Uniform result for solutions of an equation with boundary singularity.

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Uniform result for solutions of an equation with boundary singularity.

Samy Skander Bahoura

To cite this version:

Samy Skander Bahoura. Uniform result for solutions of an equation with boundary singularity.. 2018.

�hal-01250730v5�

UNIFORM RESULT FOR SOLUTIONS OF AN EQUATION WITH BOUNDARY SINGULARITY.

SAMY SKANDER BAHOURA

ABSTRACT. We consider a variational problem with boundary singularity and Dirichlet condition. We give blow-up analysis for sequences of solutions of an equation with exponential nonlinearity. Also we derive a compactness criterion for this problem under some condition.

Keywords: blow-up, boundary, singularity, a priori estimate, analytic domain, Lipschitz condition.

1. I

M

R

We set ∆ = ∂

+ ∂

on open analytic domain Ω of R

. We consider the following equation:

(P )

( −∆u = |x|

V e

in Ω ⊂ R

,

u = 0 in ∂Ω.

Here:

β ∈ (0, +∞), 0 ∈ ∂Ω, and,

u ∈ W

(Ω), |x|

e

∈ L

(Ω) and 0 ≤ V ≤ b.

We can see in [8] a nice formulation of this problem (P ) in the sens of the distributions. This Problem arises from geometrical and physical problems, see for example [1, 3, 21, 25]. The above equation was studied by many authors, with or without the boundary condition, also for Riemannian surfaces, see [1- 25], where one can find some existence and compactness results. In [7] we have the following important Theorem (β = 0):

Theorem A(Brezis-Merle [7]).For (u

)

and (V

)

two sequences of functions relative to (P ) with, 0 < a ≤ V

≤ b < +∞

then for all compact set K of Ω it holds,

sup

K

u

≤ c,

1

with c depending on a, b, K and Ω.

One can find in [7] an interior estimate if we assume a = 0, but we need an assumption on the integral of e

, namely, we have (β = 0):

Theorem B(Brezis-Merle [7]).For (u

)

and (V

)

two sequences of functions relative to the problem (P ) with,

0 ≤ V

≤ b < +∞ and Z

Ω

e

dy ≤ C, then for all compact set K of Ω it holds;

sup

K

u

≤ c, with c depending on b, C, K and Ω.

We look to the uniform boundedness in all Ω ¯ of the solutions of the Problem (P ). When a = 0, the boundedness of R

Ω

e

is a necessary condition in the problem (P ) as showed in [7] by the following coun- terexample (β = 0):

Theorem C(Brezis-Merle [7]).There are two sequences (u

)

and (V

)

of the problem (P ) with, 0 ≤ V

≤ b < +∞ and

Z

Ω

e

dy ≤ C,

such that,

sup

Ω

u

→ +∞.

To obtain the two first previous results (Theorems A and B) Brezis and Merle used an inequality (Theo- rem 1 of [7]) obtained by an approximation argument and used Fatou’s lemma and applied the maximum principle in W

(Ω) which arises from Kato’s inequality. Also this weak form of the maximum principle is used to prove the local uniform boundedness result by comparing a certain function and the Newtonian potential. We refer to [6] for a topic about the weak form of the maximum principle.

When β = 0, the above equation has many properties in the constant and the Lipschitzian cases:

Note that for the problem (P ) (β = 0), by using the Pohozaev identity, we can prove that R

Ω

e

is uniformly bounded when 0 < a ≤ V

≤ b < +∞ and ||∇V

||

≤ A and Ω starshaped, when a = 0 and

∇ log V

is uniformly bounded, we can bound uniformly R

Ω

V

e

. In [20], Ma-Wei have proved that those results stay true for all open sets not necessarily starshaped in the case a > 0.

In [10] (β = 0) Chen-Li have proved that if a = 0, ∇ log V

is uniformly bounded and u

is locally uniformly bounded in L

, then the functions are uniformly bounded near the boundary.

