On a Biharmonic Equation Involving Nearly Critical Exponent

0.24 MB·English

by  Mohamed Ben Ayed

On a Biharmonic Equation Involving Nearly Critical Exponent
∗
4 Mohamed Ben Ayeda† & Khalil El Mehdib,c‡
0
a : D´epartement de Math´ematiques, Facult´e des Sciences deSfax, Route Soukra,Sfax, Tunisia.
0
2 b : Facult´edes Sciences et Techniques,Universit´e deNouakchott,BP 5026, Nouakchott,Mauritania.

n c : The AbdusSalam ICTP, Mathematics Section, StradaCostiera 11, 34014 Trieste, Italy.
a
J

7
Abstract. This paper is concerned with a biharmonic equation under the Navier boundary condition (P∓ε) :
] ∆2u = unn−+44∓ε, u > 0 in Ω and u = ∆u = 0 on ∂Ω, where Ω is a smooth bounded domain in Rn, n ≥ 5, and
P
ε>0. Westudythe asymptotic behaviorof solutions of (P ) which are minimizing for theSobolev quotientas
A −ε
εgoestozero. Weshowthatsuchsolutionsconcentratearoundapointx ∈Ωasε→0,moreoverx isacritical
0 0
h. point of the Robin’s function. Conversely, we show that for any nondegenerate critical point x0 of the Robin’s
t function, there exist solutions of (P−ε) concentrating around x0 as ε → 0. Finally we prove that, in contrast
a with what happened in the subcritical equation (P ), the supercritical problem (P ) has no solutions which
m −ε +ε
concentrate around a point of Ω as ε→0.
[
2000 Mathematics Subject Classification : 35J65, 35J40, 58E05.
1
Key words and phrases : Elliptic PDE with critical Sobolev exponent,Noncompact variational problems.
v
1
6
1 Introduction and Results
0
1
0
In this paper, we are concerned with the following semilinear biharmonic equation under the
4
0 Navier boundary condition
/
h
∆2u= up∓ε, u >0 in Ω
t (P )
a ∓ε ∆u= u= 0 on ∂Ω,
m (cid:26)
: where Ω is a smooth bounded domain in Rn, n 5, ε is a small positive parameter, and
v
≥
i p+1= 2n/(n 4) isthecritical Sobolev exponentof theembeddingH2(Ω) H1(Ω) ֒ Lp+1(Ω).
X − ∩ 0 →
When the biharmonic operator in (P ) is replaced by the Laplacian operator, there are many
r ∓ε
a works devoted to the study of the contrepart of (P ), see for example [1], [2], [5], [7], [11], [13],
∓ε
[14], [17], [18], [20], [21], [23] and the references therein.
When ε (0,p), the mountain pass lemma proves the existence of solution to (P ) for any
−ε
∈
domain Ω. When ε = 0, the situation is more complex, Van Der Vorst showed in [24] that if Ω is
starshaped (P ) has no solution whereas Ebobisse and Ould Ahmedou proved in [15] that (P )
0 0
has a solution provided that some homology group of Ω is nontrivial. This topological condition
is sufficient, but not necessary, as examples of contractible domains Ω on which a solution exists
∗Work finished when the authors were visiting Mathematics Department of Roma University “La Sapienza”.
TheywouldliketothanktheMathematicsDepartmentforitswarmhospitality. TheauthorsalsothankProfessors
M. Grossi and F. Pacella for theirconstant support.
†Corresponding author. Fax : [phone], E-mail : [email protected].
‡E-mail : [email protected].
1
2 M. Ben Ayed & K. El Mehdi
show [16].
In view of this qualitative change in the situation when ε = 0, it is interesting to study the
asymptotic behavior of the subcritical solution u of (P ) as ε 0. Chou and Geng [12] made
ε −ε
→
the first study, when Ω is a convex domain. The aim of the first result of this paper is to remove
the convexity assumption on Ω. To state this result, we need to introduce some notation.
