A compactness result for solutions to an equation with C0 condition.
Samy Skander Bahoura∗
Equipe d’Analyse Complexe et Géométrie.
Université Pierre et Marie Curie, 75005 Paris, France.
Abstract
We give a blow-up behavior for solutions to a variational problem with continuous regular weight (not Lipschitz and not Hölderian in one point) and Dirichlet condition. An application, we have a compactness of the solutions to this Problem, with regular weight and Lipschitz condition.
Keywords: Blow-up, boundary, a priori estimate, regular weight,C0 condition.
MSC: 35J05, 35J60, 35B44, 35B45
1 Introduction and Main Results
We set∆ =∂11+∂22 on an analytic domainΩ⊂R2. We consider the following equation:
(P)
−∆u= 1
−log|x|
2d
V eu in Ω⊂R2,
u= 0 in ∂Ω.
Here:
d=diam(Ω), 0∈∂Ω, and,
0≤V ≤b <+∞, 1
−log|x|
2d
eu∈L1(Ω), u∈W01,1(Ω).
The previous equation was studied by many authors, with or without the boundary condition, also for Riemannian surfaces, see [1-14], we can find some existence and compactness results.
Among other results, we can see in [7] the following important Theorem, Theorem.(Brezis-Merle [7]).If(ui)iand(Vi)iare two sequences of functions relatively to the problem (P)with,0< a≤Vi ≤b <+∞, then, for all compact setK ofΩ,
∗e-mails: [email protected], [email protected]
sup
K
ui≤c=c(a, b, K,Ω).
If we assumeV with more regularity, we can have another type of estimates, asup + inftype inequalities. It was proved by Shafrir see [14], that, if(ui)i,(Vi)i
are two sequences of functions solutions of the previous equation without as- sumption on the boundary and, 0 < a ≤ Vi ≤ b < +∞, then we have the following interior estimate:
Ca b
sup
K
ui+ inf
Ω ui≤c=c(a, b, K,Ω).
Now, if we suppose(Vi)i uniformly Lipschitzian with A the Lipschitz con- stant, then,C(a/b) = 1andc=c(a, b, A, K,Ω), see [5].
Here we give the behavior of the blow-up points on the boundary and a proof of a Problem with Lipschitz condition.
Here, we write an extension of Brezis-Merle Problem (see [7]) is:
Problem. Suppose thatVi→V inC0( ¯Ω), with, 0≤Vi. Also, we consider a sequence of solutions(ui)of(P)relatively to(Vi)such that,
Z
Ω
1
−log|x|
2d
euidx≤C,
is it possible to have:
||ui||L∞ ≤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 the prescribed curvature are uniformly Lipschitzian. For the behavior of the blow-up points on the boundary, the following condition is enough,
0≤Vi≤b, The conditionVi →V inC0( ¯Ω)is not necessary.
But for the proof for the Brezis- Merle type problem we assume that:
||∇Vi||L∞ ≤A.
We have the following caracterization of the behavior of the blow-up points on the boundary.
Theorem 1.1 Assume thatmaxΩui→+∞, Where(ui)are solutions of the probleme (P) with:
0≤Vi≤b, and Z
Ω
1
−log|x|
2d
euidx≤C, ∀i,
then; after passing to a subsequence, there is a finction u, there is a number N ∈Nand there are N pointsx1, x2, . . . , xN ∈∂Ω, such that,
∂νui→∂νu+
N
X
j=1
αjδxj, αj ≥4π, weakly in the sense of measures
and,
ui→u in Cloc1 ( ¯Ω− {x1, . . . , xN}).
In the following theorem, we have a compactness result which concern the problem(P).
Theorem 1.2 Assume that (ui)are solutions of (P) relatively to(Vi) with the following conditions:
d=diam(Ω), 0∈∂Ω,
0≤Vi≤b, ||∇Vi||L∞≤A, and Z
Ω
1
−log|x|
2d
eui ≤C,
We have,
||ui||L∞≤c(b, A, C,Ω),
2 Proof of the theorems
Proof of theorem 1.1:
We have,
ui∈W01,1(Ω),
By [7],eui ∈Lk, ∀k≥1. By the elliptic estimates:
ui∈W2,k(Ω)∩C1,ǫ( ¯Ω), ǫ >0.
We denote∂νuithe inner derivative ofui. By the maximum principle∂νui≥ 0. By the Stokes formula:
Z
∂Ω
∂νuidσ≤C,
Thus, (using the weak convergence in the space of Radon measures), we have the existence of a non-negative Radon measureµsuch that,
Z
∂Ω
∂νuiφdσ→µ(φ), ∀ φ∈C0(∂Ω).
