arXiv:1405.7734v2 [math.AP] 10 Dec 2017
The effect of a discontinuous weight for a critical
Sobolev problem
Rejeb Hadiji∗ Sami Baraket† and Habib Yazidi ‡
Abstract
Z
p(x)|∇u|2 dx, u ∈ H01 (Ω), kuk
We study the minimizing problem inf
2N
L N−2 (Ω)
Ω
N
=1
where Ω is a smooth bounded domain of IR , N ≥ 3 and p a positive discontinuous
function. We prove the existence of a minimizer under some assumptions.
Keywords : Critical Sobolev exponent, Lack of compactness, Best Sobolev constant.
2010 AMS subject classifications: 35J20, 35J25, 35H30, 35J60.
1
Introduction
Let Ω be a smooth bounded domain of IRN , N ≥ 3, 2∗ = N2N
the critical exponent for
−2
Sobolev embedding. Define Ω1 and Ω2 two disjoint
domainsZ such that Ω̄ = Ω̄1 ∪ Ω̄2 .
∗
u ∈ H01 (Ω),
|u|2 dx = 1
Ω
Z
N
2∗
the barycenter function, see [7], β : V (Ω) → IR , u →
x |u| dx.
Denote Γ = ∂Ω1 ∩ ∂Ω2 . Define the set V (Ω) =
Ω
We consider the minimizing problem
S(p) =
and define
inf
u∈V (Ω), β(u)∈Γ
Z
p(x)|∇u|2dx,
(1)
Ω
where p is a discontinuous function as the following
(
α1
if x ∈ Ω1
p(x) =
α2
if x ∈ Ω̄2 ,
(2)
where αi , i = 1, 2 are some positive constants such that α1 < α2 . It’s important to remark
that without the additional condition β(u) ∈ Γ we have S(p) = α1 S, as one can verify
∗
Université Paris-Est Créteil, Laboratoire d’Analyse et de Mathématiques Appliquées, CNRS UMR
8050, UFR des Sciences et Technologie, 61, Avenue du Général de Gaulle Bât. P3, 4e étage, 94010 Créteil
Cedex, France. E-mail : hadiji@u-pec.fr
†
Département de Mathématiques, Faculté des Sciences de Tunis, Campus Universitaire, 2092 Tunis,
Université Tunis El Manar, Tunisie. E-mail: smbaraket@yahoo.fr
‡
Université de Tunis, Ecole Nationale Supérieure d’Ingénieurs de Tunis, 5 Avenue Taha Hssine, Bab
Mnar 1008 Tunis, Tunisie. E-mail : habib.yazidi@gmail.com
concentrating an extremal function for the best Sobolev constant S near a point in the
interior of the region Ω1 . In this case, the infimum S(p) is not achieved
This problem is closely related to the best constant in Sobolev inequality in IRN . It posses
many interesting properties, see Talenti [15], and arising in many areas of mathematics
and in a geometric context namely for example in the Yamabe problem and the prescribed
scalar curvature problem see Aubin [1]. It’s invariance under dilations produces a lack of
compactness.
The phenomenon of lack of compactness and the failure of the Palais Smale condition of
this type of problem has been the subject of several studies and it was analyzed in minute
detail by Struwe [14].
In the case where p is a constant, it is well known that (1) is not achieved for a general
domain Ω. Nevertheless, Brezis and Nirenberg showed in [6] that (1) has minimizer under
a linear perturbation, also, in the same sprit, Demyanov and Nazarov establish, in [9],
sufficient conditions for the existence of an extremal function in four embedding theorems
for more general Sobolev spaces. Bahri and Coron in [4] proved that the Euler equation
associated to this problem is solvable when some homology group of the domain with
coefficients in Z/2Z is nontrivial, see also the work of Coron [7]. In the case where p is a
smooth positive function, Hadiji and Yazidi proved that the study of problem (1) depends
on the behavior of the weight p near its minima, see [10], see also Hadiji, Molle, Passaseo
and Yazidi [11]) and Furtado and Souza[12].
One may ask whether the lack of compactness of the variational functional associated to