Let us define on Ω the following Robin’s function
1
ϕ(x) = H(x,x), with H(x,y) = G(x,y), for (x,y) Ω Ω,
x y n−4 − ∈ ×
| − |
where G is the Green’s function of ∆2, that is,
∆2G(x,.) =c δ in Ω
x Ω n x
∀ ∈ ∆G(x,.) = G(x,.) = 0 on ∂Ω,
(cid:26)
where δ denotes the Dirac mass at x and c = (n 4)(n 2)Sn−1 .
x n
− − | |
Let
n−4
δ (x) = c0λ 2 , c = [(n 4)(n 2)n(n+2)](n−4)/8, λ> 0, a Rn (1.1)
a,λ (1+λ2 x a2)n−24 0 − − ∈
| − |
It is well known (see [19]) that δ are the only solutions of
a,λ
∆2u= unn−+44, u> 0 in Rn, with u Lp+1(Rn) and ∆u L2(Rn)
∈ ∈
and are also the only minimizers of the Sobolev inequality on the whole space, that is
S = inf ∆u2 u−2 , s.t.∆u L2,u Ln2−n4,u= 0 . (1.2)
{| |L2(Rn)| |Ln2−n4(Rn) ∈ ∈ 6 }
We denote by Pδ the projection of the δ ’s on H2(Ω) H1(Ω), defined by
a,λ a,λ ∩ 0
∆2Pδ = ∆2δ in Ω and ∆Pδ = Pδ = 0 on ∂Ω.
a,λ a,λ a,λ a,λ
Let
1/2
u = ∆u2 , u H2(Ω) H1(Ω) (1.3)
|| || | | ∈ ∩ 0
(cid:18)ZΩ (cid:19)
(u,v) = ∆u∆v, u,v H2(Ω) H1(Ω) (1.4)
∈ ∩ 0
ZΩ
u = u . (1.5)
q Lq(Ω)
| | | |
Now we state the first result of this paper.
Theorem 1.1 Assume that n 6. Let (u ) be a solution of (P ), and assume that
ε −ε
≥
(H) u 2 u −2 S as ε 0,
|| ε|| | ε|p+1−ε → →
Biharmonic Equation 3
where S is the best Sobolev constant in Rn defined by (1.2). Then (up to a subsequence) there
exist a Ω, λ > 0, α > 0 and v such that u can be written as
ε ε ε ε ε
∈
u = α Pδ +v
ε ε aε,λε ε
with α 1, v 0, a Ω and λ d(a ,∂Ω) + as ε 0.
ε ε ε ε ε
→ || ||→ ∈ → ∞ →
In addition, a converges to a critical point x Ω of ϕ and we have
ε 0
∈
limε u 2 = (c c2/c )ϕ(x ),
ε→0 || ε||L∞(Ω) 1 0 2 0
where c = c2n/(n−4) dx , c = (n 4)c2n/(n−4) log(1+|x|2)(1−|x|2)dx and c is
1 0 Rn (1+|x|2)(n+4)/2 2 − 0 Rn (1+|x|2)n+1 0
defined in (1.1).
R R
Remark 1.2 It is important to point out that in the Laplacian case (see [17]), the method
of moving planes has been used to show that blowup points are away from the boundary of
domain. The process is standard if domains are convex. For nonconvex regions, the method of
movingplanesstillworksintheLaplacian casethrough theapplications ofKelvintransformations
[17]. For (P ), the method of moving planes also works for convex domains [12]. However,
−ε
for nonconvex domains, a Kelvin transformation does not work for (P ) because the Navier
−ε
boundary condition is not invariant under the Kelvin transformation of biharmonic operator.
Our method here is essential in overcoming the difficulty arising from the nonhomogeneity of
Navier boundary condition under the Kelvin transformation.
Our next result provides a kind of converse to Theorem 1.1.
Theorem 1.3 Assume that n 6, and x Ω is a nondegenerate critical point of ϕ. Then
0
≥ ∈
there exists an ε > 0 such that for each ε (0,ε ], (P ) has a solution of the form
0 0 −ε
∈
u = α Pδ +v
ε ε aε,λε ε
with α 1, v 0, a x and λ d(a ,∂Ω) + as ε 0.
ε ε ε 0 ε ε
→ || ||→ → → ∞ →
In view of the above results, a natural question arises: are equivalent results true for slightly
supercritical exponent?
The aim of the next result is to answer this question.
Theorem 1.4 Let Ω be any smooth bounded domain in Rn, n 5. Then (P ) has no solution
+ε
≥
u of the form
ε
u = α Pδ +v
ε ε aε,λε ε
with v 0, α 1, a Ω and λ d(a ,∂Ω) + as ε 0.
