Olivier Druet

Pierre-Damien Thizy

MULTI-BUMPS ANALYSIS FOR TRUDINGER-MOSER NONLINEARITIES I -QUANTIFICATION AND LOCATION OF CONCENTRATION POINTS

. We answer all of them, proving in particular that blow up phenomenon is very restrictive because of the strong interaction between bubbles in this equation. This work will have a sequel, giving existence results of critical points of the associated functional at all energy levels via degree theory arguments, in the spirit of what had been done for the Liouville equation in the beautiful work of Chen-Lin [8].

However, as studied by Adimurth-Prashanth [2], for β = 4π, finding critical points is more tricky since a lack of compactness appears in Palais-Smale sequences at this level of energy. Nevertheless, it has been proved by for the unit disk, by Struwe [37] for Ω

Introduction

We let Ω be a smooth bounded domain of R 2 and we consider the equation ∆u = λf ue u 2 in Ω, u > 0 in Ω, u = 0 on ∂Ω .

(

where ∆ = -∂ 2 ∂x 2 -∂ 2 ∂y 2 , λ > 0 and f is a smooth positive function in Ω. This equation is critical with respect to Trudinger-Moser inequality. Indeed, the nonlinearity in e u 2 is the best one can hope to control in dimension 2 by the L 2 -norm of the gradient. More precisely, we let H 1 0 (Ω) be the standard Sobolev space (with zero boundary condition) endowed with the norm ∇u 2 dx. Trudinger proved in [START_REF] Trudinger | On embedding into Orlicz spaces and some applications[END_REF] that Ω e u 2 dx is finite for any function u in H 1 0 (Ω). Moser was then a little bit more precise in [START_REF] Moser | A sharp form of an inequality by N. Trudinger[END_REF], proving that sup u∈H 1 0 (Ω), ∇u 2 =1 Ω e γu 2 dx < +∞ if and only if γ ≤ 4π .

(1.2)

Solutions of equation (1.1) are in fact critical points of the functional

J(u) = Ω f e u 2 dx (1.3)
under the constraint Ω |∇u| 2 dx = β for some β > 0. The λ appearing in (1.1) is then the Euler-Lagrange coefficient. This functional is well-defined on H 1 0 (Ω) thanks to Trudinger [START_REF] Trudinger | On embedding into Orlicz spaces and some applications[END_REF]. It is also easy to find a critical point of J if β < 4π in the constraint thanks to Moser's inequality (1.2) : these critical points may be found as maxima of J under the constraint

Ω |∇u| 2 dx = β < 4π.
close to the disk and by Flucher [START_REF] Flucher | Extremal functions for the Trudinger-Moser inequality in 2 dimensions[END_REF] for a general Ω that there are extremals in Moser's inequality (1.2) for γ = 4π, meaning in particular that there are always critical points of J for the critical value β = 4π. Note that existence of critical points for β slightly larger than 4π has also been proved by Struwe [START_REF] Struwe | Critical points of embeddings of H 1,n 0 into Orlicz spaces[END_REF] and Lamm-Robert-Struwe [START_REF] Lamm | The heat flow with a critical exponential nonlinearity[END_REF]. Struwe [START_REF] Struwe | Positive solutions of critical semilinear elliptic equations on non-contractible planar domains[END_REF] also found critical points of higher energy (for some values of β between 4π and 8π) when the domain contains an annulus (in the spirit of Coron [START_REF] Coron | Topologie et cas limite des injections de Sobolev[END_REF]). We refer also to the recent Mancini-Martinazzi [START_REF] Mancini | The Moser-Trudinger inequality and its extremals on a disk via energy estimates[END_REF] for an interesting new proof of the existence of extremal functions for Moser's inequality in the disk without using test-functions computations.

In the last decade, tools have been developed to study sequences of solutions of equation (1.1) and in particular to understand precisely their potential blow-up behaviour. This serie of works started in the minimal energy situation (β close to 4π) with Adimurthi-Struwe [START_REF] Adimurthi | Global compactness properties of semilinear elliptic equations with critical exponential growth[END_REF]. Then Adimurthi-Druet [START_REF] Adimurthi | Blow up analysis in dimension 2 and a sharp form of Trudinger-Moser inequality[END_REF] used this blow-up analysis to obtain an improvement of Moser's inequality (completing the result of Lions [START_REF] Lions | The concentration-compactness principle in the calculus of variations. The limit case[END_REF]). In the radial case (that is in the unit disk with f ≡ 1), such a blow-up analysis in the minimal energy case was recently used by Malchiodi-Martinazzi [START_REF] Malchiodi | Critical points of the Moser-Trudinger functional on a disk[END_REF] to prove that there is a β 0 > 4π for which there are solutions of (1.1) of energy less than or equal to β 0 but no solutions of energy greater than β 0 .

In order to get solutions of higher energies and to describe precisely the set of solutions for all β, one needs a fine analysis of blowing-up solutions. The first result in this direction is the quantification result of the first author [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] that we recall here since the questions we adress in the present work come from it : Theorem 1.1 (Druet [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF]). Let Ω be a smooth bounded domain in R 2 and let (f ε ) ε>0 be a sequence of functions of uniform critical growth in Ω. Also let (u ε ) ε>0 be a sequence of solutions of

∆u ε = f ε x, u ε (x)
verifying that ∇u ε 2 2 → β as ε → 0 for some β ∈ R. Then there exists a solution u 0 ∈ C 0 Ω of ∆u 0 = f 0 x, u 0 (x) in Ω, u 0 = 0 on ∂Ω , and there exists Ñ ∈ N such that ∇u ε 2 2 = ∇u 0 2 2 + 4π Ñ + o (1) . If Ñ = 0, the convergence of u ε to u 0 is strong in H 1 0 (Ω) and actually holds in C 0 Ω .

We do not define here sequences of functions of uniform critical growth in Ω. The only thing we need to know is that they include sequences of the form f ε (x, u) = h ε (x)ue u 2 as soon as h ε > 0 and h ε → h 0 in C 1 Ω . But they include much more general nonlinearities behaving like e u 2 at infinity. Note also that, in the litterature, the nonlinearity is sometimes written as e 4πu 2 (this is for instance the case in [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF]), hence the discrepancy of 4π in some results. This result describes precisely the lack of compactness in the energy space. Note that this result is not true for Palais-Smale sequences, as proved by Costa-Tintarev [START_REF] Costa | Concentration profiles for the Trudinger-Moser functional are shaped like toy pyramids[END_REF] : there are Palais-Smale sequences for the above equation which converge to 0 weakly in H 1 0 (Ω) and which present a lack of compactness at any level above 4π. This shows that the quantification result of Theorem 1.1 is specific to sequences of solutions of the equation and require a pointwise analysis as carried out in [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] and below (not only an analysis in energy space).

Note also that the above result is not empty since Del Pino-Musso-Ruf [START_REF] Del Pino | Beyond the Trudinger-Moser supremum[END_REF] constructed, via a Liapunov-Schmidt procedure, multi-pikes sequences of solutions of equation (1.1) (with f ≡ 1) in annuli. These sequences satisfy the hypothesis of the above theorem, converge weakly to 0 in H 1 0 (Ω) (that is u 0 ≡ 0 in the above result) and have an energy converging to 4π Ñ . They can construct such solutions for all Ñ ≥ 1. This suggests that the topology of the domain plays a crucial role in the existence of solutions of arbitrary energies.

However, if one wants to push further the existence results, we need to be more precise than Druet [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF]. In particular, we need to answer the following natural questions, left open in this work of the first author (see also Del Pino-Musso-Ruf [START_REF] Del Pino | New solutions for Trudinger-Moser critical equations in R 2[END_REF], Malchiodi-Martinazzi [START_REF] Malchiodi | Critical points of the Moser-Trudinger functional on a disk[END_REF] or Martinazzi [START_REF] Martinazzi | A threshold phenomenon for embeddings of H m 0 into Orlicz spaces[END_REF] where one can find these, or similar, questions) :

1. Is it possible to have both u 0 ≡ 0 and Ñ ≥ 1 in the above theorem ? 2. Are the concentration points appearing when Ñ ≥ 1 isolated 1 or not ? If yes, where are they ?

These questions are natural and can be motivated by analogy with Liouville type equations (see among others [START_REF] Chen | Sharp estimates for solutions of multi-bubbles in compact Riemann surfaces[END_REF][START_REF] Chen | Topological degree for a mean field equation on Riemann surfaces[END_REF][START_REF] Li | Blow-up analysis for solutions of -∆u = V e u in dimension two[END_REF][START_REF] Lin | An expository survey on the recent development of mean field equations[END_REF][START_REF] Malchiodi | Morse theory and a scalar field equation on compact surfaces[END_REF][START_REF] Takahashi | Blow up points and the Morse indices of solutions to the Liouville equation in two-dimension[END_REF]) or Yamabe type equations (see for instance [START_REF] Brendle | Blow-up phenomena for the Yamabe equation[END_REF][START_REF] Brendle | Blow-up phenomena for the Yamabe equation[END_REF][START_REF] Druet | From one bubble to several bubbles : the low-dimensional case[END_REF][START_REF] Druet | Blow-up theory for elliptic PDEs in Riemannian geometry[END_REF][START_REF] Khuri | A compactness theorem for the Yamabe problem[END_REF][START_REF] Yan | A Harnack type inequality for the Yamabe equation in low dimensions[END_REF][START_REF] Li | Compactness of solutions to the Yamabe problem[END_REF][START_REF] Li | Compactness of solutions to the Yamabe problem[END_REF][START_REF] Li | Yamabe type equations on three dimensional Riemannian manifolds[END_REF][START_REF] Marques | A priori estimates for the Yamabe problem in the non-locally conformally flat case[END_REF][START_REF] Marques | Blow-up examples for the Yamabe problem[END_REF][START_REF] Schoen | Prescribed scalar curvature on the n-sphere[END_REF]). We refer in particular to [START_REF] Druet | La notion de stabilité pour des équations aux dérivées partielles elliptiques[END_REF] for a survey on this kind of questions.

We attack in this paper the questions 1 and 2 above. Our result holds for more general nonlinearities but we restrict, for sake of clearness, to the simplest one. We consider a sequence (u ε ) of smooth positive solutions of

∆u ε = λ ε f ε u ε e u 2 ε in Ω, u ε = 0 on ∂Ω , (1.4)
for some sequence (λ ε ) of positive real numbers and some sequence (f ε ) of smooth functions in Ω which satisfies

f ε → f 0 in C 1 Ω as ε → 0 and ∇ 2 f ε L ∞ (Ω) = O(1) (1.5)
where f 0 > 0 in Ω. And we prove the following result :

Theorem 1.2.
Let Ω be a smooth bounded domain of R 2 and let (u ε ) be a sequence of smooth solutions of (1.4) which is bounded in H 1 0 (Ω). Assume that (1.5) holds. Then, after passing to a subsequence, λ ε → λ 0 as ε → 0 for some λ 0 ∈ R. . If λ 0 = 0, then there exists u 0 ∈ C 2 Ω solution of ∆u 0 = λ 0 f 0 u 0 e u 2 0 in Ω, u 0 = 0 on ∂Ω such that u ε → u 0 in C 2 Ω as ε → 0.

If λ 0 = 0, then u ε ⇀ 0 weakly in H 1 0 (Ω). Moreover there exist N ≥ 1 such that

Ω |∇u ε | 2 dx → 4πN as ε → 0
and N sequences of points (x i,ε ) in Ω such that a) x i,ε → x i as ε → 0 with x i ∈ Ω (not on the boundary), all the x i 's being distinct.

b) u ε → 0 in C 2 loc Ω \ S where S = {x i } i=1,...,N . c) for all i = 1, . . . , N , we have that γ i,ε = u ε (x i,ε ) → +∞ as ε → 0 and that

γ i,ε (u ε (x i,ε + µ i,ε x) -γ i,ε ) → U (x) = -ln 1 + 1 4 |x| 2 in C 2 loc R 2 as ε → 0 where µ -2 i,ε = λ ε f ε (x i,ε ) γ 2 i,ε e γ 2 i,ε → +∞ as ε → 0 .
1 By isolated, we mean here that the energy at any concentration point is exactly 4π. In other words, we mean that there are no bubble accumulations and we do not wish to rule out only bubbles towers.

d) for all i = 1, . . . , N , there exists m i > 0 such that

λ ε γ i,ε → 2 m i f 0 (x i ) as ε → 0 .
e) The points x i are such that

2m i ∇ y H (x i , x i ) + 4π j =i m j ∇ y G (x j , x i ) + 1 2 m i ∇f 0 (x i ) f 0 (x i ) = 0
and that

4π j =i m j G (x j , x i ) + 2m i H (x i , x i ) + m i ln f 0 (x i ) m 2 i + m i = 0
for all i = 1, . . . , N where

G(x, y) = 1 2π ln 1 |x -y| + H (x, y)
is the Green function of the Laplacian with Dirichlet boundary condition.

Note that this theorem proves that, if blow-up occurs, then the weak limit has to be zero so that lack of compactness can occur only at the levels β = 4πN for N ≥ 1. This is a key information to get general existence result via degree theory from this theorem; this will be the subject of a subsequent paper. We also obtain a precise characterisation of the location of concentration points. This answers in particular by the affirmative to the conjecture of Del Pino-Musso-Ruf [START_REF] Del Pino | New solutions for Trudinger-Moser critical equations in R 2[END_REF] (p. 425) since, in case f ≡ 1, the (x i , m i ) of Theorem 1.2 are critical points of the function

Φ (y i , α i ) = 2π i =j α i α j G (y i , y j ) + N i=1 α 2 i H (y i , y i ) + N i=1 α 2 i -α 2 i ln α i .
The paper is organized as follows. In Section 2, we recall the main results of Druet [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] and set up the proof of the theorem. Section 3 is devoted to a fine asymptotic analysis in the neighbourhood of a given concentration point while the theorem is proved in Section 4 which deals with the multi-spikes analysis. At last, we collect some useful estimates concerning the standard bubble and the Green function respectively in appendices A and B.

Previous results and sketch of the proof

We set up the proof of Theorem 1.2 and we recall some results obtained in Druet [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF]. We let Ω be a smooth bounded domain of R 2 and we consider a sequence (u ε ) of smooth positive solutions of

∆u ε = λ ε f ε u ε e u 2 ε in Ω, u ε = 0 on ∂Ω (2.
1) for some sequence (λ ε ) of positive real numbers and some sequence (f ε ) of smooth functions which satisfies (1.5). Note that we necessarily have that lim sup

ε→0 λ ε ≤ λ 1 min Ω f 0 (2.2)
where λ 1 > 0 is the first eigenvalue of the Laplacian with Dirichlet boundary condition in Ω. Indeed, let ϕ 1 ∈ C ∞ Ω be a positive (in Ω) eigenfunction associated to λ 1 and multiply equation (2.1) by ϕ 1 . After integration by parts, we get that

λ 1 Ω u ε ϕ 1 dx = λ ε Ω f ε u ε e u 2 ε ϕ 1 dx .
Since f ε becomes positive for ε small thanks to (1.5) and since u ε and ϕ 1 are positive, we can write that

λ 1 Ω u ε ϕ 1 dx ≥ λ ε min Ω f ε Ω u ε ϕ 1 dx , which leads to (2.2).
We assume in the following that there exists C > 0 such that

Ω |∇u ε | 2 dx ≤ C for all ε > 0 . (2.3)
Then we have the following :

Proposition 2.1 (Druet [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF]). After passing to a subsequence, λ ε → λ 0 as ε → 0, there exists a smooth solution u 0 of the limit equation

∆u 0 = λ 0 f 0 u 0 e u 2 0 in Ω, u 0 = 0 on ∂Ω (2.4)
and there exist N ≥ 0 and N sequences (x i,ε ) of points in Ω such that the following assertions2 hold :

a) u ε ⇀ u 0 weakly in H 1 0 (Ω). If N = 0, the convergence of u ε to u 0 holds in C 2 Ω . b) for any i ∈ {1, . . . , N }, u ε (x i,ε ) → +∞ as ε → 0 and ∇u ε (x i,ε ) = 0. c) for any i, j ∈ {1, . . . , N }, i = j, |x i,ε -x j,ε | µ i,ε → +∞ as ε → 0 where µ -2 i,ε = λ ε f ε (x i,ε ) u ε (x i,ε ) 2 e uε(xi,ε) 2 → +∞ as ε → 0 . d) for any i ∈ {1, . . . , N }, we have that u ε (x i,ε ) (u ε (x i,ε + µ i,ε x) -u ε (x i,ε )) → U (x) = -ln 1 + 1 4 |x| 2 in C 2 loc R 2 . e) there exists C 1 > 0 such that λ ε min i=1,...,N |x i,ε -x| 2 u ε (x) 2 e uε(x) 2 ≤ C 1 for all x ∈ Ω. f ) there exists C 2 > 0 such that min i=1,...,N |x i,ε -x| u ε (x) |∇u ε (x)| ≤ C 2 .
Proof -Even if this result is already contained in [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF], we shall give part of the proof here. The first reason is that it is not exactly stated in this way in [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF]. The second reason is that it is proved in greater generality in [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] and we thus give a proof which is in some sense more readable here.

First, it is clear thanks to (2.3) that, up to a subsequence, u ε ⇀ u 0 weakly in H 1 0 (Ω) where u 0 is a solution of (2.4). If u ε ∞ = O(1), then, by standard elliptic theory, this convergence holds in C 2 Ω and the proposition is true with N = 0. Let us assume from now on that sup Ω u ε → +∞ as ε → 0 .

(2.5)

Given N ≥ 1 and N sequences (x i,ε ) of points in Ω which verify that

γ i,ε = u ε (x i,ε ) → +∞ as ε → 0 and µ -2 i,ε = λ ε f ε (x i,ε ) γ 2 i,ε e γ 2 i,ε → +∞ as ε → 0 , (2.6)
we consider the following assertions :

P N 1 For any i, j ∈ {1, . . . , N }, i = j, |x i,ε -x j,ε | µ i,ε → +∞ as ε → 0. P N 2
For any i ∈ {1, . . . , N }, ∇u ε (x i,ε ) = 0 and

γ i,ε (u ε (x i,ε + µ i,ε x) -γ i,ε ) → U (x) in C 2 loc R 2 as ε → 0 where U (x) = -ln 1 + 1 4 |x| 2 is a solution of ∆U = e 2U in R 2 . P N 3
There exists C > 0 such that

λ ε min i=1,...,N |x i,ε -x| 2 u ε (x) 2 e uε(x) 2 ≤ C for all x ∈ Ω.
A first obvious remark is that

P N 1 and P N 2 =⇒ Ω |∇u ε | 2 dx ≥ 4πN + o(1) . (2.7)
Indeed, one has just to notice that

Ω |∇u ε | 2 dx = λ ε Ω f ε u 2 ε e u 2 ε dx , that D xi,ε (Rµ i,ε ) ∩ D xj,ε (Rµ j,ε ) = ∅ for ε > 0 small enough thanks to P N 1 and that lim ε→0 λ ε Dx i,ε (Rµi,ε) f ε u 2 ε e u 2 ε dx = D0(R) e 2U dx → R 2
e 2U dx = 4π as R → +∞ thanks to P N 2 . In the following, we shall say that property P N holds if there are N sequences (x i,ε ) of points in Ω which verify (2.6) such that assertions P N 1 and P N 2 hold.

Step 1 -Property P 1 holds.

Proof of Step 1 -Let x ε ∈ Ω be such that u ε (x ε ) = max Ω u ε .
By (2.5), we have that

γ ε = u ε (x ε ) → +∞ as ε → 0 . (2.8) We just have to check P 1 2 since P 1 1 is empty. We clearly have that ∇u ε (x ε ) = 0. We set 3 ũε (x) = γ ε (u ε (x ε + µ ε x) -γ ε ) (2.9) for x ∈ Ω ε where Ω ε = x ∈ R 2 s.t. x ε + µ ε x ∈ Ω and µ -2 ε = λ ε f ε (x ε ) γ 2 ε e γ 2 ε .
(2.10) 3 The fact that this rescaling is appropriate to understand the blow up behaviour of solutions of equation (1.1) was first discovered by Adimurthi-Struwe [START_REF] Adimurthi | Global compactness properties of semilinear elliptic equations with critical exponential growth[END_REF].

It is clear that µ ε → 0 as ε → 0 .