(1) can be made up by the discontinuity of the weight.
In this work, we investigate the effect of non smoothness of weight p on the existence of
solution without linear perturbation or additional conditions on the domain Ω.
2
Statements and proofs of results
Let
Sα1 , α2 = inf
(
α1
Z
IRN
+
|∇u|2dx + α2
Z
RN
−
)
|∇u|2dx, u ∈ H 1 (IRN ), u 6= 0 in IRN
± , kukL2∗ (IRN ) = 1 ,
′
′
N −1
N
where IRN
=
(x
,
x
)
∈
IR
×
[0,
∞[
and
IR
=
(x , xN ) ∈ IRN −1 ×] − ∞, 0] .
N
+
−
Set
)
(Z
N
=1 .
|∇u|2dx, u ∈ H 1 (IRN
S + = inf
+ ), u 6= 0 in IR+ , kukL2∗ (IRN
+)
IRN
+
and
S − = inf
(Z
|∇u|2dx,
IRN
−
)
N
u ∈ H 1 (IRN
=1 .
− ), u 6= 0 in IR− , kukL2∗ (IRN
−)
It is easy to verify that (see for example [8])
S+ = S− =
S
2
2N
,
(3)
where S is the best constant of the Sobolev embedding defined by
Z
|∇u|2 dx
N
IR
S=
inf
Z
2 .
u∈H 1 (IRN )\{0}
∗
|u|2 dx
2∗
IRN
We state our main results as follow
Theorem 2.1 The following equality holds
N
2
Sα1 , α2 =
N
2
α1 + α2
2
! N2
S.
Theorem 2.2 Let Ω, Ω1 , Ω2 , p be as defined in the introduction and let x0 ∈ Γ. Assume
that the following geometrical condition (g.c.) on Γ holds: in a neighborhood of x0 , Ω2
lies on one side of the tangent plane at x0 and the mean curvature with respect to the unit
inner normal of Ω2 at x0 is positive. Then S(p) is attained by some u ∈ H01 (Ω) .
The following proposition presents a strict lower bound for the minimizing problem
Proposition 2.1 The following inequality holds
α1 S < S(p).
Proof. We have S(p) ≥ α1 S. Arguing by contradiction, suppose that S(p) = α1 S.
Let {uZj } be a minimising sequence, that is, for every j ∈ N, uj ∈ V (Ω), β(uj ) ∈ Γ and
p(x)|∇uj |2 dx = α1 S.
lim
j→+∞
Since
ZΩ
Ω
2
p(x)|∇uj | dx ≥ α1 S then lim
j→+∞
2
Z
Ω
|∇uj |2 dx = S. Therefore, there exists z0 ∈ Ω̄
∗
such that, for a subsequence, |∇uj | → S δz0 and |uj |2 → δz0 , where δz0 is the Dirac mass
in z0 , see [13].
Since Zβ(uj ) ∈ Γ for every j ∈ N, it follows that z0 ∈ Γ and p(z0 ) = α2 > α1 . Therefore
p(x)|∇uj |2 dx = p(z0 ) S > α1 S, which gives a contradiction.
lim
j→+∞
✷
Ω
Remark 2.1 Let us give simple examples for which the condition (g.c.) in Theorem 2 is
fulfilled or not. Let Ω = B(0, R), R > 1 and e1 = (1, 0..., 0).
Set Ω2 = B(e1 , R) ∩ Ω, Ω1 = Ω \ Ω2 , Γ = ∂Ω1 ∩ ∂Ω2 and x0 in the interior of Γ. We
have condition (g.c.) holds.
For Ω1 = B(e1 , R) ∩ Ω, Ω2 = Ω \ Ω1 and , Γ = ∂Ω1 ∩ ∂Ω2 and x0 in the interior of
Γ. We have condition (g.c.) does not hold. More precisely, in any neighborhood of x0 , Ω2
does not lie on one side of the tangent plane at x0 and the mean curvature with respect to
the unit inner normal of Ω2 at x0 is negative.
Let Ω1 = {(x1 , ..., xN ) ∈ Ω s.t. x1 > 0} and Ω2 = {(x1 , ..., xN ) ∈ Ω s.t. x1 < 0}
and x0 = 0. We have condition (g.c.) hold, more precisely, in any neighborhood of 0, Ω2
lies on one side of the tangent plane at 0 but the mean curvature with respect to the unit
inner normal of Ω2 at 0 is 0.
If Γ is flat, that is mean that mean curvature at any point of Γ is zero, then we have
the following non-existence result:
Proposition 2.2 Let Ω = B(0, R), Γ = {x ∈ Ω \xN = 0} divided Ω into two subdomains
Ω1 and Ω2 . For i = 1, 2, let pi (x) = αi in Ωi with α1 < α2 . Then S(p) is not achieved.
Indeed, we have, if (1) is achieved by u then |u| is a minimization solution of (1) . Let
us suppose that S(p) is achieved by some positive function u ≥ 0, then there exists a
Lagrange multiplier µ ∈ R such that u satisfies the Euler equation