ε ε ε ε ε
|| || → → ∈ → ∞ →
The proofs of our results are based on the same framework and methods of [22], [23] and [7].
The next section will be devoted to prove Theorem 1.1, while Theorems 1.3 and 1.4 are proved
in sections 3 and 4 respectively.
4 M. Ben Ayed & K. El Mehdi
2 Proof of Theorem 1.1
Before starting the proof of Theorem 1.1, we need some preliminary results
Proposition 2.1 [10] Let a Ω and λ > 0 such that λd(a,∂Ω) is large enough. For θ =
(a,λ)
∈
δ Pδ , we have the following estimates
(a,λ) (a,λ)
−
H(a,.)
(a) 0 θ δ , (b) θ = c +f ,
≤ (a,λ) ≤ (a,λ) (a,λ) 0 λn−24 (a,λ)
where f satisfies
(a,λ)
1 ∂f 1 1∂f 1
(a,λ) (a,λ)
f = O , λ = O , = O ,
(a,λ) (cid:18)λn2dn−2(cid:19) ∂λ (cid:18)λn2dn−2(cid:19) λ ∂a (cid:18)λn+22dn−1(cid:19)
where d is the distance d(a,∂Ω),
1 ∂θ 1
(a,λ)
(c) θ = O , λ =O ,
| (a,λ) |Ln2−n4 (λd)n−24 | ∂λ |Ln2−n4 (λd)n−24
(cid:0) (cid:1) (cid:0) (cid:1)
1 1∂θ 1
(a,λ)
θ = O , = O .
|| (a,λ) || (λd)n−24 | λ ∂a |Ln2−n4 (λd)n−22
(cid:0) (cid:1) (cid:0) (cid:1)
Proposition 2.2 Let u be a solution of (P ) which satisfies (H). Then, we have
ε −ε
(a) u 2 Sn/4, (b) up+1−ε Sn/4.
|| ε|| → ε →
Z
Proof. Since u is a solution of (P ), then we have u 2 = up+1−ε. Thus, using the
ε −ε ε ε
|| ||
assumption (H), we derive that
R
2(p−1−ε)
||uε||2|uε|−p+21−ε = ||uε|| p+1−ε = S +o(1).
Therefore u 2 = up+1−ε = Sn/4+o(1). The result follows. 2
ε ε
|| ||
R
Proposition 2.3 Let u be a solution of (P ) which satisfies (H), and let x Ω such that
ε −ε ε
∈
u (x ) = u := M . Then, for ε small, we have
ε ε ε L∞ ε
| |
(a) Mε = 1+o(1).
ε
(b) u can be written as
ε
u = Pδ +v˜ ,
ε xε,λ˜ε ε
with v˜ 0, where λ˜ = c2/(4−n)M(p−1−ε)/4.
|| ε|| → ε 0 ε
Proof. First of all, we prove that M + as ε 0. To this end, arguing by contradiction,
ε
→ ∞ →
we suppose that M remains bounded for a sequence ε 0 as n + . Then, in view of
εn n → → ∞
elliptic regularity theory, we can extract a subsequence, still denoted by u , which converges
εn
uniformly to a limit u . By Proposition 2.2, u = 0, hence by taking limit in (H) we find that
o 0
6
Biharmonic Equation 5
u achieves a best Sobolev constant S, a contradiction to the fact that S is never achieved on a
0
bounded domain [25].
Now we define the rescaled functions
ω (y) = M−1u x +M(1+ε−p)/4y , y Ω = M(p−1−ε)/4(Ω x ), (2.1)
ε ε ε ε ε ∈ ε ε − ε
(cid:16) (cid:17)
ω satisfies
ε
∆2ω = ωp−ε, 0 < ω 1 in Ω
ε ε ε ≤ ε (2.2)
ω (0) = 1, ∆ω = ω = 0 on ∂Ω .
ε ε ε ε
(cid:26)
Following the same argument as in Lemma 2.3 [8], we have
M(p−1−ε)/4d(x ,∂Ω) + as ε 0.
ε ε → ∞ →
Then it follows from standard elliptic theory that there exists a positive function ω such that
(after passing to a subsequence) ω ω in C4 (Rn), and ω satisfies
ε → loc
∆2ω = ωp, 0 ω 1 in Rn
≤ ≤
ω(0) = 1, ω(0) = 0.