(2.11) Indeed, we can write that

λ ε f ε u ε e u 2 ε ≤ λ ε sup Ω f ε γ ε e γ 2 ε = γ -1 ε sup Ω f ε f ε (x ε ) µ -2 ε = o µ -2 ε
thanks to (1.5) and (2.8). If ever (2.11) was false, we would have that ∆u ε ∞ → 0 as ε → 0 which, together with the fact that u ε = 0 on ∂Ω, would contradict (2.8). Thus (2.11) holds. Thanks to (2.11), we know that, up to a subsequence and up to a harmless rotation,

Ω ε → R 2 or Ω ε → R × (-∞, d) as ε → 0 (2.12)
where d = lim

ε→0 d (x ε , ∂Ω) µ ε
. We also have that

∆ũ ε = f ε (x ε + µ ε x) f ε (x ε ) u ε (x ε + µ ε x) γ ε e uε(xε+µεx) 2 -γ 2 ε (2.13)
in Ω ε thanks to (2.1) and (2.10). Since 0 ≤ u ε ≤ γ ε in Ω and thanks to (1.5), this leads to ∆ũ ε L ∞ (Ωε) = O(1). Together with the fact that ũε ≤ 0 = ũε (0) and ũε = -γ 2 ε → -∞ as ε → 0 on ∂Ω ε , one can check that this implies that

Ω ε → R 2 as ε → 0 and that ũε → U in C 1 loc R 2
as ε → 0 after passing to a subsequence. We refer here the reader to [START_REF] Adimurthi | Global compactness properties of semilinear elliptic equations with critical exponential growth[END_REF] or [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] for the details of such an assertion. Moreover, we clearly have that U ≤ U (0) = 0 in R 2 . Noting that, as a consequence of the above convergence of ũε , we have that

u ε (x ε + µ ε x) 2 -γ 2 ε → 2U in C 0 loc R 2
, one can easily pass to the limit in equation (2.13) to obtain that

∆U = e 2U in R 2 .
Moreover, by standard elliptic theory, one has that

ũε → U in C 2 loc R 2 as ε → 0 . (2.14)
In order to apply the classification result of Chen-Li [START_REF] Chen | Classification of solutions of some nonlinear elliptic equations[END_REF], we need to check that e 2U ∈ L 1 R 2 . Using (2.9) together with (1.5), (2.10) and (2.14), we can write that

lim ε→0 λ ε Dx ε (Rµε) f ε u 2 ε e u 2 ε dx = D0(R) e 2U dx
for all R > 0. Thanks to (2.1) and (2.3), we know that

λ ε Dx ε (Rµε) f ε u 2 ε e u 2 ε dx ≤ λ ε Ω f ε u 2 ε e u 2 ε dx = Ω |∇u ε | 2 dx ≤ M so that e 2U ∈ L 1 R 2 .
Remembering that U ≤ U (0) = 0, we thus get by [START_REF] Chen | Classification of solutions of some nonlinear elliptic equations[END_REF] that

U (x) = -ln 1 + 1 4 |x| 2 .
This clearly ends the proof of Step 1. ♠

Step 2 -Assume that property P N holds for some N ≥ 1. Then either P N 3 holds or P N +1 holds.

Proof of

Step 2 -Assume that P N holds for some N ≥ 1 (with associated sequences (x i,ε )) and that P N 3 does not hold, meaning that

λ ε sup x∈Ω min i=1,...,N |x i,ε -x| 2 u ε (x) 2 e uε(x) 2 → +∞ as ε → 0 .
(2.15)

We let then y ε ∈ Ω be such that min i=1,...,N

|x i,ε -y ε | 2 u ε (y ε ) 2 e uε(yε) 2 = sup x∈Ω min i=1,...,N |x i,ε -x| 2 u ε (x)
i,ε -y ε | µ i,ε → +∞ as ε → 0 for all 1 ≤ i ≤ N . (2.17) We set μ-2 ε = λ ε f ε (y ε ) γ2 ε e γ2 ε so that, with (1.5), (2.15) and (2.16), με → 0 as ε → 0 and |x i,ε -y ε | με → +∞ as ε → 0 for all 1 ≤ i ≤ N . (2.18)
We set now ûε (x) = γε (u ε (y ε + με x)γε )

for x ∈ Ωε where Ωε = x ∈ R 2 s.t. y ε + με x ∈ Ω .
We are exactly in the situation of Step 1 except for one thing : we can not say that ûε ≤ 0 in Ωε . However, combining (2.16) and (2.18), we can say that ûε ≤ o(1) in K ∩ Ωε for all compact subset K of R 2 . This permits to repeat the arguments of Step 1, see [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] for the details, to obtain that ûε

→ U in C 2 loc R 2 as ε → 0 . (2.19)
Since U has a strict local maximum at 0, u ε must possess, for ε > 0 small, a local maximum

x N +1,ε in Ω such that |x N +1,ε -y ε | = o (μ ε ). Then ∇u ε (x N +1,ε ) = 0 and defining γ N +1,ε , µ N +1,ε with respect to this point x N +1,ε , it is easily checked that P N +1 2 and P N +1
1 hold with the sequences (x i,ε ) i=1,...,N +1 thanks to (2.17), (2.18) and (2.19). This proves that property P N +1 holds and ends the proof of Step 2. ♠

Starting from

Step 1, and applying by induction Step 2, using (2.3) and (2.7) to stop the process, we can easily prove the proposition except for point (f). But this point was the subject of Proposition 2 of [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] and we refer the reader to this paper for the proof. ♦

The main result of Druet [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] may be phrased as follows :

Theorem 2.1 (Druet [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF]). In the framework of Proposition 2.1, there exist moreover M ≥ 0 and M sequences of points (y i,ε ) in Ω such that the following assertions hold after passing to a subsequence : a) For any i ∈ {1, . . . , M } and any j ∈ {1, . . . , N },

|y i,ε -x j,ε | µ j,ε → +∞ as ε → 0 . b) For any i ∈ {1, . . . , M }, u ε (y i,ε ) (u ε (y i,ε + ν i,ε x) -u ε (y i,ε )) → U (x) = -ln 1 + 1 4 |x| 2 in C 2 loc R 2 \ S i where ν -2 i,ε = λ ε f ε (y i,ε ) u ε (y i,ε ) 2 e uε(yi,ε) 2 → +∞ as ε → 0
and

S i = lim ε→0 x j,ε -y i,ε ν i,ε , j = 1, . . . , N lim ε→0 y k,ε -y i,ε ν i,ε , k = 1, . . . , M, k = i . c) The Dirichlet norm of u ε is quantified by Ω |∇u ε | 2 dx = Ω |∇u 0 | 2 dx + 4π (N + M ) + o(1) .
It is the way that the main quantification result of Druet [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] is proved. Proposition 1 in Section 3 of [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] corresponds to Proposition 2.1 above (at the exception of f)). Then concentration points are added at the end of Section 3 of [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF], point f) of the above proposition is proved in Section 4 of [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] and it is proved during Sections 5 and 6 of [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF] that the quantification holds with these concentration points added.

Let us comment on this result. First, it is clear that u 0 ≡ 0 ⇒ λ 0 > 0. Second, if N = 0, then the convergence of u ε to u 0 is strong in H 1 0 (Ω) and in fact even holds in C 2 Ω . The two questions left open in this work of the first author were :

1. Is it possible to have u 0 ≡ 0 and N ≥ 1 together ? 2. Are the concentration points (x i,ε ) isolated or can there be bubbles accumulation ?

These two questions can be motivated, as explained in the introduction, by the situation in low dimensions for Yamabe type equations, as studied in [START_REF] Druet | From one bubble to several bubbles : the low-dimensional case[END_REF] (see also [START_REF] Druet | La notion de stabilité pour des équations aux dérivées partielles elliptiques[END_REF]). But they are also crucial in order to understand precisely the number of solutions of equation (1.1), a question we shall address in a subsequent paper.

Let us briefly sketch the proof of Theorem 1.2. We start from the above results of [START_REF] Druet | Multibumps analysis in dimension 2 -Quantification of blow up levels[END_REF]. We shall first give some fine pointwise estimates on the sequence (u ε ) in small (but not so small) neighbourhoods of the concentration points. This will be the subject of section 3. Then we prove Theorem 1.2 in section 4 through a serie of claims proving successively that : M = 0 in Theorem 2.1 above, λ 0 = 0 so that u 0 = 0 and, at last, the concentration points are isolated and of comparable size. All Theorem 1.2 then follows easily.

Local blow up analysis

In this section, we get some fine estimates on sequences of solutions of equations (2.1) in the neighbourhood of one of the concentration points (x i,ε ) of Theorem 2.1. During all this section, C denotes a constant which is independant of ε or variables x, y, . . . We let (ρ ε ) be a bounded sequence of positive real numbers (possibly converging to 0 as ε → 0) and we consider a sequence of smooth positive functions (v ε ) which are solutions of

∆v ε = λ ε f ε v ε e v 2 ε in D 0 (ρ ε ) (3.1)
where (λ ε ) is a bounded sequence of positive real numbers, (f ε ) is a sequence of smooth positive functions satisfying that there exists C 0 > 0 such that

1 C 0 ≤ f ε (0) ≤ C 0 , |∇f ε | ≤ C 0 and ∇ 2 f ε ≤ C 0 in D 0 (ρ ε ) . (3.2)
Here and in all what follows, D x (r) denotes the disk of center x and radius r. We assume moreover that

γ ε = v ε (0) → +∞ as ε → 0 and ∇v ε (0) = 0 , (3.3) that µ -2 ε = λ ε f ε (0) γ 2 ε e γ 2 ε → +∞ as ε → 0 with ρ ε µ ε → +∞ as ε → 0 , (3.4) that γ ε (v ε (µ ε x) -γ ε ) → U (x) = -ln 1 + 1 4 |x| 2 in C 2 loc R 2 as ε → 0 , (3.5)
that there exists C 1 > 0 such that

λ ε |x| 2 v 2 ε e v 2 ε ≤ C 1 in D 0 (ρ ε ) (3.6)
and that there exists C 2 > 0 such that

|x| |∇v ε | ≤ C 2 γ ε in D 0 (ρ ε ) . (3.7)
The aim of this section will be to compare in a suitable disk the sequence (v ε ) with the bubble B ε defined as the radial solution in R 2 of

∆B ε = λ ε f ε (0)B ε e B 2 ε with B ε (0) = γ ε . (3.8)
Thanks to the results of Appendix A, see in particular Claims 5.2 and 5.3, we know that

B ε (x) -γ ε - t ε (x) γ ε - t ε (x) γ 3 ε ≤ C 3 γ -2 ε for x s.t. t ε (x) ≤ γ 2 ε (3.9)
and that

∇B ε (x) -γ -1 ε 2 x |x| 2 + 4µ 2 ε ≤ C 4 γ -2 ε |x| |x| 2 + µ 2 ε for x s.t. t ε (x) ≤ γ 2 ε (3.10)
where C 3 > 0 and C 4 > 0 are some universal constants and

t ε (x) = ln 1 + |x| 2 4µ 2 ε . (3.11)
We prove the following :

Proposition 3.1. We have that :

a) if v ε (r) = 1 2πr ∂D0(r) v ε dσ, sup 0≤r≤ρε |v ε (r) -B ε (r)| = o γ -1 ε .
As a consequence, we have that

t ε (ρ ε ) ≤ γ 2 ε -1 + o (1) . b) There exists C > 0 such that |v ε -B ε | ≤ Cγ -1 ε and |∇ (v ε -B ε )| ≤ Cγ -1 ε ρ -1 ε in D 0 (ρ ε ). c)
After passing to a subsequence,

γ ε (v ε (ρ ε • ) -B ε (ρ ε )) → 2 ln 1 |x| + H as ε → 0 in C 1 loc (D 0 (1) \ {0}
) where H is some harmonic function in the unit disk satisfying

H(0) = 0 and ∇H(0) = - 1 2 lim ε→0 ρ ε ∇f ε (0) f ε (0) .
Proof of Proposition 3.1 -Let us first remark that we may assume without loss of generality that t ε (ρ ε ) ≤ γ 2 ε .

(3.12) Indeed, up to reduce ρ ε , this is the case and once a) is proved, we know that

t ε (ρ ε ) ≤ γ 2 ε -1+o (1)
. This will easily permit to prove that, for the original ρ ε , we had

t ε (ρ ε ) ≤ γ 2 ε since t ε (r) ≤ γ 2 ε -1 2
as long as t ε (r) ≤ γ 2 ε . Fix 0 < η < 1 and let

r ε = sup r ∈ (0, ρ ε ) s.t. |v ε (s) -B ε (s)| ≤ η γ ε for all 0 ≤ s ≤ r (3.13)
where

v ε (r) = 1 2πr ∂D0(r) v ε dσ .
Note that we know thanks to (3.4) and (3.5) that

r ε µ ε → +∞ as ε → 0 . (3.14)
We have that

|v ε (r) -B ε (r)| ≤ η γ ε for all 0 ≤ r ≤ r ε (3.15)
and that

|v ε (r ε ) -B ε (r ε )| = η γ ε if r ε < ρ ε . (3.16)
We set v ε = B ε + w ε (3.17) in D 0 (ρ ε ). Thanks to (3.7) and (3.15), we know that

|w ε | ≤ η + πC 2 γ ε in D 0 (r ε ) . (3.18) This clearly implies since |B ε | ≤ γ ε that v 2 ε -B 2 ε ≤ 3 (η + πC 2 ) in D 0 (r ε ) . (3.19)
Thanks to (3.1), we can write that

∆w ε = λ ε f ε v ε e v 2 ε -λ ε f ε (0)B ε e B 2 ε = λ ε e B 2 ε f ε v ε e v 2 ε -B 2 ε -f ε (0)B ε = λ ε e B 2 ε f ε w ε e v 2 ε -B 2 ε + f ε B ε e v 2 ε -B 2 ε -f ε (0)B ε
in D 0 (r ε ) so that, using (3.2), (3.18) and (3.19) but also (3.9), we get the existence of some C > 0 such that

|∆w ε | ≤ Cλ ε f ε (0) 1 + B 2 ε e B 2 ε |w ε | + Cλ ε |x| 2 γ ε + B ε e B 2 ε in D 0 (r ε ) . (3.20)
We let ϕ ε be such that

∆ϕ ε = 0 in D 0 (r ε ) and ϕ ε = w ε on ∂D 0 (r ε ) . (3.21)
Using (3.7) and (3.10), we know that

|∇w ε | ≤ Cγ -1 ε r -1 ε on ∂D 0 (r ε )
for some C > 0 so that

∇ϕ ε L ∞ (D0(rε)) = O 1 γ ε r ε . (3.22)
Note also that, up to a subsequence,

γ ε ϕ ε (r ε • ) → ϕ 0 in C 2 loc (D 0 (1)) as ε → 0 (3.23) since |ϕ ε (r ε )| ≤ ηγ -1
ε thanks to (3.15) and (3.17). It follows from standard elliptic theory thanks to (3.21).

Step 1 -There exists C > 0 such that

|∇ (w ε -ϕ ε ) (y)| ≤ C ∇w ε L ∞ (D0(rε)) + γ -1 ε µ ε µ ε + |y| + γ -2 ε + Cγ -2 ε + Cr -1 ε γ -3 ε for all y ∈ D 0 (r ε ).
Proof of Step 1 -Let y ε ∈ D 0 (r ε ). Using the Green representation formula and (3.20), we can write that

|∇ (w ε -ϕ ε ) (y ε )| ≤ Cλ ε f ε (0) D0(rε) 1 |x -y ε | 1 + B ε (x) 2 e Bε(x) 2 |w ε (x)| dx (3.24) +Cλ ε D0(rε) 1 |x -y ε | |x| 2 γ ε + B ε (x) e Bε(x) 2 dx .
We let in the following

t 1,ε = 1 4 γ 2 ε and t 2,ε = γ 2 ε -γ ε (3.25)
and we let

Ω 0,ε = D 0 (r ε ) ∩ {t ε (x) ≤ t 1,ε } , Ω 1,ε = D 0 (r ε ) ∩ {t 1,ε ≤ t ε (x) ≤ t 2,ε } and (3.26) Ω 2,ε = D 0 (r ε ) ∩ {t ε (x) ≥ t 2,ε } .
We also set, for i = 0, 1, 2,

I i,ε = λ ε f ε (0) Ωi,ε 1 |x -y ε | 1 + B ε (x) 2 e Bε(x) 2 |w ε (x)| dx (3.27)
and

J i,ε = λ ε Ωi,ε 1 |x -y ε | |x| 2 γ ε + B ε (x) e Bε(x) 2 dx . (3.28)
Case 1 -We assume first that |y ε | = O (µ ε ). Since w ε (0) = 0 and using (3.9), we can write that

I 0,ε ≤ Cλ ε f ε (0)γ 2 ε e γ 2 ε ∇w ε L ∞ (Ω0,ε) Ω0,ε |x| |x -y ε | e tε(x) 2 γ 2 ε -2tε(x) dx .
Thanks to (3.4), we can rewrite this as

I 0,ε ≤ Cµ -2 ε ∇w ε L ∞ (Ω0,ε) Ω0,ε |x| |x -y ε | e tε(x) 2 γ 2 ε -2tε(x) dx . Since t ε (x) 2 γ 2 ε -2t ε (x) ≤ - 7 4 t ε (x) in Ω 0,ε ,
this leads to

I 0,ε ≤ Cµ -2 ε ∇w ε L ∞ (Ω0,ε) Ω0,ε |x| |x -y ε | 1 + |x| 2 4µ 2 ε -7 4 dx ≤ C ∇w ε L ∞ (Ω0,ε) R 2 |x| x -yε µε 1 + |x| 2 4 -7 4
dx .

Since |y ε | = O (µ ε ), we obtain by the dominated convergence theorem that

I 0,ε = O ∇w ε L ∞ (Ω0,ε) . (3.29)
In Ω 1,ε , we have that |x| ≤ (1 + o(1)) |xy ε | since |y ε | = O (µ ε ) so that, we can write, as above

I 1,ε ≤ Cµ -2 ε ∇w ε L ∞ (Ω1,ε) Ω1,ε e tε (x) 2 γ 2 ε -2tε(x) dx ≤ C ∇w ε L ∞ (Ω1,ε) t2,ε t1,ε e t 2 γ 2 ε -t dt by the change of variables t = ln 1 + |x| 2 4µ 2 ε . Since t 2 γ 2 ε -t ≤ - t γ ε ≤ - 1 4 γ ε for 1 4 γ 2 ε = t 1,ε ≤ t ≤ t 2,ε = γ 2 ε -γ ε , we immediately get that I 1,ε ≤ C ∇w ε L ∞ (Ω1,ε) γ 2 ε e -1 4 γε . (3.30)
In Ω 2,ε , we have that B ε = O (1) thanks to (3.9) so that, using (3.2) and (3.18) , we can write that

I 2,ε ≤ Cλ ε γ -1 ε D0(rε) 1 |x -y ε | dx so that I 2,ε ≤ Cλ ε r ε γ -1 ε . (3.31) Now we notice that t ε (r ε ) ≤ γ 2 ε implies that r 2 ε µ 2 ε ≤ 4e γ 2 ε .
Using (3.2) and (3.4), this gives that

λ ε r 2 ε ≤ C γ 2 ε . (3.32)
Thus we get that I 2,ε ≤ Cr -1 ε γ -3 ε .

(3.33) For the second set of integrals, things are similar and easier. We write that

J 0,ε ≤ Cµ -2 ε γ -1 ε Ω0,ε |x| |x -y ε | e tε(x) 2 γ 2 ε -2tε(x) dx so that, see above, J 0,ε ≤ Cγ -1 ε . (3.34) We also have that J 1,ε ≤ Cγ ε e -1 4 γε
(3.35) in the same way than above. At last, for J 2,ε , we write that

J 2,ε ≤ Cλ ε r ε D0(rε) 1 |x -y ε | dx ≤ Cλ ε r 2 ε .
Thus we have thanks to (3.32) that

J 2,ε ≤ Cγ -2 ε . ( 3
|∇ (w ε -ϕ ε ) (y ε )| ≤ C ∇w ε L ∞ (D0(rε)) + Cγ -1 ε + Cr -1 ε γ -3 ε . (3.37)
Case 2 -We assume now that |yε| µε → +∞ as ε → 0. We follow the lines of the first case to estimate most of the integrals. Thus we only emphasize on the changes. First, we write that

I 0,ε ≤ Cµ -2 ε ∇w ε L ∞ (Ω0,ε) Ω0,ε |x| |x -y ε | 1 + |x| 2 4µ 2 ε -7 4 dx ≤ C ∇w ε L ∞ (Ω0,ε) R 2 |x| |x -µ -1 ε y ε | 1 + |x| 2 4 -7 4
dx .

