SINGLE BLOW-UP SOLUTIONS FOR A SLIGHTLY
SUBCRITICAL BIHARMONIC EQUATION
KHALIL EL MEHDI
Received 29 October 2004; Accepted 20 January 2005
We consider a biharmonic equation under the Navier boundary condition and with a
nearly critical exponent (Pε ): Δ2 u = u9−ε , u > 0 in Ω and u = Δu = 0 on ∂Ω, where Ω is a
smooth bounded domain in R5 , ε > 0. We study the asymptotic behavior of solutions of
(Pε ) which are minimizing for the Sobolev quotient as ε goes to zero. We show that such
solutions concentrate around a point x0 ∈ Ω as ε → 0, moreover x0 is a critical point of
the Robin’s function. Conversely, we show that for any nondegenerate critical point x0 of
the Robin’s function, there exist solutions of (Pε ) concentrating around x0 as ε → 0.
Copyright © 2006 Khalil El Mehdi. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction and results
Let us consider the following biharmonic equation under the Navier boundary condition
Δ2 u = u p−ε ,
Δu = u = 0
u > 0 in Ω
on ∂Ω,
(Qε )
where Ω is a smooth bounded domain in Rn , n ≥ 5, ε is a small positive parameter, and
p + 1 = 2n/(n − 4) is the critical Sobolev exponent of the embedding H 2 (Ω) ∩ H01 (Ω) L2n/(n−4) (Ω).
It is known that (Qε ) is related to the limiting problem (Q0 ) (when ε = 0) which exhibits a lack of compactness and gives rise to solutions of (Qε ) which blow up as ε → 0.
The interest of the limiting problem (Q0 ) grew from its resemblance to some geometric
equations involving Paneitz operator and which have widely been studied in these last
years (for details one can see [4, 6, 10, 12–14, 17] and references therein).
Several authors have studied the existence and behavior of blowing up solutions for
the corresponding second order elliptic problem (see, e.g., [1, 3, 9, 18, 21, 22, 24–26]
and references therein). In sharp contrast to this, very little is known for fourth order
elliptic equations. In this paper we are mainly interested in the asymptotic behavior and
Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2006, Article ID 18387, Pages 1–20
DOI 10.1155/AAA/2006/18387
2
Single blow-up solutions for a biharmonic equation
the existence of solutions of (Qε ) which blow up around one point, and the location of
this blow up point as ε → 0.
The existence of solutions of (Qε ) for all ε ∈ (0, p − 1) is well known for any domain Ω
(see, e.g., [16]). For ε = 0, the situation is more complex, Van Der Vorst showed in [28]
that if Ω is starshaped (Q0 ) has no solution whereas Ebobisse and Ould Ahmedou proved
in [15] that (Q0 ) 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 show [19].
In view of this qualitative change in the situation when ε = 0, it is interesting to study
the asymptotic behavior of the subcritical solution uε of (Qε ) as ε → 0. Chou and Geng
[11], and Geng [20] made a first study, when Ω is strictly convex. The convexity assumption was needed in their proof in order to apply the method of moving planes (MMP
for short) in proving a priori estimate near the boundary. Notice that in the Laplacian
case (see [21]), the MMP has been used to show that blow up points are away from the
boundary of the domain. The process is standard if domains are convex. For nonconvex
regions, the MMP still works in the Laplacian case through the applications of Kelvin
transformations [21]. For (Qε ), the MMP also works for convex domains [11]. However, for nonconvex domains, a Kelvin transformation does not work for (Qε ) because
the Navier boundary condition is not invariant under the Kelvin transformation of biha
rmonic operator. In [5], Ben Ayed and El Mehdi removed the convexity assumption of
Chou and Geng for higher dimensions, that is n ≥ 6. The aim of this paper is to prove
that the results of [5] are true in dimension 5. In order to state precisely our results, we
need to introduce some notations.
We consider the following problem
Δ2 u = u9−ε ,
u > 0 in Ω
Δu = u = 0 on ∂Ω,
(Pε )
where Ω is a smooth bounded domain in R5 and ε is a small positive parameter.
Let us define on Ω the following Robin’s function
ϕ(x) = H(x,x),
with H(x, y) = |x − y |−1 − G(x, y), for (x, y) ∈ Ω × Ω,
(1.1)
where G is the Green’s function of Δ2 , that is,
∀x ∈ Ω
Δ2 G(x, ·) = cδx
ΔG(x, ·) = G(x, ·) = 0
in Ω
on ∂Ω,
(1.2)
where δx denotes the Dirac mass at x and c = 3ω5 , with ω5 is the area of the unit sphere
of R5 . For λ > 0 and a ∈ R5 , let
c0 λ1/2
δa,λ (x) = 1/2 ,
1 + λ2 |x − a|2
c0 = (105)1/8 .
(1.3)
It is well known (see [23]) that δa,λ are the only solutions of
Δ2 u = u9 ,
u > 0 in R5
(1.4)
Khalil El Mehdi 3
and are also the only minimizers of the Sobolev inequality on the whole space, that is
S = inf |Δu|2L2 (R5 ) |u|−L102 (R5 ) , s.t. Δu ∈ L2 , u ∈ L10 , u = 0 .
(1.5)
We denote by Pδa,λ the projection of δa,λ on Ᏼ(Ω) := H 2 (Ω) ∩ H01 (Ω), defined by
Δ2 Pδa,λ = Δ2 δa,λ
in Ω,
ΔPδa,λ = Pδa,λ = 0 on ∂Ω.