(cid:26) ∇
It follows from [19] that ω writes as
2/(4−n)
ω(y)= δ (y), with α = c .
0,αn n 0
(p−1−ε)/4
Observe that, for y = M (x x ), we have
ε ε
−
(n−4)/2
M c α
M δ (y) = ε 0 n = Mε(n−4)/8δ (2.3)
ε 0,αn 1+α2M(p−1−ε)/2 x x 2 (n−4)/2 ε xε,λ˜ε
n ε | − ε|
(cid:16) (cid:17)
with λ˜ = α M(p−1−ε)/4. Then,
ε n ε
w (y) δ (y) = M−1 u (x) Mε(n−4)/8δ .
ε − 0,αn ε ε − ε xε,λ˜ε
(cid:16) (cid:17)
Let us define
u1(x) = u (x) Mε(n−4)/8Pδ (x),
ε ε − ε xε,λ˜ε
we need to compute
u1 2 = u 2+Mε(n−4)/4 Pδ 2 2Mε(n−4)/8(u ,Pδ ).
|| ε|| || ε|| ε || xε,λ˜ε|| − ε ε xε,λ˜ε
On one hand, we have
(n+4)/(n−4)
(u ,Pδ ) = u (x)δ (x)
ε xε,λ˜ε ZΩ ε xε,λ˜ε
= u (x +M(1+ε−p)/4y)δ(n+4)/(n−4)(x +M(1+ε−p)/4y)Mn(1+ε−p)/4dy
ZΩε ε ε ε xε,λ˜ε ε ε ε
= Mε(n−4)/8w (y)δ(n+4)/(n−4)(y)dy
ε ε 0,αn
ZΩε
= Mε(n−4)/8w (y)δ(n+4)/(n−4)(y)dy+ Mε(n−4)/8w (y)δ(n+4)/(n−4)(y)dy,
ε ε 0,αn ε ε 0,αn
ZB(0,R) ZΩε(cid:31)B(0,R)
6 M. Ben Ayed & K. El Mehdi
2n/(n−4)
where R is a large positive constant such that δ = o(1).
Rn(cid:31)B(0,R) 0,αn
Since
R
w2n/(n−4) = M−εn/4 u2n/(n−4) c,
ε ε ε ≤
ZΩε ZΩ
using Holder’s inequality we derive that
(n+4)/(n−4)
w δ = o(1).
ε 0,αn
ZΩε(cid:31)B(0,R)
Now, since w δ in C4 (Rn), we obtain
ε → 0,αn loc
(u ,Pδ ) = Mε(n−4)/8(Sn/4+o(1)).
ε xε,λ˜ε ε
On the other hand, one can easy verify that
Pδ ) 2 = Sn/4+o(1).
|| xε,λ˜ε ||
Thus
u1 2 = u 2 Mε(n−4)/4(Sn/4+o(1)), (2.4)
|| ε|| || ε|| − ε
and, using the fact that u1 2 0 and Proposition 2.2, we derive that
|| ε|| ≥
Mε(n−4)/4 1+o(1).
ε ≤
But, since M , we have Mε 1 and therefore claim (a) follows.
ε → ∞ ε ≥
Now we are going to prove claim (b). Observe that, using Proposition 2.2 and claim (a), (2.4)
becomes
u1 2 = (Sn/4+o(1)) (Sn/4+o(1)) = o(1).
|| ε|| −
Thus claim (b) follows. 2
Proposition 2.4 Let u be a solution of (P ) which satisfies (H). Then, there exist a Ω,
ε −ε ε
∈
α > 0, λ > 0 and v such that
ε ε ε
u = α Pδ +v
ε ε aε,λε ε
with α 1, a x 0, λ d(a ,∂Ω) , λ˜ /λ 1 and v 0. Furthermore, v
ε ε ε ε ε ε ε ε ε
→ | − | → → ∞ → || || →
satisfies
(V ) (v,Pδ )= (v,∂Pδ /∂λ ) = 0, (v,∂Pδ /∂a) = 0.
