Samy Skander Bahoura

A UNIFORM ESTIMATE FOR AN EQUATION WITH HOLDERIAN CONDITION AND BOUNDARY SINGULARITY

We consider the following problem on open set Ω of R 2 :

Here, x 0 ∈ ∂Ω and, α ∈ (0, 1/2).

We assume, for example that:

We give, a quantization analysis of the previous problem under the conditions:

and,

2) In addition to the previous hypothesis we assume that V i sholderian with 1/2 < s ≤ 1, then we have a compactness result, namely:

where A is the holderian constant of V i .

INTRODUCTION AND MAIN RESULTS

We set ∆ = ∂ 11 + ∂ 22 on open set Ω of R 2 with a smooth boundary.

We consider the following problem on Ω ⊂ R 2 :

(P ) -∆u i = |x -x 0 | -2α V i e ui in Ω ⊂ R 2 , u i = 0 in ∂Ω.
Here, x 0 ∈ ∂Ω and, α ∈ (0, 1/2).

We assume that,

0 ≤ V i ≤ b < +∞, Ω |x -x 0 | -2α e ui dy ≤ C, u i ∈ W 1,1 0 (Ω)
The above equation is called, the Prescribed Scalar Curvature equation in relation with conformal change of metrics. The function V i is the prescribed curvature.

Here, we try to find some a priori estimates for sequences of the previous problem.

Equations of this type (in dimension 2 and higher dimensions) were studied by many authors, see . We can see in [START_REF] Brezis | Uniform estimates and blow-up bihavior for solutions of -∆u = V e u in two dimensions[END_REF], different results for the solutions of those type of equations with or without boundaries conditions and, with minimal conditions on V , for example we suppose V i ≥ 0 and V i ∈ L p (Ω) or V i e ui ∈ L p (Ω) with p ∈ [1, +∞].

Among other results, we can see in [START_REF] Brezis | Uniform estimates and blow-up bihavior for solutions of -∆u = V e u in two dimensions[END_REF], the following important Theorem, Theorem A(Brezis-Merle [START_REF] Brezis | Uniform estimates and blow-up bihavior for solutions of -∆u = V e u in two dimensions[END_REF]).If (u i ) i and (V i ) i are two sequences of functions relatively to the previous problem (P ) with, 0 < a ≤ V i ≤ b < +∞, then, for all compact set K of Ω, sup K u i ≤ c = c(a, b, K, Ω).

A simple consequence of this theorem is that, if we assume u i = 0 on ∂Ω then, the sequence (u i ) i is locally uniformly bounded. We can find in [START_REF] Brezis | Uniform estimates and blow-up bihavior for solutions of -∆u = V e u in two dimensions[END_REF] an interior estimate if we assume a = 0, but we need an assumption on the integral of e ui . We have in [START_REF] Brezis | Uniform estimates and blow-up bihavior for solutions of -∆u = V e u in two dimensions[END_REF]:

Theorem B (Brezis-Merle [START_REF] Brezis | Uniform estimates and blow-up bihavior for solutions of -∆u = V e u in two dimensions[END_REF]).If (u i ) i and (V i ) i are two sequences of functions relatively to the previous problem (P ) with, 0 ≤ V i ≤ b < +∞, and, Ω e ui dy ≤ C, then, for all compact set K of Ω, sup

K u i ≤ c = c(b, C, K, Ω).
If, we assume V with more regularity, we can have another type of estimates, sup + inf. It was proved, by Shafrir, see [START_REF] Shafrir | A sup+inf inequality for the equation -∆u = V e u[END_REF], that, if (u i ) i , (V i ) i are two sequences of functions solutions of the previous equation without assumption on the boundary and, 0 < a ≤ V i ≤ b < +∞, then we have the following interior estimate:

C a b sup K u i + inf Ω u i ≤ c = c(a, b, K, Ω).
We can see in [START_REF] Chen | A sharp sup+inf inequality for a nonlinear elliptic equation in R 2[END_REF], an explicit value of C a b = a b . In his proof, Shafrir has used the Stokes formula and an isoperimetric inequality, see [START_REF] Bandle | Isoperimetric inequalities and Applications[END_REF]. For Chen-Lin, they have used the blowup analysis combined with some geometric type inequality for the integral curvature.