(1.6)
Let
u =
Ω
|Δu|2
θa,λ = δa,λ − Pδa,λ ,
1/2
u,v =
,
Ω
ΔuΔv,
u,v ∈ H 2 (Ω) ∩ H01 (Ω)
(1.7)
uq = |u|Lq (Ω) .
Thus we have the following result.
Theorem 1.1. Let (uε ) be a solution of (Pε ), and assume that
2 −2
u ε u ε 10−ε −→ S
as ε −→ 0,
(H)
where S is the best Sobolev constant in R5 defined by (1.5). Then (up to a subsequence) there
exist aε ∈ Ω, λε > 0, αε > 0 and vε such that uε can be written as
uε = αε Pδaε ,λε + vε
(1.8)
with αε → 1, vε → 0, aε ∈ Ω and λε d(aε ,∂Ω) → +∞ as ε → 0.
In addition, aε converges to a critical point x0 ∈ Ω of ϕ and we have
2
lim εuε L∞ (Ω) = c1 c02 /c2 ϕ x0 ,
ε→0
where c1 = c010
c0 = (105)1/8 .
R5 (dx/(1 + |x |
2 )9/2 ),
c2 = c010
R5 (log(1 + |x |
(1.9)
2 )(1 − |x |2 )/(1 + |x |2 )6 )dx
and
Our next result provides a kind of converse to Theorem 1.1.
Theorem 1.2. Assume that x0 ∈ Ω is a nondegenerate critical point of ϕ. Then there exists
an ε0 > 0 such that for each ε ∈ (0,ε0 ], (Pε ) has a solution of the form
uε = αε Pδaε ,λε + vε
(1.10)
with αε → 1, vε → 0, aε → x0 and λε d(aε ,∂Ω) → +∞ as ε → 0.
Our strategy to prove the above results is the same as in higher dimensions. However,
as usual in elliptic equations involving critical Sobolev exponent, we need more refined
estimates of the asymptotic profiles of solutions when ε → 0 to treat the lower dimensional
case. Such refined estimates, which are of self interest, are highly nontrivial and use in a
crucial way careful expansions of the Euler-Lagrange functional associated to (Pε ), and its
gradient near a small neighborhood of highly concentrated functions. To perform such
4
Single blow-up solutions for a biharmonic equation
expansions we make use of the techniques developed by Bahri [2] and Rey [25, 27] in the
framework of the Theory of critical points at infinity.
The outline of the paper is the following: in Section 2 we perform some crucial estimates needed in our proofs and Section 3 is devoted to the proof of our results.
2. Some crucial estimates
In this section, we prove some crucial estimates which will play an important role in
proving our results. We first recall some results.
Proposition 2.1 [8]. Let a ∈ Ω and λ > 0 such that λd(a,∂Ω) is large enough. For θ(a,λ) =
δ(a,λ) − Pδ(a,λ) , we have the following estimates
0 ≤ θ(a,λ) ≤ δ(a,λ) ,
θ(a,λ) = c0 λ−1/2 H(a, ·) + f(a,λ) ,
(2.1)
where f(a,λ) satisfies
f(a,λ) = O
1
λ5/2 d3
λ
,
1 ∂ f(a,λ)
1
= O 7/2 4 ,
λ ∂a
λ d
∂ f(a,λ)
1
= O 5/2 3 ,
∂λ
λ d
(2.2)
where d is the distance d(a,∂Ω),
θ(a,λ) ∂θ(a,λ) λ
∂λ L10
L10
θ(a,λ) = O (λd)−1/2 ,
1 ∂θ(a,λ) 1
=O
.
λ ∂a 10
(λd)3/2
= O (λd)−1/2 ,
=O
1
,
(λd)1/2
(2.3)
L
Proposition 2.2 [5]. Let uε be a solution of (Pε ) which satisfies (H). Then, there exist
aε ∈ Ω, αε > 0, λε > 0 and vε such that
uε = αε Pδaε ,λε + vε
(2.4)
with αε → 1, λε d(aε ,∂Ω) → ∞, c0−2 uε 2∞ /λε → 1, uε ε∞ → 1 and vε → 0.
Furthermore, vε ∈ E(aε ,λε ) which is the set of v ∈ Ᏼ(Ω) such that
v,Pδaε ,λε = v,∂Pδaε ,λε /∂λε = 0,
vε ,∂Pδaε ,λε /∂a = 0.
(V0 )
Lemma 2.3 [5]. λεε = 1 + o(1) as ε goes to zero implies that
2 −n)/2
δε−ε − c0−ε λε(4
= O ε log 1 + λ2ε x − aε ε
in Ω,
(2.5)
where δε = δaε ,λε and dε = d(aε ,∂Ω).
Proposition 2.4 [5]. Let (uε ) be a solution of (Pε ) which satisfies (H). Then vε occurring
in Proposition 2.2 satisfies
vε ≤ C ε + λε dε −1 ,
where C is a positive constant independent of ε.
(2.6)
Khalil El Mehdi 5
Now, we are going to state and prove the crucial estimates needed in the proof of our
theorems. In order to simplify the notations, we set δε = δaε ,λε , Pδε = Pδaε ,λε , θε = θaε ,λε
and dε = d(aε ,∂Ω).
Lemma
2.5. For ε small, we have the following estimates
(i) Ω δε9 (1/λε )(∂Pδε /∂a) = −(c1 /2λ2ε )(∂H/∂a)(aε ,aε ) + O(1/(λε dε )3 ),
)(∂H/∂a)(aε ,aε ) + O(1/(λε dε )3 + ε/(λε dε )2 ),
(ii) Ω Pδε9−ε (1/λε )(∂Pδε /∂a) = −(c1 /λ2+ε/2
ε
where c1 is the constant defined in Theorem 1.1.