2∗ −1
−α
∆u
=
µ
u
in Ω1 ,
1




∗


−α2 ∆u = µ u2 −1 in Ω2 ,


∂u
∂u
α1 ∂ν
+ α2 ∂ν
= 0 on Γ,
(4)
1
2




u 6= 0
on Γ




u=0
on ∂Ω,
where ν1 and ν2 are respectively the outward normal of Ω1 and Ω2 .
On one hand we multiply (4) by ∇u · x and we integrate. On the other hand we multiply
(4) by N 2−2 u and we integrate. We obtain, after some computations, the Pohozaev identity
Z Z
∂u 2
∂u 2
∂u
∂u
−
α1 (x · ν1 )|
| + α2 (x · ν2 )|
| dsx =
α1 (x·ν)| |2 +α2 (x·ν)| |2 dsx ,
∂ν1
∂ν2
∂ν
∂ν
Γ
∂Ω∩∂Ω1
where ν is the outward of ∂Ω. Since B(0, R) is star-shaped about 0 then x · ν > 0 on ∂Ω
and then
Z ∂u 2
∂u 2
| + α2 (x · ν2 )|
| dsx > 0
−
α1 (x · ν1 )|
∂ν1
∂ν2
Γ
which gives a contradiction since x · ν1 = x · ν2 = 0 for every x in Γ. Therefore S(p) is
not achieved.
Proof of Theorem 2.1 On one hand, we claim that
N
Sα1 , α2 ≥
N
α12 + α22
2
! N2
S.
(5)
Indeed, we see that, for all t ∈]0, 1[ we have
Sα1n
, α2 =
o
R
R
∗
∗
inf α1 IRN |∇u|2dx + α2 RN |∇u|2dx, u ∈ H 1 (IRN ), kuk2L2∗ (IRN ) = t, kuk2L2∗ (IRN ) = 1 − t .
+
−
−
+
(6)
Therefore
nR
o
∗
∗
Sα1 , α2 ≥ α1 inf IRN |∇u|2 dx, u ∈ H 1 (IRN ), kuk2L2∗ (IRN ) = t, kuk2L2∗ (IRN ) = 1 − t
+
+
−
o
nR
N
2∗
2
1
2∗
+α2 inf RN |∇u| dx, u ∈ H (IR ), kukL2∗ (IRN ) = t, kukL2∗ (IRN ) = 1 − t (. 7)
+
−
−
At this stage, define
o
n
2∗
1
N
2∗
At = u ∈ H (IR ), kukL2∗ (IRN ) = t, kukL2∗ (IRN ) = 1 − t ,
+
−
N
2∗
Bt = u ∈ H 1 (IRN
+ ), kukL2∗ (IR+ ) = t
and
N
2∗
Ct = u ∈ H 1 (IRN
− ), kukL2∗ (IR− ) = 1 − t .
We have
At ⊂ Bt
and At ⊂ Ct .
(8)
We rewrite (7) as
Sα1 , α2 ≥ α1 inf
At
Z
2
|∇u| dx + α2 inf
At
IRN
+
Using (8) , we see that
Z
Z
2
|∇u| dx + α2 inf
Sα1 , α2 ≥ α1 inf
At
At
IRN
+
Z
|∇u|2 dx.
RN
−
2
|∇u| dx ≥ α1 inf
Bt
RN
−
Z
IRN
+
2
|∇u| dx + α2 inf
Ct
Z
|∇u|2dx. (9)
RN
−
Or, looking at (3), direct computations give that
inf
Bt
2
Z
2
|∇u| dx =
IRN
+
t 2∗
2
2N
S
and
inf
Ct
Z
2
2
|∇u| dx =
(1 − t) 2∗
2
2N
IRN
−
S.
Then, (9) becomes
2
Sα1 , α2 ≥ α1
t 2∗
2
2N
1
2
S + α2
h
(1 − t) 2∗
2
2N
S
2
2∗
α1 t + α2 (1 − t)
2 max
2 N t∈[0, 1]
2
N
N !N
2
2
α1 + α2
=
S
2
≥
2
2∗
i
S
which gives (5).
On the other hand, we claim that
N
2
Sα1 , α2 ≤
N
2
α1 + α2
2
! N2
S.
(10)
In order to prove the previous claim, for every x ∈ IRN we denote x = (x′ , xN ) where
x′ ∈ IRN −1 .
+
Let {u+
j } be a minimizing sequence of S . We define the sequence
′
+ ′
N −1
u−