Now, if we suppose (V i ) i uniformly Lipschitzian with A the Lipschitz constant, then, C(a/b) = 1 and c = c(a, b, A, K, Ω), see Brezis-Li-Shafrir [START_REF] Brezis | A sup+inf inequality for some nonlinear elliptic equations involving exponential nonlinearities[END_REF]. This result was extended for Hölderian sequences (V i ) i by Chen-Lin, see [START_REF] Chen | A sharp sup+inf inequality for a nonlinear elliptic equation in R 2[END_REF]. Also, we can see in [START_REF] Yy | Harnack Type Inequality: the Method of Moving Planes[END_REF], an extension of the Brezis-Li-Shafrir to compact Riemann surface without boundary. We can see in [START_REF] Yy | Blow-up Analysis for Solutions of -∆u = V e u in Dimension Two[END_REF] explicit form, (8πm, m ∈ N * exactly), for the numbers in front of the Dirac masses, when the solutions blowup. Here, the notion of isolated blow-up point is used.

In [START_REF] Brezis | Uniform estimates and blow-up bihavior for solutions of -∆u = V e u in two dimensions[END_REF], Brezis and Merle proposed the following Problem: Problem (Brezis-Merle [START_REF] Brezis | Uniform estimates and blow-up bihavior for solutions of -∆u = V e u in two dimensions[END_REF]).If (u i ) i and (V i ) i are two sequences of functions relatively to the previous problem (P ) with,

0 ≤ V i → V in C 0 ( Ω). Ω e ui dy ≤ C,

Is it possible to prove that:

sup

Ω u i ≤ c = c(C, V, Ω) ?
Here, we assume more regularity on V i , we suppose that V i ≥ 0 is C s (s-holderian) 1/2 < s ≤ 1) and when we have a boundary singularity. We give the answer where bC < 16π for an equation with boundary singularity.

Our main results are:

Theorem 1.1. Assume Ω = B 1 (0), x 0 ∈ ∂Ω, α ∈ (0, 1/2), and,

B1(0) |x -x 0 | -2α V i e ui dy ≤ 16π -ǫ, ǫ > 0, u i (x i ) = sup B1(0) u i → +∞.
2

There is a sequences (x 0 i ) i , (δ 0 i ), such that:

(x 0 i ) i ≡ (x i ) i , δ 0 i = δ i = d(x i , ∂B 1 (0)) → 0, and, u i (x i ) = sup B1(0) u i → +∞, u i (x i ) + 2 log δ i -2α log d(x i , x 0 ) → +∞, ∀ ǫ > 0, lim sup i→+∞ B(xi,δiǫ) |x -x 0 | -2α V i e ui dy ≥ 4π > 0.
If we assume:

V i → V in C 0 ( B1 (0)), then, ∀ ǫ > 0, sup B1(0)-B(xi,δiǫ) u i ≤ C ǫ ∀ ǫ > 0, lim sup i→+∞ B(xi,δiǫ) |x -x 0 | -2α V i e ui dy = 8π.
And, thus, we have the following convergence in the sense of distributions:

B1(0) |x -x 0 | -2α V i e ui dy → B1(0) |x -x 0 | -2α V e u dy + 8πδ x0 .
Theorem 1.2. Assume that, V i is uniformly s-holderian with 1/2 < s ≤ 1, x 0 ∈ ∂Ω, α ∈ (0, 1/2), and,

B1(0) |x -x 0 | -2α V i e ui dy ≤ 16π -ǫ, ǫ > 0,
then we have:

sup Ω u i ≤ c = c(b, C, A, s, α, ǫ, x 0 , Ω).
where A is the hölderian constant of V i .

PROOFS OF THE RESULTS

Proofs of the theorems:

Without loss of generality, we can assume that Ω = B 1 (0) the unit ball centered on the origin.