Proof. Notice that
δε10
Ω\Bε
1
=O
.
(λε dε )5
(2.7)
Thus, we have, for 1 ≤ k ≤ 5
Ω
δε9
1 ∂Pδε
=
λε ∂ak
Ω
δε9
1 ∂δε
−
λε ∂ak
δε9
Ω
1 ∂θε
=−
λε ∂ak
Bε
δε9
1 ∂θε
1
+O
,
λε ∂ak
(λε dε )5
(2.8)
where Bε = B(aε ,dε ). Expanding ∂θε /∂ak around aε and using Proposition 2.1, we obtain
1 ∂θε
c0 ∂H aε ,aε
δε9
= 3/2
λ
∂a
∂a
2λ
Bε
ε
k
ε
⎛
⎞
1 ⎠
δε9 + O ⎝ 3 .
Bε
λε dε
(2.9)
Estimating the integral on the right-hand side in (2.9) and using (2.8), we easily derive
claim (i). To prove claim (ii), we write
Ω
Pδε9−ε
1 ∂Pδε
=
λε ∂ak
Ω
δε9−ε
1 ∂δε
−
λε ∂ak
(9 − ε)(8 − ε)
+
2
Ω
Ω
δε9−ε
1 ∂θε
− (9 − ε)
λε ∂ak
∂δε
+O
λ ∂ak
1
δε7−ε θε2
Ω
Ω
δε8−ε θε
1 ∂δε
λ ∂ak
1 ∂θε + δ 7−ε θ 3
ε
ε
λ ∂a δε8−ε θε ε
k
(2.10)
and we have to estimate each term on the right-hand side of (2.10).
Using Proposition 2.1 and Lemma 2.3, we have
Ω
Ω
δε7−ε θε3
3
≤ cθε ∞
δε7
1
=O
,
(λε dε )3
1 ∂θε 1
8
≤ cθε 1 ∂θε δ =O
.
∞ λε ∂ak λε ∂ak ∞ Ω ε
(λε dε )3
(2.11)
δε8−ε θε We also have
Ω
δε9−ε
1 ∂δε
=
λε ∂ak
Ω\Bε
δε9−ε
1 ∂δε
1
=O
.
λε ∂ak
(λε dε )5
(2.12)
6
Single blow-up solutions for a biharmonic equation
Expanding θε around aε and using Proposition 2.1 and Lemma 2.3, we obtain
9
Bε
δε8−ε θε
ε
1 ∂δε
c1 ∂H(aε ,aε )
1
=
+
,
+O
3
2
λε ∂ak 2λ2+ε/2
∂a
(λ
d
)
(λ
d
ε ε
ε ε)
ε
δε7−ε θε2
Bε
1 ∂δε
1
=O
.
λε ∂ak
(λε dε )3
(2.13)
In the same way, we find
⎛
⎞
ε ⎠
1 ∂θε
c1 ∂H aε ,aε
1
δε9−ε
= 2+ε/2
+ O⎝
3 + 2 .
λ
∂a
∂a
2λε
Ω
ε
k
λε dε
λε dε
(2.14)
Combining (2.10)–(2.14), we obtain claim (ii).
To improve the estimates of the integrals involving vε , we use an idea of Rey [27],
namely we write
vε = Πvε + wε ,
(2.15)
where Πvε denotes the projection of vε onto H 2 ∩ H01 (Bε ), that is
Δ2 Πvε = Δ2 vε
ΔΠvε = Πvε = 0
in Bε ;
on ∂Bε ,
(2.16)
where Bε = B(aε ,dε ). We split Πvε in an even part Πvεe and an odd part Πvεo with respect
to (x − aε )k , thus we have
vε = Πvεe + Πvεo + wε
in Bε with Δ2 wε = 0 in Bε .
(2.17)
Notice that it is difficult to improve the estimate (2.6) of the vε -part of solutions. However,
it is sufficient to improve the integrals involving the odd part of vε with respect to (x −
aε )k , for 1 ≤ k ≤ 5 and to know the exact contribution of the integrals containing the
wε -part of vε . Let us start by the terms involving wε .
Lemma 2.6. For ε small, we have that
Bε
δε8
1
δε−ε − ε ε/2
c0 λε
⎛
⎞
εvε ⎠
1 ∂δε
wε = O ⎝ 1/2 .
λε ∂ak
λε dε
(2.18)
Proof. Let ψ be the solution of
Δψ
2
= δε8
−ε
δε
1
− ε ε/2
c0 λε
1 ∂δε
λε ∂ak
in Bε ;
Δψ = ψ = 0 on ∂Bε .
(2.19)
Thus we have
Iε :=
Bε
Δ2 ψwε =
∂Bε
∂ψ
Δwε +
∂ν
∂Bε
∂Δψ
wε .
∂ν
(2.20)
Khalil El Mehdi 7
Let Gε be the Green’s function for the biharmonic operator on Bε with the Navier boundary conditions, that is,
Δ2 Gε (x, ·) = cδx
ΔGε (x, ·) = G(x, ·) = 0 on ∂Bε ,
in Bε ;
(2.21)
where c = 3w5 . Therefore ψ is given by
ψ(y) =
Bε
Gε (x, y)δε8 δε−ε −
1
c0ε λε/2
ε
1 ∂δε
,
λε ∂ak
y ∈ Bε
(2.22)
and its normal derivative by
∂ψ
(y) =
∂ν
Bε
∂Gε
1
(x, y)δε8 δε−ε − ε ε/2
∂ν
c0 λε
1 ∂δε
,
λε ∂ak
y ∈ ∂Bε .