In [10] (β = 0) Chen-Li have proved that if a = 0 and R

Ω

e

is uniformly bounded and ∇ log V

is uniformly bounded, then we have the compactness result directly. Ma-Wei in [20], extend this result in the case where a > 0.

If we assume V more regular, we can have another type of estimates, a sup + inf type inequalities. It was proved by Shafrir see [23], that, if (u

)

, (V

)

are two sequences of functions solutions of the previous

2

equation without assumption on the boundary and, 0 < a ≤ V

≤ b < +∞, then we have the following interior estimate:

C a b

sup

K

u

+ inf

Ω

u

≤ c = c(a, b, K, Ω).

One can see in [11] an explicit value of C a b

= r a

b . In his proof, Shafrir has used a blow-up function, the Stokes formula and an isoperimetric inequality, see [3]. For Chen-Lin, they have used the blow-up analysis combined with some geometric type inequality for the integral curvature.

Now, if we suppose (V

)

uniformly Lipschitzian with A the Lipschitz constant, then, C(a/b) = 1 and c = c(a, b, A, K, Ω), see Brezis-Li-Shafrir [5]. This result was extended for Hölderian sequences (V

)

by Chen-Lin, see [11]. Also, one can see in [18], an extension of the Brezis-Li-Shafrir result to compact Riemannian surfaces without boundary. One can see in [19] explicit form, (8πm, m ∈ N

exactly), for the numbers in front of the Dirac masses when the solutions blow-up. Here, the notion of isolated blow-up point is used.

In [9] we have some a priori estimates on the 2 and 3-spheres S

, S

.

Here we give the behavior of the blow-up points on the boundary and a proof of a Brezis-Merle’s type Problem when β ≥ 0.

We have the following similar problem to Brezis and Merle problem:

Problem. Suppose that V

→ V in C

( ¯ Ω), with, 0 ≤ V

. Also, we consider a sequence of solutions (u

) of (P ) relatively to (V

) such that,

Z

Ω

|x|

e

dx ≤ C,

is it possible to have:

||u

||

≤ C?

Here, we give a caracterization of the behavior of the blow-up points on the boundary and also a proof of the compactness theorem when V

are uniformly Lipschitzian. For the behavior of the blow-up points on the boundary, the following condition is enough,

0 ≤ V

≤ b,

The condition V

→ V in C

( ¯ Ω) is not necessary.

But for the proof of the compactness result (for the Brezis-Merle type problem) we assume that:

||∇V

||

≤ A.

3

We have the following caracterization of the behavior of the blow-up points on the boundary.

Theorem 1.1. Assume that max

u

→ +∞, where (u

) are solutions of the probleme (P ) with:

β ∈ (0, +∞), 0 ≤ V

≤ b, and Z

Ω

|x|

e

dx ≤ C, ∀ i,

then; after passing to a subsequence, there is a finction u, there is a number N ∈ N and N points x

, x

, . . . , x

∈ ∂Ω, such that,

∂

u

→ ∂

u +

N

X

j=1

α

δ

, α

≥ 4π weakly in the sens of measures.

and,

u

→ u in C

( ¯ Ω − {x

, . . . , x

}).

In the following theorem, we have a proof of a global a priori estimate which concern the problem (P ).

Theorem 1.2. Assume that (u

) are solutions of (P) relative to (V

) with the following conditions:

β ∈ (0, +∞), 0 ∈ ∂Ω, and,

0 ≤ V

≤ b, ||∇V

||

≤ A, and Z

Ω

|x|

e

≤ C, We have,

||u

||

≤ c(b, β, A, C, Ω),

2. P

Proof of theorem 1.1:

We have,

u

∈ W

(Ω), Since R

Ω

|x|

e

≤ C, we have, by the Brezis-Merle result see [7], e

∈ L

(Ω) for k > 2 and the elliptic estimates imply that

u

∈ W

(Ω) ∩ C

( ¯ Ω).

We denote by ∂

u

the inner normal derivative of u

. By the maximum principle, ∂

u

≥ 0.

By the Stokes formula, we obtain

Z

∂Ω

∂

u

dσ ≤ C.

