ) and 0 < r < r0 , we define the set Υ(x,r) = {ν ∈ S N −1 : Γ(u,v) ∩ Br (x) ⊆ S(x, ν, εr)} 6= ∅. Lemma 3.13. If there exists r¯ 6 r0 , x¯ ∈ Γ(u,v) ∩ BR0 (x0 ), and ν¯ ∈ Υ(¯x,¯r) such that k X

u2i = 0

in Br¯(¯ x) \ S(¯ x, ν¯, ε¯ r),

i=1

then k X

u2i = 0

i=1

in Br (x) \ S(x, ν, εr) ∀x ∈ Γ(u,v) ∩ BR0 (x0 ), r 6 r0 , ν ∈ Υ(x,r) .

P Proof. Take x ∈ Γ(u,v) ∩ BR0 , r < r0 . Observe that if ki=1 u2i = 0 in Br (x) \ S(x, ν, εr) for Pk P some ν ∈ Υ(x,r), then i=1 vi2 > 0 in such set. Moreover, if ki=1 u2i = 0 in Br (x) \ S(x, ν, εr) for some ν ∈ Υ(x,r) , then the same is true for all vectors in the latter set. Pk Thus the map ϕ : BR0 (x0 )×(0, r0 ) → {0, 1} defined as ϕ(x, r) = 0 if i=1 u2i = 0 on Br (x)\ S(x, ν, εr) for some ν ∈ Υ(x,r), ϕ(x, r) = 1 otherwise, is continuous, hence constant. Combining this result with Lemma 3.12, we have the following. Lemma 3.14. Suppose there exists r¯ 6 r0 , x¯ ∈ Γ(u,v) ∩ BR0 (x0 ), and ν¯ ∈ Υ(¯x,¯r) such that k X

u2i = 0

in Br¯(¯ x) \ S(¯ x, ν¯, ε¯ r),

i=1

Then sup

k X

Br/2 (x) i=1

u2i 6 e−C/ε sup

k X

Br (x) i=1

u2i ,

∀x ∈ Γ(u,v) ∩ BR0 (x0 ), r 6 r0 .

Proof. Take x ∈ Γ(u,v) ∩ BR0 (x0 ), r 6 r0 and ν ∈ Υ(x,r). From the previous lemma we know Pk Pk that u ˜ := i=1 u2i = 0 in Br (x) \ S(x, ν, εr) . Moreover, −∆˜ u 6 2( i=1 λ2i )˜ u in Br (x), with λi = µi /ai . Hence, rescaling u ˜: u ˜1 (y) = u ˜(x + ry), we have −∆˜ u1 6 a˜ u1 in B1 (0) for a = 2R02 Then, by Lemma 3.12,

k X

λ2i ,

and

i=1

u˜1 = 0 in B1 (0) \ S(0, ν, ε).

sup u˜1 6 e−C/ε sup u ˜1 , B1/2 (0)

and the lemma follows going back to u˜.

B1 (0)

44


Proposition 3.15 (Clean-Up). Take x ¯ ∈ Γ(u,v) ∩ BR0 (x0 ) and suppose that, for small r, k X i=1

u2i ≡ 0

in Br (¯ x) \ S(¯ x, ν(¯ x, r), εr)

for some ν(¯ x, r) ∈ Υ(¯x,r) . Then

The same results holds for

Pk

k X i=1

i=1

u2i ≡ 0

in BR0 (x0 ).

vi2 in the place of

Pk

i=1

u2i .

Proof. We want to check that the measures ∆ui + λi ui over BR0 (x0 ) is zero for each i = 1, . . . , k, doing this by covering this set with balls of radius 1/2n+1 (here, λi = µi /ai ). Take n ¯ such that 1/2n¯ 6 r0 . Then sup

k X

B1/2n+1 (¯ x) i=1

u2i 6 e−C/ε

sup B1/2n (¯ x)

We can then iterate, obtaining k X

sup

B1/2n (¯ x) i=1

k X

u2i ,

¯. ∀¯ x ∈ Γ(u,v) ∩ BR0 (x0 ), n > n

i=1

C

u2i 6 C1 e− ε (n−¯n) ,

∀n > n ¯ + 1,

for C1 depending on kuk∞ . Observe that if B1/2n+1 (x) ∩ Γ(u,v) = ∅, then ∆u1 + λ1 u1 = 0 on such a ball. Thus (in the sense of measures) Z Z Z Z − + (∆u− + λ u ) dx − (∆ui + λi ui ) dx = (∆ui + λi ui ) dx = (∆u+ i i i + λi ui ) dx i A

A

A

BR0 (x0 )

for

[

A :=

B1/2n+1 (¯ x).

x ¯∈Γ(u,v) ∩BR0 (x0 )

On the other hand, using a cut-off function argument, it is straightforward to show that v u k Z Z Z uX ± ± ± 2n 2n t u2 dx (∆u + λ u ) dx 6 C 2 u dx 6 C 2 06 B1/2n+1 (¯ x)

i

i i

2

B1/2n (¯ x)

i

2

B1/2n (¯ x)

i

i=1

22n − C (n−¯n) C e 2ε 6 C3 e− 2ε n , ∀n > n ¯ , N > 2. 2nN Since we can cover BR0 (x0 ) with ((int(R0 ) + 1)2n+2 + 2)N balls of radius 1/2n+1, we deduce that Z XZ + + + 0 6 (∆ui + λi ui ) dx 6 (∆u+ i + λi ui ) dx 6 C2

A

x ¯∈A

6 C˜4 2

B1/2n+1 (¯ x)

C n nN − 2ε

e

C

= C4 en(N log 2− ε ) → 0

as n → ∞,

C if we can choose right from the start ε < N log 2 . Since the same result clearly holds also for − ui , we conclude that −∆ui = λi ui in BR0 (x0 ), and by the unique continuation property, since ui vanishes in a set with nonempty interior, then ui ≡ 0.

This result has many strong and important consequences.


45

Theorem 3.16. Take x0 ∈ R(u,v) and (¯ u, v¯) ∈ BUx0 . Then u¯, v¯ 6≡ 0. Moreover, Γ(¯u,¯v) is a hyperplane, and there exist ν ∈ S N −1 and α1 , . . . , αk , β1 , . . . , βk ∈ R such that v¯i = βi (x · ν)−

u ¯i = αi (x · ν)+ ,

(3.20)

in RN .

Furthermore, k X

(3.21)

ai α2i =

so that the function

i=1

ai αi u¯i −

bi βi2 ,

i=1

i=1

Pk

k X

Pk

¯i i=1 bi βi v

is harmonic in RN .

Proof. 1. Let (¯ u, v¯) ∈ BUx0 with x0 ∈ R(u,v) and suppose, in view of a contradiction, that u¯ ≡ 0. Take xn ∈ Γ(u,v) with xn → x0 and tn → 0 such that u(xn + tn x) un (x) = p →u ¯, H(xn , (u, v), tn )

v(xn + tn x) vn (x) = p → v¯. H(xn , (u, v), tn )

As Γ(¯u,¯v) = Γv¯ is a vector space of dimension at most N − 1 (cf. Remark 3.8), then there exists ν¯ ∈ S N −1 such that, for every ε > 0, u ¯ ≡ 0 and v¯ 6≡ 0 in B1 (0) \ S(0, ν¯, ε). For ε > 0 fixed let γ > 0 be such that k X

v¯i2 > 2γ

in B1 (0) \ S(0, ν¯, ε).

2 vi,n >γ

in B1 (0) \ S(0, ν¯, ε),

i=1

Then, for sufficiently large n, k X i=1

and so k X

vi2

> 0,

i=1

k X i=1

u2i ≡ 0

in Btn (xn ) \ S(xn , ν¯, tn ε).

As tn < r0 , xn ∈ Γ(u,v) ∩ BR0 (x0 ) for large n, and ν¯ ∈ Υ(xn ,tn ) , then by Proposition 3.15 we obtain u ≡ 0 in Br (x0 ) for small r > 0. However, this contradicts the fact that x0 ∈ Γ(u,v) = ∂ωu ∩ Ω = ωv ∩ Ω (Corollary 2.24). 2. Next, taking into account the proof of Proposition 3.7 (see also Remark 3.8), we have that Γ(¯u,¯v) is a hyperplane, S N −1 splits into two half spheres S + , S − , and u¯i = rgi (θ),

v¯i = rhi (θ),

where gi , hi are all first eigenfunctions respectively on S + and S − . Thus there exists ν so that (3.20) holds. 3. As +

∆¯ ui = 0 in Γ := {x · ν > 0},

−

∆¯ vi = 0 in Γ := {x · ν < 0},

and Γ(¯u,¯v) = {x · ν = 0}, given a ball Br (x0 ) we have, by using the Rellich-type formula div((x − x0 )|∇¯ ui |2 − 2hx − x0 , ∇¯ ui i∇¯ ui = (N − 2)|∇¯ ui |2 − 2hx − x0 , ∇¯ ui i∆¯ ui

46

M. RAMOS, H. TAVARES, AND S. TERRACINI +

in Br (x0 ) ∩ Γ , Z r

2

+

∂Br (x0 )∩Γ

2

(|∇¯ ui | − 2(∂n u ¯i ) )dσ = (N − 2) Z −

Z

Br (x0 )∩Γ(u,¯ ¯ v)

Analogously, Z r

(|∇¯ vi |2 − 2(∂n v¯i )2 )dσ = (N − 2) − ∂Br (x0 )∩Γ Z +

i=1

and hence

Br (x0 )∩Γ(u,¯ ¯ v)

Pk

i=1

ai α2i =

|∇¯ ui |2 dx

|∇¯ ui |2 hν, x − x0 i dσ

Z

−

Br (x0 )∩Γ

Br (x0 )∩Γ(u,¯ ¯ v)

Therefore k Z X

+

Br (x0 )∩Γ

|∇¯ vi |2 dx

|∇¯ vi |2 hν, x − x0 i dσ.

ai |∇¯ ui |2 − bi |∇¯ vi |2 hν, x − x0 i dσ = 0

∀x0 ∈ RN , r > 0,

Pk

2 i=1 bi βi .