We take anx0 ∈∂Ωsuch that,µ(x0)<4π. Without loss of generality, we can assume that the following curve,B(x0, ǫ)∩∂Ω :=Iǫ is an interval.(In this
case, it is more simple to construct the following test functionηǫ). We choose a function ηǫ such that,
ηǫ≡1, on Iǫ, 0< ǫ < δ/2, ηǫ≡0, outside I2ǫ,
0≤ηǫ≤1,
||∇ηǫ||L∞(I2ǫ)≤C0(Ω, x0)
ǫ .
We take aη˜ǫsuch that,
(−∆˜ηǫ= 0 inΩ
˜
ηǫ=ηǫ on∂Ω.
We take a cutoff functionη0 inB(0,2)or B(x0,2):
We setηǫ(x) =η0(|x−x0|/ǫ)in the case of the unit disk it is sufficient.
We use a chart[f, B1(0)]withf(0) =x0.
We can take: µǫ(x) = η0(x/ǫ) and ηǫ(y) = µǫ(f−1(y)), we extend it by 0 outside f(B1(0)). We have f(B1(0)) = D1(x0), f(Bǫ(0)) = Dǫ(x0) and f(Bǫ+) =D+ǫ(x0)withf andf−1smooth diffeomorphism.
ηǫ≡1, on a the connected set Jǫ=f(Iǫ), 0< ǫ < δ/2, ηǫ≡0, outside Jǫ′=f(I2ǫ),
0≤ηǫ≤1,
||∇ηǫ||L∞(Jǫ′)≤C0(Ω, x0)
ǫ .
And, H1(Jǫ′) ≤ C1H1(I2ǫ) = C14ǫ, since f is Lipschitz. Here H1 is the Hausdorff measure.
We solve the Dirichlet Problem:
(∆¯ηǫ= ∆ηǫ in Ω⊂R2,
¯
ηǫ= 0 in ∂Ω.
and finaly we setη˜ǫ =−¯ηǫ+ηǫ. Also, by the maximum principle and the elliptic estimates we have :
||∇˜ηǫ||L∞ ≤C(||ηǫ||L∞+||∇ηǫ||L∞+||∆ηǫ||L∞)≤ C1
ǫ2, withC1 depends onΩ.
We use the following estimate, see [8],
||∇ui||Lq ≤Cq, ∀iand 1< q <2.
We deduce from the last estimate that, (ui) converge weakly in W01,q(Ω), almost everywhere to a function u ≥ 0 and R
Ω
1
−log|x|
2d
eu < +∞ (by Fatou lemma). Also, Vi weakly converge to a nonnegative function V in L∞. The function uis inW01,q(Ω)solution of :
−∆u= 1
−log|x|
2d
V eu∈L1(Ω) in Ω
u= 0 on∂Ω,
As in the corollary 1 of Brezis-Merle result, see [7], we haveeku∈L1(Ω), k >
1. By the elliptic estimates, we haveu∈C1( ¯Ω).
We can write,
−∆((ui−u)˜ηǫ) = 1
−log|x|
2d
(Vieui−V eu)˜ηǫ+ 2<∇(ui−u)|∇˜ηǫ> . (1)
We use the interior esimate of Brezis-Merle, see [7],
Step 1: Estimate of the integral of the first term of the right hand side of (1).
We use the Green formula betweenη˜ǫ andu, we obtain, Z
Ω
1
−log|x|
2d
V euη˜ǫdx= Z
∂Ω
∂νuηǫ≤C14ǫ||∂νu||L∞ =Cǫ (2) We have,
−∆ui= 1
−log|x|
2d
Vieui in Ω
ui = 0 on∂Ω,
We use the Green formula betweenui andη˜ǫ to have:
Z
Ω
1
−log|x|
2d
Vieuiη˜ǫdx= Z
∂Ω
∂νuiηǫdσ→µ(ηǫ)≤µ(Jǫ′)≤4π−ǫ0, ǫ0>0 (3)
From (2) and (3) we have for all ǫ > 0 there is i0 = i0(ǫ) such that, for i≥i0,
Z
Ω
1
−log|x|
2d
|(Vieui−V eu)˜ηǫ|dx≤4π−ǫ0+Cǫ (4)
Step 2: Estimate of integral of the second term of the right hand side of(1).