Now we can write that

R 2 |x| |x -µ -1 ε y ε | 1 + |x| 2 4 -7 4 dx = |y ε | µ ε -3 2 R 2 |x| |x -|y ε | -1 y ε | µ 2 ε |y ε | 2 + |x| 2 4 -7 4 dx ≤ C |y ε | µ ε -3 2 + 2 |y ε | µ ε -3 2 D0( 1 2 ) |x| µ 2 ε |y ε | 2 + |x| 2 4 -7 4 dx ≤ C |y ε | µ ε -3 2 + 2 µ ε |y ε | R 2 |x| 1 + |x| 2 4 -7 4 dx so that I 0,ε ≤ C ∇w ε L ∞ (Ω0,ε) µ ε |y ε | . (3.38)
Let us write once again that

I 1,ε ≤ Cµ -2 ε ∇w ε L ∞ (Ω1,ε) Ω1,ε |x| |x -y ε | e tε(x) 2 γ 2 ε -2tε(x) dx .
Let us split this integral into two parts. First,

µ -2 ε Ω1,ε\Dy ε ( 1 2 |yε|) |x| |x -y ε | e tε(x) 2 γ 2 ε -2tε(x) dx ≤ 3µ -2 ε Ω1,ε e tε (x) 2 γ 2 ε -2tε(x) dx ≤ C t2,ε t1,ε e t 2 γ 2 ε -t dt ≤ Cγ 2 ε e -1 4 γε
as in Case 1. Second,

µ -2 ε Ω1,ε∩Dy ε ( 1 2 |yε|) |x| |x -y ε | e tε(x) 2 γ 2 ε -2tε(x) dx ≤ 3 2 µ -2 ε |y ε | e s 2 ε γ 2 ε -2sε Dy ε ( 1 2 |yε|) 1 |x -y ε | dx
where

s ε = t ε y ε 2 .
Thus we have that

µ -2 ε Ω1,ε∩Dy ε ( 1 2 |yε|) |x| |x -y ε | e tε(x) 2 γ 2 ε -2tε(x) dx ≤ C |y ε | 2 µ 2 ε e s 2 ε γ 2 ε -2sε Note that Ω 1,ε ∩ D yε 1 2 |y ε | = ∅ if t ε 3 2 |y ε | = ln 1 + 9 |y ε | 2 16µ 2 ε ≤ t 1,ε = 1 4 γ 2
ε so that we may assume that

ln 1 + 9 |y ε | 2 16µ 2 ε > 1 4 γ 2 ε .
Thus

s ε = ln 1 + |y ε | 2 16µ 2 ε ≥ 1 4 γ 2 ε -ln 9 . It is also clear that if Ω 1,ε ∩ D yε 1 2 |y ε | = ∅, s ε ≤ t 2,ε = γ 2 ε -γ ε . Thus we have that s 2 ε γ 2 ε -s ε ≤ - 1 4 γ ε + O(1) .
We deduce that, if not zero,

µ -2 ε Ω1,ε∩Dy ε ( 1 2 |yε|) |x| |x -y ε | e tε(x) 2 γ 2 ε -2tε(x) dx ≤ C |y ε | 2 µ 2 ε e -1 4 γε e -sε ≤ C |y ε | 2 µ 2 ε e -1 4 γε 1 + |y ε | 2 16µ 2 ε -1 ≤ Ce -1 4 γε . Thus we arrive to I 1,ε ≤ Cγ 2 ε e -1 4 γε ∇w ε L ∞ (Ω1,ε) . (3.39)
At last, for I 2,ε , we have nothing to change to get that

I 2,ε ≤ Cr -1 ε γ -3 ε . (3.40)
For J 0,ε , J 1,ε and J 2,ε , we proceed as above or as in Case 1 to get that

J 0,ε ≤ Cγ -1 ε µ ε |y ε | , J 1,ε ≤ Cγ ε e -1 4 γε and J 2,ε ≤ C γ 2 ε .
Thus, in this second case, we obtain coming back to (3.24) with (3.38), (3.39), (3.40) and these last estimates that

|∇ (w ε -ϕ ε ) (y ε )| ≤ C ∇w ε L ∞ (D0(rε)) + γ -1 ε µ ε |y ε | + γ 2 ε e -1 4 γε +Cγ -2 ε +Cr -1 ε γ -3 ε . (3.41)
The study of these two cases clearly permits to conclude Step 1. ♠

Step 2 -We have that

∇ (w ε -ϕ ε ) L ∞ (D0(rε)) = o γ -1 ε r -1 ε + O γ -1 ε and that w ε -ϕ ε L ∞ (D0(rε)) = o γ -1 ε .
Moreover, if r ε → 0 as ε → 0, we have that

lim ε→0 ∇f ε (0) f ε (0) = -2 lim ε→0 1 r ε ∇ϕ 0 (0) . Proof of Step 2 -Let y ε ∈ D 0 (r ε ) be such that |∇ (w ε -ϕ ε ) (y ε )| = ∇ (w ε -ϕ ε ) L ∞ (D0(rε)) (3.42)
and let us assume that

α ε = |∇ (w ε -ϕ ε ) (y ε )| ≥ δ r ε γ ε + 1 δγ ε (3.43)
for some δ > 0. Thanks to (3.22), we have that

∇w ε L ∞ (D0(rε)) ≤ α ε + Cr -1 ε γ -1 ε ≤ α ε 1 + C δ . ( 3

.44)

Applying Step 1 to this sequence (y ε ), we get thanks to (3.42), (3.43) and (3.44) that

1 δ + δ r ε γ -1 ε ≤ α ε = |∇ (w ε -ϕ ε ) (y ε )| ≤ C δ α ε µ ε µ ε + |y ε | + γ -2 ε + Cγ -2 ε + Cr -1 ε γ -3 ε .
This proves that

y ε µ ε → y 0 ∈ R 2 as ε → 0 (3.45)
after passing to a subsequence and, thanks to Step 1 and (3.43), that

|∇ ( wε -φε ) (x)| ≤ C δ 1 + |x| + o(1) for all x ∈ R 2 (3.46)
where C δ depends only on δ and

wε (x) = 1 µ ε α ε w ε (µ ε x) , φε (x) = 1 µ ε α ε ϕ ε (µ ε x) . (3.47)
We know that wε (0) = 0, ∇ wε (0) = 0 and ∇ ( wεφε )

y ε µ ε = 1 . (3.48)
We also know thanks to (3.23) and (3.43) that, after passing to a subsequence,

∇ φε (x) → lim ε→0 1 γ ε r ε α ε ∇ϕ 0 (0) = A in C 1 loc R 2 as ε → 0 . (3.49)
Using (3.20), we can write that

|∆ wε | ≤ Cλ ε µ 2 ε f ε (0) 1 + B ε (µ ε x) 2 e Bε(µεx) 2 | wε | + Cλ ε α -1 ε µ 2 ε |x| 2 γ ε + B ε (µ ε x) e Bε(µεx) 2 .
Noting thanks to (3.46), (3.48) and (3.49) that

| wε (x)| ≤ C δ ln (1 + |x|) + A |x| + o (|x|)
and is thus uniformly bounded on any compact subset of R 2 , we easily deduce from the above estimate together with the definition (3.4) of µ ε and (3.43) that (∆ wε ) is uniformly bounded in any compact subset of R 2 . Thus, by standard elliptic theory, we have that, after passing to a subsequence, wε

→ w 0 in C 1,η loc R 2 as ε → 0 . ( 3
(x) -A ≤ C δ 1 + |x| in R 2 . (3.51) Thus w 0 ≡ 0. Since we know that γ ε w ε (µ ε x) → 0 in C 1 loc R 2 as ε → 0 thanks to (3.5), we deduce that γ ε µ ε α ε → 0 as ε → 0 . (3.52)
Thanks to (3.1), (3.4), (3.8), (3.17) and (3.47), we can write that

∆ wε (x) = 1 α ε µ ε λ ε f ε (µ ε x) (B ε (µ ε x) + w ε (µ ε x)) e (Bε(µεx)+wε(µεx)) 2 -f ε (0)B ε (µ ε x) e Bε(µεx) 2 . = B ε (µ ε x) γ ε e Bε(µεx) 2 -γ 2 ε 1 α ε µ ε γ ε f ε (µ ε x) f ε (0) e 2Bε(µεx)wε(µεx)+wε(µεx) 2 -1 +γ -2 ε f ε (µ ε x) f ε (0) wε (x)e Bε(µεx) 2 -γ 2 ε +2Bε(µεx)wε(µεx)+wε(µεx) 2 .
Let us write now that

γ -1 ε B ε (µ ε x) → 1 in C 0 loc R 2 as ε → 0 , that B ε (µ ε x) 2 -γ 2 ε → 2U (x) in C 0 loc R 2 as ε → 0
where U (x) =ln 1 + |x| 2 4 thanks to (3.9) and (3.11). We can also write that

f ε (µ ε x) f ε (0) = 1 + f ε (0) -1 µ ε x i ∂ i f ε (0) + O µ 2 ε |x| 2 thanks to (3.2) and that 2B ε (µ ε x) w ε (µ ε x) + w ε (µ ε x) 2 = 2µ ε α ε γ ε (w 0 + o(1)) = o(1)
thanks to (3.50) and (3.52). Thus we can write that

f ε (µ ε x) f ε (0) e 2Bε(µεx)wε(µεx)+wε(µεx) 2 -1 = 2µ ε α ε γ ε w 0 + µ ε f ε (0) -1 x i ∂ i f ε (0) + o (µ ε α ε γ ε )
Thus we obtain that

∆ wε (x) = e 2U(x) 2w 0 (x) + 1 α ε γ ε f ε (0) -1 x i ∂ i f ε (0) + o(1) .
Thanks to (3.43), we know that, after passing to a subsequence,

1 α ε γ ε ∂ i f ε (0) f ε (0) → X i as ε → 0 . (3.53)
Note that we have, again thanks to (3.43), that

X = 0 if r ε → 0 as ε → 0 . (3.54)
Then we can write that ∆ wε (x) = e 2U(x) 2w 0 (x) + X i x i + o(1)

so that ∆w 0 = e 2U 2w 0 + X i x i in R 2 . (3.55)
Now, thanks to [START_REF] Chen | Sharp estimates for solutions of multi-bubbles in compact Riemann surfaces[END_REF], lemma 2.3 or [START_REF] Laurain | Concentration of CM C surfaces in a 3-manifold[END_REF], lemma C.1, we know that the only solution of this equation satisfying (3.51) is

w 0 (x) = |x| 2 4 + |x| 2 A i x i (3.56)
and, moreover, we must have

A = - 1 2 X . (3.57)
Since w 0 ≡ 0, we must have A = 0 and thus X = 0.

This permits to prove the step. Indeed, if r ε → 0, then we have that X = 0 by (3.54), which is a contradiction. Thus, if r ε → 0, we get that (3.43) is impossible so that

α ε = o γ -1 ε r -1
ε in this case. This proves the first estimate of the step in the case r ε → 0 as ε → 0. If r ε → 0 as ε → 0, we know thanks to the fact that X = 0 and to (3.53

) that α ε = O γ -1 ε if (3.

43) holds and if it does not hold, we again have that α

ε = O γ -1 ε
. Thus we also have that the first estimate of the step holds if r ε → 0 as ε → 0. Moreover, in this second case, we know that

lim ε→0 ∂ i f ε (0) f ε (0) = -2 lim ε→0 1 r ε ∇ϕ 0 (0)
thanks to (3.49), (3.53) and (3.57). This proves the last part of the step.

It remains to notice that the second estimate of the step is a simple consequence of the first. Indeed, coming back to the estimate of Step 1 with the estimate on the gradient just proved, we have that

|∇ (w ε -ϕ ε ) (y)| ≤ Cγ -1 ε 1 + O r -1 ε µ ε µ ε + |y| + γ -2 ε + Cγ -2 ε + Cr -1 ε γ -3 ε Since w ε -ϕ ε = 0 on ∂D 0 (r ε ), this leads after integration to |w ε (y) -ϕ ε (y)| ≤ Cγ -1 ε 1 + O r -1 ε µ ε ln µ ε + r ε µ ε + |y| + O γ -2 ε
for all y ∈ D 0 (r ε ). This leads to

w ε -ϕ ε L ∞ (D0(rε)) = O γ -1 ε µ ε ln 1 + r ε µ ε + O γ -1 ε µ ε r ε ln 1 + r ε µ ε + O γ -2 ε = o γ -1 ε thanks to (3.14).
This ends the proof of Step 2. ♠

We are now in position to prove Proposition 3.1. First, since w ε (0) = 0 and ∇w ε (0) = 0, we get with Step 2 that

ϕ ε (0) = o γ -1 ε (3.58) and that |∇ϕ ε (0)| = o γ -1 ε r -1 ε + O γ -1 ε . (3.59) Since ϕ ε is harmonic, (3.58) gives that γ ε ϕ ε (0) = γ ε 2πr ε ∂D0(rε) ϕ ε dσ → 0 as ε → 0 . Since v ε -B ε = w ε = ϕ ε on ∂D 0 (r ε ), this leads to γ ε |v ε (r ε ) -B ε (r ε )| → 0 as ε → 0 ,
which is impossible if r ε < ρ ε thanks to (3.16). Thus we have proved that

r ε = ρ ε , (3.60)
for any choice of η ∈ (0, 1). This proves the first part of a). The second part of a) is then just a consequence of (3.9). Indeed,

γ ε |v ε (ρ ε ) -B ε (ρ ε )| = o(1) implies that γ ε B ε (ρ ε ) ≥ γ ε v ε (ρ ε ) + o(1)
. And (3.9) gives that

γ 2 ε -t ε (ρ ε ) -γ -2 ε t ε (ρ ε ) ≥ γ ε v ε (ρ ε ) + o(1) . which leads to t ε (ρ ε ) ≤ γ 2 ε -1+o (1) since v ε (ρ ε ) ≥ 0. Point b)
of the proposition is a consequence of Step 2 together with (3.22). It remains to prove c). Let us write that

γ ε (v ε (ρ ε x) -B ε (ρ ε )) = γ ε w ε (ρ ε x) + γ ε (B ε (ρ ε x) -B ε (ρ ε )) .
We write that

γ ε (B ε (ρ ε x) -B ε (ρ ε )) → 2 ln 1 |x| in C 1 loc (D 0 (1) \ {0}) as ε → 0
thanks to (3.9) and (3.10). Moreover, thanks to Step 2, we know that

γ ε w ε (ρ ε x) -ϕ ε (ρ ε x) L ∞ (D0(1)) = o(1)
and, combining Steps 1 and 2, that

γ ε ρ ε |∇w ε (ρ ε x) -∇ϕ ε (ρ ε x)| ≤ C µ ε µ ε + ρ ε |x| + γ -2 ε + Cρ ε γ -1 ε + Cγ -2 ε in D 0 (1)
. Thus we have that (3.23). We thus have obtained that

γ ε w ε (ρ ε x) → ϕ 0 in C 1 loc (D 0 (1) \ {0}) as ε → 0 thanks to
γ ε (v ε (ρ ε • ) -B ε (ρ ε )) → 2 ln 1 |x| + ϕ 0 in C 1 loc (D 0 (1) \ {0}) as ε → 0. Moreover, we have thanks to Step 2 that ϕ 0 (0) = 0 and ∇ϕ 0 (0) = - 1 2 lim ε→0 ρ ε ∇f ε (0) f ε (0) .
This ends the proof of the proposition. ♦ 4. Proof of Theorem 1.2

Let (u ε ) be a sequence of smooth positive solutions of

∆u ε = λ ε f ε u ε e u 2 ε in Ω, u ε = 0 on ∂Ω (4.1)
for some sequence (λ ε ) of positive real numbers and some sequence (f ε ) of functions in C 1 Ω which satisfies (1.5). We assume that there exists C > 0 such that

Ω |∇u ε | 2 dx ≤ C . (4.2)
We consider the concentration points (x i,ε ) i=1,...,N given by Proposition 2.1 together with the γ i,ε 's and µ i,ε 's. For any i ∈ {1, . . . , N }, we let

r i,ε = 1 2 min min j∈{1,...,N },j =i |x i,ε -x j,ε | , d (x i,ε , ∂Ω) . (4.3)
Note that we have

λ ε |x -x i,ε | 2 u ε (x) 2 e uε(x) 2 ≤ C 1 in D xi,ε (r i,ε ) (4.4) and |x -x i,ε | u ε (x) |∇u ε (x)| ≤ C 2 in D xi,ε (r i,ε ) (4.5)
thanks to assertions e) and f) of Proposition 2.1. We let, for i ∈ {1, . . . , N }, B i,ε be the radial solution, studied in Appendix A, of

∆B i,ε = λ ε f ε (x i,ε ) B i,ε e B 2 i,ε and B i,ε (0) = γ i,ε
and we shall write, by an obvious and not misleading abuse of notation,

B i,ε (x) = B i,ε (|x -x i,ε |) . (4.6)
We let also

t i,ε (r) = ln 1 + r 2 4µ 2 i,ε and t i,ε (x) = t i,ε (|x -x i,ε |) . (4.7)
At last, we define for i = 1, . . . , N

d i,ε = d (x i,ε , ∂Ω) . (4.8)
Let us first state a claim which explains how we shall use the results of Section 3 for the multibumps analysis :

Claim 4.1. Assume that (u ε ) satisfies equation (4.1) with (f ε ) satisfying (1.5). Assume also that (4.2) holds so that we have concentration points (x i,ε ) satisfying (4.4) and (4.5). Let 0 ≤ r ε ≤ r i,ε be such that there exists

C 3 > 0 such that |x -x i,ε | |∇u ε (x)| ≤ C 3 γ -1 i,ε in D xi,ε (r ε ) . Then we have that : a) t i,ε (r ε ) ≤ γ 2 i,ε -1 + o (1) and 1 2πr ε ∂Dx i,ε (rε) u ε dσ = B i,ε (r ε ) + o γ -1 i,ε . b) There exists C > 0 such that |u ε -B i,ε | ≤ Cγ -1 i,ε and |∇ (u ε -B i,ε )| ≤ Cγ -1 i,ε r -1 ε in D xi,ε (r ε ). c) If r ε = r i,ε
, after passing to a subsequence,

γ i,ε (u ε (x i,ε + r i,ε • ) -B i,ε (r i,ε )) → 2 ln 1 |x| + H i as ε → 0 in C 1 loc (D 0 (1) \ {0}
) where H i is some harmonic function in the unit disk satisfying

H i (0) = 0 and ∇H i (0) = - 1 2 lim ε→0 r i,ε ∇f ε (x i,ε ) f ε (x i,ε
) .

Let us start with a simple consequence of this claim :

Claim 4.2. For any i ∈ {1, . . . , N } and any sequence (r ε ) of positive real numbers such that D xi,ε (r ε ) ⊂ Ω, we have that : a) If r ε ≤ r i,ε and B i,ε (r ε ) ≥ δγ i,ε for some δ > 0, there exists C > 0 such that

|u ε -B i,ε | ≤ C γ i,ε in D xi,ε (r ε ) .
Moreover, we have that

1 2πr ε ∂Dx i,ε (rε) u ε dσ = B i,ε (r ε ) + o γ -1 i,ε . b) If lim sup ε→0 γ -1 i,ε B i,ε (r ε ) ≤ 0 and lim sup ε→0 γ -1 i,ε B i,ε (r i,ε ) ≤ 0, then we have that inf ∂Dx i,ε (rε) u ε ≤ B i,ε (r ε ) + o γ -1 i,ε . c) If lim sup ε→0 γ -1 i,ε B i,ε (r i,ε ) ≤ 0, we have that t i,ε (d i,ε ) ≤ γ 2 i,ε for ε > 0 small enough.
In other words, we have that

λ ε f ε (x i,ε ) γ 2 i,ε d 2 i,ε ≤ 4 for ε small enough. Here, d i,ε is as in (4.8).
Proof of Claim 4.2 -We first prove a). We assume that B i,ε (r ε ) ≥ δγ i,ε for some δ > 0. Define 0 ≤ s ε ≤ r ε as

s ε = max 0 ≤ s ≤ r ε s.t. u ε ≥ 1 2 δγ i,ε in D xi,ε (s) .
Thanks to (4.5), we have that

|x -x i,ε | |∇u ε | ≤ Cγ -1 i,ε in D xi,ε (s ε )
for some C > 0 so that we can apply Claim 4.1. Assertion b) of this claim gives that Let us now prove b). Let us assume first that 1 +

|u ε -B i,ε | ≤ Cγ -1 i,ε in D xi,ε (s ε ) for some C > 0. Since B i,ε (s ε ) ≥ B i,ε (r ε ) ≥ δγ i,ε , we obtain in particular that s ε = r ε . Indeed, if s ε < r ε , there would exist some x ε ∈ ∂D xi,ε (s ε ) such that u ε (x ε ) = δ 2 γ i,ε ,
r 2 ε 4µ 2 i,ε ≤ e γ 2 i,ε , that lim sup ε→0 γ -1 i,ε B i,ε (r ε ) ≤ 0 and that lim sup ε→0 γ -1 i,ε B i,ε (r i,ε
) ≤ 0 and assume by contradiction that there exists 0 < η < 1 such that inf

Dx i,ε (rε) u ε ≥ B i,ε (r ε ) + ηγ -1 i,ε . (4.9)
We claim that

u ε ≥ γ i,ε + 1 γ i,ε ln 4µ 2 i,ε |x i,ε -x| 2 - 1 -η γ i,ε + o γ -1 i,ε in D xi,ε (r ε ) \ D xi,ε (R 0 µ i,ε ) (4.10)
where

R 0 = 4 √ e 1-η -1 .
The right-hand side of (4.10) being harmonic and u ε being super-harmonic, it is sufficient to check the inequality on ∂D xi,ε (r ε ) and on ∂D xi,ε (R 0 µ i,ε ). For that purpose, let us write that

B i,ε (r ε ) = γ i,ε -γ -1 i,ε t i,ε (r ε ) -γ -3 i,ε t i,ε (r ε ) + O γ -2 i,ε
as proved in Appendix A, Claim 5.2, since we assumed for the moment that t i,ε (r ε ) ≤ γ 2 i,ε . Since we assumed that lim sup

ε→0 γ -1 i,ε B i,ε (r ε ) ≤ 0, this gives that t i,ε (r ε ) γ 2 i,ε → 1 as ε → 0 so that B i,ε (r ε ) = γ i,ε -γ -1 i,ε ln 1 + r 2 ε 4µ 2 i,ε -γ -1 i,ε + o γ -1 i,ε = γ i,ε + 1 γ i,ε ln 4µ 2 i,ε r 2 ε -γ -1 i,ε + o γ -1 i,ε
.