Here, G is the Green function of the Laplacian with Dirichlet condition on B 1 (0). We have (in complex notation):

G(x, y) = 1 2π log |1 -xy| |x -y| ,
Since u i ∈ W 1,1 0 (Ω) and α ∈ (0, 1/2), we have by the Brezis-Merle result and the elliptic estimates, (see [START_REF] Aubin | Some Nonlinear Problems in Riemannian Geometry[END_REF]):

u i ∈ C 2 (Ω) ∩ W 2,p (Ω) ∩ C 1,ǫ ( Ω) for all 2 < p < +∞. Set, v i (x) = B1(0) G(x, y)V i (y)|x -x 0 | -2α e ui(y) dy.
We decompose v i in two terms (Newtionian potential):

v 1 i (x) = B1(0) - 1 2π log |x -y|V i (y)|x -x 0 | -2α e ui(y) dy,
and,

v 2 i (x) = B1(0) 1 2π log |1 -xy|V i (y)|x -x 0 | -2α e ui(y) dy,
According to the proof in the book of Gilbarg-Trudinger see [START_REF] Gilbarg | Second elliptic equations[END_REF], v 1 i , v 2 i and thus v i are C 1 ( Ω). Indeed, we use the same proof as in [START_REF] Gilbarg | Second elliptic equations[END_REF] (Chapter 4, Newtonian potential), we have for the approximate function ∂v i,ǫ terms of type O(ǫ 1-2α log ǫ) + O(ǫ 1-2α ). Since α < 1/2, this term tends to 0.

We use this fact and the maximum principle to have v i = u i . Also, we can use integration by part (the Green representation formula, see its proof in the first chapter of [START_REF] Gilbarg | Second elliptic equations[END_REF]) to have in Ω (and not Ω):

u i (x) = - B1(0) G(x, y)∆u i (y)dy = B1(0) G(x, y)V i (y)|x -x 0 | -2α e ui(y) dy.
We write,

u i (x i ) = Ω G(x i , y)|x-x 0 | -2α V i (y)e ui(y) dx = Ω-B(xi,δi/2) G(x i , y)|x-x 0 | -2α V i e ui(y) dy+ + B(xi,δi/2) G(x i , y)|x -x 0 | -2α V i e ui(y) dy
According to the maximum principle, the harmonic function G(x i , .) on Ω -B(x i , δ i /2) take its maximum on the boundary of B(x i , δ i /2), we can compute this maximum:

G(x i , y i ) = 1 2π log |1 -xi y i | |x i -y i | = 1 2π log |1 -xi (x i + δ i θ i )| |δ i /2| = 1 2π log(2|(1+|x i |)+θ i |) < +∞ with |θ i | = 1/2.
Thus,

u i (x i ) ≤ C+ B(xi,δi/2) G(x i , y)|x-x 0 | -2α V i e ui(y) dy ≤ C+e ui(xi)-2α log d(xi,x0) B(xi,δi/2)

G(x i , y)dy

Now, we compute B(xi,δi/2) G(x i , y)dy we set in polar coordinates, y = x i + δ i tθ we find:

B(xi,δi/2) G(x i , y)dy = B(xi,δi/2) 1 2π log |1 -xi y| |x i -y| = 1 2π 2π 0 1/2 0 δ 2 i log |1 -xi (x i + δ i tθ)| δ i t tdtdθ = = 1 2π 2π 0 1/2 0 δ 2 i (log(|1 + |x i | + tθ|) -log t)tdtdθ ≤ Cδ 2 i .
Thus,

u i (x i ) ≤ C + Cδ 2 i e ui(xi)-2α log d(xi,x0
) , which we can write, because u i (x i ) → +∞,

u i (x i ) ≤ C ′ δ 2 i e ui(xi)-2α log d(xi,x0
) , We can conclude that:

u i (x i ) + 2 log δ i -2α log d(x i , x 0 ) → +∞.
Since in B(x i , δ i ǫ), d(x, x 0 ) is equivalent to d(x i , x 0 ) we can consider the following functions:

v i (y) = u i (x i + δ i y) + 2 log δ i -2α log d(x i , x 0 ), y ∈ B(0, 1/2)
The function satisfies all conditions of the Brezis-Merle hypothesis, we can conclude that, on each compact set:

v i → -∞
we can assume, without loss of generality that for 1/2 > ǫ > 0, we have:

v i → -∞, y ∈ B(0, 2ǫ) -B(0, ǫ), Lemma 2.1. For all 1/4 > ǫ > 0, we have: sup B(xi,(3/2)δiǫ)-B(xi,δiǫ) u i ≤ C ǫ . Proof of the lemma Let t ′ i and t i the points of B(x i , 2δ i ǫ) -B(x i , (1/2)δ i ǫ) and B(x i , (3/2)δ i ǫ) -B(x i , δ i ǫ)
respectively where u i takes its maximum.