0 aε,λε aε,λε ε aε,λε
Proof. ByProposition2.3,u canbewrittenasu = Pδ +v˜ with v˜ 0,λ˜ d(x ,∂Ω)
ε ε xε,λ˜ε ε || ε|| → ε ε →
as ε 0. Thus, the following minimization problem
∞ →
min u αPδ ,α > 0,a Ω,λ > 0
ε a,λ
{|| − || ∈ }
has a unique solution (α ,a ,λ ). Then, for v = u α Pδ , we have v satisfies (V ). From
ε ε ε ε ε− ε aε,λε ε 0
the two forms of u , one can easy verify that
ε
Pδ Pδ = o(1).
|| xε,λ˜ε − aε,λε||
Therefore, we derive that a x = o(1) and λ˜ /λ = 1+o(1)). The result follows. 2
ε ε ε ε
| − |
Next, we state a result which its proof is similar to the proof of Lemma 2.3 of [7], so we will
omit it.
Biharmonic Equation 7
Lemma 2.5 λε = 1+o(1) as ε goes to zero implies that
ε
1
δ−ε = O εLog(1+λ2 x a 2) in Ω.
ε − cελε(n−4)/2 ε| − ε|
0 ε
(cid:0) (cid:1)
We are now able to study the v -part of u solution of (P ).
ε ε −ε
Proposition 2.6 Let (u ) be a solution of (P ) which satisfies (H). Then v occuring in
ε −ε ε
Proposition 2.4 satisfies
1 1
v Cε+C (if n < 12)+ (if n 12) ,
|| ε|| ≤ (λεdε)n−4 (λεdε)n+24−ε(n−4) ≥ !
where C is a positive constant independent of ε.
Proof. Multiplying (P ) by v and integrating on Ω, we obtain
−ε ε
∆u .∆v up−εv = 0.
ε ε− ε ε
ZΩ ZΩ
Thus
∆v 2 (α Pδ )p−ε+(p ε)(α Pδ )p−1−εv +O δp−2−εv2χ + v p−ε v = 0.
| ε| − ε ε − ε ε ε ε ε |vε|<δε | ε| ε
ZΩ ZΩ
(cid:0) (cid:0) (cid:1)(cid:1)
Using Lemma 2.5, we find
Q (v ,v ) f (v )+o( v 2) = 0, (2.5)
ε ε ε ε ε ε
− || ||
with
Q (v,v) = v 2 (p ε) (α Pδ )p−1−εv2
ε ε ε
|| || − −
Z
and
f (v) = (α Pδ )p−εv.
ε ε ε
Z
We observe that
Q (v,v) = v 2 p (α Pδ )p−1−εv2+O ε v 2
ε ε ε
|| || − || ||
ZΩ
(cid:0) (cid:1)
= v 2 pαp−1−ε δp−1−ε+O δp−2−εθ v2+o v 2
|| || − ε ε ε ε || ||
ZΩ
p−1−ε (cid:0) (cid:0) (cid:1)(cid:1) (cid:0) (cid:1)
pα 1
= v 2 ε δp−1v2+O δ−ε δp−1 v 2 +o v 2
|| || − cε0λεε(n−4)/2 ZΩ ε ZΩ(cid:12)(cid:12) ε − cε0λεε(n−4)/2(cid:12)(cid:12) ε | | ! || ||
(cid:12) (cid:12) (cid:0) (cid:1)
(cid:12) (cid:12)
Using Lemma 2.5 and the fact that α 1, we find
ε (cid:12) (cid:12)
→
Q (v,v) = Q (v,v)+o v 2 ,
ε 0
|| ||
(cid:0) (cid:1)
8 M. Ben Ayed & K. El Mehdi
with
Q (v,v) = v 2 p δp−1v2.
0 || || − ε
ZΩ
According to [6], Q is coercive, that is, there exists some constant c > 0 independent of ε, for
0
ε small enough, such that
Q (v,v) c v 2 v E , (2.6)
0 ≥ || || ∀ ∈ (aε,λε)
where
E = v E/v satisfies (V ) , (2.7)
(aε,λε) { ∈ 0 }
(V ) is the condition defined in Proposition 2.4.
0
We also observe that
f (v) = αp−ε δp−ε +O δp−1−εθ v
ε ε ε ε ε
ZΩ
(cid:0) (cid:0) (cid:1)(cid:1)
1
= αp−ε δpv+O ε Log(1+λ2 x a 2)δp v + δp−1−εθ v .