This implies with (4.9) that

u ε ≥ γ i,ε + 1 γ i,ε ln 4µ 2 i,ε |x i,ε -x| 2 - 1 -η γ i,ε + o γ -1 i,ε on ∂D xi,ε (r ε ) . (4.11)
Let us write now that

u ε -B i,ε = o γ -1 i,ε on ∂D xi,ε (R 0 µ i,ε ) thanks to d) of Proposition 2.1. Since B i,ε (R 0 µ i,ε ) = γ i,ε -γ -1 i,ε ln 1 + R 2 0 4 + o γ -1 i,ε
, we obtain that

u ε ≥ γ i,ε + 1 γ i,ε ln 4µ 2 i,ε |x i,ε -x| 2 - 1 -η γ i,ε + o γ -1 i,ε on ∂D xi,ε (R 0 µ i,ε ) (4.12) provided that ln 1 + 4 R 2 0 < 1 -η ,
which is the case with our choice of R 0 . Thus (4.10) is proved. Now there exists

R 0 µ i,ε ≤ s ε ≤ min {r ε , r i,ε } such that B i,ε (s ε ) = η 2 γ i,ε since lim sup ε→0 γ -1 i,ε B i,ε (r ε ) ≤ 0 and lim sup ε→0 γ -1 i,ε B i,ε (r i,ε ) ≤ 0.
We can apply a) of the claim to get that

1 2πs ε ∂Dx i,ε (sε) u ε dσ = η 2 γ i,ε + o γ -1 i,ε
.

Applying (4.10), this leads to

γ i,ε + 1 γ i,ε ln 4µ 2 i,ε s 2 ε - 1 -η γ i,ε + o γ -1 i,ε ≤ η 2 γ i,ε + o γ -1 i,ε . (4.13) Since B i,ε (s ε ) = η 2 γ i,ε , it is not difficult to check thanks to Claim 5.2 of Appendix A that t i,ε (s ε ) = 1 - η 2 γ 2 i,ε -1 + O γ -1 i,ε so that, since sε µi,ε → +∞ as ε → 0, ln 4µ 2 i,ε s 2 ε = -γ 2 i,ε -1 1 - η 2 + o (1) .
Coming back to (4.13) with this leads to a contradiction. This proves that (4.9) is absurd for any 0 < η < 1. Thus we have proved assertion b) as long as t i,ε (r ε ) ≤ γ 2 i,ε . We shall now prove c), which will by the way prove that b) holds since the condition t i,ε (r ε ) ≤ γ 2 i,ε will always be satisfied. Let us assume by contradiction that t i,ε

(d i,ε ) ≥ γ 2 i,ε . Then D xi,ε (r ε ) ⊂ Ω for ε > 0 small where 1 + r 2 ε 4µ 2 ε = e γ 2 ε -1 2 .
We can apply b) in this case since t i,ε (r ε ) ≤ γ 2 i,ε and

B i,ε (r ε ) = - 1 2 γ -1 i,ε + O γ -2 i,ε
by Claim 5.2 of Appendix A. This leads to a contradiction since u ε ≥ 0 in Ω. Thus c) is proved thanks to the definition of µ i,ε and b) is also proved. This ends the proof of this claim. ♦ Claim 4.3. For any i ∈ {1, . . . , N }, we have that

lim sup ε→0 γ -1 i,ε B i,ε (r i,ε ) ≤ 0 .
Proof of Claim 4.3 -Let us reorder for this proof the concentration points in such a way that

r 1,ε ≤ r 2,ε ≤ • • • ≤ r N,ε . (4.14)
We prove the assertion by induction on i. Let i ∈ {1, . . . , N } and let us assume that lim sup

ε→0 γ -1 j,ε B j,ε (r j,ε ) ≤ 0 for all 1 ≤ j ≤ i -1 . (4.15)
Note that we do not assume anything if i = 1. We proceed by contradiction, assuming that, after passing to a subsequence,

γ -1 i,ε B i,ε (r i,ε ) ≥ 2ε 0 (4.16)
for some ε 0 > 0.

Step 1 -If (4.16) holds, then

d(xi,ε,∂Ω) ri,ε
→ +∞ as ε → 0. In particular, this implies that r i,ε → 0 as ε → 0.

Proof of Step 1 -For any η > 0 small enough, there exists a path of length less than or equal to Cd (x i,ε , ∂Ω) joining the boundary of Ω and the boundary of the disk D xi,ε (ηd (x i,ε , ∂Ω)), and avoiding all the disks D xj,ε (ηd (x i,ε , ∂Ω)) for j = 1, . . . , N . Using f) of Proposition 2.1, we deduce that, for any η > 0, there exists C > 0 such that

u ε ≤ C on ∂D xi,ε (ηd (x i,ε , ∂Ω)) . If d (x i,ε , ∂Ω) = O (r i,ε ), we can find η > 0 small enough such that ηd (x i,ε , ∂Ω) ≤ r i,
D i = {j ∈ {1, . . . , N } , j = i s.t. |x j,ε -x i,ε | = O (r i,ε )} = ∅ . (4.17)
There exists 0 < δ < 1 such that, for any j ∈ D i , any point of ∂D xj,ε (δr i,ε ) can be joined to a point of ∂D xi,ε (δr i,ε ) by a path

γ ε : [0, 1] → Ω such that |γ ε (t) -x k,ε | ≥ δr i,ε for all k = 1, . . . , N and all 0 ≤ t ≤ 1 and such that |γ ′ ε (t)| ≤ δ -1 r i,ε . Thanks to assertion f) of Proposition 2.1, the existence of such paths give that inf ∂Dx j,ε (δri,ε) u 2 ε ≥ inf ∂Dx i,ε (δri,ε) u 2 ε -2C 2 δ -2 for all j ∈ D i .
Thanks to (4.16), we can apply a) of Claim 4.2 to obtain also that

u ε ≥ B i,ε (δr i,ε ) -Cγ -1 i,ε on ∂D xi,ε (δr i,ε ) for some C > 0. Since B i,ε (δr i,ε ) = B i,ε (r i,ε ) + O γ -1 i,ε
, the two previous estimates , together with (4.16), lead to the existence of some C > 0 such that

u ε ≥ B i,ε (r i,ε ) -Cγ -1 i,ε on ∂D xj,ε (δr i,ε ) for all j ∈ D i . (4.18)
Step 2 -If (4.16) holds, then for any j ∈ D i , we have that

lim inf ε→0 γ -1 j,ε B j,ε (r j,ε ) > 0 .
In particular, we have that j ≥ i + 1.

Proof of

Step 2 -Assume on the contrary that there exists j ∈ D i such that, after passing to a subsequence, lim sup

ε→0 γ -1 j,ε B j,ε (r j,ε ) ≤ 0 . (4.19)
Since j ∈ D i , we also know that

r j,ε ≤ 1 2 |x i,ε -x j,ε | ≤ Cr i,ε . (4.20)
Thus we also have that lim sup

ε→0 γ -1 j,ε B j,ε (δr i,ε ) ≤ 0 . (4.21)
We can apply b) of Claim 4.2 with r ε = δr i,ε to obtain that

B i,ε (r i,ε ) -Cγ -1 i,ε ≤ B j,ε (δr i,ε ) + o γ -1 j,ε (4.22)
thanks to (4.18). Combining (4.16) and (4.21), we get that

γ i,ε = o (γ j,ε ) . (4.23)
Thus we also have that µ j,ε ≤ µ i,ε . Let us write now thanks to Claim 5.2 of Appendix A that

B j,ε (δr i,ε ) = -γ -1 j,ε ln r 2 i,ε 4 -γ -1 j,ε ln λ ε γ 2 j,ε + O γ -1 j,ε
and that

B i,ε (r i,ε ) = -γ -1 i,ε ln r 2 i,ε 4 -γ -1 i,ε ln λ ε γ 2 i,ε + O γ -1 i,ε
to obtain that

B j,ε (δr i,ε ) = γ i,ε γ j,ε B i,ε (r i,ε ) + γ -1 j,ε ln γ 2 i,ε γ 2 j,ε + O γ -1 j,ε
.

Coming back to (4.22) with this, (4.16) and (4.23), we obtain that

(2ε 0 + o(1)) γ i,ε ≤ γ -1 j,ε ln γ 2 i,ε γ 2 j,ε + O γ -1 i,ε ≤ O γ -1 i,ε
, which is a clear contradiction.

Step 2 is proved. ♠

We can now conclude the proof of the claim by proving that (4.16) is absurd if (4.15) holds. Continue to assume that (4.16) holds. Then we know thanks to Step 2 that for any j ∈ D i , j ≥ i + 1 so that r j,ε ≥ r i,ε . We set, for j ∈ D i , and up to a subsequence,

xj = lim ε→0 x j,ε -x i,ε r i,ε (4.24)
and we let Ŝ = {x j , j ∈ D i } . Let K be a compact subset of R 2 \ Ŝ. We can use assertion f) of Proposition 2.1 to write 4 that

γ i,ε |u ε (x i,ε + r i,ε x) -B i,ε (r i,ε )| ≤ C K in K . (4.29)
Thanks to (4.1), we can write that

∆û ε = λ ε r 2 i,ε γ i,ε f ε (x i,ε + r i,ε x) u ε (x i,ε + r i,ε x) 2 e uε(xi,ε+ri,εx) 2 where ûε = γ i,ε (u ε (x i,ε + r i,ε x) -B i,ε (r i,ε )) .
4 see the argument between Steps 1 and 2.

Using (4.29), we can write that

|∆û ε | ≤ C K µ -2 i,ε r 2 i,ε e Bi,ε(ri,ε) 2 -γ 2 i,ε in K for any compact subset K of R 2 \ Ŝ.
Thanks to (4.16), we have that

e Bi,ε(ri,ε) 2 -γ 2 i,ε ≤ C 1 + r 2 i,ε 4µ 2 i,ε -1-2ε0 so that |∆û ε | ≤ C K µ i,ε r i,ε 4ε0 → 0 uniformly in K .
By standard elliptic theory, we thus have that

ûε = γ i,ε (u ε (x i,ε + r i,ε x) -B i,ε (r i,ε )) → û0 in C 1 loc R 2 \ Ŝ as ε → 0 (4. 30
)
where

∆û 0 = 0 in R 2 \ Ŝ . (4.31)
Since r j,ε ≥ r i,ε for j ∈ D i , (4.28) permits to apply a) of Claim 4.2, which in turn implies thanks to (4.5) that we can apply Claim 4.1 for all j ∈ D i with r ε = r i,ε . Assertion c) of this claim gives that

γ j,ε (u ε (x j,ε + r i,ε x) -B j,ε (r i,ε )) → 2 ln 1 |x| + H j (4.32)
in C 1 loc (D 0 (1) \ {0}) as ε → 0 where H j is harmonic in the unit disk and satifies H j (0) = 0 and ∇H j (0) = 0 (note here that we know thanks to Step 1 that r i,ε → 0 as ε → 0). This gives that

γ j,ε γ i,ε ûε + γ j,ε (B i,ε (r i,ε ) -B j,ε (r i,ε )) → 2 ln 1 |x -xj | + Ĥj (4.33)
in C 1 loc D xj (1) \ {x j } as ε → 0 for all j ∈ D i (and also for j = i if we set xi = 0). It remains to write thanks to Claim 5.2 of Appendix A and to (1.5) that

B i,ε (r i,ε ) -B j,ε (r i,ε ) = 1 - γ i,ε γ j,ε B i,ε (r i,ε ) + γ -1 j,ε ln γ 2 j,ε γ 2 i,ε + O γ -1 j,ε
to deduce from (4.16), (4.30) and (4.33) that in D xj (1) \ {x j } where ϕ j is smooth and harmonic and satisfies ∇ϕ j (x j ) = 0. Thus we can write that

γ j,ε γ i,ε → 1 as ε → 0 and γ i,ε |γ i,ε -γ j,ε | = O (1) . ( 4
û0 = 2 ln 1 |x| + 2 j∈Di ln 1 |x -xj | + ϕ (4.36)
where ϕ is a smooth harmonic function in R 2 . Thanks to assertion f) of Proposition 2.1, we also know that

|∇ϕ(x)| ≤ C 1 + |x| in R 2
for some C > 0 so that ϕ ≡ Cst. Now this gives that for any k ∈ D i ,

∇   ln 1 |x| + j∈Di,j =k ln 1 |x -xj |   (x k ) = 0 . Let k ∈ D i be such that |x k | ≥ |x j | for all j ∈ D i . Then ∇   ln 1 |x| + j∈Di, j =k ln 1 |x -xj |   (x k ) , xk = -|x k | - j∈Di, j =k |x k | 2 -xk , xj |x k -xj | ≤ -|x k | < 0 ,
which gives the desired contradiction. This proves that (4.16) is absurd as soon as (4.15) holds. And this ends the proof of the claim by an induction argument. ♦ Claim 4.4. For any i = 1, . . . , N , we have that

λ ε f ε (x i,ε ) γ 2 i,ε d 2 i,ε ≤ 4 for ε small enough. Proof of Claim 4.4 -It is a direct consequence of c) of Claim 4.2 together with Claim 4.3. ♦ Claim 4.5. We have that Ω |∇u ε | 2 dx = Ω |∇u 0 | 2 dx + 4πN + o(1) .
In other words, M = 0 in Theorem 2.1.

Proof of Claim 4.5 -We prove that M = 0 in Theorem 2.1. Assume on the contrary that there exists some sequence (y 1,ε ) such that the assertion b) of Theorem 2.1 holds. This means that ν -2

1,ε = λ ε f ε (y 1,ε ) u ε (y 1,ε ) 2 e uε(y1,ε) 2 → +∞ as ε → 0 .
By e) of Proposition 2.1, we know that

ν -2 1,ε min i=1,...,N |x i,ε -y 1,ε | 2 = min i=1,...,N |x i,ε -y 1,ε | 2 λ ε f ε (y 1,ε ) u ε (y 1,ε ) 2 e uε(y1,ε) 2 ≤ C 1 f ε (y 1,ε ) .
This proves that there exists i ∈ {1, . . . , N } such that

|x i,ε -y 1,ε | = O (ν 1,ε ) . Since |x i,ε -y 1,ε | µ i,ε → +∞ as ε → 0 by a) of Theorem 2.1, we have that ν 1,ε µ i,ε → +∞ as ε → 0 .
Thanks to the definition of ν 1,ε and µ i,ε , this leads to

e γ 2 i,ε -uε(y1,ε) 2 γ 2 i,ε u ε (y 1,ε ) 2 → +∞ as ε → 0 , which implies that γ 2 i,ε -u ε (y 1,ε ) 2 → +∞ as ε → 0 . (4.37)
Now, by the convergence of b) of Theorem 2.1, we know that

u ε ≥ u ε (y 1,ε ) -Cu ε (y 1,ε ) -1 on ∂D xi,ε (Rν 1,ε )
for some R > 0 and C > 0. Thanks to Claim 4.3, we can use assertion b) of Claim 4.2 to deduce that

u ε (y 1,ε ) -Cu ε (y 1,ε ) -1 ≤ B i,ε (Rν 1,ε ) + o γ -1 i,ε
. This leads after some simple computations, using Claim 5.2 of Appendix A, to

u ε (y 1,ε ) -Cu ε (y 1,ε ) -1 ≤ γ -1 i,ε ln u ε (y 1,ε ) 2 γ 2 i,ε + u ε (y 1,ε ) 2 γ i,ε + Cγ -1 i,ε
so that, thanks to (4.37),

u ε (y 1,ε ) 2 1 - u ε (y 1,ε ) γ i,ε ≤ C .
This clearly implies that

u ε (y 1,ε ) γ i,ε → 1 as ε → 0 and then that u ε (y 1,ε ) ≥ γ i,ε -Cγ -1 i,ε
for some C > 0. This contradicts (4.37). Thus we have proved that M = 0 in Theorem 2.1 and the claim follows. ♦

For any i ∈ {1, . . . , N }, thanks to Claim 4.3 and a) of Claim (4.2), there exists 0

≤ s i,ε ≤ r i,ε such that lim sup ε→0 γ -1 i,ε B i,ε (s i,ε ) ≤ 0 and |u ε -B i,ε | ≤ D i γ i,ε in D xi,ε (s i,ε ) (4.38)
for some D i > 0.

Claim 4.6. We have that lim inf

ε→0 Dx i,ε (si,ε) |∇u ε | 2 dx ≥ 4π .
Proof of Claim 4.6 -Let δ > 0. Let us write thanks to (4.38) that

Dx i,ε (si,ε) |∇u ε | 2 dx ≥ Dx i,ε (si,ε) ∇ (u ε -δγ i,ε ) + 2 dx = Dx i,ε (si,ε) (u ε -δγ i,ε ) + ∆u ε dx .
Thanks to (4.1), this leads to

Dx i,ε (si,ε) |∇u ε | 2 dx ≥ λ ε Dx i,ε (si,ε) f ε (u ε -δγ i,ε ) + u ε e u 2 ε dx ≥ λ ε Dx i,ε (Rµi,ε) f ε (u ε -δγ i,ε ) + u ε e u 2 ε dx for all R > 0. Now we have that lim ε→0 λ ε Dx i,ε (Rµi,ε) f ε (u ε -δγ i,ε ) + u ε e u 2 ε dx = (1 -δ) D0(R)
Ω ε = Ω \ N i=1 D xi,ε (s i,ε ) (4.39)
where s i,ε is as in (4.38) and

w ε =    u ε in Ω ε min u ε , B i,ε (s i,ε ) + 2 D i γ i,ε in D xi,ε (s i,ε ) for i = 1, . . . , N (4.40)
Claim 4.7. We have that

Ω |∇ (w ε -u 0 )| 2 dx → 0 as ε → 0 .
Proof of Claim 4.7 -Let us write that

Ω |∇ (w ε -u 0 )| 2 dx = Ω |∇w ε | 2 dx -2 Ω ∇w ε , ∇u 0 dx + Ω |∇u 0 | 2 dx = Ω |∇u ε | 2 dx -2 Ω ∇u ε , ∇u 0 dx + Ω |∇u 0 | 2 dx + Ω ∇ (w ε -u ε ) , ∇u ε + ∇w ε -2∇u 0 dx = 4πN + o(1) + Ω ∇ (w ε -u ε ) , ∇u ε + ∇w ε -2∇u 0 dx
thanks to the weak convergence of u ε to u 0 in H 1 and to Claim 4.5. Let us remark now that ∇ (w εu ε ) ≡ 0 in Ω ε and that ∇ (w εu ε ) , ∇w ε = 0 a.e. Thus we can write that

Ω |∇ (w ε -u 0 )| 2 dx = 4πN + o(1) + N i=1 Dx i,ε (si,ε) ∇ (w ε -u ε ) , ∇u ε -2∇u 0 dx . (4.41)
Since w εu ε is null on the boundary of D xi,ε (s i,ε ), we can proceed as in the proof of Claim 4.6 to get that

Dx i,ε (si,ε) ∇ (w ε -u ε ) , ∇u ε -2∇u 0 dx = Dx i,ε (si,ε) (w ε -u ε ) (∆u ε -2∆u 0 ) dx ≤ -4π + o(1) + O γ i,ε Dx i,ε (si,ε) |∆u 0 | dx .
Here we used the fact that w ε ≤ u ε and

|w ε | = o (γ i,ε ) in D xi,ε (s i,ε
). If u 0 ≡ 0, the last term disappears. If u 0 ≡ 0, then λ ε → λ 0 with λ 0 > 0 and Claim 4.4 gives that γ i,ε s 2 i,ε = o(1). Thus, in any case, we have that

Dx i,ε (si,ε) ∇ (w ε -u ε ) , ∇u ε -2∇u 0 dx ≤ -4π + o(1) .
Coming back to (4.41) with this proves the claim. ♦

The next two claims are devoted to obtaining good pointwise estimates on u ε and ∇u ε .