According to the Brezis-Merle work, we have:

u i (t ′ i ) + 2 log δ i -2α log d(x i , x 0 ) → -∞ We write, u i (t i ) = Ω G(t i , y)|x-x 0 | -2α V i (y)e ui(y) dx = Ω-B(xi,2δiǫ) G(t i , y)|x-x 0 | -2α V i e ui(y) dy+ + B(xi,2δiǫ)-B(xi,(1/2)δiǫ) G(t i , y)|x -x 0 | -2α V i e ui(y) dy+ + B(xi,(1/2)δiǫ) G(t i , y)|x -x 0 | -2α V i e ui(y) dy
But, in the first and the third integrale, the point t i is far from the singularity x i and we know that the Green function is bounded. For the second integrale, after a change of variable, we can see that this integale is bounded by (we take the supremum in the annulus and use Brezis-Merle theorem)

δ 2 i e ui(t ′ i )-2α log d(xi,x0) × I j
where I j is a Jensen integrale (of the form

1 0 2π 0 (log(|1 + |x i | + tθ) -log |θ i -tθ|)tdtdθ which is bounded ).
we conclude the lemma.

From the lemma, we see that far from the singularity the sequence is bounded, thus if we take the supremum on the set B 1 (0) -B(x i , δ i ǫ) we can see that this supremum is bounded and thus the sequence of functions is uniformly bounded or tends to infinity and we use the same arguments as for x i to conclude that around this point and far from the singularity, the seqence is bounded.

The process will be finished , because, according to Brezis-Merle estimate, around each supremum constructed and tending to infinity, we have:

∀ ǫ > 0, lim sup i→+∞ B(xi,δiǫ) |x -x 0 | -2α V i e ui dy ≥ 4π > 0.
Finaly, with this construction, we have a finite number of "exterior "blow-up points and outside the singularities the sequence is bounded uniformly, for example, in the case of one "exterior" blow-up point, we have:

u i (x i ) → +∞ ∀ ǫ > 0, sup B1(0)-B(xi,δiǫ) u i ≤ C ǫ ∀ ǫ > 0, lim sup i→+∞ B(xi,δiǫ) |x -x 0 | -2α V i e ui dy ≥ 4π > 0.
x i → x 0 ∈ ∂B 1 (0). Remark: For the general case, the process of quantization can be extended to more than one blow-up points.

We have the following lemma:

Lemma 2.2. Each δ k i is of order d(x k i , ∂B 1 (0)). Namely: there is a positive constant C > 0 such that for ǫ > 0 small enough:

δ k i ≤ d(x k i , ∂B 1 (0)) ≤ (2 + C ǫ )δ k i .
Proof of the lemma Now, if we suppose that there is another "exterior" blow-up (t i ) i , we have, because (u i ) i is uniformly bounded in a neighborhood of ∂B(x i , δ i ǫ), we have :

d(t i , ∂B(x i , δ i ǫ)) ≥ δ i ǫ If we set, δ ′ i = d(t i , ∂(B 1 (0) -B(x i , δ i ǫ))) = inf{d(t i , ∂B(x i , δ i ǫ)), d(t i , ∂(B 1 (0)))} then, δ ′ i is of order d(t i , ∂B 1 (0)).
To see this, we write:

d(t i , ∂B 1 (0)) ≤ d(t i , ∂B(x i , δ i ǫ)) + d(∂B(x i , δ i ǫ), x i ) + d(x i , ∂B 1 (0)), Thus, d(t i , ∂B 1 (0)) d(t i , ∂B(x i , δ i ǫ)) ≤ 2 + 1 ǫ ,
Thus,

δ ′ i ≤ d(t i , ∂B 1 (0)) ≤ δ ′ i (2 + 1 ǫ ).
Now, the general case follow by induction. We use the same argument for three, four,..., n blow-up points.