(2.23)
Notice that for y ∈ ∂Bε we have the following estimates: for x ∈ Bε \ B(y,dε /2), we have
1
∂Gε
(x, y) = O 2 ;
∂ν
dε
∂ΔGε
1
(x, y) = O 4
∂ν
dε
(2.24)
for x ∈ Bε ∩ B(y,dε /2), we have
∂Gε
c
≤
(x,
y)
∂ν
|x − y |2 ;
∂ΔGε
c
≤
(x,
y)
∂ν
|x − y |4
(2.25)
for x ∈ Bε ∩ B(y,dε /2), we have
δε8 δε−ε −
ε logλε dε
1 ∂δε
=O
,
λε ∂ak
(λε dε )9
(2.26)
2 1 ∂δε
= O δε9 ε log 1 + λ2ε x − aε .
λε ∂ak
(2.27)
1
c0ε λε/2
ε
for x ∈ Bε \ B(y,dε /2), we have
δε8 δε−ε −
1
c0ε λε/2
ε
Therefore
∂ψ
∂ν (y) = O
ε
.
λ1/2
ε dε2
(2.28)
In the same way, we have
∂Δψ
∂ν (y) = O
ε
λ1/2
ε dε4
.
(2.29)
Using (2.20), (2.28), (2.29), we obtain
Iε = O
ε
λ1/2
ε dε2
∂Bε
Δwε +
ε
λ1/2
ε dε4
∂Bε
wε .
(2.30)
8
Single blow-up solutions for a biharmonic equation
To estimate the right-hand side of (2.30), we introduce the following function
w̄(X) = dε1/2 wε aε + dε X ,
v̄(X) = dε1/2 vε aε + dε X
for X ∈ B(0,1).
(2.31)
w̄ satisfies
Δ2 w̄ = 0
Δw̄ = Δv̄,
in B := B(0,1);
w̄ = v̄ on ∂B.
(2.32)
We deduce that
∂B
|Δw̄ | +
∂B
|Δw̄ | ≤ C
1/2
|Δv̄ |
2
B
=C
Bε
Δvε 2
1/2
.
(2.33)
But, we have
1
|Δw̄ | +
|Δw̄ | =
d
∂B
∂B
ε
3/2 ∂Bε
Δwε +
1
dε
7/2 ∂Bε
wε .
(2.34)
Using (2.30), (2.33) and (2.34), the lemma follows.
Lemma
2.7. For ε small, we have
(i) Bε Δ((1/λε )(∂Πδε /∂ak ))Δwε = O(vε /(λε dε )3/2 ),
(ii) Bε δε8−ε Πvεo wε = O(vε Πvεo /(λε dε )1/2 ).
Proof. Using (2.17), we obtain
1 ∂Πδε
Δ
Δwε =
λε ∂ak
Bε
∂ψk
Δwε ,
∂ν
∂Bε
with ψk =
1 ∂Πδε
.
λε ∂ak
(2.35)
Using an integral representation for ψk as in (2.23), we obtain for y ∈ ∂Bε ,
∂ψ
(y) =
∂ν
Bε
∂Gε
(x, y)Δ2 ψk ,
∂ν
(2.36)
where Gε is the Green’s function defined in (2.21). Clearly, we have
c0 λ1/2
ε
Πδε (x) = δε (x) − 1/2
1 + λ2ε dε2
cε aε ,dε x − aε 2 − d 2 ,
−
ε
10
(2.37)
with cε (aε ,dε ) = Δδε|∂Bε . Thus we deduce that
∂ψ
(y) = 9
∂ν
Bε
∂Gε
1 ∂δε
.
(x, y)δε8
∂ν
λε ∂ak
(2.38)
In Bε \ B(aε ,dε /2), we argue as in (2.28) and (2.25), we obtain
Bε
∂Gε
1 ∂δε
1
= O 9/2
.
(x, y)δε8
∂ν
λε ∂ak
λε dε6
(2.39)
Khalil El Mehdi 9
Furthermore, since
∂Gε
1
∇
∂ν (x, y) = O d 3
for (x, y) ∈ B aε ,dε /2 × ∂Bε ,
ε
(2.40)
we obtain
∂Gε
1 ∂δε ≤ c
(x, y)δε8
d3
∂ν
λ
∂a
B(aε ,dε /2)
ε
k
ε
B(aε ,dε /2)
δε9 x − aε = O
1
λ3/2
ε dε3
,
(2.41)
where we have used the evenness of δε and the oddness of its derivative. Thus
∂ψk
1
(y) = O 3/2 3 .
∂ν
λε dε
(2.42)
Using (2.35) and (2.42), we obtain
Bε
Δ
1 ∂Πδε
c
Δwε ≤ 3/2 3
λε ∂ak
λε dε
∂Bε
Δwε .
(2.43)
Arguing as in (2.34), claim (i) follows. To prove claim (ii), let ψ be such that
Δ2 ψ = δε8−ε Πvεo
in Bε ;
Δψ = ψ = 0 on ∂Bε .
(2.44)
We have
Bε
δε8−ε Πvεo wε =
∂Bε
∂Δψ
wε +
∂ν
∂Bε
∂ψ
Δwε .
∂ν
(2.45)
As before, we prove that, for y ∈ ∂Bε
o
Πv ∂ψ
(y) = O 1/2 ε 2 ,
∂ν
λε dε
o
Πv ∂Δψ
(y) = O 1/2 ε 4 .