Claim 4.8. For any sequence

(x ε ) of points in Ω such that |x ε -x i,ε | µ i,ε → +∞ as ε → 0 for i = 1, . . . , N , we have that a) if d ε = d (x ε , ∂Ω) → 0 as ε → 0, then u ε (x ε ) = ψ ε (x ε ) + N i=1 4π + o(1) γ i,ε G (x i,ε , x ε ) +O N i=1 γ -1 i,ε µ i,ε |x i,ε -x ε | + γ -1 i,ε ln s i,ε |x i,ε -x ε | + 2 , b) if d ε → 0 as ε → 0, then u ε (x ε ) = ψ ε (x ε ) + N i=1 4π + o(1) γ i,ε G (x i,ε , x ε ) +O i∈A γ -1 i,ε µ i,ε |x i,ε -x ε | + γ -1 i,ε ln s i,ε |x i,ε -x ε | + 2 +O i∈B d ε d ε + d i,ε γ -1 i,ε µ i,ε + γ -1 i,ε s i,ε
where G is the Green function of the Laplacian with Dirichlet boundary condition in Ω and ψ ε is a solution of

∆ψ ε = λ ε f ε w ε e w 2 ε in Ω and ψ ε = 0 on ∂Ω . In b), A is defined as the set of i ∈ {1, . . . , N } such that |x i,ε -x ε | ≤ s i,ε + o (d ε ) and B as its complementary.
Proof of Claim 4.8 -We let G be the Green function of the Laplacian with Dirichlet boundary condition in Ω. We let (x ε ) be a sequence of points in Ω such that

|x ε -x i,ε | µ i,ε → +∞ as ε → 0 for i = 1, . . . , N . (4.42)
Then we have thanks to (4.1) and to the definition of ψ ε that

u ε (x ε ) -ψ ε (x ε ) = λ ε Ω G (x ε , x) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx .
Using the definition (4.40) of w ε , this gives that

u ε (x ε ) -ψ ε (x ε ) (4.43) = N i=1 λ ε G (x ε , x i,ε ) Dx i,ε (si,ε) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx + N i=1 A i,ε
where

A i,ε = λ ε Dx i,ε (si,ε) (G (x ε , x) -G (x ε , x i,ε )) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx . (4.44)
We fix i ∈ {1, . . . , N } in the following and we let

Ω 0,ε = D xi,ε (s i,ε ) ∩ {t ε (x) ≤ t 1,ε } , Ω 1,ε = D xi,ε (s i,ε ) ∩ {t 1,ε ≤ t ε (x) ≤ t 2,ε } and (4.45) Ω 2,ε = D xi,ε (s i,ε ) ∩ {t ε (x) ≥ t 2,ε } . where t ε (x) = ln 1 + |x-xi,ε| 2 4µ 2 i,ε , t 1,ε = 1 4 γ 2 i,ε and t 2,ε = γ 2 i,ε -γ i,ε .
Step 1 -We have that

λ ε Dx i,ε (si,ε) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx = 4πγ -1 i,ε + o γ -1 i,ε . Proof of Step 1 -We write that λ ε Dx i,ε (si,ε) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx = λ ε Dx i,ε (Rεµi,ε) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx +λ ε Ω0,ε\Dx i,ε (Rεµi,ε) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx +λ ε Ω1,ε f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx +λ ε Ω2,ε f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx where R ε → +∞ is such that |u ε -B i,ε | = o γ -1 i,ε and γ -1 i,ε B i,ε (x) = 1 + o(1) in D xi,ε (R ε µ i,ε
). Such a R ε does exist thanks to d) of Proposition 2.1. Then we have, using also (4.38), that

λ ε Dx i,ε (Rεµi,ε) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx = λ ε (f ε (x i,ε ) + o(1)) Dx i,ε (Rεµi,ε) B i,ε (x)e Bi,ε(x) 2 dx +O λ ε B i,ε (s i,ε ) e Bi,ε(si,ε) 2 R 2 ε µ 2 i,ε = γ -1 i,ε D0(Rε) e 2U dx + o(1) + o γ -1 i,ε = 4πγ -1 i,ε + o γ -1 i,ε
.

In Ω 0,ε , we write that

B i,ε (x) 2 = γ 2 i,ε -2t ε (x) + t ε (x) 2 γ 2 i,ε + O(1) ≤ γ 2 i,ε - 7 4 t ε (x) + O(1)
so that e Bi,ε(x) 2 ≤ e γ 2 i,ε

1 + |x -x i,ε | 2 4µ 2 i,ε -7 4
.

Thus we can write that

0 ≤ λ ε Ω0,ε\Dx i,ε (Rεµi,ε) f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx ≤ Cγ -1 i,ε µ -2 i,ε Ω\Dx i,ε (Rεµi,ε) 1 + |x -x i,ε | 2 4µ 2 i,ε -7 4 dx = o γ -1 i,ε
.

In Ω 1,ε , we write that

e Bi,ε(x) 2 ≤ e γ 2 i,ε -1 so that 0 ≤ λ ε Ω1,ε f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx ≤ Cµ -2 i,ε e -1 4 γi,ε γ -1 i,ε Ω1,ε 1 + |x -x i,ε | 2 4µ 2 i,ε -1 dx ≤ Ce -1 4 γi,ε γ -1 i,ε ln s i,ε µ i,ε = o γ -2 i,ε since 2 ln si,ε µi,ε = γ 2 i,ε + O(1) thanks to Claim 4.4. At last, in Ω 2,ε , we have that B i,ε = O(1) so that 0 ≤ λ ε Ω2,ε f ε (x) u ε (x)e uε(x) 2 -w ε (x)e wε(x) 2 dx ≤ Cλ ε s 2 i,ε = O γ -2 i,ε
thanks to Claim 4.4. Combining all these estimates clearly proves Step 1. ♠

We shall now estimate the A i 's involved in (4.43) and defined in (4.44). We write since u ε ≥ w ε and thanks to (4.38) that

|A i,ε | ≤ Cλ ε Dx i,ε (si,ε) |G (x ε , x) -G (x ε , x i,ε )| B i,ε + Cγ -1 i,ε e Bi,ε(x) 2 dx . (4.46)
Step 2 -Assume that d ε = d (x ε , ∂Ω) ≥ d for some d > 0. Then we have that

|A i,ε | ≤ Cγ -1 i,ε µ i,ε |x i,ε -x ε | + γ -1 i,ε ln s i,ε |x i,ε -x ε | + 2 .

Proof of

Step 2 -We use (6.1) to write that

|G (x ε , x) -G (x ε , x i,ε )| ≤ 1 2π ln |x i,ε -x ε | |x ε -x| + C |x -x i,ε | .
Thus we have thanks to (4.46) that

|A i,ε | ≤ Cλ ε Dx i,ε (si,ε) ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| B i,ε + Cγ -1 i,ε e Bi,ε(x) 2 dx .
In Ω 0,ε , we have that

B i,ε (x) ≤ γ 2 i,ε - 7 4 ln 1 + |x i,ε -x| 2 4µ 2 i,ε so that λ ε Ω0,ε ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| B i,ε + Cγ -1 i,ε e Bi,ε(x) 2 dx ≤ Cµ -2 i,ε γ -1 i,ε Ω0,ε ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| 1 + |x i,ε -x| 2 4µ 2 i,ε -7 4 dx .
This leads after simple computations, since

|xi,ε-xε| µi,ε → +∞, as ε → 0 to λ ε Ω0,ε ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| B i,ε + Cγ -1 i,ε e Bi,ε(x) 2 dx ≤ Cγ -1 i,ε µ i,ε |x i,ε -x ε | .
In Ω 1,ε , we can write that

e Bi,ε(x) 2 ≤ e γ 2 i,ε -1 4 γi,ε 1 + |x -x i,ε | 2 4µ 2 i,ε -1 so that λ ε Ω1,ε ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| B i,ε + Cγ -1 i,ε e Bi,ε(x) 2 dx ≤ Cγ -1 i,ε e -1 4 γi,ε Ω1,ε ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| |x -x i,ε | -2 dx ≤ Cγ -1 i,ε e -1 4 γi,ε ln r 2,ε µ i,ε 2
where t i,ε (r 2,ε ) = t 2,ε . We have that ln r 2,ε µ i,ε ≤ Cγ 2 ε so that λ ε Ω1,ε ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| B i,ε + Cγ -1 i,ε e Bi,ε(x) 2 dx = O γ 3 i,ε e -1 4 γi,ε = o γ -2 i,ε . At last, in Ω 2,ε , we have that B i,ε = O(1) so that λ ε Ω2,ε ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| B i,ε + Cγ -1 i,ε e Bi,ε(x) 2 dx ≤ λ ε Ω2,ε ln |x i,ε -x ε | |x ε -x| + |x i,ε -x| dx ≤ λ ε s 2 i,ε ln s i,ε |x i,ε -x ε | + 2 ≤ Cγ -2 i,ε ln s i,ε |x i,ε -x ε | + 2
by direct computations and Claim 4.4. Combining the above estimates gives Step 2. ♠

Step 3 -Assume now that d ε = d (x ε , ∂Ω) → 0 as ε → 0 and that |x i,εx ε | ≥ s i,ε + δd ε for some δ > 0. Then we have that

|A i,ε | ≤ C d ε d ε + d i,ε γ -1 i,ε µ i,ε + γ -2 i,ε s i,ε .
Proof of Step 3 -In this setting, we can apply (6.12) to write that

|G (x ε , x) -G (x ε , x i,ε )| ≤ C d ε d ε + d i,ε |x -x i,ε | so that |A i,ε | ≤ Cλ ε d ε d ε + d i,ε Dx i,ε (si,ε) |x -x i,ε | B i,ε + Cγ -1 i,ε e B 2 i,ε dx
thanks to (4.44). In Ω 0,ε , we have that

B i,ε + Cγ -1 i,ε e B 2 i,ε ≤ Cγ i,ε e γ 2 i,ε 1 + |x -x i,ε | 2 4µ 2 i,ε -7 4 so that λ ε Ω0,ε |x -x i,ε | B i,ε + Cγ -1 i,ε e B 2 i,ε dx ≤ Cµ -2 i,ε γ -1 i,ε Dx i,ε (si,ε) |x -x i,ε | 1 + |x -x i,ε | 2 4µ 2 i,ε -7 4 dx ≤ Cµ i,ε γ -1 i,ε .
In Ω 1,ε , we have that

B i,ε + Cγ -1 i,ε e Bi,ε(x) 2 ≤ γ i,ε e γ 2 i,ε -1 4 γi,ε 1 + |x -x i,ε | 2 4µ 2 i,ε -1 so that λ ε Ω1,ε |x -x i,ε | B i,ε + Cγ -1 i,ε e B 2 i,ε dx ≤ Cµ -2 i,ε γ -1 i,ε e -1 4 γi,ε Dx i,ε (si,ε) |x -x i,ε | 1 + |x -x i,ε | 2 4µ 2 i,ε -1 dx ≤ γ -1 i,ε e -1 4 γi,ε s i,ε . At last, in Ω 2,ε , we have that B i,ε = O(1) so that λ ε Ω2,ε |x -x i,ε | B i,ε + Cγ -1 i,ε e B 2 i,ε dx ≤ λ ε s 3 i,ε .
Combining the above estimates with Claim 4.4, we get the estimate of Step 3. ♠

Step 4 -Assume now that

d ε = d (x ε , ∂Ω) → 0 as ε → 0 and that |x i,ε -x ε | ≤ s i,ε + o (d ε ). Then we have that |A i,ε | ≤ Cγ -1 i,ε µ i,ε |x i,ε -x ε | + γ -1 i,ε ln s i,ε |x i,ε -x ε | + 2 .
Proof of Step 4 -Let us remark that in this case, we necessarily have that

d ε ≤ |x i,ε -x ε | + d i,ε ≤ s i,ε + d i,ε + o (d ε ) ≤ 3 2 d i,ε + o (d ε ) so that d ε = O (d i,ε ). This leads in turn to |x i,ε -x ε | ≤ s i,ε + o (d i,ε
). And then we can write that

d i,ε ≤ d ε + |x i,ε -x ε | ≤ s i,ε + o (d i,ε ) + d ε ≤ 1 2 d i,ε + d ε + o (d i,ε ) so that d i,ε = O (d ε )
. Thanks to (6.12), we can write that

|G (x ε , x) -G (x ε , x i,ε )| ≤ C |x -x i,ε | d i,ε + C ln |x i,ε -x ε | |x ε -x| so that
|∇ (u ε -ψ ε ) (x)| ≤ C N i=1 γ -1 i,ε (µ i,ε + |x -x i,ε |) -1
where ψ ε is as in Claim 4.8.

Proof of Claim 4.9 -We use again the Green representation formula with equation (4.1) (together with the equation satisfied by ψ ε , see Claim 4.8) to write that

|∇ (u ε -ψ ε ) (x)| ≤ λ ε Ω |∇G(x, y)| f ε (y) u ε (y)e uε(y) 2 -w ε (y)e wε(y) 2 dy .
Thanks to standard estimates on the Green function and to the definition (4.40), this leads to |x -y| -1 u ε (y)e uε(y) 2 dy (4.48)

|∇ (u ε -ψ ε ) (x)| ≤ C N i=1 λ ε Dx i,ε (si,ε) |x -y| -1 u ε (y)e uε(
≤ Cλ ε k=0,1,2 Ω k,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy
where the Ω α,ε 's are as in (4.45). In Ω 0,ε , we write that

B i,ε + C i γ -1 i,ε (y) e Bi,ε(y) 2 ≤ Cγ i,ε e γ 2 i,ε 1 + |y -x i,ε | 2 4µ 2 i,ε -7 4 so that λ ε Ω0,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy ≤ Cµ -2 i,ε γ -1 i,ε Ω0,ε |x -y| -1 1 + |y -x i,ε | 2 4µ 2 i,ε -7 4 dy .
Direct computations give that

λ ε Ω0,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy ≤ Cγ -1 i,ε (µ i,ε + |x -x i,ε |) -1 . ( 4

.49)

In Ω 1,ε , we write that

B i,ε + C i γ -1 i,ε (y) e Bi,ε(y) 2 ≤ Cγ i,ε e γ 2 i,ε e tε(y) 2 γ 2 i,ε -2tε(y) so that λ ε Ω1,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy ≤ Cγ -1 i,ε µ -2 i,ε Ω1,ε |x -y| -1 e tε(y) 2 γ 2 i,ε -2tε(y)
.

In Ω 1,ε , we have that

t ε (y) 2 γ 2 i,ε -2t ε (y) ≤ -t ε (y) - 1 4 γ i,ε so that λ ε Ω1,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy ≤ Cγ -1 i,ε e -1 4 γi,ε Ω1,ε |x -y| -1 µ 2 i,ε + |y -x i,ε | 2 -1
dy .

In Ω 1,ε we have that |yx i,ε | ≥ µ i,ε so that

λ ε Ω1,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy ≤ Cγ -1 i,ε e -1 4 γi,ε Ω1,ε |x -y| -1 |y -x i,ε | -2 dy .
Noting that D xi,ε (r 1,ε ) ∩ Ω 1,ε = ∅ for ε small where

r 1,ε = µ i,ε e 1 8 γ 2
i,ε , we get by direct computations that

λ ε Ω1,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy ≤ Cγ -1 i,ε e -1 4 γi,ε (|x -x i,ε | + r 1,ε ) -1 ln 2 + |x -x i,ε | r 1,ε .
Thanks to the value of r 1,ε , this leads to

λ ε Ω1,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy = o γ -1 i,ε (µ i,ε + |x -x i,ε |) -1 . (4.50) At last, in Ω 2,ε , we have that B ε (y) = O(1) so that λ ε Ω2,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy ≤ Cλ ε Ω2,ε |x -y| -1 dy ≤ Cλ ε s 2 i,ε s i,ε + |x -x i,ε | .
Thanks to Claim 4.4, this leads to Let us reorder the concentration points in a suitable way. For this purpose, we notice that, up to a subsequence, for any i, j ∈ {1, . . . , N }, there exists C i,j , possibly 0 or +∞ (but nonnegative) such that lim

λ ε Ω2,ε |x -y| -1 B i,ε (y) + C i γ -1 i,ε e Bi,ε(y) 2 dy ≤ Cγ -2 i,ε (s i,ε + |x -x i,ε |) -1 . ( 4
ε→0 γ i,ε γ j,ε = C i,j . (4.52)
Note that C i,j = C -1 j,i (with obvious conventions when C i,j = 0 or +∞). Then there exists C ≥ 1 such that for any i, j ∈ {1, . . . , N } , either C i,j = 0 or

C i,j = +∞ or 1 C ≤ C i,j ≤ C . (4.53)
It is then easily checked that we can order the concentration points in such a way that for any i, j ∈ {1, . . . , N } , i < j ⇒ C i,j < +∞ (4.54) and for any i, j ∈ {1, . . . , N } , i < j and

C i,j > 0 ⇒ r i,ε ≤ r j,ε . (4.55)
Let us give some estimates on ψ ε , involved in Claims 4.8 and 4.9. Using Claim 4.7, we clearly have that λ -1 ε ∆ψ ε is uniformly bounded in any L p (Ω) thanks to Trudinger-Moser inequality. Thus we know that there exists C > 0 such that

ψ ε C 1,α (Ω) ≤ Cλ ε (4.56)
for 0 < α < 1 by standard elliptic theory. Now, if λ ε → 0, we know that u 0 ≡ 0 and we can be a little bit more precise. Indeed,

∆ψ ε L p (Ω) ≤ λ ε f ε L ∞ (Ω) w ε e w 2 ε L p (Ω) ≤ λ ε f ε L ∞ (Ω) u ε L 2p (Ω) e w 2 ε L 2p (Ω)
.

Since u 0 ≡ 0, we know thanks to Claim 4.7 and to Trudinger-Moser inequality that e w 2 ε is bounded in any L q . Thus we have that

∆ψ ε L p (Ω) ≤ Cλ ε u ε L 2p (Ω)
thanks to (1.5). Using Claim (4.9), we get that

∇ (u ε -ψ ε ) L q (Ω) ≤ C q γ 1,ε
for some C q > 0 for all 1 ≤ q < 2. Remember that concentration points are ordered such that (4.54) holds. This gives that

u ε L 2p (Ω) ≤ C p γ -1 1,ε + ∇ψ ε C 1 (Ω) so that ∆ψ ε L p (Ω) ≤ Cλ ε γ -1 1,ε + ∇ψ ε C 1 (Ω)
. By standard elliptic theory and since we assumed that λ ε → 0, we finally obtain that

if λ ε → 0 as ε → 0, then ψ ε C 1,α (Ω) ≤ C λ ε γ 1,ε . (4.57)
Claim 4.10. We have that r 1,ε ≥ δ 0 for some δ 0 > 0.

Proof of Claim 4.10 -We assume by contradiction that r 1,ε → 0 as ε → 0. We let in the following

D ⋆ 1 = {i ∈ {2, . . . , N } s.t. |x i,ε -x 1,ε | = O (r 1,ε )} and D 1 = D ⋆ 1 ∪ {1} . (4.58)
After passing to a subsequence, we let

S ⋆ 1 = xi = lim ε→0 x i,ε -x 1,ε r 1,ε , i ∈ D ⋆ 1 and S 1 = S ⋆ 1 ∪ {x 1 = 0} . (4.59)
We also let

Ω 1,ε = y ∈ R 2 s.t. x 1,ε + r 1,ε y ∈ Ω . ( 4
.60) Note that, after passing to a subsequence (and up to a harmless rotation if necessary), we have that

Ω 1,ε → Ω 0 as ε → 0 where          Ω 0 = R 2 if d 1,ε r 1,ε → +∞ as ε → 0 Ω 0 = R × (-∞, L) if d 1,ε r 1,ε → L as ε → 0 (4.61)
Here d 1,ε = d (x 1,ε , ∂Ω), as defined in (4.8). For R > 0, we shall also let

Ω R 0 = (Ω 0 ∩ D 0 (R)) \ i∈D1 D xi 1 R . ( 4

.62)

We shall distinguish three cases, depending on the behaviour of d 1,ε = d (x 1,ε , ∂Ω) and r 1,ε . Case 1 -We assume that d 1,ε → 0 as ε → 0, meaning that, after passing to a subsequence, x 1,ε → x 1 as ε → 0 with x 1 ∈ Ω. We let y ∈ Ω R 0 for some R > 0 and we set x ε = x 1,ε + r 1,ε y. Since d 1,ε → 0 and r 1,ε → 0, we are in situation a) of Claim 4.8. Note indeed that

|x ε -x i,ε | µ i,ε → +∞ as ε → 0 for all i = 1, . . . , N .
It is obvious if i ∈ D 1 since we clearly have in this case

|x ε -x i,ε | µ i,ε = |x ε -x i,ε | r 1,ε r 1,ε r i,ε r i,ε µ i,ε with |x ε -x i,ε | r 1,ε ≥ R -1 + o(1), r 1,ε r i,ε ≥ 2 |x i | -1 + o(1) for i ∈ D ⋆ 1 and equal to 1 if i = 1, and r i,ε µ i,ε → +∞ as ε → 0 thanks to assertion c) of Proposition 2.1. While, if i ∈ D 1 , we can write that |x ε -x i,ε | µ i,ε ≥ (1 + o(1)) |x i,ε -x 1,ε | µ i,ε ≥ (2 + o(1)) r i,ε µ i,ε → +∞ as ε → 0 .
Thus, applying a) of Claim 4.8, we can write that

u ε (x ε ) = ψ ε (x ε ) + N i=1 (4π + o(1)) γ -1 i,ε G (x i,ε , x ε ) +O N i=1 γ -1 i,ε µ i,ε |x i,ε -x ε | + γ -1 i,ε ln s i,ε |x i,ε -x ε | + 2 .
Now, for any i ∈ {1, . . . , N },

γ -1 i,ε µ i,ε |x i,ε -x ε | = o γ -1 i,ε = o γ -1 1,ε
thanks to (4.54) and

γ -2 i,ε ln s i,ε |x i,ε -x ε | + 2 = o γ -1 1,ε thanks to the fact that s i,ε ≤ r i,ε = O (|x i,ε -x ε |). Note that Claim 4.4 implies that λ ε = O γ -2
1,ε in our case so that (4.57) gives that

ψ ε (x ε ) = O γ -3 1,ε = o γ -1 1,ε . Thus we have that u ε (x ε ) = N i=1 (4π + o(1)) γ -1 i,ε G (x i,ε , x ε ) + o γ -1 1,ε . (4.63)
We can now use (6.3) to write that

G (x i,ε , x ε ) = 1 2π ln 1 r 1,ε + O (1) if i ∈ D 1 and that G (x i,ε , x ε ) = G (x i,ε , x 1,ε ) + O(1) if i ∈ D 1 .
Thus we have that

u ε (x ε ) = (2 + o(1)) γ -1 1,ε ln 1 r 1,ε   1 + i∈D ⋆ 1 C 1,i   (4.64) + i ∈D1 (4π + o(1)) γ -1 i,ε G (x i,ε , x 1,ε ) + O γ -1 1,ε .
Note that C 1,i ≤ C for all i > 1 thanks to (4.53). Thus we have in particular that

γ -1 1,ε ln 1 r 1,ε   1 + i∈D ⋆ 1 C 1,i   ≤ 1 2 + o(1) u ε (x ε ) ≤ γ -1 1,ε ln 1 r 1,ε 1 + (N -1) C . (4.65)
Note that we also have thanks to Claim 4.9 and to (4.57) that

|∇u ε (x)| ≤ Cγ -1 1,ε |x 1,ε -x| -1 for all x ∈ D x1,ε (r 1,ε ) . (4.66)
We are thus in position to apply Claim 4.