4

Thus, (using the weak convergence in the space of Radon measures), we have the existence of a nonegative Radon measure µ such that

Z

∂Ω

∂

u

dσ ≤ C,

Without loss of generality, we can assume that ∂

u

> 0. Thus, (using the weak convergence in the space of Radon measures), we have the existence of a positive Radon measure µ such that,

Z

∂Ω

∂

u

ϕdσ → µ(ϕ), ∀ ϕ ∈ C

(∂Ω).

We take an x

∈ ∂Ω such that, µ(x

) < 4π. Without loss of generality, we can assume that the following curve, B(x

, ǫ) ∩ ∂Ω := I

is an interval.

We take a test function:



 

 



 

 



η

≡ 1, on I

, 0 < ǫ < δ/2, η

≡ 0, outside I

,

0 ≤ η

≤ 1,

||∇η

||

≤ C

(Ω, x

)

ǫ .

We take a η ˜

such that,

−∆˜ η

= 0 in Ω

˜

η

= η

on ∂Ω.

Remark: We use the following steps in the construction of η ˜

: We take a cutoff function η

in B (0, 2) or B(x

, 2):

1- We set η

(x) = η

(|x − x

|/ǫ) in the case of the unit disk it is sufficient.

2- Or, in the general case: we use a chart (f, Ω) ˜ with f (0) = x

and we take µ

(x) = η

(f (|x|/ǫ)) to have connected sets I

and we take η

(y) = µ

(f

(y)). Because f, f

are Lipschitz, |f(x) − x

| ≤ k

|x| ≤ 1 for |x| ≤ 1/k

and |f(x) − x

| ≥ k

|x| ≥ 2 for |x| ≥ 2/k

> 1/k

, the support of η is in I

.



 

 



 

 



η

≡ 1, on f(I

), 0 < ǫ < δ/2, η

≡ 0, outside f (I

),

0 ≤ η

≤ 1,

||∇η

||

1)ǫ)

≤ C

(Ω, x

)

ǫ .

5

3- Also, we can take: µ

(x) = η

(x/ǫ) and η

(y) = µ

(f

(y)), we extend it by 0 outside f (B

(0)). We have f (B

(0)) = D

(x

), f (B

(0)) = D

(x

) and f (B

) = D

(x

) with f and f

smooth diffeomor- phism.



 

 



 

 



η

≡ 1, on a the connected set J

= f(I

), 0 < ǫ < δ/2, η

≡ 0, outside J

= f (I

),

0 ≤ η

≤ 1,

||∇η

||

≤ C

(Ω, x

)

ǫ .

And, H

(J

) ≤ C

H

(I

) = C

4ǫ, since f is Lipschitz. Here H

is the Hausdorff measure.

We solve the Dirichlet Problem:

( ∆¯ η

= ∆η

in Ω ⊂ R

,

¯

η

= 0 in ∂Ω.

and finaly we set η ˜

= −¯ η

+ η

. Also, by the maximum principle and the elliptic estimates we have :

||∇˜ η

||

≤ C(||η

||

+ ||∇η

||

+ ||∆η

||

) ≤ C

ǫ

, with C

depends on Ω.

We use the following estimate, see [4,8,14,25]

||∇u

||

≤ C

, ∀ i and 1 < q < 2.

We deduce from the last estimate that, (u

) converge weakly in W

(Ω), almost everywhere to a function u ≥ 0 and R

Ω

|x|

e

< +∞ (by Fatou lemma). Also, V

weakly converge to a nonnegative function V in L

. The function u is in W

(Ω) solution of :

( −∆u = |x|

V e

∈ L

(Ω) in Ω

u = 0 on ∂Ω,

As in the corollary 1 of Brezis-Merle result, see [7], we have e

∈ L

(Ω), k > 1. By the elliptic estimates, we have u ∈ C

( ¯ Ω).

We can write,

−∆((u

− u)˜ η

) = |x|

(V

e

− V e

)˜ η

+ 2 < ∇(u

− u)|∇˜ η

> . (1) We use the interior esimate of Brezis-Merle, see [7],

6

Step 1: Estimate of the integral of the first term of the right hand side of (1).