We are now ready to improve the result of Lemma 3.11, showing that Γ(u,v) ∩ BR0 (x0 ) verifies the so called (N − 1)–dimensional (δ, R0 )– Reifenberg flatness condition for every small ε and some r0 = r0 (ε) > 0. Corollary 3.17. Within the previous framework, for any given δ > 0 there exists R0 > 0 such that for every x ∈ Γ(u,v) ∩ BR0 (x0 ) and 0 < r < R0 there exists a hyperplane H = Hx,r containing x such that 5 dH (ΓU ∩ Br (x), H ∩ Br (x)) 6 δr. Proof. In the ground of Theorem 3.16, the proof follows exactly as the one of [33, Lemma 5.3]. The Reifenberg flat condition yields a local separation result. Corollary 3.18. Given x0 ∈ R(u,v) , there exists R0 such that Γ(u,v) ∩ BR0 (x0 ) = R(u,v) ∩ BR0 (x0 ), and Moreover, k X i=1

BR0 (x0 ) \ Γ(u,v) u2i > 0,

k X i=1

has exactly two connected components Ω1 , Ω2 .

vi2 ≡ 0 on Ω1

and

k X i=1

u2i = 0,

k X

vi2 > 0 on Ω2 .

i=1

Finally, given y ∈ Γ(u,v) ∩ BR0 (x0 ) and 0 < r < R0 there exists a hyperplane H = Hy,r containing y and a unitary vector ν = νy,r orthogonal to H such that (3.22) {x + tν ∈ Br (y) : x ∈ H, t > δr} ⊂ Ω1 ,

{x − tν ∈ Br (y) : x ∈ H, t > δr} ⊂ Ω2 .

5Here d (A, B) := max{sup H a∈A dist(a, B), supb∈B dist(A, b)} denotes the Hausdorff distance. Notice that dH (A, B) 6 δ if and only if A ⊆ Nδ (B) and B ⊆ Nδ (A), where Nδ (·) is the closed δ–neighborhood of a set. Observe moreover that when H is a hyperplane, Nδ (H) is a strip S(x, ν, δ) for x ∈ H and ν ∈ S N−1 a vector orthogonal to H.


47

Proof. See [33, Proposition 5.4], or [32, Proposition 3.38] for a detailed proof. For a result in the same direction, check [24, Theorem 4.1] Combining (3.22) with Corollary 3.17, we see that both Ω1 and Ω2 as defined above are (δ, R0 )– Reifenberg flat domain. We suppose that δ > 0 is small enough so that Ω1 and Ω2 are also NTA domains (check Definitions A.1 and A.4 ahead, and [27, Theorem 3.1] for the proof of this result). 3.4. The set R(u,v) is locally a regular hypersurface. Define p p √ √ V (x) := ( b1 v1 (x), . . . , bk vk (x)), U (x) := ( a1 u1 (x), . . . , ak uk (x)),

and observe that locally R(u,v) is characterized as being the set of zeros of |U (x)| − |V (x)|. We will prove the regularity of such set by proving by definition that |U | − |V | is differentiable at each point of R(u,v) , having nonzero gradient (Theorem 3.27). We fix x0 ∈ R(u,v) and work in BR0 (x0 ) as in the previous subsection. We also recall that Ω1 = {x ∈ BR0 (x0 ) :

k X

u2i > 0},

and

i=1

Ω2 = {x ∈ BR0 (x0 ) :

k X

vi2 > 0}.

i=1

To simplify the notation, in this subsection we denote Γ := Γ(u,v) ∩ BR0 (x0 ). Given x ∈ Ω1 , define √ √ ( a1 u1 (x), . . . , ak uk (x)) U (x) U(x) = = p |U (x)| a1 u21 (x) + . . . + ak u2k (x) and, for x ∈ Ω2 ,

V(x) =

√ √ ( b1 v1 (x), . . . , bk vk (x)) V (x) = p 2 . |V (x)| b1 v1 (x) + . . . + bk vk2 (x)

Our main interest will be to evaluate U, V at Γ. By Corollary 2.24 (or Proposition A.5 combined with Theorem 3.16) we have that at least one component of u and of v is signed. Suppose from now on, without loss of generality, that u1 > 0 in Ω1 and v1 > 0 in Ω2 . This crucial fact allows us to apply the regularity results shown in Appendix A. More precisely, (u, v, int(Ω1 ∪ Ω2 )) ∈ G (cf. Definition A.6) and so Proposition A.7 implies that U and V can be uniquely extended up to Γ, as soon as one observes that for instance U can be rewritten as √ √ √ ( a1 , a2 uu21 . . . , ak uuk1 ) U=r 2 2 a1 + a2 uu21 . . . + ak uuk1 Moreover, from this proposition we also deduce that (3.23)

|U(x) − U(x0 )| 6 Crα ,

|V(x) − V(x0 )| 6 Crα ,

∀x ∈ Br (x0 ).

Observe that, formally, U(x0 ) = lim

x→x0

−∂n U (x0 ) U (x)/dist(x, ∂Ω) = , |U (x)/dist(x, ∂Ω)| |∂n U (x0 )|

V(x0 ) =

−∂n V (x0 ) |∂n V (x0 )|

for x0 ∈ Γ.

Rigorously, however, we can only for the moment observe that U (y), for y ∈ Γ, is related to the blowup limits centered at y.

48


Lemma 3.19. Take y ∈ Γ and let

(¯ u, v¯) = (α1 (x · ν)+ , . . . , αk (x · ν)+ , β1 (x · ν)− , . . . , βk (x · ν)− ) ∈ BUy .

Then

√ √ √ √ ¯ 1 , . . . , ak u ¯k ) ( a1 u ( a1 α1 , . . . , ak αk ) p U(y) = p = a1 u ¯21 + . . . + ak u ¯2k a1 α21 + . . . + ak α2k

and

√ √ √ √ ( b 1 β1 , . . . , b k βk ) ( b1 v¯1 , . . . , bk v¯k ) p p = . V(y) = b1 v¯12 + . . . + bk v¯k2 b1 β12 + . . . + bk βk2

Proof. Let (¯ u, v¯) ∈ BUy be as in the statement, limit of a blowup sequence centered at xn → y and with parameter tn ↓ 0. Take z¯ ∈ RN such that |(¯ u1 (¯ z ), . . . , u ¯k (¯ z ))| 6= 0 and (for large n) xn + tn z¯ ∈ Ω1 . Then √ √ ( a1 u¯1 , . . . , ak u¯k ) U (xn + tn z¯) U (xn + tn z¯)/ρn U(y) = lim = lim = p , n→∞ |U (xn + tn z ¯)| n→∞ |U (xn + tn z¯)/ρn | a1 u ¯21 + . . . ak u ¯2k and a similar property holds for V.

We now define two important auxiliary functions. Definition 3.20. Given x0 ∈ Γ we define ux0 (x) = U(x0 ) · U (x),

vx0 (x) = V(x0 ) · V (x).

The ideia for considering these functions is the following: given a blowup limit (¯ u, v¯), the Almgren’s monotonicity formula tells us that a certain linear combination of u ¯i extends in an harmonic way with a linear combination of v¯i (Theorem 3.16). However, the coefficients of each linear combination, in general, may vary from point to point (namely the αi and βi ). Fixing U, V at x0 together with estimate (3.23) will allow us to control in some sense this variation. Let us see that ux0 , vx0 are positive close to x0 , and check which problem they solve. Lemma 3.21. There exists r0 such that ux0 > 0, vx0 > 0 in Br0 (x0 ). In particular, there exist positive Radon measures Mx0 , Nx0 , both concentrated on Γ, such that −∆ux0 =

k X µi i=1

ai

Ui (x0 )ui − Mx0 ,

−∆vx0 =

k X νi i=1

bi

Vi (x0 )vi − Nx0

in Br0 (x0 ).

Proof. By using (3.23) we see that ux0 (x) = U(x) · U (x) + (U(x0 ) − U(x)) · U (x) > (1 − Cr0α )|U (x)| > 0

∀x ∈ Br0 (x0 ),

whenever r0 > 0 is small enough. The rest of the proof follows easily.

Our aim now it to analyze the difference ϕx0 := ux0 − vx0 , as a first step before analysing |U | − |V | at x0 . For that, let us take a blowup sequence centered at y ∈ Γ ∩ Br0 (x0 ). Define, given tn → 0+ , ux0 (y + tn x) n , uy,t x0 (x) := p H(y, (u, v), tn )

vx (y + tn x) n (x) := p 0 vxy,t . 0 H(y, (u, v), tn )


49

Up to a subsequence, there exists (¯ u, v¯) ∈ BUy such that √ √ y,tn ¯ =: u¯yx ¯ 1 , . . . , ak u ¯k ) = U(x0 ) · U ux0 → U(x0 ) · ( a1 u 0 p p y,tn y ¯ vx0 → V(x0 ) · ( b1 v¯1 , . . . , bk v¯k ) = V(x0 ) · V =: v¯x0 .

Observe that, if y = x0 , then by Lemma 3.19 one has q q ¯2k , v¯yy = b1 v¯12 + . . . + bk v¯k2 u ¯yy = a1 u¯21 + . . . + ak u

for some (¯ u, v¯) ∈ BUy .

In such a case, by Theorem 3.16 we have with

Pk

(3.24)

i=1

v¯i = βi (x · ν)−

u¯i = αi (x · ν)+ ,

Pk

so that u ¯yy − v¯yy is harmonic. In particular, v v u k u k uX uX y t 2 ¯ |∇uy (0)| = |∇U (0)| = vyy (0)| ai αi = t bi βi2 = |∇V¯ (0)| = |∇¯

ai α2i =

2 i=1 bi βi ,

i=1

i=1

Lemma 3.22. Under the previous notations, let 1 1 n n My,t (·) := Mx0 (y + tn ·), Nxy,t N (y + tn ·) x0 (·) := 0 N −2 −2 x0 ρ n tn ρ n tN n with ρ2n = H(y, (u, v), tn ), so that in Br0 /tn (x0 − y) n −∆uy,t x0

=

Then there exists and moreover Z y Mx0 (E) =

E∩Γ

t2n

k X µi

i=1 y Mx0 ,

ai

Ui (x0 )ui,n −

t2n

=

y

n ⇀ Mx0 , Mxy,t 0

=

|∇¯ uyx0 (0)||E∩Γ|,

k X νi i=1

y

N x0 such that

|∇¯ uyx0 | dσ

−∆vxy,n 0

n My,t x0 ,

bi

Vi (x0 )vi,n − Nxy,n 0

y

n ⇀ N x0 Nxy,t 0

y N x0 (E)

=

Z

E∩Γ

vxy0 (0)||E∩Γ|. |∇¯ vxy0 | dσ = |∇¯

In particular, when y = x0 , one has that the limiting measures coincide, that is y

y

My ≡ N y . Proof. The proof is similar to the one of Theorem 3.1 and of (3.21) in Theorem 3.16, and so we omit it here. This result, together with (3.23) and (3.24), allows us to compare Mx0 and Nx0 close to x0 . Lemma 3.23. We have Mx0 6 (1 + Crα )Nx0

Nx0 6 (1 + Crα )Mx0

and

in Br (x0 )

for some C > 0 independent of x0 and r > 0. In particular, the new functions ϕx0 ,r := (1 + Crα )ux0 − vx0 ,

and

ϕx

0 ,r

:= ux0 − (1 + Crα )vx0

satisfy −∆ϕx0 ,r 6 (1 + Crα )

k X µi i=1

ai

Ui (x0 )ui −

k X νi i=1

bi

Vi (x0 )vi

50


and −∆ϕx in Br (x0 ).