1 for i = 1 to write that, if |x| = 1 2 , u ε (x 1,ε + r 1,ε x) = B 1,ε (r 1,ε ) + O γ -1
1,ε . Combined with (4.65), this gives that

(2 + o(1)) γ -1 1,ε ln 1 r 1,ε   1 + i∈D ⋆ 1 C 1,i   ≤ B 1,ε (r 1,ε ) ≤ (2 + o(1)) γ -1 1,ε ln 1 r 1,ε 1 + (N -1) C .
(4.67) We write now thanks to Claim 5.2 of Appendix A that

B 1,ε (r 1,ε ) = 2γ -1 1,ε ln 1 r 1,ε -γ -1 1,ε ln λ ε γ 2 1,ε + O γ -1 1,ε (4.68)
to deduce that

(2 + o(1)) γ -1 1,ε ln 1 r 1,ε i∈D ⋆ 1 C 1,i ≤ -γ -1 1,ε ln λ ε γ 2 1,ε ≤ 2(N -1) C + o(1) γ -1 1,ε ln 1 r 1,ε . (4.69) Fix now i ∈ D ⋆ 1 . It is clear that there exists δ > 0 such that ∂D xi,ε (δr 1,ε ) ⊂ x 1,ε + r 1,ε y, y ∈ Ω R 0
for some R > 0. Thus we can write that inf ∂Dx i,ε (δr1,ε) (4.65). We can also apply b) of Claim 4.2 with r ε = δr 1,ε thanks to Claim 4.3 and to the fact that

u ε ≥ (2 + o(1)) γ -1 1,ε ln 1 r 1,ε   1 + i∈D ⋆ 1 C 1,i   thanks to
r 1,ε r i,ε ≥ 2 |x i | -1 + o(1)
. This leads to

(2 + o(1)) γ -1 1,ε ln 1 r 1,ε   1 + i∈D ⋆ 1 C 1,i   ≤ B i,ε (δr 1,ε ) + o γ -1 i,ε .
We have that

B i,ε (δr 1,ε ) = 2γ -1 i,ε ln 1 r 1,ε -γ -1 i,ε ln λ ε γ 2 i,ε + O γ -1 i,ε
.

This leads together with (4.69) to

(2 + o(1)) γ -1 1,ε ln 1 r 1,ε   1 + i∈D ⋆ 1 C 1,i   ≤ 2(N -1) C + 2 + o(1) γ -1 i,ε ln 1 r 1,ε -γ -1 i,ε ln γ 2 i,ε γ 2 1,ε
. This is clearly impossible if C 1,i = 0. Thus we have proved that

for any i ∈ D ⋆ 1 , C 1,i > 0 . (4.70)
This implies thanks to (4.55) that r i,ε ≥ r 1,ε for all i ∈ D ⋆ 1 . Then we can apply Claim 4.1 to all i ∈ D ⋆ 1 thanks to Claim 4.9 and to what we just said to get that, for any xi ∈ D 1 ,

γ i,ε (u ε (x 1,ε + r 1,ε x) -B i,ε (r 1,ε )) → 2 ln 1 |x -xi | + H i in C 1 loc (D xi (1) \ {x i }) as ε → 0 (4.71)
where H i is some harmonic function in D xi (1) satisfying H i (x i ) = 0 and ∇H i (x i ) = 0 (note here that we assumed that r 1,ε → 0 as ε → 0). Let us set now

v ε (x) = γ 1,ε (u ε (x 1,ε + r 1,ε x) -B 1,ε (r 1,ε )) .
Thanks to Claim 4.9, we have that

|∇v ε | ≤ C R in Ω R 0 for all R > 0. This clearly proves that (v ε ) is uniformly bounded in any Ω R 0 . Since ∆v ε = λ ε r 2 1,ε γ 1,ε f ε (x 1,ε + r 1,ε x) u ε (x 1,ε + r 1,ε x) e uε(x1,ε+r1,εx) 2 in Ω R 0 , we have that |∆v ε | = O λ ε r 2 1,ε γ 1,ε B 1,ε (r 1,ε ) + γ -1 1,ε e B1,ε(r1,ε) 2 in Ω R 0 .
Thanks to (4.68), we know that

λ ε r 2 1,ε ≤ Cγ -2 1,ε e -γ1,εB1,ε(r1,ε) so that |∆v ε | = O γ -1 1,ε B 1,ε (r 1,ε ) + γ -1 1,ε e B1,ε(r1,ε) 2 -γ1,εB1,ε(r1,ε) = o(1) in Ω R 0 thanks to Claim 4.
3. Thus we have by standard elliptic theory that

v ε → v 0 in C 1 loc R 2 \ S 1 as ε → 0 (4.72)
where v 0 is some harmonic funtion in R 2 \ S 1 which satisfies, thanks to Claim 4.9,

|∇v 0 | ≤ C |x| for |x| large. (4.73)
Thanks to (4.71), we know that

v 0 (x) = 2C 1,i ln 1 |x -xi | + C 1,i H i + B i in D xi (1)
for all i ∈ D 1 where B i is a constant given by

B i = (1 -C 1,i ) 1 + ln f 0 (x 1 ) 4 + 2C 1,i ln C 1,i + lim ε→0 1 - γ 1,ε γ i,ε ln λ ε γ 2 1,ε r 2 1,ε
.

Thus we have that

v 0 (x) = 2 ln 1 |x| + 2 i∈D ⋆ 1 C 1,i ln 1 |x -xi | + w 0
where w 0 is harmonic in R 2 and satisfies thanks to (4.73) that |∇w 0 | ≤ C|x| -1 for |x| large. This implies that w 0 ≡ A 0 for some constant A 0 . Thus we have that

v 0 (x) = 2 ln 1 |x| + 2 i∈D ⋆ 1 C 1,i ln 1 |x -xi | + A 0 . (4.74)
Moreover, the H i 's of (4.71) are given by

H i (x) = 2 ln 1 |x| + 2 j∈D ⋆ 1 , j =i C 1,j ln 1 |x -xj | + A 0
and they satisfy ∇H i (x i ) = 0 for all i ∈ D 1 . Note that, by the definition of r 1,ε and since we assumed that

r1,ε d1,ε → +∞ as ε → 0, we know that D ⋆ 1 = ∅. Let us pick up i ∈ D ⋆ 1 such that |x i | ≥ |x j | for all j ∈ D ⋆ 1 . It is then clear that ∇H i (x i ) , xi = -2 -2 j∈D ⋆ 1 , j =i C 1,i xi -xj , xi |x i -xj | 2 ≤ -2 ,
which contradicts the fact that ∇H i (x i ) = 0. This is the contradiction we were looking for and this proves that, if r 1,ε → 0 as ε → 0, this first case can not happen, that is we must have d 1,ε → 0 as ε → 0. ♠ Case 2 -We assume that d 1,ε → 0 and that r1,ε dε → 0 as ε → 0. We let y ∈ Ω R 0 for some R > 0 and we set x ε = x 1,ε + r 1,ε x. Since d 1,ε → 0 and r 1,ε → 0, we are in situation b) of Claim 4.8. Indeed, as in Case 1, we have that

|x ε -x i,ε | µ i,ε → +∞ as ε → 0 for all i = 1, . . . , N . u ε (x ε ) = ψ ε (x ε ) + N i=1 4π + o(1) γ i,ε G (x i,ε , x ε ) +O i∈A γ -1 i,ε µ i,ε |x i,ε -x ε | + γ -1 i,ε ln s i,ε |x i,ε -x ε | + 2 +O i∈B d ε d ε + d i,ε γ -1 i,ε µ i,ε + γ -2 i,ε s i,ε
where A is defined as the set of i ∈ {1, . . . , N } such that |x i,ε -

x ε | ≤ s i,ε + o (d ε
) and B as its complementary. Noting that |x i,εx ε | ≥ Cr i,ε for all i ∈ {1, . . . , N }, we have that for any i ∈ A,

γ -1 i,ε µ i,ε |x i,ε -x ε | + γ -2 i,ε ln s i,ε |x i,ε -x ε | + 2 = o γ -1 i,ε
and, for any i ∈ B,

d ε d ε + d i,ε γ -1 i,ε µ i,ε + γ -2 i,ε s i,ε = o γ -1 i,ε
.

Thus we have that

u ε (x ε ) = ψ ε (x ε ) + N i=1 4π + o(1) γ i,ε G (x i,ε , x ε ) + o γ -1 1,ε thanks to (4.54). For i ∈ D 1 , we have that |x i,ε -x ε | = o (d 1,ε
) so that, thanks to (6.12),

G (x i,ε , x ε ) = 1 2π ln 2d 1,ε r 1,ε + O(1) .
For any i ∈ D 1 , we know that

G (x i,ε , x ε ) = G (x i,ε , x 1,ε ) + o(1)
thanks to (6.12). Thus we can write that

u ε (x ε ) = ψ ε (x ε ) + i∈D1 2 + o(1) γ i,ε ln 2d 1,ε r 1,ε + i ∈D1 γ -1 i,ε G (x i,ε , x 1,ε ) + O γ -1 1,ε .
If λ 0 = 0, then we can write thanks to the fact that ψ ε = 0 on ∂Ω and to (4.56) that ψ ε (x ε ) = O (d 1,ε ). This leads with Claim 4.4 to ψ ε (x ε ) = O γ -1 1,ε . If λ 0 = 0, then we can use (4.57) to arrive to the same result. Thus we finally get that

u ε (x ε ) = 2 γ 1,ε   1 + i∈D ⋆ 1 C 1,i   ln d 1,ε r 1,ε + i ∈D1 γ -1 i,ε G (x i,ε , x 1,ε ) + o γ -1 1,ε ln d 1,ε r 1,ε . (4.75)
Note that C 1,i ≤ C for all i > 1 thanks to (4.53). Thus we have in particular that 2γ -1 1,ε ln

d 1,ε r 1,ε   1 + i∈D ⋆ 1 C 1,i   ≤ u ε (x ε ) + o γ -1 1,ε ln d 1,ε r 1,ε ≤ 2γ -1 1,ε ln d 1,ε r 1,ε 1 + (N -1) C .
(4.76) Here we used (6.12) to estimate G (x i,ε , x 1,ε ) for i ∈ D 1 . Note that we also have thanks to Claim 4.9 and to (4.57) that

|∇u ε (x)| ≤ Cγ -1 1,ε |x 1,ε -x| for all x ∈ D x1,ε (r 1,ε ) . (4.77)
The proof now follows exactly Case 1, from (4.66) to the end. We will not repeat it here. ♠ Case 3 -We assume that d 1,ε → 0 as ε → 0 and that

d 1,ε r 1,ε → L as ε → 0 where L ≥ 2.
We are thus in the case where, after some harmless rotation,

Ω 0 = R × (-∞, L) .
We let y ∈ Ω R 0 for some R > 0 and we set x ε = x 1,ε + r 1,ε y. Since d 1,ε → 0 and r 1,ε → 0, we are in situation b) of Claim 4.8. Indeed, as in Case 1, we have that

|x ε -x i,ε | µ i,ε → +∞ as ε → 0 for all i = 1, . . . , N .
Thus we can write that

u ε (x ε ) = ψ ε (x ε ) + N i=1 4π + o(1) γ i,ε G (x i,ε , x ε ) +O i∈A γ -1 i,ε µ i,ε |x i,ε -x ε | + γ -1 i,ε ln s i,ε |x i,ε -x ε | + 2 +O i∈B d ε d ε + d i,ε γ -1 i,ε µ i,ε + γ -2 i,ε s i,ε
where A is defined as the set of i ∈ {1, . . . , N } such that |x i,εx ε | ≤ s i,ε + o (d ε ) and B as its complementary. As in Case 2, we have that

γ -1 i,ε µ i,ε |x i,ε -x ε | + γ -2 i,ε ln s i,ε |x i,ε -x ε | + 2 = o γ -1 i,ε for all i ∈ A while d ε d ε + d i,ε γ -1 i,ε µ i,ε + γ -2 i,ε s i,ε = o γ -1 i,ε
for all i ∈ B. Thus we have that

u ε (x ε ) = ψ ε (x ε ) + N i=1 4π + o(1) γ i,ε G (x i,ε , x ε ) + o γ -1 1,ε . If i ∈ D 1 , we have that G (x i,ε , x ε ) = 1 2π ln |ỹ i -y| |x i -y| + o(1)
where ỹi = R (x i ) , R being the reflection with respect to the straight line R × {L}. Here we used (6.12). If i ∈ D 1 , we have that

G (x i,ε , x ε ) = o(1)
thanks to (6.12). Thus we can write, remembering (4.54), that

u ε (x ε ) = ψ ε (x ε ) + 2γ -1 1,ε i∈D1 C 1,i ln |ỹ i -y| |x i -y| + o γ -1 1,ε .
Using (4.56), we know that

ψ ε (x ε ) r 1,ε → A (L -y 2 )
where y = (y 1 , y 2 ) for some A independent of y. Moreover, we have that A ≥ 0 by the maximum principle since ∆ψ ε ≥ 0 in Ω and ψ ε = 0 on ∂Ω. If λ 0 = 0, we can use Claim 4.4 to deduce that

γ 1,ε ψ ε (x ε ) → B (L -y 2 )
for some B > 0, independent of y. If λ 0 = 0, then (4.57) implies that

γ 1,ε ψ 1,ε (x ε ) = O (λ ε r 1,ε ) = o(1) .
Thus, up to change the B above, we can write that

γ 1,ε u ε (x ε ) → B (L -y 2 ) + 2 i∈D1 C 1,i ln |ỹ i -y| |x i -y| as ε → 0 . (4.78)
Then, by the equation satisfied by u ε , it is clear that

v ε (x) = γ 1,ε u ε (x 1,ε + r 1,ε x)
has a Laplacian uniformly converging to 0 in any Ω R 0 . Thus, by standard elliptic theory, we can conclude that

γ 1,ε u ε (x 1,ε + r 1,ε y) → B (L -y 2 ) + 2 i∈D1 C 1,i ln |ỹ i -y| |x i -y| in C 1 loc (Ω 0 \ S 1 ) as ε → 0 . (4.79) Writing that |∇ψ ε | ≤ Cλ ε in D x1,ε (r 1,ε )
thanks to (4.56), we get with Claim 4.4 that

|∇ψ ε | ≤ C λ ε d -1 1,ε γ -1 1,ε in D x1,ε (r 1,ε
) so that we can use Claim 4.9 and (4.54) to obtain that

|∇u ε | ≤ Cγ -1 1,ε |x 1,ε -x ε | -1 in D x1,ε (r 1,ε ) .
We are thus in position to apply Claim 4.1 to i = 1. In particular, combined with (4.79), we get that

γ 1,ε B 1,ε (r 1,ε ) = O (1)
. This leads with Claim 5.2 of Appendix A to

ln λ ε r 2 1,ε γ 2 1,ε = O (1) .
Thus we have that, up to a subsequence,

λ ε f ε (x 1,ε ) r 2 1,ε γ 2 1,ε → α 0 as ε → 0 (4.80)
for some α 0 > 0. Let now i ∈ D 1 be such that the second coordinate of xi satisfies (x i ) 2 < L and (x i ) 2 ≥ (x j ) 2 or (x j ) 2 = L for all j ∈ D 1 .

Note that such a i does exists since 1 ∈ D 1 . Moreover, we have that

L > (x i ) 2 ≥ (x 1 ) 2 = 0 . Note also that d i,ε ≥ (L -(x i ) 2 + o(1)) r 1,ε so that Claim 4.4 implies that O(1) = λ ε f ε (x 1,ε ) γ 2 i,ε d 2 i,ε = γ 2 i,ε γ 2 1,ε d 2 i,ε r 2 1,ε λ ε f ε (x 1,ε ) γ 2 1,ε r 2 1,ε ≥ (L -(x i ) 2 ) 2 α 0 + o(1) γ 2 i,ε γ 2 1,ε
thanks to (4.80). This implies that C 1,i = 0. Thanks to (4.55), we then have that r i,ε ≥ r 1,ε . Once again, thanks to Claim 4.9 and to (4.54), we see now that

|∇u ε | ≤ Cγ -1 i,ε |x i,ε -x ε | -1 in D xi,ε (r 1,ε
) and that we can apply Claim 4.1. In particular, using (4.80), we get that

γ i,ε u ε (x 1,ε + r 1,ε x) → 2 ln 1 |x -xi | + H i -ln α 0 4C 2 1,i in C 1 loc (D xi (1) \ {x i }) as ε → 0
where H i is harmonic in D xi (1) and satisfies ∇H i (x i ) = 0 (since r 1,ε → 0 as ε → 0 by assumption). Now, combining this with (4.79), we know that

H i = B C 1,i (L -x 2 ) + 2 j∈D1,j =i C 1,j C 1,i ln |ỹ j -x| |x j -x| + 2 ln |ỹ i -x| + ln α 0 4C 2 1,i .
The derivative of H i with respect to the second coordinate at xi is

∂H i ∂x 2 (x i ) = - B C 1,i + 2 j∈D1,j =i C 1,j C 1,i (x i ) 2 -(ỹ j ) 2 |ỹ j -xi | 2 - (x i ) 2 -(x j ) 2 |x j -xi | 2 + 2 (x i ) 2 -(ỹ i ) 2 |ỹ i -xi | 2 . Note now that (ỹ j ) 2 = 2L -(x j ) 2 so that ∂H i ∂x 2 (x i ) = - B C 1,i + 2 j∈D1,j =i C 1,j C 1,i (x i ) 2 + (x j ) 2 -2L |ỹ j -xi | 2 - (x i ) 2 -(x j ) 2 |x j -xi | 2 + 4 (x i ) 2 -L |ỹ i -xi | 2 .
We claim that

(x i ) 2 + (x j ) 2 -2L |ỹ j -xi | 2 ≤ (x i ) 2 -(x j ) 2 |x j -xi | 2 (4.81)
for all j ∈ D 1 with j = i. This will imply that i ∂x 2 (x i ) < 0 , all the terms above being nonpositive, the last one being negative. This will give a contradiction with the fact that ∇H i (x i ) = 0, thus proving that this last case is not possible either. In order to prove (4.81), we first notice that

|x i -ỹj | 2 = |x i -xj | 2 + 4 (L -(x i ) 2 ) L -(x j ) 2 .
Thus we can write that

(x i ) 2 + (x j ) 2 -2L |x i -xj | 2 -(x i ) 2 -(x j ) 2 |x i -ỹj | 2 = 2 |x i -xj | 2 (x j ) 2 -L -4 (L -(x i ) 2 ) L -(x j ) 2 (x i ) 2 -(x j ) 2 = 2 (x j ) 2 -L |x i -xj | 2 + 2 (L -(x i ) 2 ) (x i ) 2 -(x j ) 2 ≤ 0 since (x j ) 2 -L ≤ 0 and if (x j ) 2 -L = 0, (x i ) 2 -(x j ) 2 ≥ 0.
This clearly proves (4.81) and, as already said, proves that this last case is not possible. ♠

The study of these three cases proves that the assumption r 1,ε → 0 is absurd and thus proves the claim. ♦

Note that this claim implies that

x 1,ε → x 1 as ε → 0 with x 1 ∈ Ω . (4.82)
We also have thanks to Claims 4.4 and 4.10 that

λ ε = O γ -2
1,ε so that λ 0 = 0 and u 0 ≡ 0. Moreover, we can transform (4.57) into

∇ψ ε C 1,α (Ω) = O γ -3 1,ε . (4.83)
Let us now give a simple consequence of the previous claim :

Claim 4.11. After passing to a subsequence,

λ ε γ 2 1,ε → α 0 as ε → 0 for some 0 < α 0 ≤ 4 f 0 (x 1 ) d (x 1 , ∂Ω) 2 .
Proof of Claim 4.11 -We already said that λ ε = O γ -2 1,ε . Claim 4.9 with (4.83) gives that

|∇u ε | ≤ C N i=1 γ -1 i,ε (µ i,ε + |x -x i,ε |) -1 in Ω .
This gives in particular that

|∇u ε | ≤ Cγ -1 1,ε |x -x 1,ε | -1
in D x1,ε (δ 0 ) where δ 0 is as in Claim 4.10. Thus we are in position to apply Claim 4.1 to i = 1. This gives in particular that

γ 1,ε (u ε (x) -B 1,ε (δ 0 )) = O(1)
for all |xx 1,ε | = δ0 2 . Now Claim 4.8 combined with (4.83) gives that

γ 1,ε u ε (x) = O(1) on ∂D x1,ε δ 0 2 so that the above leads to γ 1,ε B 1,ε (δ 0 ) = O(1) . Since γ 1,ε B 1,ε (δ 0 ) = -ln λ ε γ 2 1,ε + O(1) , we obtain that ln λ ε γ 2 1,ε = O(1) .
This clearly permits to prove the claim. ♦ Claim 4.12. We have that r i,ε ≥ δ 1 for some δ 1 > 0 for all i = 1, . . . , N .