ε "cε0λεε(n−4)/2 ZΩ ε (cid:18) ZΩ ε| − ε| ε| | ZΩ ε ε| |(cid:19)#
The last equality follows from Lemma 2.5. Therefore we can write, with B = B(a ,d )
ε ε
f (v) c ε v + δp−1−εθ v + δp v
ε ≤ || || ε ε| | ε| |
(cid:18) ZB ZRn(cid:31)B (cid:19)
n+4 n+4
(p−1−ε) 2n 2n 2n 2n
c v ε+ θ δ n+4 + δn−4 .
ε L∞ ε ε
≤ || || | |
(cid:18)ZB (cid:19) (cid:18)ZRn(cid:31)B (cid:19) !
We notice that
1
δ2n/(n−4) = O (2.8)
ε (λ d )n
ZRn(cid:31)B (cid:18) ε ε (cid:19)
and
n+4
2n(p−1−ε) 2n c c
θ δ n+4 ( if n 12)+ ( if n < 12). (2.9)
| ε|L∞(cid:18)ZB ε (cid:19) ≤ (λεdε)n+24−ε(n−4) ≥ (λεdε)n−4
Thus we obtain
1 1
f (v) C v ε+ ( if n <12)+ ( if n 12) (2.10)
| ε | ≤ || || (λεdε)n−4 (λεdε)n+24−ε(n−4) ≥ !
Combining (2.5), (2.6) and (2.10), we obtain the desired estimate. 2
Next we prove the following crucial result :
Biharmonic Equation 9
Proposition 2.7 For u = α Pδ + v solution of (P ) with λε = 1 +o(1) as ε goes to
ε ε aε,λε ε −ε ε
zero, we have the following estimate
H(a ,a ) 1
(a) c ε+O(ε2) c ε ε +o = 0.
2 − 1 λεn−4 (cid:18)(λεdε)n−4(cid:19)
and for n 6, we also have
≥
c ∂H 1
(b) 3 (a ,a )+O(ε2)+o = 0,
λεn−3∂aε ε ε (cid:18)(λεdε)n−3(cid:19)
where c , c are the constants defined in Theorem 1.1, and where c is a positive constant.
1 2 3
Proof. We start by giving the proof of Claim (a). Multiplying the equation (P ) by
−ε
λ (∂Pδ )/(∂λ ) and integrating on Ω, we obtain
ε ε ε
∂Pδ ∂Pδ
0= ∆2u λ ε up−ελ ε
ε ε ∂λ − ε ε ∂λ
ZΩ ZΩ
∂Pδ
= α δpλ ε (α Pδ )p−ε+(p ε)(α Pδ )p−1−εv
ε ε ε ∂λ − ε ε − ε ε ε
ZΩ ZΩ
(cid:2) ∂Pδ
+O δp−2−ε v 2+ v p−εχ λ ε. (2.11)
ε | ε| | ε| δε≤|vε| ε ∂λ
(cid:0) (cid:1)(cid:3)
We estimate each term of the right-hand side in (2.11). First, using Proposition 2.1, we have
∂Pδ 1
δpλ ε c δp+1 = O( )
ε ε ∂λ ≤ ε (λ d )n
ZBc ε ZBc ε ε
∂Pδ ∂δ (n 4)c ∂f
δpλ ε = δpλ ε + − 0 δpH δpλ ε,
ZB ε ε ∂λ ZB ε ε ∂λ 2λ(εn−4)/2 ZB ε −ZB ε ε ∂λ
with B = B(a ,d ). Expanding H(a ,.) around a and using Proposition 2.1, we obtain
ε ε ε ε
∂Pδ 1 (n 4)c 1
δpλ ε = O + − 0H(a ,a ) δp +O .
ZB ε ε ∂λ (cid:18)(λεdε)n(cid:19) 2λ(εn−4)/2 ε ε ZB ε (cid:18)(λεdε)n−2(cid:19)
Therefore, estimating the integral, we obtain
∂Pδ n 4 H(a ,a ) 1
δpλ ε = − c ε ε +O (2.12)
ZΩ ε ε ∂λ 2 1 λεn−4 (cid:18)(λεdε)n−2(cid:19)
with c = c2n/(n−4) dx .
