A compactness result for an elliptic equation in dimension 2.

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A compactness result for an elliptic equation in dimension 2.

Samy Skander Bahoura

To cite this version:

Samy Skander Bahoura. A compactness result for an elliptic equation in dimension 2.. 2018. �hal-

01101650v8�

A COMPACTNESS RESULT FOR AN ELLIPTIC EQUATION IN DIMENSION 2.

SAMY SKANDER BAHOURA

ABSTRACT. We give a blow-up behavior for the solutions of an elliptic equation under some conditions. We also derive a compactness creterion for this equation.

Mathematics Subject Classification: 35J60 35B45 35B50

Keywords: blow-up, boundary, elliptic equation, a priori estimate, Lipschitz condition, starshaped domains.

1. I

M

R

Let us consider the following operator:

L

:= ∆ + ǫ(x

∂

+ x

∂

) = div[a

(x)∇]

a

(x) with a

(x) = e

2 2

. We consider the following equation:

(P

)

( −∆u − ǫ(x

∂

u + x

∂

u) = −L

u = V e

in Ω ⊂ R

, u = 0 in ∂Ω.

Here, we assume that:

Ω starshaped, and,

u ∈ W

(Ω), e

∈ L

(Ω), 0 ≤ V ≤ b, 1 ≥ ǫ ≥ 0.

When ǫ = 0 the previous equation was studied by many authors with or without the boundary condition, also for Riemann surfaces see [1-20] where one can find some existence and compactness results. Also we have a nice formulation in the sens of the distributions of this Problem in [7].

Among other results, we can see in [6] the following important Theorem,

Theorem A (Brezis-Merle [6]) If (u

)

and (V

)

are two sequences of functions relative to the problem (P

) with ǫ = 0 and,

0 < a ≤ V

≤ b < +∞

then it holds,

sup

K

u

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

We can find in [6] an interior estimate if we assume a = 0 but we need an assumption on the integral of e

, namely:

1

Theorem B(Brezis-Merle [6]).For (u

)

and (V

)

two sequences of functions relative to the problem (P

) with,

0 ≤ V

≤ b < +∞ and Z

Ω

e

dy ≤ C, then it holds;

sup

K

u

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

The condition R

Ω

e

dy ≤ C is a necessary condition in the Problem (P

) as showed by the following counterexample for ǫ = 0:

Theorem C (Brezis-Merle [6]).There are two sequences (u

)

and (V

)

of the problem (P

) with;

0 ≤ V

≤ b < +∞, 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 (Theorem 1 of [6]) obtained by an approximation argument and they 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 [5] for a topic about the weak form of the maximum principle.

Note that for the problem (P

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

In [8] Chen-Li have proved that if a = 0 and R

Ω

e

is uniformly bounded and ∇ log V

is uniformly bounded then (u

)

is bounded near the boundary and we have directly the compactness result for the prob- lem (P

). Ma-Wei in [17] extend this result in the case where a > 0.

When ǫ = 0 and if we assume V more regular we can have another type of estimates called sup + inf type inequalities. It was proved by Shafrir see [19] that, if (u

)

, (V

)

are two sequences of functions solutions of the Problem (P

) without assumption on the boundary and 0 < a ≤ V

≤ b < +∞ then it holds:

C a b

sup

K

u

+ inf

Ω

u

≤ c = c(a, b, K, Ω).

We can see in [9] an explicit value of C a b

= r a

b . In his proof, Shafrir has used the blow-up function, the Stokes formula and an isoperimetric inequality see [2]. For Chen-Lin, they have used the blow-up analysis combined with some geometric type inequality for the integral curvature see [9].

Now, if we suppose (V

)

uniformly Lipschitzian with A its Lipschitz constant then C(a/b) = 1 and c = c(a, b, A, K, Ω) see Brezis-Li-Shafrir [4]. This result was extended for H ¨olderian sequences (V

)

by

2

Chen-Lin see [9]. Also have in [15], an extension of the Brezis-Li-Shafrir result to compact Riemannian surfaces without boundary. One can see in [16] 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.