Proof of Claim 4.12 -We shall prove it by induction on i. This is already proved for i = 1 in the previous claim. Fix 2 ≤ i ≤ N and assume that r j,ε ≥ δ 1 > 0 for all j < i .

(4.84)

In particular, after passing to a subsequence, we have that

x j,ε → x j as ε → 0 with x j ∈ Ω . By (4.55), this implies that C j,i = 0 for all j < i so that γ j,ε = o (γ i,ε ) for all j < i . (4.87)

We shall now proceed as in the proof of Claim 4.10 and distinguish three cases. We let in the following

D ⋆ i = {j > i s.t. |x i,ε -x j,ε | = O (r i,ε )} and D i = D ⋆ i ∪ {i} . (4.88)
After passing to a subsequence, we let

S ⋆ i = xj = lim ε→0 x j,ε -x i,ε r i,ε , j ∈ D ⋆ i and S i = S ⋆ i ∪ {x i = 0} . (4.89)
We also let

Ω i,ε = y ∈ R 2 s.t. x i,ε + r i,ε y ∈ Ω . (4.90)
Note that, after passing to a subsequence (and up to a harmless rotation if necessary), we have that Ω i,ε → Ω 0 as ε → 0 where

         Ω 0 = R 2 if d i,ε r i,ε → +∞ as ε → 0 Ω 0 = R × (-∞, L) if d i,ε r i,ε → L as ε → 0 (4.91)
Here d i,ε = d (x i,ε , ∂Ω), as defined in (4.8). For R > 0, we shall also let

Ω R 0 = (Ω 0 ∩ D 0 (R)) \ j∈Di D xj 1 R . ( 4

.92)

Case 1 -We assume that d i,ε → 0 as ε → 0, meaning that, after passing to a subsequence, x i,ε → x i as ε → 0 with x i ∈ Ω. We let y ∈ Ω R 0 for some R > 0 and we set x ε = x i,ε + r i,ε y. Since d i,ε → 0 and r i,ε → 0 as ε → 0, we are in situation a) of Claim 4.8. Note indeed that

|x ε -x j,ε | µ j,ε → +∞ as ε → 0 for all j = 1, . . . , N .
It is obvious if j < i since r j,ε ≥ δ 1 > 0 and r i,ε → 0 as ε → 0. It is also obvious if j ∈ D i since we clearly have in this case 1) for j ∈ D ⋆ i and equal to 1 if j = i, and r j,ε µ j,ε → +∞ as ε → 0 thanks to assertion c) of Proposition 2.1. While, if j > i and j ∈ D i , we can write that

|x ε -x j,ε | µ j,ε = |x ε -x j,ε | r i,ε r i,ε r j,ε r j,ε µ j,ε with |x ε -x j,ε | r i,ε ≥ R -1 + o(1), r i,ε r j,ε ≥ 2 |x j | -1 + o(
|x ε -x j,ε | µ i,ε ≥ (1 + o(1)) |x i,ε -x j,ε | µ j,ε ≥ (2 + o(1)) r j,ε µ j,ε → +∞ as ε → 0 .
Thus, applying a) of Claim 4.8, we can write that

u ε (x ε ) = ψ ε (x ε ) + N j=1 (4π + o(1)) γ -1 j,ε G (x j,ε , x ε ) +O N i=1 γ -1 j,ε µ j,ε |x j,ε -x ε | + γ -2 j,ε ln s j,ε |x j,ε -x ε | + 2 .
For j < i, we have that

(4π + o(1)) γ -1 j,ε G (x j,ε , x ε ) = 4πγ -1 1,ε C 1,j G (x j , x i ) + o γ -1 1,ε
thanks to the assumption that x i,ε → x i with x i ∈ Ω, to (4.85) and to (4.52), (4.54). We also obviously have that

γ -1 j,ε µ j,ε |x j,ε -x ε | + γ -2 j,ε ln s j,ε |x j,ε -x ε | + 2 = o γ -1 1,ε .
We also know thanks to (4.83) that ψ ε (x ε ) = o γ -1 1,ε . For j > i, we can proceed exactly as in Case 1 of Claim 4.10 to finally obtain that

u ε (x ε ) = 4πγ -1 1,ε i-1 j=1 C 1,j G (x j , x i ) + o γ -1 1,ε + (2 + o(1)) γ -1 i,ε ln 1 r i,ε   1 + j∈D ⋆ i C i,j   (4.93) + j>i, j ∈Di (4π + o(1)) γ -1 j,ε G (x i,ε , x j,ε ) + O γ -1 i,ε
This gives in particular that u ε ≥ Cγ -1 1,ε on ∂D xi,ε (r i,ε ) for some C > 0. Using b) of Claim 4.2, we deduce that

Cγ -1 1,ε ≤ B i,ε (r i,ε ) + O γ -1 i,ε = -γ -1 i,ε ln λ ε γ 2 i,ε r 2 i,ε + O γ -1 i,ε . Since γ 1,ε = o (γ i,ε
), see (4.87), we deduce that λ ε γ 2 i,ε r 2 i,ε → 0 as ε → 0 . Thanks to Claim 4.11, this gives that

γ i,ε r i,ε = o (γ 1,ε ) . (4.94)
We apply now Claim 4.9 combined with (4.83) and this last estimate to write that

|∇u ε (x)| ≤ Cγ -1 i,ε |x i,ε -x| -1 + Cγ -1 1,ε ≤ C ′ γ -1 i,ε |x i,ε -x| -1 in D xi,ε (r i,ε
). Thus we can apply Claim 4.1 to i : this gives that, if

|x| = 1 2 , u ε (x i,ε + r i,ε x) = B i,ε (r i,ε ) + O γ -1 i,ε .
Combined with (4.93), this leads to

B i,ε (r i,ε ) = 4πγ -1 1,ε i-1 j=1 C 1,j G (x j , x i ) + o γ -1 1,ε + (2 + o(1)) γ -1 i,ε ln 1 r i,ε   1 + j∈D ⋆ i C i,j   + j>i, j ∈Di (4π + o(1)) γ -1 j,ε G (x i,ε , x j,ε ) + O γ -1 i,ε . Since B i,ε (r i,ε ) = -γ -1 i,ε ln λ ε γ 2 i,ε r 2 i,ε + O γ -1 i,ε , this leads to -ln λ ε γ 2 i,ε r 2 i,ε = γ i,ε γ -1 1,ε   4π i-1 j=1 C 1,j G (x j , x i ) + o(1)   +2 ln 1 r i,ε   1 + j∈D ⋆ i C i,j   (4.95) + j>i, j ∈Di (4π + o(1)) C i,j G (x i,ε , x j,ε ) + o ln 1 r i,ε
.

Thanks to Claim 4.11 and (4.87), we deduce that

4π i-1 j=1 C 1,j G (x j , x i ) + o(1) + γ 1,ε γ i,ε ln 1 r i,ε   2 j∈D ⋆ 1 C i,j + o(1)   ≤ 0 . (4.96) Let k ∈ D ⋆ i .
It is clear that there exists δ > 0 such that ∂D x k,ε (δr i,ε ) ⊂ x i,ε + r i,ε y, y ∈ Ω R 0 for some R > 0. Thus we can write that inf ∂Dx k,ε (δri,ε) (4.93). We can also apply b) of Claim 4.2 with r ε = δr i,ε thanks to Claim 4.3 and to the fact that

u ε ≥ Cγ -1 1,ε + 2γ -1 i,ε ln 1 r i,ε   1 + j∈D ⋆ i C i,j + o(1)   thanks to
r i,ε r k,ε ≥ 2 |x k | -1 + o(1)
. This leads to

Cγ -1 1,ε + 2γ -1 i,ε ln 1 r i,ε   1 + j∈D ⋆ i C i,j + o(1)   ≤ B k,ε (δr i,ε ) + o γ -1 k,ε = -γ -1 k,ε ln λ ε r 2 i,ε γ 2 k,ε + O γ -1 k,ε
.

Combined with (4.95), this gives that

Cγ -1 1,ε + 2γ -1 i,ε ln 1 r i,ε   1 + j∈D ⋆ i C i,j + o(1)   ≤ γ -1 k,ε γ i,ε γ -1 1,ε   4π i-1 j=1 C 1,j G (x j , x i ) + o(1)   +O γ -1 k,ε ln 1 r i,ε -γ -1 k,ε ln γ 2 k,ε γ 2 i,ε
.

Assume by contradiction that γ i,ε = o (γ k,ε ). We then have that

Cγ -1 1,ε + 2γ -1 i,ε ln 1 r i,ε   1 + j∈D ⋆ i C i,j + o(1)   ≤ o γ -1 1,ε + o γ -1 i,ε ln 1 r i,ε
, which is absurd. Thus we have proved that → +∞ for all j < i, we are sure that D ⋆ i = ∅ and, with (4.97), that

C i,j > 0 for all j ∈ D ⋆ i . ( 4
j∈D ⋆ i C i,j > 0 .
Then (4.96) leads to a contradiction. This proves that this first case is absurd. ♠ Case 2 -We assume that d i,ε → 0 and that ri,ε di,ε → 0 as ε → 0. We let x i = lim ε→0 x i,ε . Note that x i ∈ ∂Ω. We let y ∈ Ω R 0 for some R > 0 and we set x ε = x i,ε + r i,ε y. Since d i,ε → 0 and r i,ε → 0, we are in situation b) of Claim 4.8. Indeed, as in Case 1, we have that |x εx j,ε | µ j,ε → +∞ as ε → 0 for all j = 1, . . . , N .

Thus we have that

u ε (x ε ) = ψ ε (x ε ) + N j=1 4π + o(1) γ j,ε G (x j,ε , x ε ) +O   j∈A γ -1 j,ε µ j,ε |x j,ε -x ε | + γ -2 j,ε ln s j,ε |x j,ε -x ε | + 2   +O   j∈B d ε d ε + d j,ε γ -1 j,ε µ j,ε + γ -2 j,ε s j,ε  
where A is defined as the set of j ∈ {1, . . . , N } such that |x j,εx ε | ≤ s j,ε + o (d ε ) and B as its complementary. Noting that |x j,εx ε | ≥ Cr j,ε for all j ∈ {1, . . . , N }, we have that for any j ∈ A,

γ -1 j,ε µ j,ε |x j,ε -x ε | + γ -2 j,ε ln s j,ε |x j,ε -x ε | + 2 = o γ -1 j,ε
.

And, for any j ∈ B,

d ε d ε + d j,ε γ -1 j,ε µ j,ε + γ -2 j,ε s j,ε = o γ -1 j,ε
.

Note also that, if j < i, we have that j ∈ B thanks to (4.84) and that

d ε d ε + d j,ε γ -1 j,ε µ j,ε + γ -2 j,ε s j,ε = o d ε γ -1 j,ε
.

Thus we have that

u ε (x ε ) = ψ ε (x ε ) + N j=1 4π + o(1) γ j,ε G (x j,ε , x ε ) + o γ -1 i,ε + o d ε γ -1 1,ε .
We can write thanks to (4.83) and since ψ ε = 0 on ∂Ω that

ψ ε (x ε ) = O d ε γ -3 1,ε .
Then we have that, for any j < i,

G (x j,ε , x ε ) = -d ε ∂ ν G (x j , x i ) + o (d ε ) .
And, for j ≥ i, we have thanks to (6.12) that

G (x j,ε , x ε ) = 1 2π ln 2d i,ε r i,ε + 1 2π ln r i,ε |x j,ε -x ε | + O r i,ε d i,ε + O (d i,ε ) if j ∈ D i and that G (x j,ε , x ε ) = G (x i,ε , x j,ε ) + o(1)
if j ∈ D i . We thus arrive to

u ε (x ε ) = 4π d i,ε γ 1,ε   i-1 j=1 C 1,j (-∂ ν G (x j , x i ))   +2   j∈D1 C i,j + o(1)   γ -1 i,ε ln 2d i,ε r i,ε (4.98)
+ j>i, j ∈Di

(4π + o(1)) γ -1 j,ε G (x i,ε , x j,ε ) + o d i,ε γ -1 1,ε .
This gives in particular that u ε ≥ Cd i,ε γ -1 1,ε on ∂D xi,ε (r i,ε ) for some C > 0. Using b) of Claim 4.2, we deduce that

Cd i,ε γ -1 1,ε ≤ B i,ε (r i,ε ) + O γ -1 i,ε = -γ -1 i,ε ln λ ε γ 2 i,ε r 2 i,ε + O γ -1 i,ε
.

Thanks to Claim 4.11, this gives that

C d i,ε γ i,ε γ 1,ε ≤ -ln γ 2 i,ε r 2 i,ε γ 2 1,ε + O(1) .
Since di,ε ri,ε → +∞ as ε → 0 in our case, this implies that γ i,ε r i,ε = o (γ 1,ε ) . (4.99)

We apply now Claim 4.9 combined with (4.83), (4.84) and this last estimate to write that

|∇u ε (x)| ≤ Cγ -1 i,ε |x i,ε -x| -1 + Cγ -1 1,ε ≤ C ′ γ -1 i,ε |x i,ε -x| -1 in D xi,ε (r i,ε
). Thus we can apply Claim 4.1 to i : this gives that

, if |x| = 1 2 , u ε (x i,ε + r i,ε x) = B i,ε (r i,ε ) + O γ -1 i,ε .
Combined with (4.98) and (4.99), this leads to

B i,ε (r i,ε ) = 4π d i,ε γ 1,ε   i-1 j=1 C 1,j (-∂ ν G (x j , x i )) + o(1)   +2   j∈Di C i,j + o(1)   γ -1 i,ε ln 2d i,ε r i,ε (4.100) + j>i, j ∈Di (4π + o(1)) γ -1 j,ε G (x i,ε , x j,ε ) . Since B i,ε (r i,ε ) = -γ -1 i,ε ln λ ε γ 2 i,ε r 2 i,ε + O γ -1 i,ε = -γ -1 i,ε ln γ 2 i,ε d 2 i,ε γ 2 1,ε -2γ -1 i,ε ln r i,ε d i,ε + O γ -1 i,ε
thanks to Claim 4.11, this leads to

-ln γ 2 i,ε d 2 i,ε γ 2 1,ε = 4π d i,ε γ i,ε γ 1,ε   i-1 j=1 C 1,j (-∂ ν G (x j , x i )) + o(1)   +2   j∈D ⋆ i C i,j + o(1)   ln d i,ε r i,ε (4.101) + j>i, j ∈Di (4π + o(1)) γ i,ε γ j,ε G (x i,ε , x j,ε ) ,
from which we can infer that, for ε small,

2 γ 1,ε d i,ε γ i,ε ln γ 1,ε d i,ε γ i,ε ≥ 2π i-1 j=1 C 1,j (-∂ ν G (x j , x i )) (4.102) +2   j∈D ⋆ i C i,j + o(1)   γ 1,ε d i,ε γ i,ε ln d i,ε r i,ε . Let j ∈ D ⋆ i . Note that, since di,ε ri,ε → +∞ as ε → 0, we know that D ⋆ i = ∅. There exists δ > 0 such that ∂D xj,ε (δr i,ε ) ⊂ Ω R
0 for some R > 0. Thus we can write that inf ∂Dx j,ε (δri,ε)

u ε ≥ 1 + o(1) B i,ε (r i,ε )
thanks to (4.98) and (4.100). We can also apply b) of Claim 4.2 with r ε = δr i,ε thanks to Claim 4.3 and to the fact that r i,ε r j,ε ≥ 2 |x j | -1 + o(1). This leads to

B j,ε (δr i,ε ) ≥ 1 + o(1) B i,ε (r i,ε ) . Since B j,ε (δr i,ε ) = -γ -1 j,ε ln λ ε γ 2 j,ε r 2 i,ε + O γ -1 j,ε and B i,ε (r i,ε ) = -γ -1 i,ε ln λ ε γ 2 i,ε r 2 i,ε + O γ -1 i,ε
thanks to Claim 4.11, we obtain that

-γ -1 j,ε ln λ ε γ 2 j,ε r 2 i,ε + O γ -1 j,ε ≥ -1 + o(1) γ -1 i,ε ln λ ε γ 2 i,ε r 2 i,ε + O γ -1 i,ε . This implies since γ i,ε = O (γ j,ε ), see (4.54), that ln λ ε γ 2 i,ε r 2 i,ε 1 + o(1) - γ i,ε γ j,ε ≥ -C
for some C > 0. Since ri,ε di,ε → 0 as ε → 0, we get with Claim 4.4 that λ ε r 2 i,ε γ 2 i,ε → 0 as ε → 0 and the above implies that C i,j ≥ 1. Thus we have obtained that

C i,j ≥ 1 for all j ∈ D ⋆ i . (4.103)
Thanks to (4.55), we know that r j,ε ≥ r i,ε for all i ∈ D ⋆ i . Using Claim 4.9, (4.83) and (4.84), we thus obtain that

|∇u ε | ≤ C   γ -1 1,ε + γ -1 i,ε j∈Di |x j,ε -x| -1   in D xi,ε (Rr i,ε
) for all R > 0. Thanks to (4.99), this leads to

|∇u ε | ≤ Cγ -1 i,ε j∈Di |x j,ε -x| -1 in D xi,ε (Rr i,ε
) for all R > 0. We are now in position to follow exactly the end of the proof of Case 2 of Claim 4.10. We can prove that

γ i,ε (u ε (x i,ε + r i,ε x) -B i,ε (r i,ε )) → 2 ln 1 |x| + 2 j∈D ⋆ i C i,j ln 1 |x -xj | + A 0 in C 1 loc R 2 \ S i
as ε → 0 for some constant A 0 and then get a contradiction with Claim 4.1 for j ∈ D ⋆ i (which is non-empty) such that |x j | ≥ |x k | for all k ∈ D ⋆ i . Note here that we assumed that r i,ε → 0 as ε → 0, see (4.86). This proves that this second case can not happen either. ♠ Case 3 -We assume that d i,ε → 0 as ε → 0 and that

d i,ε r i,ε → L as ε → 0 where L ≥ 2.
We are thus in the case where, after some harmless rotation,

Ω 0 = R × (-∞, L) .
We let y ∈ Ω R 0 for some R > 0 and we set x ε = x i,ε + r i,ε x. Since d i,ε → 0 and r i,ε → 0, we are in situation b) of Claim 4.8. Indeed, as in Case 1, we have that

|x ε -x j,ε | µ j,ε → +∞ as ε → 0 for all j = 1, . . . , N .
Thus we can write that

u ε (x ε ) = ψ ε (x ε ) + N j=1 4π + o(1) γ j,ε G (x j,ε , x ε ) +O   j∈A γ -1 j,ε µ j,ε |x j,ε -x ε | + γ -2 j,ε ln s j,ε |x j,ε -x ε | + 2   +O   j∈B d ε d ε + d j,ε γ -1 j,ε µ j,ε + γ -2 j,ε s j,ε  
where A is defined as the set of j ∈ {1, . . . , N } such that |x j,εx ε | ≤ s j,ε + o (d ε ) and B as its complementary. As in Case 2, we have that

γ -1 j,ε µ j,ε |x j,ε -x ε | + γ -2 j,ε ln s j,ε |x j,ε -x ε | + 2 = o γ -1 j,ε for all j ∈ A while d ε d ε + d j,ε γ -1 j,ε µ j,ε + γ -2 j,ε s j,ε = o γ -1 j,ε
for all j ∈ B. Note also that, if j < i, we have that j ∈ B thanks to (4.84) and that

d ε d ε + d j,ε γ -1 j,ε µ j,ε + γ -2 j,ε s j,ε = o d ε γ -1 j,ε = o r i,ε γ -1 j,ε .
Thus we have that

u ε (x ε ) = ψ ε (x ε ) + N i=1 4π + o(1) γ i,ε G (x i,ε , x ε ) + o γ -1 i,ε + o r i,ε γ -1 1,ε .
We can write thanks to (4.83) and since ψ ε = 0 on ∂Ω that

ψ ε (x ε ) = O r i,ε γ -3 1,ε .
Then we have that, for any j < i,

G (x j,ε , x ε ) = -d ε ∂ ν G (x j , x i ) + o (r i,ε ) .
And, for j ≥ i, we have that

G (x j,ε , x ε ) = 1 2π ln |ỹ j -y| |x j -y| + o(1)
if j ∈ D i where ỹj = R (x j ) , R being the reflection with respect to the straight line R × {L}. Here we used (6.12). At last, for j ≥ i and j ∈ D i , we have that G (x j,ε , x ε ) = o(1) thanks to (6.12). This leads to

u ε (x ε ) = 4πd ε γ -1 1,ε i-1 j=1 (-C 1,j ∂ ν G (x j , x i )) + j∈Di 2 + o(1) γ j,ε ln |ỹ j -y| |x j -y| + o γ -1 i,ε + o r i,ε γ -1 1,ε .
This gives in particular that u ε ≥ Cr i,ε γ -1 1,ε on ∂D xi,ε (r i,ε ) for some C > 0. Using b) of Claim 4.2, we deduce that

Cr i,ε γ -1 1,ε ≤ B i,ε (r i,ε ) + O γ -1 i,ε = -γ -1 i,ε ln λ ε γ 2 i,ε r 2 i,ε + O γ -1 i,ε
.