×] − ∞, 0] and for all j ∈ N.
j (x , xN ) = uj (x , −xN ) for all x ∈ IR
−
Easily we see that {u−
j } is a minimizing sequence of S .
N
( αα21 ) 2
There exists t0 =
N such that
1 + ( αα21 ) 2
sup
t∈[0, 1]
(α1 t
2
2∗
2
2∗
+ α2 (1 − t) ) S
2
2
N
2
2∗
2
2∗
(α1 t0 + α2 (1 − t0 ) ) S
=
2
2
N
N
2
=
N
2
α1 + α2
2
! N2
S.
We define the following functions :
where θ =
′
θ
vj+ (x′ , xN ) = t0 u+
j (x , t0 xN )
for all x ∈ IRN −1 ×]0, +∞]
′
θ
vj− (x′ , xN ) = t0 u−
j (x , (1 − t0 ) xN )
for all x ∈ IRN −1 ×] − ∞, 0],
2(2∗ − 1)
. Now, consider
2∗
( + ′
vj (x , xN )
wj (x′ , xN ) =
vj− (x′ , xN )
for all x′ ∈ IRN −1 and for all xN ≥ 0
for all x′ ∈ IRN −1 and for all xN ≤ 0
(11)
An easy computation yields that, for large j, wj is a testing function for Sα1 , α2 defined
by (6).
Therefore
Z
Z
+ 2
|∇vj− |2 dx.
|∇vj | dx + α2
Sα1 , α2 ≤ α1
IRN
−
IRN
+
Using the definitions of vj+ and vj− , we obtain
2
Sα1 , α2 ≤ α1
t02∗
2
2N
2
S + α2
(1 − t0 ) 2∗
2
2N
S + o(1).
Then, using the definition of t0 and letting j → +∞, we obtain
N
2
Sα1 , α2 ≤
N
2
α1 + α2
2
! N2
S,
which gives (10).
Finally, (5) and (10) give the conclusion of Theorem 2.1.
✷
The proof of Theorem 2.2 follows from the following two Lemmas.
Lemma 2.1 Following the hypothesis of Theorem 2.2, we have if α1 S < S(p) < Sα1 , α2
then the infimum in (1) is achieved.
Proof.
We follow the arguments of Brezis-Nirenberg ([6], proof of Lemma 2.1) and we rely on
some idea of Demyanov-Nazarov ([9], proof of Proposition 1.1). Let {uj } ⊂ H01 (Ω) be a
minimizing sequence for (1) that is,
Z
p(x)|∇uj |2 dx = S(p) + o(1),
(12)
Ω
kuj kL2∗ = 1,
(13)
β(uj ) ∈ Γ.
(14)
and
Easily we see that {uj } is bounded in H01 (Ω), we may extract a subsequence still denoted
by uj , such that
uj ⇀ u weakly in H01 (Ω),
uj → u strongly in L2 (Ω),
uj → u
a.e. on Ω,
with kukL2∗ ≤ 1. Set vj = uj − u, so that
weakly in H01 (Ω)
vj ⇀ 0
vj → 0 strongly in L2 (Ω),
vj → 0
Using (12) we write
Z
2
p(x)|∇u(x)| dx +
Ω
a.e. on Ω.
Z
p(x)|∇vj (x)|2 dx = S(p) + o(1),
(15)
Ω
since vj ⇀ 0 weakly in H01 (Ω).
On the other hand, it follows from a result of Brezis-Lieb ([5], relation (1)) that
∗
∗
∗
ku + vj k2L2∗ = kuk2L2∗ + kvj k2L2∗ + o(1),
∗
(which holds since vj is bounded in L2 and vj → 0 a.e.). Thus, by (13), we have
∗
∗
1 = kuk2L2∗ + kvj k2L2∗ + o(1)
(16)
1 ≤ kuk2L2∗ + kvj k2L2∗ + o(1).
(17)