0 ,r

>

k X µi i=1

ai

Ui (x0 )ui − (1 + Crα )

k X νi i=1

bi

Vi (x0 )vi

Proof. 1. We claim that (3.25)

lim

t→0

Mx0 (B t (y)) > 1 − Crα Nx0 (B t (y))

for every y ∈ Γ(u,v) ∩ Br (x0 ).

Fix y ∈ Γ(u,v) ∩ Br (x0 ) and consider any arbitrary sequence tn ↓ 0. Using the previous y,tn y,tn n n notations, take uy,t and the rescaled measures My,t x0 (x), vx0 x0 , Nx0 ; defined the blowup y y y y limits by u¯x0 , v¯x0 , Mx0 , N x0 . One has (by [18, Sect. 1.6, Th. 1]) y

M (B 1 (0)) |∇¯ uyx0 (0)| My,n Mx0 (B tn (y)) x0 (B 1 (0)) . = lim y,n = yx0 = (3.26) lim n Nx (B 1 (0)) k Nx (B t (y)) |∇¯ vxy0 (0)| N x0 (B 1 (0)) 0 0 n √ √ √ ¯ := (√a1 u ¯1 , . . . , ak u¯k ), V¯ := ( b1 v¯1 , . . . , bk v¯k ), Now, for some (¯ u, v¯) ∈ BUy , and U |U(x0 ) · ∇U ¯ (0)| − |U(y) · ∇U ¯ (0)| 6 |(U(x0 ) − U(y)) · ∇U¯ (0)| 6 Crα |∇U ¯ (0)| = Crα |∇¯ uyy (0)|,

so

|∇¯ uyy (0)| 6 Crα |∇¯ uyy (0)| uyx0 (0)| − |∇¯

where we have used (3.23) and (3.24). Analogously, one has |∇¯ vyy (0)| 6 Crα |∇¯ vyy (0)| vxy (0)| − |∇¯ 0

Combining these two inequalities with |∇¯ uyy (0)| = |∇¯ vyy (0)| (cf. (3.24)) and (3.26) yields the validity of our initial claim (3.25). 2. By the previous point, we have DNx0 Mx0 (y) 6 (1 + Crα ) for Nx0 –a.e. y ∈ Br0 (x0 ) and hence the Radon-Nikodym Decomposition Theorem (check for instance [18, Sect. 1.6, Th. 3]) yields that for every Borel set E ⊆ Br0 (x0 ) Z DNx0 Mx0 dNx0 6 (1 + Crα )Nx0 (E). Mx0 (E) = (Mx0 )s (E) + E

(where (Mx0 )s > 0 represents the singular part of Mx0 with respect to Nx0 ), and this proves the first inequality in the proposition. The proof of the other inequality follows in the exact same way. Recall that ϕx0 := ux0 − vx0 and let ψx0 ,r , for each small r > 0, be the solution of ( P P −∆ψx0 ,r = ki=1 µaii Ui (x0 )ui − ki=1 νbii Vi (x0 )vi in Br (x0 ) on ∂Br (x0 ). ψx0 ,r = ϕx0 = ux0 − vx0 As we will see ahead, ∇ψx0 ,r (x0 ) will be a good approximation, for r small, to a normal vector of R(u,v) at x0 . Having that in mind, our next aim is to estimate the difference ϕx0 − ψx0 ,r in terms of r, in each ball Br (x0 ). Introduce the auxiliary functions ψ x0 ,r , ψ x ,r defined in 0 Br (x0 ) by the conditions: ( Pk Pk −∆ψx0 ,r = (1 + Crα ) i=1 µaii Ui (x0 )ui − i=1 νbii Vi (x0 )vi in Br (x0 ) on ∂Br (x0 ). ψ x0 ,r = ϕx0 ,r = (1 + Crα )ux0 − vx0


and

(

−∆ψ x

ψx

0 ,r

0 ,r

=

= ϕx

Pk

µi i=1 ai Ui (x0 )ui

0 ,r

− (1 + Crα ) α

= ux0 − (1 + Cr )vx0

Pk

νi i=1 bi Vi (x0 )vi

51

in Br (x0 ) on ∂Br (x0 ).

By the comparison principle and since ux0 , vx0 > 0, we have that in Br (x0 ), 6 ϕx ,r 6 ϕx0 6 ϕx0 ,r 6 ψ x0 ,r 0 Pk Pk and moreover (since also i=1 µaii Ui (x0 )ui , i=1 νbii Vi (x0 )vi > 0) ψx

0 ,r

ψx

0 ,r

6 ψx0 ,r 6 ψ x0 ,r

in Br (x0 ).

Now, since ψ x0 ,r − ψ x ,r satisfies the problem 0  Pk µi α −∆(ψ ) = Cr − ψ x0 ,r i=1 ai Ui (x0 )ui + x0 ,r α ψ x ,r − ψ = Cr (ux0 + vx0 )

νi bi Vi (x0 )vi

x0 ,r

0

in Br (x0 ) on ∂Br (x0 ).

we have, by using also the maximum principle [19, Theorem 3.7] and the fact that ui , vi are Lipschitz continuous and zero at x0 , kϕx0 − ψx0 ,r kL∞ (Br (x0 )) 6 kψ x0 ,r − ψ x

0 ,r

kL∞ (Br (x0 )) ′ α

′ α

6 C r kux0 + vx0 kL∞ (∂Br (x0 )) + C r 6 C ′′ r1+α

(3.27)

k X i=1

∀x ∈ Br (x0 ).

kui + vi kL∞ (Br (x0 ))

Proposition 3.24. There exists ν(x0 ) := lim ∇ψx0 ,r (x0 ). r→0

N

Moreover, the map Γ → R , x0 7→ ν(x0 ) is H¨ older continuous of order α. Proof. 1. Fix r > 0 and let us prove that {∇ψx0 ,r/2k (x0 )}k is a Cauchy sequence. One has kϕx0 − ψx0 ,r kL∞ (Br (x0 )) 6 Cr1+α ,

kϕx0 − ψx0 ,r/2 kL∞ (Br/2 (x0 )) 6

C 1+α r , 21+α

which implies that kψx0 ,r − ψx0 ,r/2 kL∞ (Br/2 (x0 )) 6 kψx0 ,r − ϕx0 kL∞ (Br/2 (x0 )) + kψx0 ,r/2 − ϕx0 kL∞ (Br/2 (x0 )) C 1+α 1 1+α 6 Cr + 1+α r = C 1 + 1+α r1+α 2 2

As ψx0 ,r − ψx0 ,r/2 is harmonic in Br/2 (x0 ), then by using classical estimates we have |∇ψx0 ,r (x0 ) − ∇ψx0 ,r/2 (x0 )| 6 C ′ (1 +

1

21+α

)rα .

Iterating this procedure, we obtain (3.28)

|∇ψx0 ,r/2n+m (x0 ) − ∇ψx0 ,r/2m (x0 )| 6 C ′ (1 +

1 21+α

)

m+n−1 X i=m

1 2αi

!

rα .

Thus, for each r > 0, {∇ψx0 ,r/2k (x0 )}k is a Cauchy sequence, and thus ν(x0 ) := limk→∞ ∇ψx0 ,r/2k (x0 ) exists.

52


2. Now let rn → 0 be an arbitrary sequence. Suppose, without loss of generality, that r r n 2n+1 < rn 6 2n (if not, one can pass to a subsequence of {r/2 }n . Then kϕx0 − ψx0 ,rn kL∞ (Br/2n+1 (x0 )) 6 Crn1+α ,

kϕx0 − ψr/2n+1 kL∞ (Br/2n+1 (x0 )) 6

C r1+α . 2(n+1)(1+α)

and so kψx0 ,rn − ψx0 ,r/2n+1 kL∞ (Br/2n+1 (x0 )) 6 Crn1+α + 6

C 2n(1+α)

Once again by classical estimates, |∇ψx0 ,rn (x0 ) − ∇ψx0 ,r/2n+1 (x0 )| 6 C and thus

C 2(n+1)(1+α)

r1+α +

1 2n(1+α)

+

r1+α

C 2(n+1)(1+α) 1

2(n+1)(1+α)

r1+α .

rα

|∇ψx0 ,rn (x0 ) − ν(x0 )| 6 |∇ψx0 ,rn (x0 ) − ∇ψx0 ,r/2n+1 (x0 )| + |∇ψx0 ,r/2n+1 (x0 ) − ν(x0 )|

and now the conclusion follows by taking the limit as n → ∞.

3. Finally, let us prove that x0 7→ ν(x0 ) is H¨ older continuous of order α. Take x0 , x1 ∈ Γ and take r := |x0 − x1 |. From (3.28) we have show that there exists C > 0 (independent of the point and of the radius) such that |∇ψxi ,r − ν(xi )| 6 Crα .

On the other hand, the H¨ older continuity of U, V one can easily show that kϕx1 − ϕx0 kL∞ (Br (x0 )∩Br (x1 )) 6 Cr1+α . Moreover, kϕxi − ψr,xi kL∞ (Br (x0 )) 6 Cr1+α . By combining all this we get |∇ψx0 ,r (x0 ) − ∇ψx1 ,r (x0 )| 6 Crα , and the proof is complete. Remark 3.25. For future reference, we observe that from the previous proof it is also clear that, given xr ∈ Br/2 (x0 ), we also have ∇ψx0 ,r (xr ) → ν(x0 )

as r → 0.