∂ν
λε dε
(2.46)
Therefore
Bε
δε8−ε Πvεo wε
cΠvo 1
≤ 1/2 ε
λε dε4 δε3/2
∂Bε
wε + 1
δε7/2
∂Bε
o
Δwε ≤ c vε Πvε .
1/2
The proof of the lemma is completed.
λε dε
(2.47)
Lemma
2.8. For ε small, we have
1/2
(i) Bε δε7−ε vε (1/λε )(∂δε /∂ak ) = O(Πvεo /λ1/2
ε + vε /λε dε ),
(ii) Bε δε7−ε θε vε (1/λε )(∂δε /∂ak ) = O(Πvεo /λε dε + vε /(λε dε )3/2 ).
Proof. Claim (i) can be proved in the same way as Lemma 2.6, so we omit its proof. Claim
(ii) follows from Proposition 2.1 and claim (i).
10
Single blow-up solutions for a biharmonic equation
Let us now compute the contribution of the following integral which involves vε2 .
Lemma 2.9. For ε small, we have
⎛
2 ⎞
vε 1 ∂δε
δε7−ε vε2
= O ⎝Πvεo vε + 1/2 ⎠ .
λε ∂ak
Bε
λε dε
(2.48)
Proof. Using (2.17) and the fact that the even part of vε2 has no contribution to the integrals, we obtain
Bε
δε7−ε vε2
1 ∂δε
=
λε ∂ak
Bε
δε7−ε
1 ∂δε 2vε − wε wε + O Πvεo vε .
λε ∂ak
(2.49)
Let Ψ be the solution of
Δ2 Ψ = δε7−ε
1 ∂δε 2vε − wε in Bε ;
λε ∂ak
ΔΨ = Ψ = 0 on ∂Bε .
(2.50)
Thus, as in the proof of Lemma 2.6, we obtain for y ∈ ∂Bε
vε ∂Ψ
(y) = O 1/2 2 ,
∂ν
λε dε
vε ∂ΔΨ
(y) = O 1/2 4
∂ν
λε dε
(2.51)
and therefore
⎛ ⎞
2
Bε
δε7−ε
vε 1 ∂δε 2vε − wε wε = O ⎝ 1/2 ⎠ .
λε ∂ak
λε dε
(2.52)
Thus our lemma follows.
Next we are going to estimate the integrals involving the odd part of vε with respect to
(x − aε )k , for 1 ≤ k ≤ 5.
Lemma 2.10. For ε small, we have
Bε
u9ε −ε Πvεo = 9
8
Bε
o 2
δε Πvε
⎛
⎛
o 3/2
o 2 + o Πv + O ⎝Πv ⎝ε + ε
⎞⎞
1
3/2 ⎠⎠ .
λε dε
ε
(2.53)
Proof. We have
Bε
u9ε −ε Πvεo = α9ε −ε
Bε
Pδε9−ε Πvεo + (9 − ε)α8ε −ε
+O
= α9ε −ε
Bε
Bε
2
Pδε7−ε vε Πvεo +
Bε
Bε
Pδε9−ε Πvεo + (9 − ε)α8ε −ε
Pδε8−ε vε Πvεo
9−ε o vε Πv ε
Bε
2 Pδε8−ε vε Πvεo + O vε Πvεo .
(2.54)
Khalil El Mehdi 11
We estimate the two integrals on the right-hand side in (2.54). First, using Proposition 2.1
and the Holder inequality, we have
Bε
Pδε8−ε vε Πvεo
=
δε8−ε vε Πvεo + O
Bε
vε Πv o ε
λε dε
=
Bε
δε8−ε
2
Πvεo +
Bε
δε8−ε Πvεo wε ,
(2.55)
where we have used in the last equality the evenness of δε and Πvεe and the oddness of
Πvεo . By Lemmas 2.3 and 2.7 we obtain
Bε
Pδε8−ε vε Πvεo =
Bε
δε8 Πvεo
2
+O
vε Πv o ε
(λε dε )1/2
.
(2.56)
Secondly, we write
Bε
Pδε9−ε Πvεo =
Bε
δε9−ε Πvεo − (9 − ε)
Bε
δε8−ε θε Πvεo + O
Bε
δε7−ε θε2 Πvεo .
(2.57)
Thus, using the evenness of δε , the oddness of Πvεo and Holder inequality, we obtain
Bε
⎛
⎞
Πv o Pδε9−ε Πvεo = O ⎝ ε2 ⎠ .
λε dε
(2.58)
Using (2.54), (2.56), (2.58) and Propositions 2.2 and 2.4, we easily derive our lemma.
Lemma 2.11. For ε small, we have
⎛
o
Πv = O ⎝ε3/2 + ⎞
1
3/2 ⎠ .
λε dε
ε
(2.59)
Proof. We write
o
ε + αΠδε + βλε
Πvεo = Πv
∂Πδε 1 ∂Πδε
+ γr
∂λ
λε ∂ar
r =1
5
(2.60)
with
εo ,
εo ,Πδε = Πv
Πv
∂Πδε
∂λ
εo , ∂Πδε = 0
= Πv
∂ar
for each r ∈ {1,2,3,4,5}.