and therefore
Using the definition of Sα1 , α2 , extending vj by 0 in IRN (still denoted by vj ) we obtain
#
" Z
Z
1
kvj k2L2∗ ≤
|∇vj (x)|2 dx
|∇vj (x)|2 dx + α2
α1
N
N
Sα1 , α2
IR−, x
IR+, x
0
0
#
" Z
Z
1
≤
|∇vj (x)|2 dx
|∇vj (x)|2 dx + α2
α1
N
N
Sα1 , α2
Ω∩IR−, x
Ω∩IR+, x
0
0
Z
Z
1
2
2
≤
α1 |∇vj (x)| dx + α2 |∇vj (x)| dx
Sα1 , α2
Ω
Ω
Z
1
p(x)|∇vj (x)|2 dx.
(18)
kvj k2L2∗ ≤
Sα1 , α2 Ω
N
N −1
′
′
where IRN
, xN > x0N } and IRN
+, x0 = {x = (x , xN ) ∈ IR , \ x ∈ IR
−, x0 = {x =
N
N −1
′
′
′
(x , xN ) ∈ IR , \ x ∈ IR
, xN < x0N } with xON is such that x0 = (x , x0N ).
We claim that u 6≡ 0.
Indeed, suppose that u ≡ 0. From (15) we obtain
Z
p(x)|∇vj |2 dx = S(p) + o(1),
Ω
then
lim
j→+∞
From (16) we see that
Z
p(x)|∇vj |2 dx = S(p).
Ω
lim kvj kL2∗ = 1.
j→+∞
Or (18) gives that
kvj k2L2∗ Sα1 , α2 ≤
Z
p(x)|∇vj |2 dx.
Ω
Passing to limit in the previous inequality we obtain Sα1 , α2 ≤ S(p). This contradicts the
hypothesis S(p) < Sα1 , α2 . Consequently u 6≡ 0.
Now, we deduce from (17) and (18) that
Z
S(p)
2
p(x)|∇vj (x)|2 dx + o(1).
(19)
S(p) ≤ S(p)kukL2∗ +
Sα1 , α2 Ω
Combining (15) and (19) we obtain
Z
Z
Z
S(p)
2
2
2
p(x)|∇u(x)| + p(x)|∇vj (x)| dx ≤ S(p)kukL2∗ +
p(x)|∇vj (x)|2 dx + o(1).
S
α1 , α2 Ω
Ω
Ω
Thus
Z
Ω
2
p(x)|∇u(x)| dx ≤ S(p)kuk2L2∗ +
S(p)
−1
Sα1 , α2
Z
p(x)|∇vj (x)|2 dx + o(1).
Ω
Since S(p) < Sα1 , α2 , we deduce
Z
p(x)|∇u(x)|2 dx ≤ S(p)kuk2L2∗ ,
(20)
Ω
Therefore
Z
Ω
p(x)|∇u(x)|2 dx = S(p)kuk2L2∗ .
∗
It follows that uj → u strongly in L2 (Ω) and β(u) ∈ Γ. This means that u is a minimum
of S(p).
✷
Lemma 2.2 Assume that there exists x0 in the interior of Γ such that the (g.c.) holds.
Then
S(p) < Sα1 , α2 .
N −1
1 X
Proof. Let {λi (x0 )}1≤i≤N −1 , denote the principal curvatures and H(x0 ) =
λi (x0 )
N − 1 i=1
the mean curvature at x0 with respect to the unit normal.
For the simplicity, we suppose that x0 = 0. Therefore we note {λi }1≤i≤N −1 the principal
N −1
1 X
λi . Let R > 0, such that
curvatures at 0 and H(0) =
N − 1 i=1
B(R) ∩ Ω1 = {(x′ , xN ) ∈ B(R); xN > ρ(x′ )}
B(R) ∩ Ω2 = {(x′ , xN ) ∈ B(R); xN < ρ(x′ )}
B(R) ∩ Γ = {(x′ , xN ) ∈ B(R); xN = ρ(x′ )}
where x′ = (x1 , x2 , ..., xN −1 ) and ρ(x′ ) is defined by
N −1
ρ(x′ ) = Σi=1
λi x2i + O(|x′ |3 ).
We note that the condition (g.c.) implies that ρ(x′ ) ≥ 0.
Let us define, for ε > 0 and for t ∈]0, 1[ the function