The following lemma relates ν(x0 ) some blowup limits, providing at the same time that ν(x0 ) 6= 0. Lemma 3.26. Given x0 ∈ Γ(u,v) . Then for every (¯ u, v¯) ∈ BUx0 with fixed center x0 we have that + − u(x0 ) ν(x0 ) and v¯i = βi x · , u ¯i = αi x · |ν(x0 )| |ν(x0 )| for some αi , βi ∈ R (not all zero). In particular, ν(x0 ) 6= 0. Proof. Take u ¯i , v¯i coming from a limit of tn → 0 with fixed center x0 , and define ux (x0 + tn x) vx (xn + tn x) ϕx0 ,n (x) := p 0 −p 0 in B1 (0), H(x0 , (u, v), tn ) H(x0 , (u, v), tn )

which converges to

v v u k u k u uX X √ √ + t 2 ¯k ) − V(x0 ) · (b1 v¯1 , . . . , bk v¯k ) = ai αi (x · ν˜) = t bi βi2 (x · ν˜)− U(x0 ) · ( a1 u¯1 , . . . , ak u i=1

i=1


53

for some αi , βi (i = 1, . . . , k) and ν˜ ∈ S N −1 . On the other hand, we have that ψx ,t (x0 + tn x) , ψex0 ,tn (x) := p 0 n H(x0 , (u, v), tn )

defined in B1 (0),

2,γ coincides at the boundary (for qP ∂B1 (0) with ϕx0 ,n (x), and hence it converges strongly C k 2 2 every γ ∈ (0, 1)) to ˜) . In particular, i=1 ai αi (x · ν v u k uX tn t ai α2i ν˜ = lim ∇ψex0 ,tn (0) = lim p ∇ψx0 ,tn (xn ) = Aν(x0 ), n n H(x , 0 (u, v), tn ) i=1 p with A 6= 0 being the limit of tn / H(x0 , (u, v), tn ). Hence ν(x0 ) k ν˜.

We are now ready to prove the regularity of the set R(u,v) .

Theorem 3.27. The map |U (x)| − |V (x)| =

q q a1 u21 + . . . + ak u2k − b1 v12 + . . . + bk vk2

is differentiable at each x0 ∈ R(u,v) , with

∇ (|U | − |V |) (x0 ) = ν(x0 ).

(3.29)

In particular, the set R(u,v) is locally a C 1,α –hypersurface, for some α ∈ (0, 1). Proof. We prove this by definition, namely we show that |U (x)| − |V (x)| − ν(x0 ) · (x − x0 ) →0 |x − x0 |

as x → x0 .

In the following, take r := 2|x − x0 |. We have

|ϕx0 (x) − ψx0 ,r (x)| 6 Crα → 0, |x − x0 |

and

by (3.27). Moreover,

|ψx0 ,r (x0 )| |ψx0 ,r (x0 ) − ϕx0 (x0 )| = 6 Crα → 0. |x − x0 | |x − x0 |

ψx0 ,r (x) − ψx0 ,r (x0 ) − ν(x0 ) · (x − x0 ) x − x0 = (∇ψx0 ,r (xr ) − ν(x0 )) · →0 |x − x0 | |x − x0 |

by Remark 3.25. Finally, (3.23) and the Lipschitz continuity of U and V yield

and so

(U(x) − U(x0 )) · U (x) →0 |x − x0 |

(V(x) − V(x0 )) · V (x) → 0, |x − x0 |

| U(x) · U (x) − V(x) · V (x) − ν(x0 ) · (x − x0 )| | |U (x)| − |V (x)| − ν(x0 ) · (x − x0 )| = |x − x0 | |x − x0 | |ϕx0 (x) − ν(x0 ) · (x − x0 )| + o(1) 6 |x − x0 | |ϕx0 (x) − ψx0 ,r (x)| |ψx0 ,r (x0 )| ψx0 ,r (x) − ψx0 ,r (x0 ) − ν(x0 ) · (x − x0 ) + + + o(1) 6 |x − x0 | |x − x0 | |x − x0 |

54


converges to 0. The last statement of the theorem is now a simple consequence of the Implicit Function Theorem, since ν(x0 ) 6= 0. Proof of Theorem 1.6 completed. Items (i), (ii) and (iii) are a consequence of Theorem 2.8 and Corollaries 2.12, 2.23 and 2.24. In item (iv), the fact that (¯ ω1 , . . . , ω ¯ m ) solves c∞ is a consequence of Corollary 2.12, while its regularity is provided by Theorem 3.27 (for N = 2, the more precise statement for Γ \ R follows also from Theorem 3.10, while the equal angle property, at this point, follows exactly as in and [33, p. 305]). Finally, the extremality condition in (v) comes out from (3.29). 4. The limit of the approximating solutions as p → ∞. Existence and regularity results for the initial problem 4.1. Properties of the approximating solutions up , vp . Proof of Theorems 1.3 and 1.5. We are now ready to prove our first two main results, Theorems 1.3 and 1.5. Until now, we have proved existence and regularity for the approximate problem (1.6) and so, in particular, for (1.5). As explained in the introduction, a solution of our initial model problem (1.1) will be obtained by taking p → +∞ in (1.5). Let us summarize what we have proved so far regarding the latter problem. Given p ∈ N, we have shown the existence of up = (u1p , . . . , ukp ), vp = (u1p , . . . , ukp ), with components in H01 (Ω) ∩ C 0,1 (Ω), such that up and vp are segregated (in the sense that uip · vjp ≡ 0 for every i, j), they are L2 –normalized Z Z 2 u2ip dx = vip dx = 1 ∀i = 1, . . . , k, Ω

Ω

and mutually H01 –orthogonal, Z Z ∇vip · ∇vjp dx = 0, ∇uip · ∇ujp dx = Ω

Ω

∀i 6= j.

We may assume that the components are ordered in the following way Z Z Z Z |∇vkp |2 dx. |∇v1,p |2 dx 6 . . . 6 |∇ukp |2 dx, |∇u1,p |2 dx 6 . . . 6 (4.1) Ω

Ω

Ω

Ω

Moreover, there exist coefficients a1p , . . . , akp , b1p , . . . , bkp > 0 and µ1p , . . . , µkp , ν1p , . . . , νkp > 0 such that −bip ∆vip = νip vip in ωvp −aip ∆uip = µip uip in ωup , with ωup = {|up | > 0} = Ω \ {|vp | > 0}, ωvp = {|vp | > 0} = Ω \ {|up | > 0},

while the coefficients aip and bip are given by the following formulas: 6 p−1 R p−1 R |∇uip |2 dx |∇vip |2 dx Ω Ω aip = , bip = p p−1 p−1 R R Pk Pk p p 2 2 dx p |∇u | dx |∇v | jp jp j=1 j=1 Ω Ω From (4.1), we have the ordering

0 < a1p 6 . . . 6 akp , 6Recall that a ip = P p 1/p k . i=1 ξi

∂ψ (ku1p k2 , . . . , kukp k2 ) ∂ξi

0 < b1p 6 . . . 6 bkp .

and bip =

∂ψ (kv1p k2 , . . . , kvkp k2 ), ∂ξi

for ψ(ξ1 , . . . , ξk ) =


The partition (ωup , ωvp ) solves (1.5), and p !1/p p !1/p k Z k Z X X |∇uip |2 dx + |∇vip |2 dx i=1

Ω

Ω

i=1

(4.2)

=

k X

(λj (ωup ))p

j=1

=

55

inf

(ω1 ,ω2 )∈P2 (Ω)

1/p

+

k 2 X X

k X

(λj (ωvp ))p

j=1

(λj (ωi ))p

j=1

i=1

1/p

1/p

=: cp .

Finally, the following Pohozaev–type formula holds: given x0 ∈ Ω and r ∈ (0, dist(x0 , ∂Ω)), we have that k Z X (4.3) (aip |∇uip |2 + bip |∇vip |2 ) dx (2 − N ) Br (x0 )

i=1

=

k Z X i=1

∂Br (x0 )

aip r(2(∂n uip )2 − |∇uip |2 ) dσ + +

k Z X i=1

∂Br (x0 )

r(µip u2ip

+

k Z X

∂Br (x0 )

i=1

2 νip vip ) dσ

−

k Z X i=1

bip r(2(∂n vip )2 − |∇vip |2 ) dσ

Br (x0 )

2 N (µip u2ip + νip vip ) dx.

We have also proved that, up to a set of Hausdorff dimension at most N − 2 (corresponding to Σ(up ,vp ) ) the common boundary ∂ωup ∩ Ω = ∂ωvp ∩ Ω is locally a hypersurface of class C 1,α , for some 0 < α < 1. Lemma 4.1. We have cp 6 k 1/p c∞ . In particular, kup kH01 , kvp kH01 are bounded independently of p ∈ N. Proof. The comparison between the levels follows easily from the estimate 1/p  k X  ξjp  6 k 1/p max{ξ1 , . . . , ξk } for ξ1 , . . . , ξk > 0. j=1

The H01 –boundedness of up and vp then follows directly from (4.2).

Thus, we have guaranteed the existence of u˜ = (˜ u1 , . . . , u˜k ) and v˜ = (˜ v1 , . . . , v˜k ) such that (up to a subsequence) up ⇀ u˜, vp ⇀ v˜ in H01 (Ω; Rk ), and up → u ˜, vp → v˜ in L2 (Ω; Rk ). In particular, u˜i · v˜j = 0 a.e. in Ω ∀i, j, and Z Z (4.4) u ˜i u ˜j dx = v˜i v˜j dx = δij . Ω

Moreover, it is easy to check that R p−1 2 ΩR|∇uip | dx 6 aip 6 1, k 1/p Ω |∇ukp |2 dx

Ω

R

p−1 2 ΩR|∇vip | dx k 1/p Ω |∇vkp |2 dx

6 bip 6 1.

56


Therefore, there exists 1 6 l 6 k such that (up to a subsequence) (4.5)

aip , bip → 0,

∀i = 1, . . . , l − 1;

aip → a ˜i 6= 0,

bip → ˜bi 6= 0

∀i = l, . . . , k

(observe that it is not necessarily true that the same number of components ai and bi vanish, and here we just suppose it without loss of generality to make the notations simpler). As Z Z aip |∇uip |2 dx = µip , bip |∇vip |2 dx = νip , Ω

Ω

also

µip , νip → 0,

∀i = 1, . . . , l − 1;

µip → µ ˜i 6= 0,

νip → ν˜i 6= 0

∀i = l, . . . , k.