Also one can see in [10] refined estimates near the isolated blow-up points and the bubbling behavior of the blow-up sequences.

Here we give the behavior of the blow-up points on the boundary and a proof of a compactness result with Lipschitz condition. Note that our problem is an extension of the Brezis-Merle Problem.

The Brezis-Merle Problem (see [6]) is:

Problem. Suppose that V

→ V in C

( ¯ Ω) with 0 ≤ V

. Also, we consider a sequence of solutions (u

) of (P

) relative to (V

) such that,

Z

Ω

e

dx ≤ C, is it possible to have:

||u

||

≤ C = C(b, C, V, Ω)?

Here we give blow-up analysis on the boundary when V (similar to the prescribed curvature when ǫ = 0) are nonegative and bounded, and on the other hand, if we add the assumption that these functions (similar to the prescribed cruvature) are uniformly Lipschitzian, we have a compactness of the solutions of the problem (P

) for ǫ small enough. (In particular we can take a sequence of ǫ

tending to 0):

For the behavior of the blow-up points on the boundary, the following condition is sufficient, 0 ≤ V

≤ b,

The condition V

→ V in C

( ¯ Ω) is not necessary. But for the compactness of the solutions we add the following condition:

||∇V

||

≤ A.

Our main results are:

Theorem 1.1. Assume that max

u

→ +∞, where (u

) are solutions of the probleme (P

) with:

0 ≤ V

≤ b, and Z

Ω

e

dx ≤ C, ǫ

→ 0,

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

, . . . , x

∈ ∂Ω, such that,

∂

u

→ ∂

u +

N

X

j=1

α

δ

, α

≥ 4π, in the sens of measures on ∂Ω.

u

→ u in C

( ¯ Ω − {x

, . . . , x

}).

Theorem 1.2. Assume that (u

) are solutions of (P

) relative to (V

) with the following conditions:

0 ≤ V

≤ b, ||∇V

||

≤ A and Z

Ω

e

≤ C, ǫ

→ 0.

Then we have:

||u

||

≤ c(b, A, C, Ω),

3

2. P

Proof of theorem 1.1:

First remark that:

( −∆u

= ǫ

(x

∂

u

+ x

∂

u

) + V

e

∈ L

(Ω) in Ω ⊂ R

,

u

= 0 in ∂Ω.

and,

u

∈ W

(Ω).

By the corollary 1 of Brezis-Merle see [6] we have e

∈ L

(Ω) for all k > 2 and the elliptic estimates of Agmon and the Sobolev embedding see [1] imply that:

u

∈ W

(Ω) ∩ C

( ¯ Ω).

Also remark that, we have for two positive constants C

= C(q, Ω) and C

= C

(Ω) (see [7]) :

||∇u

||

≤ C

||∆u

||

≤ (C

+ ǫC

||∇u

||

), ∀ i and 1 < q < 2.

Thus, if ǫ > 0 is small enough and by the Holder inequality, we have the following estimate:

||∇u

||

≤ C

, ∀ i and 1 < q < 2.

Step 1: interior estimate

First remark that, if we consider the following equation:

( −∆w

= ǫ

(x

∂

u

+ x

∂

u

) ∈ L

, 1 < q < 2 in Ω ⊂ R

,

w

= 0 in ∂Ω.

If we consider v

the Newtonnian potential of ǫ

(x

∂

u

+ x

∂

u

), we have:

v

∈ C

( ¯ Ω), ∆(w

− v

) = 0.

By the maximum principle w

− v

∈ C

( ¯ Ω) and thus w

∈ C

( ¯ Ω).

Also we have by the elliptic estimates that w

∈ W

⊂ L

, and we can write the equation of the Problem as:

4

( −∆(u

− w

) = ˜ V

e

in Ω ⊂ R

, u

− w

= 0 in ∂Ω.

with,

0 ≤ V ˜

= V

e

≤ ˜ b, Z

Ω

e

≤ C. ˜ We apply the Brezis-Merle theorem to u

− w

to have:

u

− w

∈ L

(Ω), and, thus:

u

∈ L

(Ω).

Step2: boundary estimate

Set ∂

u

the inner derivative of u

. By the maximum principle ∂

u

≥ 0.