We use the Green formula between η ˜

and u, we obtain,

Z

Ω

|x|

V e

η ˜

dx = Z

∂Ω

∂

uη

≤ 4ǫ||∂

u||

= Cǫ (2)

We have,

( −∆u

= |x|

V

e

in Ω

u

= 0 on ∂Ω,

We use the Green formula between u

and η ˜

to have:

Z

Ω

|x|

V

e

η ˜

dx = Z

∂Ω

∂

u

η

dσ → µ(η

) ≤ µ(I

) ≤ 4π − ǫ

, ǫ

> 0 (3)

From (2) and (3) we have for all ǫ > 0 there is i

= i

(ǫ) such that, for i ≥ i

,

Z

Ω

|x|

|(V

e

− V e

)˜ η

|dx ≤ 4π − ǫ

+ Cǫ (4)

Step 2: Estimate of integral of the second term of the right hand side of (1).

Let Σ

= {x ∈ Ω, d(x, ∂Ω) = ǫ

} and Ω

= {x ∈ Ω, d(x, ∂Ω) ≥ ǫ

}, ǫ > 0. Then, for ǫ small enough, Σ

is hypersurface.

The measure of Ω − Ω

is k

ǫ

≤ µ

(Ω − Ω

) ≤ k

ǫ

.

Remark: for the unit ball B(0, ¯ 1), our new manifold is B(0, ¯ 1 − ǫ

).

(Proof of this fact; let’s consider d(x, ∂Ω) = d(x, z

), z

∈ ∂Ω, this imply that (d(x, z

))

≤ (d(x, z))

for all z ∈ ∂Ω which it is equivalent to (z − z

) · (2x − z − z

) ≤ 0 for all z ∈ ∂Ω, let’s consider a chart around z

and γ(t) a curve in ∂Ω, we have;

(γ(t) − γ(t

) · (2x − γ(t) − γ(t

)) ≤ 0 if we divide by (t − t

) (with the sign and tend t to t

), we have γ

(t

) · (x − γ(t

)) = 0, this imply that x = z

− sν

where ν

is the outward normal of ∂Ω at z

))

With this fact, we can say that S = {x, d(x, ∂Ω) ≤ ǫ} = {x = z

− sν

, z

∈ ∂Ω, −ǫ ≤ s ≤ ǫ}. It is sufficient to work on ∂Ω. Let’s consider a charts (z, D = B (z, 4ǫ

), γ

) with z ∈ ∂Ω such that ∪

B (z, ǫ

) is cover of ∂Ω . One can extract a finite cover (B(z

, ǫ

)), k = 1, ..., m, by the area formula the measure of S ∩ B(z

, ǫ

) is less than a kǫ (a ǫ-rectangle). For the reverse inequality, it is sufficient to consider one chart around one point of the boundary).

We write,

7

Z

Ω

| < ∇(u

− u)|∇˜ η

> |dx = Z

Ω_ǫ3

| < ∇(u

− u)|∇˜ η

> |dx+

+ Z

Ω−Ω_ǫ3

< ∇(u

− u)|∇ η ˜

> |dx. (5)

Step 2.1: Estimate of R

Ω−Ω_ǫ3

| < ∇(u

− u)|∇˜ η

> |dx.

First, we know from the elliptic estimates that ||∇˜ η

||

≤ C

/ǫ

, C

depends on Ω.

We know that (|∇u

|)

is bounded in L

, 1 < q < 2, we can extract from this sequence a subsequence which converge weakly to h ∈ L

. But, we know that we have locally the uniform convergence to |∇u| (by Brezis-Merle theorem), then, h = |∇u| a.e. Let q

be the conjugate of q.

We have, ∀f ∈ L

(Ω)

Z

Ω

|∇u

|f dx → Z

Ω

|∇u|f dx

If we take f = 1

, we have:

for ǫ > 0 ∃ i

= i

(ǫ) ∈ N , i ≥ i

, Z

Ω−Ω_ǫ3

|∇u

| ≤ Z

Ω−Ω_ǫ3

|∇u| + ǫ

.