We have, by induction and, here we use the fact that u i is uniformly bounded outside a small ball centered at x j i , j = 0, . . . , k -1:

δ j i ≤ d(x j i , ∂B 1 (0)) ≤ C 1 δ j i , j = 0, . . . , k -1, . d(x k i , ∂B(x j i , δ j i ǫ/2)) ≥ ǫδ j i , ǫ > 0, j = 0, . . . , k -1, .
and let's consider x k i such that:

u i (x k i ) = sup B1(0)-∪ k-1 j=0 B(x j i ,δ j i ǫ) u i → +∞, take ,
δ k i = inf{d(x k i , ∂B 1 (0)), d(x k i , ∂(B 1 (0) -∪ k-1 j=0 B(x j i , δ j i ǫ/2))}, if, we have, δ k i = d(x k i , ∂B(x j i , δ j i ǫ/2)), j ∈ {0, . . . , k -1}. Then, δ k i ≤ d(x k i , ∂B 1 (0)) ≤ ≤ d(x k i , ∂B(x j i , δ j i ǫ/2)) + d(∂B(x j i , δ j i ǫ/2), x j i ) + d(x j i , ∂B 1 (0)) ≤ (2 + C 1 ǫ )δ k i .
To apply lemma 2.1 for m blow-up points, we use an induction:

We do directly the same approch for t i as x i by using directly the Green function of the unit ball.

If we look to the blow-up points, we can see, with this work that, after finite steps, the sequence will be bounded outside a finite number of balls , because of Brezis-Merle estimate:

∀ ǫ > 0, lim sup i→+∞ B(x k i ,δ k i ǫ) |x -x 0 | -2α V i e ui dy ≥ 4π > 0.
Here, we can take the functions:

u k i (y) = u i (x k i + δ k i y) + 2 log δ k i -2α log d(x i , x 0
). Indeed, by corollary 4 of the paper of Brezis-Merle, if we have:

lim sup i→+∞ B(x k i ,δ k i ǫ) |x -x 0 | -2α V i e ui dy ≤ 4π -ǫ 0 < 4π,
then, (u k i ) + would be bounded and this contradict the fact that u k i (0) → +∞. Finaly, we can say that, there is a finite number of sequences andeach δ k i is of order d(x k i , ∂B 1 (0)). and,

(x k i ) i , (δ k i ), 0 ≤ k ≤ m, such that: (x 0 i ) i ≡ (x i ) i , δ 0 i = δ i = d(x i , ∂B 1 (0)), (x 1 i ) i ≡ (t i ) i , δ 1 i = δ ′ i = d(t i , ∂(B 1 (0) -B(x i , δ i ǫ)),
u i (x k i ) = sup B1(0)-∪ k-1 j=0 B(x j i ,δ j i ǫ) u i → +∞, u i (x k i ) + 2 log δ k i -2α log d(x k i , x 0 ) → +∞, ∀ ǫ > 0, sup B1(0)-∪ m j=0 B(x j i ,δ j i ǫ) u i ≤ C ǫ ∀ ǫ > 0, lim sup i→+∞ B(x k i ,δ k i ǫ) |x -x 0 | -2α V i e ui dy ≥ 4π > 0.
The work of YY.Li-I.Shafrir Since in B(x i , δ i ǫ), d(x, x 0 ) is equivalent to d(x i , x 0 ) we can consider the following functions:

v i (y) = u i (x i + δ i y) + 2 log δ i -2α log d(x i , x 0 ).
With the previous method, we have a finite number of "exterior" blow-up points (perhaps the same) and the sequences tend to the boundary. With the aid of proposition 1 of the paper of Li-Shafrir, we see that around each exterior blow-up, we have a finite number of "interior" blow-ups. Around, each exterior blow-up, we have after rescaling with δ k i , the same situation as around a fixed ball with positive radius. If we assume:

V i → V in C 0 ( B1 (0)), then, ∀ ǫ > 0, lim sup i→+∞ B(x k i ,δ k i ǫ) |x -x 0 | -2α V i e ui dy = 8πm k , m k ∈ N * .
And, thus, we have the following convergence in the sense of distributions:

B1(0) |x-x 0 | -2α V i e ui dy → B1(0) |x-x 0 | -2α V e u dy+ m k=0 8πm ′ k δ x k 0 , m ′ k ∈ N * , x k 0 ∈ ∂B 1 (0).
Consequence: using a Pohozaev-type identity, proof of theorem 2

By a conformal transformation, we can assume that our domain Ω = B + is a half ball centered at the origin, B + = {x, |x| ≤ 1, x 1 ≥ 0}, and, x 0 = 0. In this case the normal at the boundary is ν = (-1, 0) and u i (0, x 2 ) ≡ 0. Also, we set x i the blow-up point and x 2 i = (0, x 2 i ) and x 1 i = (x 1 i , 0) respectevely the second and the first part of x i . Let ∂B + the part of the boundary for which u i and its derivatives are uniformly bounded and thus converge to the corresponding function.