Thanks to Claim 4.11, this gives that

C r i,ε γ i,ε γ 1,ε ≤ -ln γ 2 i,ε r 2 i,ε γ 2 1,ε + O(1) .
This proves that r i,ε γ i,ε = O (γ 1,ε ) so that, up to a subsequence,

r i,ε γ i,ε γ 1,ε → B 0 as ε → 0 . (4.104)
Then, by the equation satified by u ε , it is clear that

v ε (x) = γ i,ε u ε (x 1,ε + r i,ε x)
has a Laplacian uniformly converging to 0 in any Ω R 0 . Thus, by standard elliptic theory, we can conclude that

γ i,ε u ε (x i,ε + r i,ε x) → B 1 (L -y 2 ) + 2 j∈Di C i,j ln |ỹ j -y| |x j -y| in C 1 loc (Ω 0 \ S 1 ) as ε → 0 . (4.105)
Using (4.83), (4.104), (4.105) and Claim 4.9, we have that

|∇u ε | ≤ Cγ -1 i,ε |x i,ε -x| -1 in D xi,ε (r i,ε ) .
We are thus in position to apply the results of Section 3 to u ε (x i,ε + •) in the disk D 0 (r i,ε ). In particular, applying c) of Proposition 3.1 and combining it with (4.105), we get that

γ i,ε B i,ε (r i,ε ) = O (1) .
This leads with Claim 5.2 of Appendix A to ln λ ε r 2 i,ε γ 2 i,ε = O (1) . Thanks to Claim 4.11, we thus have that B 0 > 0 in (4.104) and B 1 > 0 in (4.105). We can then proceed exactly as in Case 3 of Claim 4.10 to get a contradiction in this last case. ♠

The study of these three cases, all leading to a contradiction, proves that (4.87) is absurd when we assume (4.84). As already said, this permits to prove the claim by induction on i. ♦

We are now in position to prove Theorem 1.2. We know thanks to Claim 4.12 that

x i,ε → x i as ε → 0 where x i ∈ Ω . ( 4
u ε √ λ ε → 4π N i=1 m i G (x i , x) in C 1 loc (Ω \ S) (4.108)
where S = {x i } i=1,...,N . Moreover, using again (4.83) this time together with Claim 4.9, we know that

|∇u ε | ≤ C λ ε N i=1 |x i,ε -x| -1
in Ω. We are thus in position to apply Claim 4.1 for all i = 1, . . . , N . This gives that

γ i,ε (u ε (x i,ε + δx) -B i,ε (δ)) → 2 ln 1 |x| + H i (x) in C 1 loc (D 0 (1) \ {0}) as ε → 0 (4.109)
where H i (0) = 0 and ∇H i (0) = -1 2 δ ∇f0(xi) f0(xi) . Let us write thanks to Claim 5.2 that

B i,ε (δ) = γ i,ε -γ -1 i,ε 1 + γ -2 i,ε ln 1 + δ 2 4µ 2 i,ε + O γ -2 i,ε = γ i,ε -γ -1 i,ε 1 + γ -2 i,ε ln 1 µ 2 i,ε -γ -1 i,ε ln δ 2 4 + o γ -1 i,ε = -γ -1 i,ε -γ -1 i,ε ln f 0 (x i ) λ ε γ 2 i,ε -γ -1 i,ε ln δ 2 4 + o γ -1 i,ε
so that, thanks to (4.107),

γ i,ε B i,ε (δ) → -ln δ 2 f 0 (x i ) 4m 2 i -1 .
Coming back to (4.109) with this, we get that

γ i,ε u ε (x) → 2 ln 1 |x -x i | + H i x -x i δ -ln f 0 (x i ) 4m 2 i -1 in C 1 loc (D xi (δ) \ {x i }) as ε → 0 . (4.110)
On the other hand, using (4.107) and (4.108), we also have that

γ i,ε u ε (x) → 4π m i N j=1 m j G (x j , x) in C 1 loc (D xi (δ) \ {x i }) as ε → 0 . ( 4
m i H i x -x i δ = 4π N j=1 m j G (x j , x) -2m i ln 1 |x -x i | + m i ln f 0 (x i ) 4m 2 i + m i . Writing G (x, y) = 1 2π ln 1 |x -y| + H (x, y) ,
this leads to

m i H i x -x i δ = 4π j =i m j G (x j , x) + 2m i H (x i , x) + m i ln f 0 (x i ) 4m 2 i + m i .
The conditions that H i (0) = 0 and

∇H i (0) = -1 2 δ ∇f0(xi) f0(xi) read as 4π j =i m j G (x j , x i ) + 2m i H (x i , x i ) + m i ln f 0 (x i ) 4m 2 i + m i = 0 (4.112) and 4π j =i m j ∇ y G (x j , x i ) + 2m i ∇ y H (x i , x i ) = - 1 2 m i ∇f 0 (x i ) f 0 (x i ) . (4.113)
This ends the proof of Theorem 1.2, up to change the m i 's as in the statement of the theorem. ♦

Appendix A -The standard bubble

In this appendix, we develop the exact form of the standard bubble B ε which is defined as the radial solution of

∆B ε = µ -2 ε γ -2 ε B ε e B 2 ε -γ 2 ε in R 2 with B ε (0) = γ ε (5.1)
where γ ε → +∞ and µ ε → 0 as ε → 0. Note that, by standard ordinary differential equations theory, this function is defined on [0, +∞) and is decreasing. We perform the change of variables t = ln 1 + r 2 4µ 2 ε (5.2) so that we can rewrite equation (5.1) as

e t 1 -e -t B ′ ε ′ = - B ε γ 2 ε e 2t+B 2 ε -γ 2 ε . (5.3)
We shall need the following lemma which can be proved by direct computations : e -s F (s) 1 -2e -t 1 -2e -s ln e t -1 e s -1 + 4 e -se -t ds .

Proof of Lemma 5.1 -We clearly have that ϕ(0) = 0 so that we just have to check that ϕ satisfies the given differential equation. Let us differentiate to obtain that ϕ ′ (t) = t 0 e -s F (s) 2e This proves the second part of (5.6) by passing to the limit t → +∞ and the first part by integration.

We set now

B ε (t) = γ ε - t γ ε + γ -3 ε ϕ 0 + R ε .
(5.7)

Claim 5.1. There exists D 0 > 0 such that

|R ′ ε (t)| ≤ D 0 γ -5
ε for all 0 ≤ t ≤ γ 2 ε -T ε where T ε is any sequence such that T ε = o (γ ε ) and γ k ε e -Tε → 0 as ε → 0 for all k.

Proof of Claim 5.1 -Fix such a sequence T ε . Let D 0 > 0 that we shall choose later. Since R ′ ε (0) = 0, there exists 0 < t ε ≤ γ 2 ε -T ε such that |R ′ ε (t)| ≤ D 0 γ -5 ε for all 0 ≤ t ≤ t ε .

(5.8)

Note that this implies since R ε (0) = 0 that |R ε (t)| ≤ D 0 γ -5 ε t for all 0 ≤ t ≤ t ε .

(5.9)

We will prove that, for some choice of D 0 , this t ε may be chosen equal to γ 2 ε -T ε , which will prove the claim. Now, assume this is not the case, then, for the maximal t ε such that (5.8) holds, we have that |R ′ ε (t ε )| = D 0 γ -5 ε .

(5.10) This is the statement we will contradict by an appropriate choice of D 0 . Let us use (5.3), (5.5) and (5.7) to write that L (R ε ) = F ε where

F ε = 1 γ ε - B ε γ 2 ε e 2t+B 2 ε -γ 2 ε + 2R ε -γ -3 ε t -t 2 -2ϕ 0 .
For 0 ≤ t ≤ min {t ε , T ε }, we have that

2t + B 2 ε -γ 2 ε = t 2 γ 2 ε + 2γ ε R ε + 2γ -2 ε 1 - t γ 2 ε ϕ 0 + o γ -4 ε and that B ε γ 2 ε = γ -1 ε -γ -2 ε t + γ -5 ε ϕ 0 + o γ -6
ε thanks to (5.6) and (5.9). Thus we have in particular that

2t + B 2 ε -γ 2 ε ≤ 2t 2 γ 2 ε + 2D 0 γ -4 ε t + 2γ -2 ε (C 0 + 1) + o γ -4 ε = o(1)
again with (5.6) and (5.9). We can write that Combined with (5.11), this gives that

|R ′ ε (t)| ≤ D 2 γ -5 ε + o γ -5 ε .
This proves that (5.10) is impossible, up to choose D 0 ≥ 2D 2 . This ends the proof of this claim.

♦.

If we want to push a little bit further the estimates, we can get Claim 5.2. There exists C 0 > 0 such that

B ε -γ ε + t γ ε + t γ 3 ε ≤ C 0 γ -2
ε for all 0 ≤ t ≤ γ 2 ε .

Proof of Claim 5.2 -It is clear that it holds for any 0 ≤ t ≤ γ 2 ε -T ε for T ε as in Claim 5.1. This is a consequence of Claim 5.1 and of (5.6). We also know that

B ε γ 2 ε -T ε = T ε γ ε - 1 γ ε + T ε γ 3 ε + O γ -3 ε .
(5.12)

and that

B ′ ε γ 2 ε -T ε = - 1 γ ε - 1 γ 3 ε + O γ -5 ε . (5.13)
Let us integrate twice the equation (5.3) between γ 2 ε -T ε and t ε = γ 2 εα ε for 0 ≤ α ε ≤ T ε to write that Using (5.12) and (5.13), and remembering that α ε ≤ T ε = o (γ ε ), we obtain that -s ds ≤ Cγ ε for all 0 ≤ t ≤ γ 2 ε . This permits to end the proof of the claim. ♦

B ε (t ε ) = B ε γ 2 ε -T ε + B ′ ε γ 2 ε -T ε 1 -e Tε-
B ε (t ε ) = γ ε - t ε γ ε - t ε γ 3 ε + O γ -3 ε - 1 γ 2 ε γ 2 ε -αε γ 2 ε -
B ′ ε (t) + γ -1 ε ≤ C 1 γ -2

Appendix B -Estimates on the Green function

We list and prove some useful estimates on the Green function of the Laplacian with Dirichlet boundary condition in some smooth domain Ω. We fix such a two-dimensional domain and we let G (x, y) be such that ∆ x G(x, y) = δ y with G (x, y) = 0 if x ∈ ∂Ω .

It is well known that G is symmetric and smooth outside of the diagonal. Except on the disk of radius R where G is explicitly given by G (x, y) = 1 4π ln We let now (y ε ) be a sequence of points in Ω such that d ε = d (y ε , ∂Ω) → 0 as ε → 0 . (6.4)

We let now ỹε ∈ R 2 be such that ỹε = 2π (y ε )y ε (6.5) where π is the projection on the boundary of Ω. Note that π (y ε ) is unique thanks to (6.4) and to the fact that Ω is smooth. Moreover, we have that ỹε = y ε + 2d ε ν ε (6.6) where ν ε is the unit outer normal of ∂Ω at π (y ε ). We let now

  .50) Moreover, we have thanks to (3.45), (3.46), (3.48) and (3.50) that w 0 (0) = 0, ∇w 0 (0) = 0, ∇w 0 (y 0 ) -A = 1 and ∇w 0

(4. 25 )

 25 We know thanks to Step 1 that there exists j ∈ D i such that|x j | = 2 (4.26)and that|x kxl | ≥ 2 for all k, l ∈ D i , k = l .(4.27) Since r j,ε and r i,ε are comparable, we also have thanks to Step 2 that lim inf ε→0 γ -1 j,ε B j,ε (r j,ε ) > 0 .(4.28)

2 e

 2 2U dx = 4π , the result follows by letting R go to +∞ and δ go to 0. ♦ Let us set now

Claim 4 . 9 .

 49 the computations of Step 2 lead to the result of Step 4. ♠ Of course, the combination of Steps 1 to 4 gives the estimate of the claim.♦ There exists C > 0 such that

  that r i,ε → 0 as ε → 0 .(4.86)

  → +∞ as ε → 0 and since |xi,ε-xj,ε| ri,ε

Lemma 5 . 1 .

 51 The solution ϕ ofL (ϕ) = e t 1e -t ϕ ′ ′ + 2ϕ = F with ϕ(0) = 0 and F smooth is ϕ(t) = t 0

e 2t+B 2 ε -γ 2 ε - 1 -2t + B 2 ε -γ 2 ε≤ 2 2t + B 2 ε -γ 2 ε 2 ≤ 20γ - 4 ε t 4 + 5 ε 5 ε t 0 e -s 1 +

 22122222445501 (C 0 + 1)2 for all 0 ≤ t ≤ min {t ε , T ε } for ε small. Coming back to F ε , this leads to|F ε | ≤ D 1 1 + t 4 γ -for all 0 ≤ t ≤ min {t ε , T ε } where D 1 depends on C 0 but not on D 0 . We can use the representation formula of Lemma 5.1 to deduce that|R ′ ε (t)| ≤ D 1 γ -s 4 2e -t

=

  O γ ε e -1 2 Tε = o γ -5 ε .

γ 2 ε ln e γ 2 ε -αε -1 e γ 2 ε -Tε - 1 ( 5 ln e tε -1 e t - 1 B

 22151 ε (t)e t+Bε(t) 2 -γ 2 ε dt .

ε 2 ε e t e t - 1 t 0 B 1 ε≤ γ -2 ε e t e t - 1 t 0 Be s 2 γ 2 ε 1 ε 2 ε 2 ε

 2101102122 for all 0 ≤ t ≤ γ 2 ε .Proof of Claim 5.3 -Let us start from the fact thatB ′ ε (t) = -γ -ε (s)e s+Bε(s) 2 -γ 2 ε dsobtained by integrating(5.3). This leads toB ′ ε (t) + γ -ε (s)e 2s+Bε(s) 2 -γ 2 εγ ε e -s ds .Let us use Claim 5.2 to write thatB ε (s)e 2s+Bε(s) 2 -γ 2 εγ ε ≤ Cγ ε -1 e -s + C s + γ ε γ ε e some C > 0 independent of ε and of 0 ≤ s ≤ γ 2 ε . Thus we get that B ′ ε (t) + γ --1 e -s ds + Cγ -3 -1 e -s ds ≤ Cγ -2

First, if 3 )

 3 y ∈ K for some compact subset K of Ω, we clearly have that|H y (x)| ≤ C K and |∇H y (x)| ≤ C K (6.2)for some C K > 0 for all x ∈ Ω so thatG (x, y) -|∇ x G (x, y)| ≤ C K |x -y| -1 , G (x, y) -G (z, y) -1 2π ln |z -y| |x -y| ≤ C K |x -z|for all x, y, z ∈ K ⊂⊂ Ω (distinct points).

  which is impossible by what we just proved. Thus a) is clearly proved, applying again Claim 4.1 this time with r ε .

  ε . Then the above estimate would clearly contradict a) of Claim 4.2 together with (4.16). Thus Step 1 is proved.

♠

Thanks to

Step 1, we know that, if (4.16) holds, then

  = 1, . . . , N with m i > 0. Thanks to Claim 4.8, to (4.83) and to the equation satisfied by u ε , by standard elliptic theory, we obtain that

				.106)
	Claim 4.4 then gives that λ ε γ 2 i,ε = O(1) for all i = 1, . . . , N . Thanks to Claim 4.11 and (4.54),
	this implies that, up to a subsequence
	√	1 λ ε γ i,ε	→ m i as ε → 0	(4.107)
	for all i

  -t 1 -2e -s ln e t -1 e s -1 + e t -2 e t -1 1 -2e -s + 4e -t ds F (s) -2e -t 1 -2e -t 1 -2e -s ln e t -1 e s -1 + 8e -2t -8e -t e -s ds = e -t F (t) -2e -t ϕ(t) ,

	so that
	1 -e -t ϕ ′ (t) = Differentiating again, we get that t 0 e -s F (s) 2 e t -1 e 2t 1 -2e -s ln	e t -1 e s -1	+	e t -2 e t	1 -2e -s + 4	e t -1 e 2t	ds .
	1 -e -t ϕ ′ ′ (t) = e -t F (t)	e t -2 e t	1 -2e -t + 4	e t -1 e 2t
	t
	+ e -s which proves the lemma. 0							♦
	Let us define
	ϕ 0 (t) =

t 0 e -s ss

2

1 -2e -t 1 -2e -s ln e t -1 e s -1 + 4 e -se -t ds (5.4) so that, by lemma 5.1,

L (ϕ 0 ) (t) = tt 2 .

(5.5)

We claim now that

|ϕ 0 (t) + t| ≤ C 0 and ϕ ′ 0 (t) → 1 as t → +∞ (5.6)

for some C 0 > 0. Let us write that

ϕ 0 (t) = t 0 e -s ss 2 2e -t

1 -2e -s ln e t -1 e s -1 + e t -2 e t -1 1 -2e -s + 4e -t ds = 2e -t ln e t -1 + e t -2 e t -1 t 0 e -s ss 2 1 -2e -s ds +e -t t 0 e -s ss 2 2 1 -2e -s ln 1 e s -1 + 4 ds = 2e -t ln e t -1 + e t -2 e t -1 1 + t + t 2 e -tt 2 e -2t -1 +O e -t = -1 + O 1 + t 2 e -t .

  1 -2e -s ln e t -1 e s -1 + e t -2 e t -1 1 -2e -s + 4e -t ds ≤ D 2 γ -5 ε for all 0 ≤ t ≤ min {T ε , t ε } where D 2 depends only on C 0 , not on D 0 . Up to choose D 0 > 2D 2 ,we get that t ε > T ε thanks to (5.10). Moreover we have that|R ′ ε (T ε )| ≤ D 2 γ -5 ε .(5.11)From now on, we assume that t ε ≥ T ε . For all T ε ≤ t ≤ γ 2 ε -T ε , we can write that|F ε (t)| ≤ Cγ ε efor some C > 0, depending on D 0 and C 0 . Then we write that|R ′ ε (t) -R ′ ε (T ε )| ≤ Cγ ε -s 2e -t1 -2e -s ln e t -1 e s -1 + e t -2 e t -1 1 -2e -s + 4e -t ds

		t	e	s 2 γ 2 ε
		Tε
		t
	≤ Cγ ε	Tε
				t 2 γ 2 ε

  Assume that the statement of the Claim holds up to t ε . If we are able to prove that, under this condition, ≤ t ≤ t ε , then we can write that ln e tε -1e t -1 |B ε (t)| e t+Bε(t) 2 -γ 2 ε = O γ -1 ε 1 + s 2 e -sin γ 2 ε -T ε , t ε with t = γ 2 εs so that it is easily checked that

	γ 2 ε -αε γ 2 ε -Tε then the argument already used in the previous claim will conclude. ln e tε -1 B ε (t)e t+Bε(t) 2 -γ 2 ε dt = o(1) , e t -1	(5.16)
	If
			B ε -γ ε +	t γ ε	+	t γ 3 ε	≤ C 0 γ -2 ε
	for all 0 γ 2 ε -αε γ 2 ε -Tε	ln	e tε -1 e t -1	B
						Tε	ln	e tε -1 e t -1	B ε (t)e t+Bε(t) 2 -γ 2 ε dt .	(5.15)

ε (t)e t+Bε(t) 2 -γ 2 ε dt = O γ -1 ε ,

which ends the proof of this claim. ♦ Claim 5.3. There exists C 1 > 0 such that

  and so where all the estimates below follow from explicit computations, we need to be a little bit careful to estimate the Green function for various x and y.

		|y| R x -Ry |y| |x -y| 2	2
	We know that where	G (x, y) = ∆ x H y (x) = 0 in Ω and H y (x) = -1 2π ln 1 |x -y|	+ H y (x) 1 2π ln 1 |x -y|	on ∂Ω .	(6.1)

We assume for assertions b) to f) that N ≥ 1.

γi,ε e -tε(x)

Acknowledgments

The authors wish to thank Luca Martinazzi to have drawn their attention again on this question and for the fruitful discussions they had during the preparation of this work.

It is easily checked since Ω ∈ C 2 that Hε (x) ≤ C Ω d ε for some C Ω > 0 independent of ε and for all x ∈ ∂Ω. Thus we have that

It is also easily checked that ∇ T Hε (x) ≤ C Ω (6.10) for all x ∈ ∂Ω where ∇ T denotes the tangential derivative. Thus we have that

Let us give some useful consequences of (6.9) and (6.11). Let y ε be such that d ε = d (y ε , ∂Ω) → 0 as ε → 0, then we have that for any sequence (x ε ) in Ω

These are the only estimates which were used in this paper.