We have:

Z

∂Ω

∂

u

dσ ≤ C.

We have the existence of a nonnegative Radon measure µ such that, Z

∂Ω

∂

u

ϕdσ → µ(ϕ), ∀ ϕ ∈ C

(∂Ω).

We take an x

∈ ∂Ω such that, µ(x

) < 4π. Set B (x

, ǫ) ∩ ∂Ω := I

. We choose a function η

such that,



 

 



 

 



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 ⊂ R

,

˜

η

= η

in ∂Ω.

5

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 .

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 diffeo- morphism.



 

 



 

 



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, because 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.

As we said in the beguening, see also [3, 7, 13, 20], we have:

||∇u

||

≤ C

, ∀ i and 1 < q < 2.

6

We deduce from the last estimate that, (u

) converge weakly in W

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

Ω

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 = V e

∈ L

(Ω) in Ω ⊂ R

,

u = 0 in ∂Ω.

According to the corollary 1 of Brezis-Merle result, see [6], we have e

∈ L

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

(Ω) ∩ C

( ¯ Ω).

We denote by f · g the inner product of any two vectors f and g of R

. We can write,

−∆((u

− u)˜ η

) = (V

e

− V e

)˜ η

− 2∇(u

− u) · ∇˜ η

+ ǫ

(∇u

· x)˜ η

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

V e

η ˜

dx = Z

∂Ω

∂

uη

≤ Cǫ = O(ǫ) (2)

We have,

( −∆u

− ǫ

∇u

· x = V

e

in Ω ⊂ R

, u = 0 in ∂Ω.

We use the Green formula between u

and η ˜

to have:

Z

Ω

V

e

η ˜

dx = Z

∂Ω

∂

u

η

dσ − ǫ

Z

Ω

(∇u

· x)˜ η

=

= Z

∂Ω

∂

u

η

dσ + o(1) → µ(η

) ≤ µ(J

) ≤ 4π − ǫ

, ǫ

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

such that, for i ≥ i

,

Z

Ω

|(V

e

− V e

)˜ η

|dx ≤ 4π − ǫ

+ Cǫ (4) Step 2.1: Estimate of integral of the second term of the right hand side of (1).

7

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 an hypersurface.

The measure of Ω − Ω

is k

ǫ

≤ meas(Ω − Ω

) = 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,

Z

Ω

|∇(u

− u) · ∇ η ˜

|dx = Z

Ω_ǫ3

|∇(u

− u) · ∇˜ η

|dx + Z

Ω−Ω_ǫ3

|∇(u

− u) · ∇˜ η

|dx. (5)

Step 2.1.1: 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 the Brezis-Merle’s 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

ǫ3

, 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),

8

Z

Ω−Ω_ǫ3

|∇u

| ≤ meas(Ω − Ω

)||∇u||

+ ǫ

= ǫ

(k

||∇u||

+ 1) = O(ǫ

).

Thus, we obtain, Z

Ω−Ω_ǫ3

|∇(u

− u) · ∇˜ η

|dx ≤ ǫC

(2k

||∇u||

+ 1) = O(ǫ) (6) The constant C

does not depend on ǫ but on Ω.

Step 2.1.2: 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

. We write,

Z

Ω_ǫ3

|∇(u

− u) · ∇˜ η

|dx ≤ ||∇(u

− u)||

||∇˜ η

||

= C

ǫ = O(ǫ) for i ≥ i

,

For ǫ > 0, we have for i ∈ N , i ≥ i

, Z

Ω

|∇(u

− u) · ∇˜ η

|dx ≤ ǫC

(2k

||∇u||

+ 2) = O(ǫ) (7) From (4) and (7), we have, for ǫ > 0, there is i

such that, i ≥ i

,

Z

Ω

|∆[(u

− u)˜ η

]|dx ≤ 4π − ǫ

+ ǫ2C

(2k

||∇u||

+ 2 + C) = 4π − ǫ

+ O(ǫ) (8) We choose ǫ > 0 small enough to have a good estimate of (1).