The case of one blow-up point:

Theorem 2.3. If V i is s-Holderian with 1/2 < s ≤ 1 and, Ω |x| -2α V i e ui dy ≤ 16π -ǫ, ǫ > 0,
we have :

2(1 -2α)V i (x i ) B(xi,δiǫ)
|x| -2α e ui dy = o(1), which means that there is no blow-up points.

Proof of the theorem

In order to use the Pohozaev identity we need to have a good function u i , since α ∈ (0, 1/2), we have a function u i such that:

u i ∈ C 2 (Ω) ∩ W 2,p (Ω) ∩ C 1 ( Ω) Thus, ∂ j u i , ∂ k u i ∈ W 1,p (Ω) ∩ C 0 ( Ω).
Thus, we can use integration by parts, in fact we have for the following product (here "." is the usual product of function):

∂ j u i .∂ k u i ∈ W 1,p (Ω) ∩ C 0 ( Ω).
The Pohozaev identity gives us the following formula:

Ω < (x -x i 2 )|∇u i > (-∆u i )dy = Ω < (x -x i 2 )|∇u i > |x| -2α V i e ui dy = A i A i = ∂B + < (x -x i 2 )|∇u i >< ν|∇u i > dσ + ∂B + < (x -x i 2 )|ν > |∇u i | 2 dσ
We can write it as:

Ω < (x-x i 2 )|∇u i > (V i -V i (x i ))|x| -2α e ui dy = A i +V i (x i ) Ω < (x-x i 2 )|∇u i > |x| -2α e ui dy = = A i + V i (x i ) Ω < (x -x i 2 )|x| -2α |∇(e ui ) > dy
And, if we integrate by part the second term, we have (because x 1 = 0 on the boundary and ν 2 = 0):

Ω < (x -x i 2 )|∇u i > (V i -V i (x i ))|x| -2α e ui dy = -2(1 -α)V i (x i ) Ω |x| -2α e ui dy+ +2αV i (x i ) B(xi,δiǫ)
x 2 x i 2 |x| -2α-2 e ui dy + 2αV i (x i )

Ω-B(xi,δiǫ)

x 2 x i 2 |x| -2α-2 e ui dy + B i where B i is,

B i = V i (x i ) ∂B + < (x -x i 2 )|ν > |x| -2α e ui dy
applying the same procedure to u, we can write:

Ω < (x -x i 2 )|∇u > (V -V (0))|x| -2α e u dy = -2(1 -α)V (0) Ω |x| -2α e u dy+ +2αV (0)
B(xi,δiǫ)

x 2 x i 2 |x| -2α-2 e u dy + 2αV (0)

Ω-B(xi,δiǫ)
x 2 x i 2 |x| -2α-2 e u dy + B, with,

B = V (0) ∂B + < (x -x i 2 )|ν > |x| -2α e u dy
we use the fact that, u i is bounded outside B(x i , δ i ǫ) and the convergence of u i to u on compact set of Ω -{0}, and the fact that α ∈ (0, 1/2), to write the following:

2(1 -2α)V i (x i ) B(xi,δiǫ) |x| -2α e ui dy + o(1) = = Ω < (x-x i 2 )|∇u i > (V i -V i (x i ))|x| -2α e ui dy- Ω < (x-x i 2 )|∇u > (V -V (0))|x| -2α e u dy+ +(A i -A) + (B i -B)
, where A and B are,

A = ∂B + < (x -x i 2 )|∇u >< ν|∇u > dσ + ∂B + < (x -x i 2 )|ν > |∇u| 2 dσ B = V (0) ∂B + < (x -x i 2 )|ν > |x| -2α e u dy
and, because of the uniform convergence of u i and its derivatives on ∂B + , we have: 1) and B i -B = o(1) which we can write as:

A i -A = o(
2(1 -2α)V i (x i ) B(xi,δiǫ) |x| -2α e ui dy + o(1) = = Ω < (x -x i 2 )|∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy+ + Ω < (x -x i 2 )|∇u > (V i -V i (x i ))|x| -2α (e ui -e u )dy+ + Ω < (x -x i 2 )|∇u > (V i -V i (x i ) -(V -V (0)))|x| -2α e u dy + o(1)
We can write the second term as:

Ω < (x-x i 2 )|∇u > (V i -V i (x i ))|x| -2α (e ui -e u )dy = Ω-B(0,ǫ) < (x-x i 2 )|∇u > (V i -V i (x i ))(e ui -e u )|x| -2α dy+ + B(0,ǫ) < (x -x i 2 )|∇u > (V i -V i (x i ))(e ui -e u )|x| -2α dy = o(1),
because of the uniform convergence of u i to u outside a region which contain the blow-up and the uniform convergence of V i . For the third integral we have the same result:

Ω < (x -x i 2 )|∇u > (V i -V i (x i ) -(V -V (0)))|x| -2α e u dy = o(1),
because of the uniform convergence of V i to V . Now, we look to the first integral:

Ω < (x -x i 2 )|∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy,
we can write it as:

Ω < (x-x i 2 )|∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy = Ω < (x-x i )|∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy+ + Ω < x i 1 |∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy,
Thus, we have proved by using the Pohozaev identity the following equality:

Ω < (x -x i )|∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy+ + Ω < x i 1 |∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy = = 2(1 -2α)V i (x i ) B(xi,δiǫ) |x| -2α e ui dy + o(1)
We can see, because of the uniform boundedness of u i outside B(x i , δ i ǫ) and the fact that :

||∇(u i -u)|| 1 = o(1),
it is sufficient to look to the integral on B(x i , δ i ǫ).

Assume that we are in the case of one blow-up, it must be (x i ) and isolated, we can write the following inequality as a consequence of YY.Li-I.Shafrir result:

u i (x) + 2 log |x -x i | -2α log d(x, 0) ≤ C,
We use this fact and the fact that V i is s-holderian to have that, on B(x i , δ i ǫ), (1-s) , ∀ ǫ ′ > 0, and, we use the fact that: ||∇(u i -u)|| q = o(1), ∀ 1 ≤ q < 2 to conclude by the Holder inequality that:

|(x -x i )(V i -V i (x i ))|x| -2α e ui | ≤ C |x -x i | 1-s ∈ L (2-ǫ ′ )/
B(xi,δiǫ) < (x -x i 2 )|∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy = o(1),
For the other integral, namely:

B(xi,δiǫ) < x i 1 |∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy,
We use the fact that, because our domain is a half ball, and the sup + inf inequality to have: But in the second member, for 1/2 < s ≤ 1, we have q s = 1/(3/4 -s/2) > 2 and thus q ′ s < 2 and, (|x| -2α e ui ) 3/4-s/2 ∈ L qs |x| -2α(3/4-s/2) e ((3/4)-(s/2))ui ∈ L qs ||∇(u i -u)|| q ′ s = o(1), ∀ 1 ≤ q ′ s < 2, one conclude that:

x i 1 = δ i ,
B(xi,δiǫ) < x i 1 |∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy = o(1)
Finaly, with this method, we conclude that, in the case of one blow-up point and V i is s-Holderian with 1/2 < s ≤ 1 :

2(1 -2α)V i (x i ) B(xi,δiǫ)
|x| -2α e ui dy = o(1) which means that there is no blow-up, which is a contradiction.

Finaly, for one blow-up point and V i is is s-Holderian with 1/2 < s ≤ 1, the sequence (u i ) is uniformly bounded on Ω.

  u i (x) + 4 log δ i -4α log d(x, 0) ≤ C and, |x| -sα e (s/2)ui(x) ≤ |x -x i | -s , |V i -V i (x i )| ≤ |x -x i | s ,Finaly, we have:| B(xi,δiǫ) < x i 1 |∇(u i -u) > (V i -V i (x i ))|x| -2α e ui dy| ≤ ≤ C B(xi,δiǫ) |∇(u i -u)|(|x| -2α e ui ) (3/4-s/2) ,