ϕ(x)

If xN > 0
N−2
 (ε+|x′|2 +t− N−2
2 xN 2 ) 2
u0,ε,t(x) =
ϕ(x)


If xN < 0
N−2
− N−2
2
′ 2
(ε+|x | +(1−t)
where ϕ is a radial C ∞ -function such that
(
ϕ(x) =
2
xN )
2
1 if |x| ≤ R4
0 if |x| ≥ R2 .
N
There exists t0 =
sup
t∈[0, 1]
(α1 t
2
2∗
( αα21 ) 2
N
1 + ( αα21 ) 2
such that
2
2∗
+ α2 (1 − t) ) S
2
2
N
2
2∗
=
2
2∗
(α1 t0 + α2 (1 − t0 ) ) S
2
2
N
We note u0,ε (x) = u0,ε,t0 (x). Set, for i ∈ {1, 2},
Z
αi |∇u|2dx
Qi (u) = ZΩi
2
∗
|u|2 dx
2∗
Ω
and
Q(u) = Q1 (u) + Q2 (u).
N
2
=
N
2
α1 + α2
2
! N2
S.
In order to obtain the result of Lemma 2.2, we use u0,ε as a test function for S(p).
From ([2], page 13), direct computation gives

2
2∗


 α1 t02 S + α1 SH(0) A(N) ε 21 | ln(ε)| + O(ε 21 )
if N = 3
2N
Q1 (u0, ε ) =
(21)
2
∗

1
α1 t02 S


+ α1 SH(0) A(N) ε 2 + O(ε| ln(ε)|)
if N ≥ 4
2
2N
and
Q2 (u0, ε ) =

2
∗
1
1

 α2 (1−t20 ) 2 S − α1 SH(0) A(N) ε 2 | ln(ε)| + O(ε 2 )
2N
2

 α2 (1−t20 ) 2∗ S − α2 SH(0) A(N) ε 21 + O(ε| ln(ε)|)
2N
if N = 3
(22)
if N ≥ 4
where A(N) is a positive constant.
Combining (21) and (22) we see that,

2
2
∗

1
1
(α1 t02 +α2 (1−t0 ) 2∗ ) S


− (α2 − α1 )H(0) S A(N) ε 2 | ln(ε)| + O(ε 2 ) if N = 3
2
2N
Q(u0, ε ) =
2
2
∗

2
∗

 (α1 t0 +α2 2(1−t0 ) 2 ) S − (α2 − α1 )H(0) S A(N) ε 21 + O(ε| ln(ε)|) if N ≥ 4.
2N
Therefore, using the definition of t0 , we obtain

N N N2


1
1
α12 +α22


S − (α2 − α1 )H(0) S A(N) ε 2 | ln(ε)| + O(ε 2 ) if N = 3

2

Q(u0, ε ) ≤
N N N2

1

α12 +α22


S − (α2 − α1 )H(0) S A(N) ε 2 + O(ε| ln(ε)|) if N ≥ 4.

2

Finally, since α1 < α2 then we obtain the desired result.
✷
Remark 2.2 (see [2]): By looking at the previous proof, it follows that we can relax the
condition (g.c.) by allowing some of the λi ’s to be negative with mean curvature positive.
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