LetΣǫ={x∈Ω, d(x, ∂Ω) =ǫ3} and Ωǫ3 ={x∈Ω, d(x, ∂Ω)≥ǫ3},ǫ >0.
Then, forǫsmall enough,Σǫ is hypersurface.
The measure ofΩ−Ωǫ3 isk2ǫ3≤µL(Ω−Ωǫ3)≤k1ǫ3.
Remark: for the unit ballB(0,¯ 1), our new manifold isB(0,¯ 1−ǫ3).
(Proof; let’s considerd(x, ∂Ω) =d(x, z0), z0∈∂Ω, this imply that(d(x, z0))2≤ (d(x, z))2 for allz∈∂Ωwhich it is equivalent to(z−z0)·(2x−z−z0)≤0 for allz∈∂Ω, let’s consider a chart aroundz0 andγ(t)a curve in∂Ω, we have;
(γ(t)−γ(t0)·(2x−γ(t)−γ(t0))≤0 if we divide by(t−t0)(with the sign and tendt tot0), we haveγ′(t0)·(x−γ(t0)) = 0, this imply thatx=z0−sν0
whereν0is the outward normal of∂Ωatz0))
With this fact, we can say thatS ={x, d(x, ∂Ω)≤ǫ}={x=z0−sνz0, z0∈
∂Ω, −ǫ ≤ s ≤ ǫ}. It is sufficient to work on ∂Ω. Let’s consider a charts (z, D=B(z,4ǫz), γz)withz∈∂Ωsuch that∪zB(z, ǫz)is cover of∂Ω. One can extract a finite cover(B(zk, ǫk)), k= 1, ..., m, by the area formula the measure ofS∩B(zk, ǫk)is less than akǫ(aǫ-rectangle). For the reverse inequality, it is sufficient to consider one chart around one point of the boundary).
We write, Z
Ω
|<∇(ui−u)|∇˜ηǫ>|dx= Z
Ωǫ3
|<∇(ui−u)|∇˜ηǫ>|dx+
+ Z
Ω−Ωǫ3
<∇(ui−u)|∇˜ηǫ>|dx. (5)
Step 2.1:Estimate of R
Ω−Ωǫ3|<∇(ui−u)|∇˜ηǫ>|dx.
First, we know from the elliptic estimates that ||∇˜ηǫ||L∞ ≤ C1/ǫ2, C1 de- pends onΩ
We know that(|∇ui|)iis bounded inLq,1< q <2, we can extract from this sequence a subsequence which converge weakly toh∈Lq. But, we know that we have locally the uniform convergence to|∇u|(by Brezis-Merle theorem), then, h=|∇u|a.e. Letq′ be the conjugate ofq.
We have,∀f ∈Lq′(Ω) Z
Ω
|∇ui|f dx→ Z
Ω
|∇u|f dx
If we takef = 1Ω−Ωǫ3, we have:
forǫ >0∃i1=i1(ǫ)∈N, i≥i1, Z
Ω−Ωǫ3
|∇ui| ≤ Z
Ω−Ωǫ3
|∇u|+ǫ3.
Then, fori≥i1(ǫ), Z
Ω−Ωǫ3
|∇ui| ≤meas(Ω−Ωǫ3)||∇u||L∞+ǫ3=Cǫ3. Thus, we obtain,
Z
Ω−Ωǫ3
|<∇(ui−u)|∇˜ηǫ>|dx≤ǫC1(2k1||∇u||L∞+ 1) (6) The constantC1 does not depend onǫbut onΩ.
Step 2.2:Estimate of R
Ωǫ3|<∇(ui−u)|∇˜ηǫ>|dx.
We know that,Ωǫ⊂⊂Ω, and ( because of Brezis-Merle’s interior estimates) ui→uinC1(Ωǫ3). We have,
||∇(ui−u)||L∞(Ωǫ3)≤ǫ3,fori≥i3=i3(ǫ).
We write,
Z
Ωǫ3
|<∇(ui−u)|∇˜ηǫ>|dx≤ ||∇(ui−u)||L∞(Ωǫ3)||∇˜ηǫ||L∞ ≤C1ǫfori≥i3,
Forǫ >0, we have fori∈N,i≥max{i1, i2, i3}, Z
Ω
|<∇(ui−u)|∇˜ηǫ>|dx≤ǫC1(2k1||∇u||L∞+ 2) (7) From (4) and (7), we have, for ǫ > 0, there is i3 = i3(ǫ) ∈ N, i3 = max{i0, i1, i2} such that,
Z
Ω
|∆[(ui−u)˜ηǫ]|dx≤4π−ǫ0+ǫ2C1(2k1||∇u||L∞+ 2 +C) (8) We chooseǫ >0small enough to have a good estimate of(1).