Then, for i ≥ i

(ǫ),

Z

Ω−Ω_ǫ3

|∇u

| ≤ mes(Ω − Ω

)||∇u||

+ ǫ

= ǫ

(k

||∇u||

+ 1).

Thus, we obtain,

Z

Ω−Ω_ǫ3

| < ∇(u

− u)|∇ η ˜

> |dx ≤ ǫC

(2k

||∇u||

+ 1) (6)

The constant C

does not depend on ǫ but on Ω.

Step 2.2: Estimate of R

Ω_ǫ3

| < ∇(u

− u)|∇˜ η

> |dx.

We know that, Ω

⊂⊂ Ω, and ( because of Brezis-Merle’s interior estimates) u

→ u in C

(Ω

). We have,

||∇(u

− u)||

ǫ3)

≤ ǫ

, for i ≥ i

= i

(ǫ).

8

We write,

Z

Ω_ǫ3

| < ∇(u

− u)|∇˜ η

> |dx ≤ ||∇(u

− u)||

ǫ3)

||∇˜ η

||

≤ C

ǫ for i ≥ i

,

For ǫ > 0, we have for i ∈ N , i ≥ max{i

, i

, i

},

Z

Ω

| < ∇(u

− u)|∇˜ η

> |dx ≤ ǫC

(2k

||∇u||

+ 2) (7)

From (4) and (7), we have, for ǫ > 0, there is i

= i

(ǫ) ∈ N , i

= max{i

, i

, i

} such that,

Z

Ω

|∆[(u

− u)˜ η

]|dx ≤ 4π − ǫ

+ ǫ2C

(2k

||∇u||

+ 2 + C) (8)

We choose ǫ > 0 small enough to have a good estimate of (1).

Indeed, we have:

( −∆[(u

− u)˜ η

] = g

in Ω, (u

− u)˜ η

= 0 on ∂Ω.

with ||g

||

≤ 4π − ǫ

.

We can use Theorem 1 of [7] to conclude that there is q > 1 such that:

Z

Vǫ(x0)

e

dx ≤ Z

Ω

e

dx ≤ C(ǫ, Ω).

where, V

(x

) is a neighberhood of x

in Ω. ¯

Thus, for each x

∈ ∂Ω − {¯ x

, . . . , x ¯

} there is ǫ

> 0, q

> 1 such that:

Z

B(x0,ǫx0)

e

dx ≤ C, ∀ i.

Now, we consider a cutoff function η ∈ C

( R

) such that

η ≡ 1 on B(x

, ǫ

/2) and η ≡ 0 on R

− B(x

, 2ǫ

/3).

We write

9

−∆(u

η) = |x|

V

e

η − 2∇u

· ∇η − u

∆η.

By the elliptic estimates (see [15]) (u

)

is uniformly bounded in W

(V

(x

)) and also, in C

(V

(x

)).

Finaly, we have, for some ǫ > 0 small enough,

||u

||

≤ c

∀ i.

Proof of theorem 1.2:

Without loss of generality, we can assume that 0 is a blow-up point. Since the boundary is an analytic curve γ(t), there is a neighborhood of 0 such that the curve γ can be extend to a holomorphic map such that γ

(0) 6= 0 (series) and by the inverse mapping one can assume that this map is univalent around 0. In the case when the boundary is a simple Jordan curve the domain is simply connected, see [24]. In the case that the domains has a finite number of holes it is conformally equivalent to a disk with a finite number of disks removed, see [17]. Here we consider a general domain. Without loss of generality one can assume that γ(B

) ⊂ Ω and also γ (B

) ⊂ ( ¯ Ω)

and γ (−1, 1) ⊂ ∂Ω and γ is univalent. This means that (B

, γ) is a local chart around 0 for Ω and γ univalent. (This fact holds if we assume that we have an analytic domain, in the sense of Hofmann see [16], (below a graph of an analytic function), we have necessary the condition

∂ Ω = ¯ ∂Ω and the graph is analytic, in this case γ (t) = (t, ϕ(t)) with ϕ real analytic and an example of this fact is the unit disk around the point (0, 1) for example).