In the rest of this section, our aim is to show that: (H1) uip → u˜i and vip → v˜i in C 0,γ ∩ H01 (Ω) for every 0 < γ < 1, ∀i = l, . . . , k; (H2) u ˜i and v˜i are Lipschitz continuous, for every i = l, . . . , k; (H3) The nodal set Γlu˜,˜v := {x ∈ Ω : u˜i (x) = v˜i (x) = 0, ∀i = l, . . . , k} has Hausdorff measure at most N − 1, the pair  ( ( ) )  k k X X (4.6) (e ω1 , ω e2 ) := int  u ˜2i > 0  , int  v˜i2 > 0  , i=l

i=l

belongs to P2 (Ω), and it is a regular partition in the sense of Definition 1.1. Moreover, for each regular regular point x0 ∈ R, lim

k X

x→x0 x∈e ω1 i=l

a ˜i |∇˜ ui (x)|2 = lim

k X

x→x0 x∈e ω2 i=l

˜bi |∇˜ vi (x)|2 6= 0.

Once this is proved, we can proceed as follows and end the proofs of Theorems 1.3 and 1.5, proving that the optimal partition problem (1.1) is achieved by the open regular partition (e ω1 , ω e2 ). Proof of Theorems 1.3 and 1.5. The statements for p fixed in Theorem 1.5 are a direct consequence of Theorem 1.6, while the convergences of ainp and uñp as p → ∞ follow from (4.5) and (H1). As u˜i · v˜j = 0 for every i, j, and Γl(˜u,˜v) has dimension at most N − 1, then ( k ( k ) ) X X 2 2 v˜i = 0 a.e. in u ˜i > 0 , v˜i > 0 . u ˜i = 0 a.e. in i=l

i=l

7

The sets ∂ ω e1 ∩ Ω, ∂ ω e2 ∩ Ω have zero Lebesgue measure , so u ˜i ∈ H01 (e ω1 ) and v˜i ∈ H01 (e ω2 ) for every i = 1, . . . , k. Moreover, from the strong convengence (H1) for ukp together with the weak convergence in H01 which holds for the remaining components uip , we have Z ∇˜ uk · ∇˜ ui dx = 0 ∀i = 1, . . . , k − 1. Ω

As u ˜1 , . . . , u ˜k−1 are linearly independent (because of (4.4)), we conclude that Z Z 2 |∇˜ uk | dx > λk (e ω1 ) and, analogously, |∇˜ vk |2 dx > λk (e ω2 ) . Ω

Ω

7Although this is not important at this point, we will check ahead that these sets coincide.


57

Thus (recalling Lemma 4.1) we have c∞ = lim (k 1/p c∞ ) = lim cp p→∞

p→∞

= lim

p→∞

Z

k Z X i=1

p !1/p |∇uip | dx + 2

Ω

Z

k Z X

Ω

i=1

p !1/p |∇vip | dx 2

> lim |∇ukp |2 dx + |∇vkp |2 dx p→∞ Ω Ω Z Z = |∇˜ uk |2 dx + |∇˜ vk |2 dx Ω

Ω

> λk (e ω1 ) + λk (e ω 2 ) > c∞ ,

which implies that c∞ is achieved by the open regular partition (4.6). The remaining statements now follow from (H2) and (H3). Remark 4.2. Observe that the proof for the more general case (1.2) follows exactly in the same way as the one of (1.1). In fact, in the proof of the latter case we have only P used properties (F1) and (F2) - cf. the Introduction - of the function F (x1 , . . . , xm ) = m i=1 xi , and never its specific form. Dealing with the general case, one would have to approximate it with   km k1 1/p 1/p X X , (λj (ω1 )p (λj (ω1 )p ,..., inf F (ω1 ,...,ωm )∈Pm (Ω)

j=1

j=1

instead of (1.5), which would only lead to heavier notations.

So, to conclude, we will show (H1) in the following subsection, and (H2)–(H3) in Subsection 4.3. 4.2. Proof of property (H1): Strong convergence of the last components of up , vp . We will follow the ideas of Subsection 2.2, which here apply in an easier way. First, we deduce L∞ bounds. Lemma 4.3. There exists C > 0, independent of p, such that kuip kL∞ (Ω) , kvip kL∞ (Ω) 6 C, Proof. Observe that −∆|uip | 6 and

µip |uip | aip

∀i = 1, . . . , k. in Ω,

Z µip |∇uip |2 dx 6 cp 6 k 1/p c∞ 6 kc∞ . aip = Ω

Thus the result follows from a standard Brezis–Kato type argument. This is the first step to prove uniform H¨ older bounds for the last k − l + 1 components. Lemma 4.4. For each 0 < γ < 1 there exists D > 0 such that kuip kC 0,γ (Ω) , kvip kC 0,γ (Ω) 6 D,

∀i = l, . . . , k.

58


Proof. Suppose, by contradiction, that        |uip (x) − uip (y)| |vip (x) − vip (y)|  →∞ , sup sup Lp := max  i=l,...,k  |x − y|γ |x − y|γ   x,y∈Ω  x,y∈Ω x6=y

x6=y

as p → ∞, and suppose without loss of generality that Lp is achieved for ukp at xp , yp , and that xp ∈ ωup . Since we have uniform L∞ bounds, then rp := |xp − yp | → 0. Take, for i = l, . . . , k, the rescaled functions u ¯ip (x) = which solve

1 uip (xp + rp x), Lp rpγ (

v¯ip (x) =

1 vip (xp + rp x), Lp rpγ

−aip ∆¯ uip = rp2 µip u¯ip

−bip ∆¯ vip = rp2 νip v¯ip

in in

x ∈ Ωp =

Ω − xp , rp

ωup −xp rp ωvp −xp . rp

Moreover, due to the normalization,        |¯ uip (x) − u ¯ip (y)| |¯ vip (x) − v¯ip (y)|  max sup (4.7) , sup  i=l,...,k  |x − y|γ |x − y|γ   x,y∈Ωp x,y∈Ωp  x6=y x6=y yp − xp = u ¯kp (0) − u¯kp =1 rp

Let zp ∈ Γ(up ,vp ) be such that dist(xp , Γ(up ,vp ) ) = |xp − zp |. We now have to split the proof in two cases: Case 1. Suppose that |xp − zp |/rp → ∞. ukp = rp2 µkp u¯kp in BR (0) for large p. Then, given R > 0, BR rp (xp ) ⊂ ωup , and −akp ∆¯ By taking the auxiliary function u ˜kp (x) := u ¯kp (x) − u ¯kp (0), we have that it is bounded in C 0,γ (BR (0)), and there exists v˜k such that ′

in C 0,γ (BR (0)), ∀0 < γ ′ < γ.

v˜kp → v˜k As −akp ∆˜ ukp = rp2 µkp u ¯kp =

µkp rp2−γ ukp (xp + rp ·) → 0 Lp

in BR (0), ∀R > 0,

and akp → a ˜k 6= 0, then v˜k is harmonic in RN , and (4.8)

sup x,y∈RN x6=y

|˜ vk (x) − v˜k (x)| = 1 < ∞. |x − y|γ

Thus [29, Corollary 2.3] yields that v˜k is constant, which contradicts (4.8). Case 2. Suppose that |xp − zp |/rp is bounded. In this case, observe that this information combined with (4.7) and zp − xp zp − xp u ¯ip = v¯ip =0 rp rp


59

yields boundedness of u ¯ip and v¯ip in C 0,γ (BR (0))–norm for every R > 0 sufficiently large, ′ i = l, . . . , k. Thus there exists u ¯i , v¯i (i = l, . . . , k) such that u¯ip → u ¯i , v¯ip → v¯i in C 0,γ (BR (0)) ′ for every 0 < γ < γ and R > 0. As ! ! k k k k X X X X νip µik 2 |¯ uip |, −∆ |¯ vip | in Ω, |¯ vip | 6 rp2 −∆ |¯ uip | 6 rp a bik i=1 ik i=l i=l i=l Pk Pk then the limiting functions ui | and vi | are nonnegative in RN , have i=l |¯ i=l |¯ P a com k N 0,γ N mon zero, are subharmonic in R , have finite C (R ) semi-norm and satisfy ui | · i=l |¯ P k vi | ≡ 0. Thus by the Liouville–type result [29, Proposition 2.2] we have that one of i=l |¯ Pk these functions is zero. By (4.7), we necessarily have i=l |¯ vi | ≡ 0. Now we can finish this proof exactly as in the final part of Step B of the proof of Theorem 2.11-(i) (or as in [29, p. 293]), as soon as we prove an Almgren monotonicity formula for the remaining components. From (4.3), we have k Z X (aip |∇¯ uip |2 + bip |∇¯ vip |2 ) dx (2 − N ) Br (x0 )

i=1

=

k Z X i=1

∂Br (x0 )

+

aip r(2(∂n u ¯ip )2 − |∇¯ uip |2 ) dσ +

k Z X i=1

∂Br (x0 )

k Z X

i=l

i=1

2 rp2 r(µip u ¯2ip + νip v¯ip ) dσ −

and, passing to the limit as p → ∞, k Z k Z X X 2 a ˜i |∇¯ ui | dx = (2 − N ) Br (x0 )

i=l

∂Br (x0 )

bip r(2(∂n v¯ip )2 − |∇¯ vip |2 ) dσ

k Z X i=1

Br (x0 )

2 rp2 N (µip u¯2ip + νip v¯ip ) dx.

a ˜i r(2(∂n u ¯i )2 − |∇¯ ui |2 ) dσ

∂Br (x0 )

for every x0 ∈ RN , r > 0 (observe that, as aip , bip → 0 from i = 1, . . . , l − 1, the previous sums start from i = l). Then, as in Theorem 2.21, for N (x0 , (¯ ul , . . . , u ¯k ), r) :=

E(x0 , (¯ ul , . . . , u ¯k ), r) , H(x0 , (¯ ul , . . . , u¯k ), r)

with E(x0 , (¯ ul , . . . , u ¯k ), r) =

1 rN −2

and H(x0 , (¯ ul , . . . , u ¯k ), r) =

k Z X i=l

1 rN −1

Br (x0 )

a ˜i |∇¯ ui |2 dx

k Z X i=l

∂Br (x0 )

a ˜i u ¯2i dσ,

we can prove a monotonicity result, namely that N (x0 , (¯ ul , . . . , u ¯k ), r) is a nondecreasing function in r, and 2N (x0 , (¯ ul , . . . , u ¯k ), r) d log H(x0 , (¯ ul , . . . , u ¯k ), r) = , dr r

∀x0 ∈ RN , r > 0.

We can now end the subsection with the proof of our first aim.