Indeed, we have:
(−∆[(ui−u)˜ηǫ] =gi,ǫ in Ω, (ui−u)˜ηǫ= 0 on ∂Ω.
with||gi,ǫ||L1(Ω)≤4π−ǫ0.
We can use Theorem 1 of [7] to conclude that there isq≥q >˜ 1such that:
Z
Vǫ(x0)
eq|u˜ i−u|dx≤ Z
Ω
eq|ui−u|˜ηǫdx≤C(ǫ,Ω).
where,Vǫ(x0)is a neighberhood ofx0 inΩ.¯
Thus, for eachx0∈∂Ω− {x¯1, . . . ,x¯m} there isǫx0>0, qx0>1 such that:
Z
B(x0,ǫx0)
eqx0uidx≤C, ∀ i.
Now, we consider a cutoff functionη∈C∞(R2)such that:
η≡1 on B(x0, ǫx0/2) and η≡0 on R2−B(x0,2ǫx0/3).
We write,
−∆(uiη) = 1
−log|x|
2d
Vieuiη−2<∇ui|∇η >−ui∆η.
By the elliptic estimates,(uiη)iis uniformly bounded inW2,q1(Ω)and also, in C1( ¯Ω).
Finaly, we have, for someǫ >0small enough,
||ui||C1,θ[B(x0,ǫ)]≤c3 ∀ i. (9) We have proved that, there is a finite number of pointsx¯1, . . . ,x¯msuch that the squence(ui)i is locally uniformly bounded inΩ¯− {x¯1, . . . ,x¯m}.
And, finaly, we have:
∂νui→∂νu+
N
X
j=1
αjδxj, αj ≥4πweakly in the sense of measures. (10) Proof of theorem 1.2:
Without loss of generality, we can assume that 0 = x1 is a blow-up point (either, we are in the regular case). Also, by a conformal transformationγ, we can assume that Ω =B1+, the half ball, and∂+B1+ is the exterior part, a part which not contain 0 and on which ui converge in the C1 norm to u. Let us considerB+ǫ, the half ball with radiusǫ >0.
We know that:
ui∈W2,k(Ω)∩C1,ǫ( ¯Ω).
Thus we can use integrations by parts (Gauss-Green-Riemann-Stokes for- mula). The second Pohozaev identity applied around the blow-up0 =x1gives:
Z
B+ǫ
∆ui(x· ∇ui)dx= Z
∂+B+ǫ
g(∇ui)dσ, (11)
with,
g(∇ui) = (ν· ∇ui)(x· ∇ui)−x·ν|∇ui|2 2 . After integration by parts, we obtain:
Z
Bǫ+
1
−log|x|
2d
Vieuidx+
Z
Bǫ+
< x|∇Vi > 1
−log|x|
2d
Vieuidx+
Z
∂B+ǫ
1
−log|x|
2d
Vieui(x·ν) =
= Z
∂+Bǫ+
g(∇ui)dσ, (12)
Z
Bǫ+
1
−log|x|
2d
V eudx+
Z
B+ǫ
< x|∇V > 1
−log|x|
2d
V eudx+
Z
∂Bǫ+
1
−log|x|
2d
V eu(x·ν) =
= Z
∂+B+ǫ
g(∇u)dσ, (13)
Thus,
Z
B+ǫ
1
−log|x|
2d
Vieuidx− Z
B+ǫ
1
−log|x|
2d
V eudx+
+ Z
Bǫ+
< x|∇Vi> 1
−log|x|
2d
Vieuidx− Z
B+ǫ
< x|∇V > 1
−log|x|
2d
V eudx+
+ Z
∂B+ǫ
1
−log|x|
2d
Vieui(x·ν)− Z
∂B+ǫ
1
−log|x|
2d
V eu(x·ν) =
= Z
∂+B+ǫ
g(∇ui)−g(∇u)dσ=o(1),
First, we tendito infinity afterǫto 0, we obtain:
ǫ→0lim lim
i→+∞
Z
B+ǫ
1
−log|x|
2d
Vieuidx= 0, (14)
But,
Z
γ(Bǫ+)
1
−log|x|
2d
Vieuidx= Z
∂γ(Bǫ+)
∂νui+o(ǫ) +o(1)→α1>0.
A contradiction.
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