1 0 Rn (1+|x|2)(n+4)/2
Secondly, we compute
R
∂Pδ ∂Pδ
(Pδ )p−ελ ε = δp−ε (p ε)δp−1−εθ +O θ2δp−2−ε+θp−ε λ ε
ε ε ∂λ ε − − ε ε ε ε ε ε ∂λ
ZΩ ZΩ
(cid:2) ∂δ ∂θ (cid:0) (cid:1)(cid:3) ∂δ
= δp−ελ ε δp−ελ ε (p ε) δp−1−εθ λ ε (2.13)
ε ε ∂λ − ε ε ∂λ − − ε ε ∂λ
ZB ZB ZB
∂θ 1
+O δp−1−εθ λ ε + θ2δp−1−ε+ θp−εδ +
ZΩ ε ε| ε ∂λ| ZΩ ε ε ZΩ ε ε (λεdε)n−εn−24!
10 M. Ben Ayed & K. El Mehdi
and we have to estimate each term of the right hand-side of (2.13).
Using the fact that λ ∂δε = n−4 1−λ2ε|x−aε|2 δ , we derive that
ε ∂λ 2 1+λ2ε|x−aε|2 ε
(cid:16) (cid:17)
∂δ n 4 cp+1−ε 1 1 x2 1
δp−ελ ε = − 0 −| | dx+O
ZB ε ε ∂λ 2 λεε(n2−4) ZRn (1+|x|2)n−ε(n2−4) 1+|x|2 (cid:18)(λεdε)n−ε(n−4)(cid:19)
n 4 1
= − c ε+O(ε2) +O (2.14)
2
2λεε(n−4)/2 − (cid:18)(λεdε)n−ε(n−4)(cid:19)
(cid:0) (cid:1)
with c = n−4cn2−n4 Log(1+|x|2)|x|2−1dx > 0.
2 2 0 Rn (1+|x|2)n |x|2+1
For the other terms in (2.13), using Proposition 2.1, we have
R
∂θ ∂ c H
δp−ελ ε = δp−ελ 0 f
ZB ε ε ∂λ ZB ε ε∂λ λ(εn−4)/2 − ε!
(p−ε)(n−4)
n 4 p+1−εH(aε,aε) λε 2 1
= − c +O
− 2 0 λ(εn−4)/2 ZB(cid:18)1+λ2ε|x−aε|2(cid:19) (cid:18)(λεdε)n−2(cid:19)
p+1−ε
n 4H(a ,a ) 1 c 1
= − ε ε 0 +O
− 2 λεn−4 λεε(n−4)/2 ZB(0,λd) (1+ x2)n+24−εn−24 (cid:18)(λεdε)n−2(cid:19)
| |
n 4 H(a ,a ) 1 ε 1
ε ε
= − c +O + (2.15)
− 2 1 λεn−4 λεε(n−4)/2 (cid:18)(λεdε)n−4 (λεdε)n−2(cid:19)
and
∂δ c ∂δ
(p ε) δp−1−εθ λ ε = (p ε) δp−1−ε 0 Hλ ε +O δp−εf
− ZB ε ε ε ∂λ − ZB ε λ(εn−4)/2 ε ∂λ (cid:18)ZB ε ε(cid:19)
c ∂δ 1
= (p ε) 0 H(a ,a ) δp−1−ελ ε +O
− λ(εn−4)/2 ε ε ZB ε ε ∂λ (cid:18)(λεdε)n−2(cid:19)
c (p ε)H(a ,a ) (n 4)(1 x2) 1
0 ε ε
= − − −| | +O
λεεn−24 λεn−4 ZB(0,λεdε) 2(1+|x|2)n+26−εn−24 (cid:18)(λεdε)n−2(cid:19)
n 4 c H(a ,a ) ε 1
1 ε ε
= − +O + .
− 2 λεε(n−4)/2 λεn−4 (cid:18)(λεdε)n−4 (λεdε)n−2(cid:19)
(2.13), (2.14), (2.15) and additional integral estimates of the same type provide us with the
expansion
∂Pδ n 4 H(a ,a )
(Pδ )p−ελ ε = − c ε+O(ε2)+2c ε ε (2.16)
ZΩ ε ε ∂λ 2λεε(n−4)/2 (cid:20)− 2 1 λεn−4
1 1
+O +(if n = 5) .
(λ d )n−2 (λ d )2
(cid:18) ε ε ε ε (cid:19)(cid:21)