60


Proof of property (H1). The previous lemma provides the existence of u ˜i , v˜i (i = l, . . . , k) such that uip → u ˜i , vip → v˜i in C 0,γ (Ω), ∀0 < γ < 1. Moreover, it is now clear that ( P −˜ ai ∆˜ ui = µ ˜i u˜i in { ki=l u ˜2i > 0}, P k ˜ −bi ∆˜ vi = ν˜i v˜i in { i=l v˜i2 > 0}.

The strong H01 –convergence now follows in a standard way, combining the previous system with the one for uip , vip .

4.3. Proof of properties (H2) and (H3): Regularity of the last components of u ˜, v˜, and of the partition (e ω1 , ω e2 ). From property (H1), proved before, we garantee the existence of u ˜i , v˜i ∈ H01 (Ω) ∩ C 0,γ (Ω) for every i = l, . . . , k, which satisfy ( k ) X 2 −˜ ai ∆˜ ui = µ ˜i u˜i in the open set u ˜i > 0 , i=l

while

(

−˜bi ∆˜ vi = ν˜i v˜i in the open set

k X

v˜i2

i=l

)

>0 .

Moreover, by passing to the limit p → ∞ in (4.3), we get the limiting local Pohozaev–type formula: k Z X (˜ ai |∇˜ ui |2 + ˜bi |∇˜ vi |2 ) dx (2 − N ) i=l

=

k Z X i=l

∂Br (x0 )

a ˜i r(2(∂n u ˜i )2 − |∇˜ ui |2 ) dσ + +

k Z X i=l

∂Br (x0 )

r(˜ µi u ˜2i

+

k Z X i=l

ν˜i v˜i2 ) dσ

Br (x0 )

∂Br (x0 )

−

k Z X i=l

˜bi r(2(∂n v˜i )2 − |∇˜ vi |2 ) dσ

Br (x0 )

N (˜ µi u ˜2i + ν˜i v˜i2 ) dx,

for every x0 ∈ Ω and r ∈ (0, dist(x0 , ∂Ω)). Having these properties at hand, the regularity of the pair (e ω1 , ω e2 ) follows almost exactly as in Section 3. In this subsection we will simply go through the proof, highlighting the similarities and the differences with respect to what we already presented. One difference is that one needs to work with the components (˜ ul , . . . , u ˜k , v˜l , . . . , v˜k ) instead of with the whole vector (˜ u, v˜). Another delicate point is that although some results apply to the common nodal set Γlu˜,˜v = {x ∈ Ω : u ˜i (x) = v˜i (x) = 0, ∀i = l, . . . , k}, the regularity is proved only for its subsets ∂ ω e1 , ∂ ω e2 . A first simple by crucial observation is the following.

Lemma 4.5. We have ∂ ω e1 ∩ Ω = ∂ ω e2 ∩ Ω ⊆ Γl(˜u,˜v) . Pk Proof. Take x0 ∈ ∂ ω e1 ∩ Ω, so that in particular i=l u˜2i (x0 ) = 0. Suppose, in view of a P contradiction, that x0 6∈ ∂ ω e2 ∩ Ω. Then there exists δ > 0 so that ki=l v˜i2 ≡ 0 in Bδ (x0 ). For every x ∈ Bδ (x0 ), as Γl(˜u,˜v ) has non empty interior, then in each neighborhood of x


61

Pk Pk ˜2i > 0, whence x ∈ { i=l u˜2i > 0}. We have concluded that there exists a point where i=l u P e1 , a contradiction. The other inclusion is proved in ˜2i > 0}, therefore x0 ∈ ω Bδ (x0 ) ⊆ { ki=l u a similar way.

Let us describe the proofs of (H2) and (H3). From the system and the local Pohozaev– type identities, we have a version of the Almgren’s Monotonicity Formula (Theorem 2.21) for (¯ ul , . . . , u ¯k ), (¯ vl , . . . , v¯k ). Theorem 4.6. Take e 0 , r) = E(x

and

1 rN −2

e 0 , r) = H(x

k Z X i=l

1 rN −1

Br (x0 )

k Z X i=l

(˜ ai |∇˜ ui |2 + ˜bi |∇˜ vi |2 ) dx,

∂Br (x0 )

(˜ ai u ˜2i + ˜bi v˜i2 ) dσ,

e e (x0 , r) = E(x0 , r) . N e 0 , r) H(x

˜ ⋐ Ω there exists C > 0 and r˜ > 0 such that, for every x0 ∈ Ω, ˜ Then given Ω Cr 2 e (N (x0 , (u, v), r) + 1) is a non decreasing function for r ∈ (0, r˜], and the limit N (x0 , (u, v), 0+ ) := limr→0+ N (x0 , (u, v), r) exists and is finite. Also, d 2 log(H(x0 , (u, v), r)) = N (x0 , (u, v), r) dr r

∀r ∈ (0, r˜)

and Γl(˜u,˜v) has no interior points. This implies versions of Corollaries 2.22 and 2.23; in particular, u ˜i , v˜i are Lipschitz continuous for i = l, . . . , k (hence (H2) is proved), and N (x0 , 0+ ) > 1 whenever x0 ∈ Γl(˜u,˜v) . fi , N ei supported Similarly to Proposition 2.25, we prove that for i > l there exist measures M l on Γ(˜u,˜v) , such that fi , −˜ ai ∆˜ ui = µ ˜i u˜i − M

ei −˜b∆˜ vi = ν˜i v˜i − N

in D ′ (Ω).

Let us now pass through the results of Section 3. From the monotonicity formula and the equations for u ˜i , v˜i , we can repeat the arguments of Subsection 3.1, obtaining that, given ˜ ⋐ Ω and sequences xn ∈ Ω, e tn → 0, then Ω u ¯i (xn + tn x) u ˜i,n (x) := q → u¯i , e n , tn ) H(x

v¯i (xn + tn x) v˜i,n (x) := q → v¯i e n , tn ) H(x

0,γ 1 in Cloc ∩ Hloc (RN ), for every l 6 i 6 k, 0 < γ < 1. Moreover, (¯ ul , . . . , u ¯k , v¯l , . . . , v¯k ) 6= 0, u¯i · v¯j = 0 ∀i, j = l, . . . , k, and

¯ i, −˜ ai ∆¯ ui = −M

¯ i, N ¯i are measures concentrated on where M

¯i , −˜bi ∆¯ vi = −N

¯i (x) = v¯i (x) = 0 ∀i = l, . . . , k}. Γl(¯u,¯v) := {x ∈ Ω : u

62


Finally, we also have a local Pohozaev type identity, which yields an Almgren monotonicity formula similar to Corollary 3.3, this time for the limiting Almgren’s quotient Pk R ¯ 0 , r) r i=l Br (x0 ) (˜ ai |∇¯ ui |2 + ˜bi |∇¯ vi |2 ) dx E(x ¯ . = N (x0 , r) := ¯ Pk R H(x0 , r) ai u¯2i + ˜bi v¯i2 ) dσ i=l ∂Br (x0 ) (˜ In particular, Γl(˜u,˜v) has an empty interior, and if xn is either a constant sequence, or xn ∈ ¯ (x0 , 0+ ) = 1, then Γl(˜u,˜v) satisfies xn → x0 with N ¯ (0, r) ≡ N e (x0 , r) =: α N

for every r > 0

and the limiting functions in polar coordinates write as u ¯i = rα gi (θ),

v¯i = rα hi (θ),

with −∆S N −1 gi = λgi

in S N −1 \ Γl(¯u,¯v) ,

− ∆S N −1 hi = λhi

λ = α(α + N − 2).

Following now Subsection 3.2, we define e (x, 0+ ) = 1} Rl(˜u,˜v) = {x ∈ Γl(˜u,˜v) : N

and

l l + e S(˜ u,˜ v ) = {x ∈ Γ(˜ u,˜ v ) : N (x, 0 ) > 1}.

Following word by word the proofs presented there, we get the existence of δN > 0 such that ∀x0 ∈ Γl(˜u,˜v) ,

either

e (x0 , 0+ ) = 1 or N e (x0 , 0+ ) > 1 + δN N

(as in Proposition 3.7). This implies (Theorem 3.10) that

l Hdim (S(˜ u,˜ v ) ) 6 N − 2,

Hdim (Γl(˜u,˜v) ) 6 N − 1,

which yields the first part of property (G2). Moreover, as observed in Remark 3.8, given a blowup limit (¯ u, v¯) ∈ BUx0 with x0 ∈ Rl(˜u,˜v) , the nodal set Γl(˜u,˜v) is a vector space which is at most a hyperplane; if u ¯, v¯ 6≡ 0, then it is actually an hyperplane. By using a Clean-Up result (Proposition 3.15), we know that if v¯ ≡ 0 then also v˜ ≡ 0 in a neighborhood of x0 . However, this can well be the case for some x0 ∈ Γl(˜u,˜v) . This is the reason why, when going through Subsections 3.3 and 3.4, we need to restrict our attention to e := ∂ ω Γ e1 ∩ Ω = ∂ ω e2 ∩ Ω,

e := Γ e ∩ Rl R (˜ u,˜ v) .

and

e and (¯ Given x0 ∈ R u, v¯) ∈ BUx0 , exactly as in Theorem 3.16 one can now prove that u¯, v¯ 6≡ 0, and show the existence of ν ∈ S N −1 and αi , βi ∈ R (l 6 i 6 k) such that u ¯i = αi (x · ν)+ , v¯i = βi (x · ν)− ,

(4.9)

and

k X

a ˜i α2i =

i=l

k X

˜bi β 2 . i

i=l

e there exists R0 > 0 such that As a corollary, we obtain that given x0 ∈ R BR0 (x0 ) \ Γl(˜u,˜v) = Ω1 ∪ Ω2 ,

with Ω1 ∩ Ω2 = ∅, and Ω1 , Ω2 are (δ, R0 )–Reifenberg flat and NTA domains. Moreover, k X i=l

u ˜2i > 0,

k X i=1

v˜i2 ≡ 0 on Ω1

and

k X i=1

u ˜2i = 0,

k X i=1

v˜i2 > 0 on Ω2 .


63

Define p p e (x) = ( a ˜l u ˜l (x), . . . , a ˜k u ˜k (x)), U

and

q q Ve (x) = ( ˜bl v˜l (x), . . . , ˜bk v˜k (x)),

U(x) :=

V(x) :=

e (x) U , e (x)| |U

Ve (x) , |Ve (x)|

for x ∈ Ω1

for x ∈ Ω2 .