(2.61)
12
Single blow-up solutions for a biharmonic equation
Taking the scalar product in H 2 ∩ H01 (Bε ) of (2.60) with Πδε , λε ∂Πδε /∂λ, λ−ε 1 ∂Πδε /∂ar , 1 ≤
r ≤ 5, provides us with the following invertible linear system in α, β, γr (with 1 ≤ r ≤ 5)
λε
∂Πδε
= α C + o(1) + β Πδε ,λε
∂λ
Πδε ,Πvεo
∂Πδε
∂Πδε
,Πvεo = α Πδε ,λε
∂λ
∂λ
1 ∂Πδε
1 ∂Πδε
,Πvεo = α Πδε ,
λε ∂ak
λε ∂ak
+
5
γr
r =1
+
5
γr
r =1
5
+ β C + o(1) +
γr λ ε
r =1
1 ∂Πδε
Πδε ,
λε ∂ar
∂Πδε 1 ∂Πδε
+ β λε
,
∂λ λε ∂ak
∂Πδε 1 ∂Πδε
,
∂λ λε ∂ar
(S)
1 ∂Πδε 1 ∂Πδε
,
.
λε ∂ak λε ∂ar
Observe that
∂Πδε
Πδε ,λε
∂λ
∂Πδε 1 ∂Πδε
λε
,
∂λ λε ∂ar
1 ∂Πδε
Πδε ,
λε ∂ar
1 ∂Πδε 1 ∂Πδε
,
λε ∂ak λε ∂ar
1
;
=O
λε dε
⎛
⎞
1 ⎠
= O⎝
2 ;
λε dε
⎛
⎞
(2.62)
1 ⎠
= O⎝
2 ;
λε dε
⎛
⎞
1 ⎠
= C + o(1) δkr + O ⎝ 2 ,
λε dε
where δkr denotes the Kronecker symbol.
Now, because of the evenness of δε and the oddness of Πvεo with respect to (x − aε )k
we obtain
Πδε ,Πvεo =
Bε
ΔΠδε · ΔΠvεo =
Bε
δε9 Πvεo = 0.
(2.63)
In the same way we have
λε
∂Πδε
,Πvεo =
∂λ
1 ∂Πδε
,Πvεo = 0 for each r = k.
λε ∂ar
(2.64)
We also have
1 ∂Πδε
,Πvεo =
λε ∂ak
1 ∂Πδε
Δ
· Δ vε − Πvεe − wε
λε ∂ak
Bε
1 ∂Πδε
Δ
· Δvε −
=
λ
Bε
ε ∂ak
1 ∂Πδε
Δ
· Δwε ,
λ
Bε
ε ∂ak
(2.65)
Khalil El Mehdi 13
where we have used in the last equality the fact that Πvεe is even with respect to (x − aε )k .
Using (2.37) and Holder inequality, we obtain
1 ∂Πδε
Δ
· Δvε ≤ cvε λ
∂a
Bε
ε
k
Ω\Bε
⎛ ⎞
1/2
vε 1 ∂δε 2
Δ
= O⎝
3/2 ⎠ .
λ ∂a ε
λε dε
k
(2.66)
Equation (2.66) and Lemma 2.7 imply that
⎛
⎞
vε 1 ∂Πδε
,Πvεo = O ⎝ 3/2 ⎠ .
λε ∂ak
λε dε
(2.67)
Inverting the linear system (S), we deduce from the above estimates
⎛
⎞
vε 7/2 ⎠ ,
α = O⎝
⎛
⎛
β = O⎝
λε dε
⎞
vε γk = O ⎝ 3/2 ⎠ ,
⎛
⎞
vε 7/2 ⎠ ,
λε dε
⎞
vε γr = O ⎝ 7/2 ⎠ ,
λε dε
λε dε
(2.68)
r = k.
This implies through (2.60)
⎛ ⎞
vε o
Πv − Πv
εo = O ⎝ 3/2 ⎠ ,
ε
⎛ ⎞
v 2
o 2 o 2
Πv = Πv
ε + O⎝ ε ⎠.
ε
3
λε dε
λε dε
(2.69)
εo . Since uε is a solution of
We now turn to the last step, which consists in estimating Πv
(Pε ), we have
Bε
Δ2 uε Πvεo =
Bε
u9ε −ε Πvεo .
(2.70)
Because of the evenness of δε and the oddness of Πvεo with respect to (x − aε )k , (2.70)
becomes
o 2
Πv =
ε
Bε
u9ε −ε Πvεo .
(2.71)
By (2.69), (2.71) and Lemma 2.10, we obtain
Πv
o 2 − 9
ε
Bε
δε8
⎛
2
εo 2 = O ⎝ε3 +
Πvεo + o Πv
⎞
1 ⎠
3 .
λε dε
(2.72)
Using now (2.72) and the fact that the quadratic form
v −→
Bε
|Δv |2 − 9
Bε
δi8 v2
(2.73)
14
Single blow-up solutions for a biharmonic equation
is positive definite (see [6]) on the subset [Span(Πδε ,∂Πδε /∂λ,∂Πδε /∂ak 1 ≤ k ≤
5)]⊥H 2 ∩H01 (Bε ) , we obtain
⎛
Πv
o ≤ C ⎝ ⎞
1
3/2
3/2 + ε ⎠ .
λε dε
ε
(2.74)
Our lemma follows from (2.69) and (2.74).
Before ending this section, let us prove the following estimate which will be needed
later.
Lemma 2.12. For ε small, we have
⎛
⎞
∂2 Pδε
1
3/2
,vε = O ⎝ 3/2 + ε ⎠ .
∂λ∂ak
λε dε
(2.75)
Proof. We have
∂2 Pδε
Δ
Δvε =
∂λ∂ak
Ω
Bε
Δ2
=
Δ
2
Bε
⎛
⎞
vε ∂2 Pδε
vε + O ⎝ 9/2 ⎠
∂λ∂ak
λε dε
∂2 Pδε
Πvεo +
∂λ∂ak
Δ
2
Bε
⎛
⎞
vε ∂2 Pδε
wε + O ⎝ 9/2 ⎠ .