Indeed, we have:

( −∆[(u

− u)˜ η

] = g

in Ω ⊂ R

, (u

− u)˜ η

= 0 in ∂Ω.

with ||g

||

≤ 4π − ǫ

/2.

We can use Theorem 1 of [6] to conclude that there are q ≥ q > ˜ 1 such that:

9

Z

Vǫ(x0)

e

dx ≤ Z

Ω

e

dx ≤ C(ǫ, Ω).

where, V

(x

) is a neighberhooh of x

in Ω. Here we have used that in a neighborhood of ¯ x

by the elliptic estimates, 1 − Cǫ ≤ η ˜

≤ 1.

Thus, for each x

∈ ∂Ω − {¯ x

, . . . , x ¯

} there is ǫ

> 0, q

> 1 such that:

Z

B(x0,ǫ0)

e

dx ≤ C, ∀ i.

By the elliptic estimate see [14] we have:

||u

||

≤ c

∀ i.

We have proved that, there is a finite number of points x ¯

, . . . , x ¯

such that the squence (u

)

is locally uniformly bounded in C

, (θ > 0) on Ω ¯ − {¯ x

, . . . , x ¯

}.

Proof of theorem 1.2:

The Pohozaev identity gives :

Z

∂Ω

1

2 (x · ν)(∂

u

)

dσ + ǫ Z

Ω

(x · ∇u

)

dx + Z

∂Ω

(x · ν)V

e

dσ = Z

Ω

(x · ∇V

+ 2V

)e

dx

We use the boundary condition and the fact that Ω is starshaped and the fact that ǫ > 0 to have that:

Z

∂Ω

(∂

u

)

dx ≤ c

(b, A, C, Ω). (9) Thus we can use the weak convergence in L

(∂Ω) to have a subsequence ∂

u

, such that:

Z

∂Ω

∂

u

ϕdx → Z

∂Ω

∂

uϕdx, ∀ ϕ ∈ L

(∂Ω), Thus, α

= 0, j = 1, . . . , N and (u

) is uniformly bounded.

Remark 1: Note that if we assume the open set bounded starshaped and V

uniformly Lipschitzian and between two positive constants we can bound, by using the inner normal derivative R

Ω

e

.

Remark 2: If we assume the open set bounded starshaped and ∇ log V

uniformly bounded, by the previous Pohozaev identity (we consider the inner normal derivative) one can bound R

Ω

V

e

uniformly.

10

Remark 3: One can consider the problem on the unit ball and an ellipse. These two problems are differents, because:

1) if we use a linear transformation, (y

, y

) = (x

/a, x

/b), the Laplcian is not invariant under this map.

2) If we use a conformal transformation, by a Riemann theorem, the quantity x · ∇u is not invariant under this map.

We can not use, after using those transofmations, the Pohozaev identity.

3. A

We start with the notation of the counterexample of Brezis and Merle.

The domain Ω is the unit ball centered in (1, 0).

Lets consider z

(obtained by the variational method), such that:

−∆z

− ǫ

x · ∇z

= −L

(z

) = f

. With Dirichlet condition. By the regularity theorem we have z

∈ C

( ¯ Ω).

We have:

||f

||

= 4πA.

Thus by the duality theorem of Stampacchia or Brezis-Strauss, we have:

||∇z

||

≤ C

, 1 ≤ q < 2.

We solve:

−∆w

= ǫ

x · ∇z

, With Dirichlet condition.

By the elliptic estimates, w

∈ C

( ¯ Ω) and w

∈ C

( ¯ Ω) uniformly.

By the maximum principle we have:

z

− w

≡ u

. Where u

is the function of the counterexemple of Brezis Merle.

We write:

11

−∆z

− ǫ

x · ∇z

= f

= V

e

. Thus, we have:

Z

Ω

e

≤ C

, and 0 ≤ V

≤ C

, and,

z

(a

) ≥ u

(a

) − C

→ +∞, a

→ O.

R

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DEPARTEMENT DEMATHEMATIQUES, UNIVERSITEPIERRE ETMARIECURIE, 2PLACEJUSSIEU, 75005, PARIS, FRANCE. E-mail address:[email protected], [email protected]

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