Proposition A.5 together with (4.9) implies that at least one component among (˜ ul , . . . , u ˜k ) and among (˜ vl , . . . , v˜k ) is signed in a neighborhood of x0 . Thus, since (˜ ul , . . . , u ˜k , v˜l , . . . , v˜k , int(Ω1 ∪ Ω2 )) ∈ G (check Definition A.6), as before, we can extend U, V up to Γ := ∂Ω1 ∩ ∂Ω2 ∩ Br0 (x0 ), and (3.23) holds. The rest of Subsection 3.4 can now be applied to our situation. e(˜u,˜v) is locally a C 1,α hypersurface. As the remaining In conclusion, we have shown that R e has Hausdorff dimension at most N − 2, we have thus proved property (H2), as part of Γ wanted. Having proved by now (H1), (H2) and (H3), as seen in the end of Subsection 4.1 we have everything we need in order to prove Theorems 1.3 and 1.5. Appendix A. Boundary behavior on NTA and Reifenberg flat domains A.1. Results on NTA domains. For convenience of the reader, we recall here the notion of non-tangencially accessible domain (NTA), introduced in [25] (the version we present here is taken from [28]). Definition A.1. A bounded domain Ω ⊂ RN is called an NTA if there exist M, r0 > 0 (the NTA constants) such that (1) Corkscrew condition: Given x ∈ ∂Ω and 0 < r < r0 there exist x0 ∈ Ω such that M −1 r < dist(x0 , ∂Ω) < |x − x0 | < M r.

(2) RN \ Ω satisfies the corkscrew condition (3) If w ∈ ∂Ω and w1 , w2 ∈ B(w, r0 ) ∩ Ω, then there is a rectifiable curve γ : [0, 1] → Ω with γ(0) = w1 , γ(1) = w2 , and: (a) H 1 (γ) 6 M |w1 − w2 |, (b) min{H 1 (γ([0, t])), H 1 (γ([t, 1]))} 6 M dist(γ(t), ∂Ω). Based on the seminal work of Jerison and Kenig [25] (see also the book [26]), we prove a boundary Harnack principle for solutions of −∆w = λw

(A.1) for some λ > 0, in a NTA domain.

Lemma A.2. Let Ω be an NTA domain, λ > 0, and x0 ∈ ∂Ω. Then there exist R0 , C > 0 (depending only on λ and the NTA constants) such that for every 0 < 2r < R0 and for every u, v solutions of (A.1) in Ω ∩ B2r (x0 ) with u = v on ∂Ω ∩ B2r (x0 ), and u, v > 0 in Ω, then C −1

v(x) v(y) v(y) 6 6C u(y) u(x) u(y)

for every x ∈ Ω ∩ Br (x0 ), y ∈ Ω ∩ Br (x0 ).

64


Proof. Let R0 > 0 be so that there exists ϕ0 > 0 nontrivial solution of −∆ϕ = λϕ0 in B3R0 (x0 ),

ϕ0 = 0 on ∂B3R0 (x0 ).

Then ϕ0 > 0 on B2R0 (x0 ), and w div(ϕ20 ∇( )) = div(∇wϕ0 − w∇ϕ0 ) ϕ0 = ∆wϕ0 − ∇w∇ϕ0 − ∇w∇ϕ0 − w∆ϕ0 = 0

in B2R0 (x0 )

for w ≡ u and w ≡ v. Thus we can apply [26, Lemma 1.3.7] to u/ϕ0 , v/ϕ0 , which provides the result. By using the same strategy, we can use [25, 26] to prove that v/u is α–Hölder continuous.

Lemma A.3. Let Ω be an NTA domain, λ > 0, and x0 ∈ ∂Ω. There exist R0 , C > 0 (depending only on λ and the NTA constants) such that for every 0 < 2r < R0 and for every u, v solutions of (A.1) in Ω ∩ B2r (x0 ) with u = v on ∂Ω ∩ B2r (x0 ), and u, v > 0 in Ω, then there exists α (depending only on λ and on the NTA constants) such that v is H¨ older continuous of order α on Ω ∩ Br (x0 ). u More precisely, v(x) v(z) |x − y|α v(y) ∀x, y ∈ Ω ∩ Br (x0 ), z ∈ Ω ∩ Br (x0 ). − u(x) u(y) 6 C u(z) rα ,

A.2. Results on Reifenberg flat domains. Next, we prove that if some blowup limit of a λ–harmonic function u centered at 0 converges to a nonnegative, nontrivial function, then u > 0 in a neighborhood of x0 . First, we need the following result, which holds for Reifenberg flat domains. Let us recall its definition.

Definition A.4. Let Ω ⊂ RN be an open set and take δ, R > 0. We say that Ω is a (δ, R)–Reifenberg flat domain if: - ∀x ∈ ∂Ω there exists a hyperplane H = Hx,R containing x, and ν = ν(x, R) ∈ S N −1 orthogonal to H such that {x + tν ∈ BR (x) : x ∈ H, t > 2δR} ⊂ Ω,

{x − tν ∈ BR (x) : x ∈ H, t > 2δR} ⊂ RN \ Ω.

- ∀x ∈ ∂Ω, 0 < r < R there exists a hyperplane H = Hx,r containing x such that dH (∂Ω ∩ Br (x), H ∩ Br (x)) 6 δr.

Recall that for δ = δ(N ) > 0 sufficiently small a (δ, R)–Reifenberg flat domain is NTA [27, Theorem 3.1]. We assume from now take we have such δ. Proposition A.5. Let Ω be a Reifenberg flat domain, and take u such that −∆u = λu in Ω, for some λ > 0. Take x0 ∈ ∂Ω and suppose that u = 0 on ∂Ω ∩ BR0 (x0 ). If, for some sequences tn , ρn → 0, we have u(x0 + tn x) un (x) := → u¯ > 0 in C 0,α (Ω), ρn with u¯ 6≡ 0, then u>0 in Ω ∩ Br (x0 ) for some sufficiently small r > 0.


65

Proof. Define by wn , zn the λ–harmonic extensions of u+ and u− on B2tn (x0 ) ∩ Ω, namely: −∆wn = λwn +

in B2tn (x0 ) ∩ Ω,

− ∆zn = λzn −

wn = u , zn = u on ∂(B2tn (x0 ) ∩ Ω). en , zen denote the blowups and observe that u = wn − zn in B2tn (x0 ) ∩ Ω. Let w zn (x0 + tn x) Ω − x0 wn (x0 + tn x) . and zen = , defined in B2 (0) ∩ w en = ρn ρn tn

At the limit, we find a harmonic equation in an half sphere (since Ω is a Reifenberg flat domain), and the boundary data converges to u ¯ and 0 respectively; hence we have w en → u¯i ,

Thus there exists y¯ ∈ ∂B1/2 (0) ∩

zen → 0 Ω−x0 tn

in a half-sphere of radius 2. such that

zn (¯ y) zñ (¯ y) 1 = < wn (¯ y) wn (¯ y) C for n large, where C is the constant introduced in Lemma A.2, depending only on λ and on the NTA constants of Ω. Then by this very same lemma applied to zn , wn , we have zñ (¯ y) Ω − x0 zn (x0 + tn x) , 6C 0 in Brn (x0 ) for sufficiently large n.

Throughout this article we need a version of Lemma A.3 for u and v solutions of problems like (A.1) with different λ’s. We were not able to prove this fact in general (and we are not sure it is actually true), but we were able to prove it for the situation we need. Definition A.6. We say that the triple (u, v, Ω), with (u, v) = (u1 , . . . , uk ; v1 , . . . , vk ) defined in Ω, belongs to the class G if: ˙ 2 ∪Γ, ˙ (G1) Ω = Ω1 ∪Ω where Γ = ∂Ω1 ∩ ∂Ω2 and Ω1 , Ω2 are Reifenberg flat domains (as well as NTA). Pk Pk 2 2 (G2) i=1 ui is positive in Ω1 , zero in Ω2 ; i=1 vi is zero in Ω1 , positive in Ω2 ; (G3) u, v are Lipschitz continuous in Ω, and u = v = 0 on Γ; (G4) there exists ai , bi , λi , µi > 0, i = 1, . . . , k so that −ai ∆ui = λi ui in Ω1 ,

−bi ∆vi = µi vi in Ω2 ;

(G5) Given x0 ∈ Ω and r ∈ (0, dist(x0 , ∂Ω)), we have that k Z X (ai |∇ui |2 + bi |∇vi |2 ) dx (2 − N ) i=1

=

k Z X i=1

∂Br (x0 )

ai r(2(∂n ui )2 − |∇ui |2 ) dσ + +

k Z X i=1

∂Br (x0 )

r(µi u2i

+

k Z X i=1

νi vi2 ) dσ

Br (x0 )

bi r(2(∂n vi )2 − |∇vi |2 ) dσ

∂Br (x0 )

−

k Z X i=1

Br (x0 )

N (µi u2i + νi vi2 ) dx.

Suppose without loss of generality that u1 > 0 in Ω1 , v1 > 0 in Ω2 , and take x0 ∈ Γ so that Br (x0 ) ∩ ∂Ω1 = Br (x0 ) ∩ ∂Ω2 ⊆ Γ. Our aim in the remaining part of this appendix is to prove the following result.

66


Proposition A.7. Within the previous framework, there exists C > 0 and α ∈ (0, 1) (independent of r > 0) such that, for each i 6= 1, ui (x) ui (x0 ) α ∀x ∈ Br (x0 ) ∩ Ω1 . u1 (x) − u1 (x0 ) 6 Cr In particular, ui /u1 is H¨ older continuous of order α in Br (x0 ) ∩ Ω1 . Similar results hold also for vi /v1 in Br (x0 ) ∩ Ω2 .