By this conformal transformation, we can assume that Ω = B

, the half ball, and ∂

B

is the exterior part, a part which not contain 0 and on which u

converge in the C

norm to u. Let us consider B

, the half ball with radius ǫ > 0. Also, one can consider a C

domain (a rectangle between two half disks) and by charts its image is a C

domain).

We know that:

u

∈ C

( ¯ Ω).

Thus we can use integrations by parts (Gauss-Green-Riemann-Stokes formula). The second Pohozaev identity applied around the blow-up 0 see for example [2, 20, 22] gives :

2(1 + β) Z

B⁺_ǫ

|x|

V

e

dx + Z

B⁺_ǫ

< x|∇V

> |x|

V

e

dx + Z

∂B_ǫ⁺

< ν |x > |x|

V

e

dσ =

= Z

∂⁺Bǫ⁺

g(∇u

)dσ, (9)

with,

g(∇u

) =< ν|∇u

>< x|∇u

> − < x|ν > |∇u

|

2 .

10

also,

2(1 + β) Z

Bǫ⁺

|x|

V e

dx + Z

Bǫ⁺

< x|∇V > |x|

V e

dx + Z

∂Bǫ⁺

< ν|x > |x|

V e

dσ =

= Z

∂⁺Bǫ⁺

g(∇u)dσ, (10)

Thus,

2(1 + β) Z

Bǫ⁺

|x|

V

e

dx − 2(1 + β) Z

Bǫ⁺

|x|

V e

dx+

+ Z

B⁺_ǫ

< x|∇V

> |x|

V

e

dx − Z

B⁺_ǫ

< x|∇V > |x|

V e

dx + o(1) =

= Z

∂⁺Bǫ⁺

g(∇u

) − g(∇u)dσ = o(1),

First, we tend i to infinity after ǫ to 0, we obtain:

ǫ→0

lim lim

i→+∞

2(1 + β) Z

B⁺ǫ

|x|

V

e

dx = 0, (11)

however

Z

γ(B⁺ǫ)

|x|

V

e

dx = Z

∂γ(Bǫ⁺)

∂

u

dσ = α

+ O(ǫ) + o(1) > 0. (12)

which is a contradiction.

Here we used a theorem of Hofmann see [16], which gives the fact that γ (B

) is a Lipschitz domain.

Also, we can see that γ((−ǫ, ǫ)) and γ (∂

B

) are submanifolds.

We start with a Lipschitz domain B

because it is convex and by the univalent and conformal map γ the image of this domain γ(B

) is a Lipschitz domain and thus we can apply the integration by part and here we know the explicit formula of the unit outward normal it is the usual unit outward normal (normal to the tangent space of the boundary which we know explicitly because we have two submanifolds).

In the case of the disk D = Ω, it is sufficient to consider B (0, ǫ) ∩ D which is a Lipschitz domain because it is convex (and not necessarily γ(B

)).

There is a version of the integration by part which is the Green-Riemann formula in dimension 2 on a domain Ω. This formula holds if we assume that there is a finite number of points y

, ..., y

such that

∂Ω − (y

, ..., y

) is a C

manifold and for C

tests functions, see [2], for the Gauss-Green-Riemann-Stokes formula, for C

domains with singular points (here a finite number of singular points).

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Remark 1: Note that a monograph of Droniou contain a proof of all fact about Sobolev spaces (with Strong Lipschitz property) with only weak Lipschitz property (Lipschitz-Charts), we start with Strong Lip- schitz property and by γ we have weak Lipschtz property.