∂λ∂ak
λε dε
(2.76)
For the first integral on the right-hand side in (2.76), we have
Bε
Δ2
⎛
⎞
∂2 Pδε
1
3/2
Πvεo = O Πvεo = O ⎝ 3/2 + ε ⎠ ,
∂λ∂ak
λε dε
(2.77)
where we have used in the last equality Lemma 2.11.
Now let ψ4 be the solution of
Δ2 ψ4 = Δ2
∂2 Pδε
∂λ∂ak
in Bε ,
Δψ4 = ψ4 = 0 on ∂Bε .
(2.78)
Thus, as in the proof of Lemma 2.6, we obtain for y ∈ ∂Bε
∂ψ4
1
(y) = O 1/2 2 ,
∂ν
λε dε
∂Δψ4
1
(y) = O 1/2 4
∂ν
λε dε
(2.79)
and therefore
Δ
2
Bε
⎛
⎞
vε ∂2 Pδε
wε = O ⎝ 1/2 ⎠ .
∂λ∂ak
λε dε
From (2.76), (2.77), (2.80) and Proposition 2.4, we easily deduce our lemma.
(2.80)
Khalil El Mehdi 15
3. Proof of theorems
Let us start by proving the following crucial result.
Proposition 3.1. For uε = αε Pδaε ,λε + vε solution of (Pε ) with λεε = 1 + o(1) as ε goes to
zero, we have the following estimates
(a) c2 ε + O(ε2 ) − c1 (H(aε ,aε )/λε ) + o(1/λε dε ) = 0,
(b) (c3 /λ2ε )(∂H(aε ,aε )/∂a)+o(1/(λε dε )2 )+O(ε5/2 +ε log(λε )/(λε dε )2 +1/(λε dε )5/2 ) = 0,
where c1 , c2 are the constants defined in Theorem 1.1, and where c3 > 0.
Proof. Since claim (a) was proved in [5], we only need to prove claim (b). Multiplying
(Pε ) by (1/λε )(∂Pδε /∂ak ) and integrating on Ω, we obtain for 1 ≤ k ≤ 5
1 ∂Pδε
1 ∂Pδε
− u9ε −ε
λε ∂ak
λε ∂ak
Ω
Ω
9−ε
8−ε
1 ∂Pδε
= αε δε9
−
αε Pδε
+ (9 − ε) αε Pδε
vε
λε ∂ak
Ω
Ω
7−ε 2 1 ∂Pδε
3 (9 − ε)(8 − ε) +
αε Pδε
vε
+ O vε .
2
λε ∂ak
0=
Δ2 uε
(3.1)
We estimate each term on the right-hand side in (3.1). First, by Proposition 2.1 and the
Holder inequality, we have
Ω
Pδε7−ε vε2
1 ∂Pδε
=
λε ∂ak
Ω
δε7−ε vε2
2
vε 1 ∂δε
+O
.
λε ∂ak
λε dε
(3.2)
Secondly, we compute
Ω
Pδε8−ε vε
1 ∂Pδε
1 ∂Pδε
+ (8 − ε) δε7−ε θε vε
+O
δε7−ε θε2 vε λε ∂ak
λε ∂ak
Ω
Ω
Ω
1 ∂δε
8−ε
8−ε 1 ∂θε δε vε
+O
δε vε =
λε ∂ak
λε ∂ak Ω
Ω
1 ∂δε
+ (8 − ε) δε7−ε θε vε
λε ∂ak
Ω
7−ε 1 ∂θε 7−ε 2 +O
δε θε vε +
O
δ
θ
v
.
ε
ε
ε
λε ∂ak Ω
Ω
(3.3)
1 ∂Pδε
=
λε ∂ak
δε8−ε vε
By Proposition 2.1 and the Holder inequality, we obtain
Ω
⎛ ⎞
vε δε7−ε θε2 vε = O ⎝ 2 ⎠ ,
Ω
λε dε
Ω
⎛ ⎞
1
∂θ
ε
⎝ vε ⎠ ,
δε7−ε θε vε λ ∂a = O 3
ε
k
⎛ ⎞
vε 8−ε 1 ∂θε = O⎝
δε vε 2 ⎠ .
λ ∂a
ε
k
λε dε
λε dε
(3.4)
16
Single blow-up solutions for a biharmonic equation
We also have by Proposition 2.2
Ω
δε8−ε vε
1 ∂δε
=
λε ∂ak
Ω
δε8
Bε
c −ε
1 ∂δε
− 0ε/2 vε
λ
λε
ε ∂ak
−ε
vε c −ε
1 ∂δε
− 0ε/2 vε
+ O⎝
9/2 ⎠ .
λε ∂ak
λε
λε dε
δε
=
−ε
δε8
⎛
δε
⎞
(3.5)
Using (2.17), Lemma 2.3 and the Holder inequality, we derive that
1 ∂δε
δε8−ε vε
=
λε ∂ak
Ω
Bε
δε8
⎛
⎛
⎞
vε c−ε 1 ∂δε
δε−ε − 0ε/2
wε + O ⎝εΠvεo + 9/2 ⎠
λε ∂ak
λε
λε dε
⎞
vε o
εvε = O⎝
1/2 + εΠvε + 9/2 ⎠ ,
λε dε
λε dε
(3.6)
where we have used Lemma 2.6 in the last equality.