We will present the proof for u (the one for v is analogous). The idea is to consider suitable deformations of u1 so that the result is either a sub solution or super solution of −a1 ∆u1 = λ1 u1 with comparable boundary data, considering afterwards some λ1 /a1 –harmonic extension in view of using Lemma A.3. We start by deforming u1 into a sub solution of the original equation. In order to do that, we need to prove first that u1 survives for any blowup, which already requires some work. Take ε > 0, and take g : R → R defined by g(s) = s +

s3−ε . (3 − ε)(2 − ε)

Consider any function w satisfying −∆w 6 λw having |w| small. Then

−∆(g ◦ w) = −div(g ′ (w)∇w) = −g ′ (w)∆w − g ′′ (w)|∇w|2 6 λg ′ (w)w − g ′′ (w)|∇w|2 λ w2−ε w − w1−ε |∇w|2 = w λ − |∇w|2 w−ε + w2−ε =λ 1+ 2−ε 2−ε 2 2 2−ε |∇w| d (x) w λ = w λ− (A.2) + w2−ε , w2 d2 (x) 2 − ε

where we denote d(x) := dist(x, Γ). We will apply this deformation to the auxiliary function 1/2 P k 2 ρ(x) = , which satisfies (after a direct computation): u i i=1 λi ¯ ¯ , i = 1, . . . , k . −∆ρ 6 λρ in Ω1 , for λ := max ai Lemma A.8. Given x0 ∈ Γ, there exists R0 and C > 0 such that, for every x ∈ BR0 (x0 )∩Ω1 , |∇ρ(x)|2 d2 (x) (i) C −1 6 6 C; ρ2 (x) 2−ε ρ(x) (ii) lim = +∞, for some ε > 0. x→x0 d2 (x) Proof. (i) The proof goes by contradiction. Suppose that there exists rn → 0 and xn ∈ Brn (x0 ) ∩ Ω1 such that |∇ρ(xn )|2 d2 (xn ) v 2 (xn )

converges either to 0 or to + ∞.

Take tn := d(xn ) → 0 and let x′n ∈ Γ such that d(xn ) = |xn − x′n |. Using the notations of Subsection 3.1, consider the blowup sequence ui (x′n + tn x) , ui,n (x) := p H(x′n , (u, v), tn )

x ∈ Ωn =

Ω − x′n tn


67

which (by Theorem 3.1 and Theorem 3.16) converges, without loss of generality, to u ¯i = 1/2 1/2 P P Pk k k + 2 2 2 γi xN , with , and ρ¯(x) := = ¯i i=1 γi > 0. Define ρn (x) := i=1 ui,n i=1 u 1/2 P k 2 x+ and observe that N i=1 γi x −x′

|∇ρn ( ntn n )|2 |∇ρ(xn )|2 d2 (xn ) = x −x′ ρ2 (xn ) ρ2n ( ntn n ) x −x′

x −x′

Now dist( ntn n , Γ(un ,vn ) ) = 1 and ntn n → x¯ ∈ ∂B1 (0); since the convergence ui,n → u¯i is C 1,α away from a strip around Γ(¯u,¯v) (which is a hyperplane), then we have xn −x′n 2 tn )| ′ x −x ρ2n ( ntn n )

|∇ρn (

→

|∇¯ ρ(¯ x)|2 1 = + ∈]0, +∞[, 2 ρ¯ (¯ x) x ¯N

which is a contradiction. (ii) Using the same kind of argument as before, one can check that there exists C > 0 such that, for x ∼ x0 , ρ(x) C −1 6 p 6 C. ′ H(x , (u, v), d(x))

There exists R0 > 0 such that N (x′ , (u, v), 0+ ) = 1 for x′ ∈ BR0 (x0 ) ∩ Γ. Then there exists r¯ > 0 and a small ε′ > 0 such that for r 6 r¯ N (x′ , (u, v), r) 6 1 + ε′

x′ ∈ BR0 (x0 ) ∩ Γ(u,v) .

Thus we deduce from Theorem 2.21 that, for some C > 0, H(x′ , (u, v), r) . r2(1+ε′ ) and, as x → x0 , we have, by choosing ε < 2 so that 2/(2 − ε) > 1 + ε′ , "p #2−ε H(x′ , (u, v), d(x)) ρ(x)2−ε ρ(x)2−ε = p → +∞ 2−ε d2 (x) d(x)2/(2−ε) H(x′ , (u, v), d(x)) C6

Lemma A.9. Given x0 ∈ Γ, there exists R0 > 0 such that −∆(g ◦ ρ) 6 0,

g◦ρ>0

in Ω1 ∩ BR0 (x0 )

Proof. This is now a simple consequence of the previous lemma together with (A.2).

Suppose that u1 > 0 in BR0 (x0 ) ∩ Ω1 for small R0 > 0. Our aim is to check that u1 survives for any type of blowup; for that, we will compare it with ρ. For sufficiently small r > 0, define g˜r as the λ1 /a1 –harmonic extension of g ◦ ρ in Br (x0 ) ∩ Ω1 , namely −∆˜ gr =

λ1 g˜r in Br (x0 ) ∩ Ω1 , a1

g˜r = g ◦ ρ > 0 on ∂(Br (x0 ) ∩ Ω1 ).

By the comparison principle, for r > 0 small, ρ 6 g ◦ ρ 6 g˜r

in Br (x0 ) ∩ Ω1 .

68


Lemma A.10. Suppose that u1 > 0 in BR0 (x0 ) ∩ Ω1 . Then there exists C1 > 0 (eventually depending on R0 ) such that (A.3)

u1 (x) > C1 ρ(x)

∀x ∈ BR0 /2 (x0 ) ∩ Ω1 .

In particular, the first component u1 survives for any blowup sequence at x0 , i.e., u ¯1 6≡ 0 whenever (¯ u, v¯) ∈ BUx0 . Proof. Both u1 and g˜r are positive in BR0 (x0 ) ∩ Ω1 . Thus, by Lemma A.2, we have that, fixing any y ∈ BR0 /2 (x0 ) ∩ Ω1 , C1 := C −1

u1 (y) u1 (x) 6 ρr (y) ρ(x)

∀x ∈ BR0 /2(x0 ) ∩ Ω1 ,

which wields (A.3). As for the last statement, taking xn → x0 and tn → 0, we have, for all x ∈ B1 (0) and large n, xn + tn x ∈ BR0 /2 (x0 ) ∩ Ω1 . Thus u1 (xn + tn x) ρn (xn + tn x) u1,n := p > C1 p , H(xn , (u, v), tn ) H(xn , (u, v), tn ) and u1,n 6→ 0, as claimed. After all this, we are now ready to deform u1 (or, in general, any component which does not vanish after a blowup). In fact, by using the fact (provided by the previous result) that u1 does not vanish for any blowup at x0 , we can repeat the proofs of Lemmas A.8 with u1 in the place of ρ and get the properties stated there for this function, namely: C −1 6

|∇u1 (x)|2 d2 (x) 6 C, u21 (x)

for x ∈ BR0 (x0 ) ∩ Ω1 ,

and

u2−ε 1 = +∞. x→x0 d2 (x) This allows us to repeat the proof of Lemma A.9, and hence we can deform u1 in a subharmonic positive function. lim

Lemma A.11. Suppose that u1 > 0 in BR0 (x0 ) ∩ Ω1 . Then for a sufficiently small r > 0, we have −∆(g ◦ u1 ) 6 0, g ◦ u1 > 0 in Br (x0 ) ∩ Ω1 . Now take the function h : R → R defined by h(s) = s − and observe that

s3−ε (3 − ε)(2 − ε)

−∆(h ◦ u1 ) = −div(h′ (u1 )∇u1 ) = −h′ (u1 )∆u1 − h′′ (u1 )|∇u1 |2 6 λ1 h′ (u1 )u1 − h′′ (u1 )|∇u1 |2 λ1 λ1 |∇u1 |2 d2 (x) u2−ε λ1 2−ε 1 > = u1 + u1 − u a1 u21 d2 (x) a1 (2 − ε) 1 a1

in Br (x0 ) ∩ Ω, for sufficiently small r > 0. Let u¯r and u ˜r the λi /ai –harmonic extensions in Br (x0 )∩Ω1 of g ◦u1 and h◦u1 respectively. By the comparison principle, one has u1 6 g ◦ u1 6 u ¯r ,

and

u ˜ r 6 h ◦ u1 6 u1 .


69

Moreover, on ∂(Br (x0 ) ∩ Ω), one has (by using the fact that u1 is Lipschitz continuous)   2−ε g ◦ u1 = h ◦ u1 1 +

2u1 (3−ε)(2−ε)

1−

u2−ε 1 (3−ε)(2−ε)

Thus, for C > 0 independent of r > 0, u ¯r 6 u ˜r (1 + Cr2−ε ),

hence

 6 h ◦ u1 (1 + Cr2−ε ).

u1 6 u¯r 6 u˜r (1 + Cr2−ε ) 6 u1 (1 + Cr2−ε ),

and in particular 16

u ¯r 6 (1 + Cr2−ε ) u1

in Br (x0 ) ∩ Ω1 .

Lemma A.12. Under the previous notations, there exist 0 < δ 0, such that ui (x) ui (y) − 6 Cr(1−δ)α/δ , ∀x, y ∈ Ω ∩ Br1/δ (x0 ). u ¯r (x) u ¯r (y) − Proof. Take wr and zr respectively the λi /ai –extensions of u+ i and ui in Ω1 ∩B2r (x0 ), namely

−∆wr =

λi wr , ai

−∆zr =

λi zr ai

in Ω1 ∩ B2r (x0 )

with wr = u+ i ,

zr = u− i

on ∂(Ω1 ∩ B2r (x0 )).

By applying Lemma A.3 to the quotients wr /¯ ur and zr /¯ ur , we deduce the existence of C > 0 (independent of r) such that ui (x) wr (ξ) + zr (ξ) |x − y|α ui (y) 6 C , ∀x, y ∈ Ω ∩ Br (x0 ), z ∈ Ω ∩ Br (x0 ). − ur (x) ur (y) u ¯r (ξ) rα

Reasoning as before, by choosing ξ = ξr ∈ ∂Br/2 (x0 ) ∩ Ω such that dist(ξ, ∂Ω) > rε (which r (ξ) exists since Ω is Reifenberg flat) one proves that the quotient wr (ξ)+z is bounded. Take u ¯r (ξ) δ > 0 small; then we conclude that α ui (x) rα/δ ui (y) ′ |x − y| 6 C ′ α = C ′ r(1−δ)α/δ , ∀x, y ∈ Ω ∩ Br1/δ (x0 ), − ur (x) ur (y) 6 C α r r

Proof of Proposition A.7. We will use the decomposition ui ui ur = , u1 ur u1 so ui (x) ui (x0 ) ui (x) ui (x0 ) ur (x) ur (x) ur (x0 ) ui (x0 ) − − − 6 + u1 (x) u1 (x0 ) ur (x) ur (x0 ) u1 (x) u1 (x) u1 (x0 ) ur (x0 ) 6 C ′ r(1+δ)α/δ (1 + Cr2−ε ) + C ′′ r2−ε 6 κr(1−δ)α/δ

for every x ∈ Ω ∩ Br1/δ (x0 ), and the result follows.

70


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