Remarks about the conformal map :

1-It sufficient to prove that γ

((−ǫ, ǫ)) = ∂Ω∩˜ γ

(B

) = ∂Ω∩ γ ˜

(B

)∩{|abscissa| < ǫ}, for ǫ > 0 small enough. Where ˜ γ

is the holomorphic extension of γ

(t) = t + iϕ(t). For this, we argue by contradiction, we have for z

∈ B

, γ ˜

(z

) = (t

, ϕ(t

)) for |t

| ≥ ǫ. Because γ ˜

is injective on B

and γ ˜

= γ

= t+iϕ(t) on the real axis, we have necessirely |t

| ≥ 1. But, by continuity |˜ γ

(z

)| → 0 because z

→ 0. And, we use the fact that |˜ γ

(z

)| = |(t

, ϕ(t

))| ≥ |t

| ≥ 1, to have a contradiction.) (This means that for a small radius when the graph go out from the ball, it never retruns to the ball). (This fact imply that, when we have a curve which cut ∂Ω in γ ˜

(B

) then the point have an abscissa such that |abscissa| < ǫ. This fact (by a contradiction with the fact ∂ Ω = ¯ ∂Ω and consider paths), imply that the image of the upper part of the ball is one side of the curve and the image of the lower part is in the other side of the curve.

2- Also, we can consider directly the coordinate T and change the function u(z) → u(T ) by z = T /a

+ x

.

Set: ψ(λ

, λ

) → M ∈ Ω such that, −−→

x

M = λ

i

+ λ

j

where (i

, j

) is a basis such that i

= e

i

, j

= e

j

. And, ϕ(x

, x

) → M such that, −−→

OM = x

i

+ x

i

the canonical basis (i

, j

).

Then, we have two charts ϕ and ψ and the complex affix T

= λ

+ iλ

and z

= x

+ ix

are such that (transition map):

T

/e

+ x

= z

= ϕ

oψ(λ

, λ

), We have:

∂

= cos θ∂

+ sin θ∂

,

∂

= − sin θ∂

+ cos θ∂

,

Thus, the metric in the chart ψ or coordinates (λ

, λ

) is : g

= δ

and the Laplacian in the two charts, ψ and ϕ are the usual Laplacian ∂

+ ∂

.

We write:

∆u(M) = ∆

(uoψ(λ

, λ

))

And then we apply the conformal map ˜ γ

which send the affix T

, M in a neighborhood of x

∈ ∂Ω to B

with the fact that send T

, M ∈ ∂Ω to the real axis (−ǫ, ǫ) and the other parts of Ω and Ω ¯

.

We have: ψ : (λ

, λ

) → M and ˜ γ

: (λ

, λ

) → (µ

, µ

). Then the map considerded is: ψo˜ γ

. It is a chart. Indeed the first chart is: ψog

where g is g : (λ

, λ

) → [λ

, λ

+ ϕ

(λ

)] with ϕ

the "graph map"

as in the defiinition of ∂Ω.

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Then the transition map is:

go˜ γ

is smooth from a neighborhood of 0 to a neighborhood of 0 and send the upper part of a ball (lower part) to an upper part of a domain (lower part). And then ψo˜ γ

is a chart with the property that γ ˜

is conformal.

Remark that: ψ and ϕ are charts for Ω and ψ is almost chart for ∂Ω. But ψo˜ γ

and ψog

are the "real"

charts of the boundary.

3-We can remark that a definition of C

, k ≥ 1 domain, is equivalent to a definition of a submanifold with the condition ∂ Ω = ¯ ∂Ω or Ω = Ω. ˙¯

Remark 2: about a variational formulation.

we consider a solutions in the sense of distribution. By the same argument (in the proof of the maximum principle obtained by Kato’s inequality W

is sufficient), see [6], we prove that the solutions are in the sense C

of Agmon, see [1]. Also, we have corollary 1 of [7]. We use Agmon’s regularity theorem. We return to the usual variational formulation in W

and thus we have the estimate of [8] or by Stampacchia duality theorem in W

.

R

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DEPARTEMENT DEMATHEMATIQUES, UNIVERSITEPIERRE ETMARIECURIE, 2PLACEJUSSIEU, 75005, PARIS, FRANCE. E-mail address:samybahoura@yahoo.fr, samybahoura@gmail.com

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