Using (3.2)–(3.6), Lemmas 2.5, 2.8, 2.9, Proposition 2.2 and the fact that λεε = 1 +
O(ε log λε ), we easily derive our result.
We are now able to prove Theorem 1.1.
Proof of Theorem 1.1. Let (uε ) be a solution of (Pε ) which satisfies (H). Then, using
Proposition 2.2, uε = αε Pδaε ,λε + vε with αε → 1, λεε → 1, λε d(aε ,∂Ω) → ∞, vε satisfies (V0 )
and vε → 0. Now, using claim (a) of Proposition 3.1, we derive that
c H aε ,aε
1
ε= 1
+o
c2
λε
λε dε
1
.
=O
λε dε
(3.7)
Therefore, it follows from claim (b) of Proposition 3.1 that
∂H aε ,aε
1
=o 2 .
∂a
dε
(3.8)
Using (3.8) and the fact that for a near the boundary (∂H/∂a)(aε ,aε ) ∼ cd(aε ,∂Ω)−2 , we
derive that aε is away from the boundary and it converges to a critical point x0 of ϕ.
Finally, using (3.7), we obtain
ελε −→
c1 ϕ x0
c2
as ε −→ 0.
(3.9)
By Proposition 2.2, we have
2
u ε ∞ ∼ c 2 λ ε
0
L
as ε −→ 0.
This concludes the proof of Theorem 1.1.
The sequel of this section is devoted to the proof of Theorem 1.2.
(3.10)
Khalil El Mehdi 17
Proof of Theorem 1.2. Let x0 be a nondegenerate critical point of ϕ. It is easy to see that
d(a,∂Ω) > d0 > 0 for a near x0 . We will take a function u = αPδ(a,λ) + v where (α − α0 )
is very small, λ is large enough, v is very small, a is close to x0 and α0 = S−5/8 and we
will prove that we can choose the variables (α,λ,a,v) so that u is a critical point of Jε with
u = 1. Here Jε denotes the functional corresponding to problem (Pε ) defined by
Jε (u) =
Ω
|Δu|2
Ω
|u|10−ε
−2/(10−ε)
.
(3.11)
Let
Mε = (α,λ,a,v) ∈ R∗+ × R∗+ × Ω × Ᏼ(Ω)/ α − α0 < ν0 ,
da > d0 , λ > ν−0 1 , ε log λ < ν0 , v < ν0 and v ∈ E(a,λ) ,
(3.12)
where ν0 and d0 are two suitable positive constants and where da = d(a,∂Ω).
Let us define the functional
Kε (α,a,λ,v) = Jε αPδ(a,λ) + v .
Kε : Mε −→ R,
(3.13)
It is known that (α,λ,a,v) is a critical point of Kε if and only if u = αPδ(a,λ) + v is a critical
point of Jε on Ᏼ(Ω). So this fact allows us to look for critical points of Jε by successive
optimizations with respect to the different parameters on Mε .
First, arguing as in [25, Proposition 4] we see that the following problem
min Jε αPδ(a,λ) + v , v satisfying V0 and v < ν0
(3.14)
is achieved by a unique function v which satisfies the estimate of Proposition 2.4. This
implies that there exist A, B and Ci ’s such that
∂
∂Kε
∂
Pδ(a,λ) ,
(α,λ,a,v) = ∇Jε αPδ(a,λ) + v = APδ(a,λ) + B Pδ(a,λ) + Ci
∂v
∂λ
∂a
i
i =1
5
(3.15)
where ai is the ith component of a.
According to [5], we have that
A = O ε log λ + |β| +
1
,
λ
B = O λε + 1 ,
Cj = O
ε2 1
.
+
λ λ3
(3.16)
To find critical points of Kε , we have to solve the following system
∂Kε
=0
∂α
2
2
5
∂Kε
∂ Pδ
∂ Pδ
=B
,
v̄
+
C
,
v̄
i
∂λ
∂λ2
∂λ∂ai
i=1
∂Kε
∂2 Pδ
∂2 Pδ
=B
, v̄ + Ci
, v̄ ,
∂a j
∂λ∂a j
∂ai ∂a j
i=1
5
for each j = 1,...,5.
(E1 )
18
Single blow-up solutions for a biharmonic equation
Observe that for ψ = Pδ(a,λ) , ∂Pδ(a,λ) /∂λ, ∂Pδ(a,λ) /∂ai with i = 1,...,5 and for u = αPδ(a,λ) +
v, we have
∇Jε (u),ψ = 2Jε (u) α Pδ(a,λ) ,ψ − Jε (u)5−ε/2 |u|8−ε uψ .
Ω
(3.17)
We also have (see [5])
Jε αPδ(a,λ) + v = S + O ε log λ +
1
,
λ
(3.18)
∂Kε 1
= ∇Jε (αPδ + v),Pδ = 2Jε (u) αS5/4 1 − α8 S5 + O ε logλ +
,
∂α
λ
∂K
∂Pδ
H(a,a) λ ε = ∇Jε (αPδ + v),λ
1 − 2α8 S5 + c2 S5 α9 ε
= Jε (u) αc1
∂λ
∂λ
λ
ε logλ 1
+ 3 .
+ O ε2 log λ +
λ
λ
(3.19)
(3.20)
Following the proof of claim (b) of Proposition 3.1, we obtain, for each j = 1,...,5,
1 ∂Pδ
1 ∂Kε
= ∇Jε (αPδ + v),
λ ∂a j
λ ∂a j
ε logλ
1
cα ∂H(a,a) 1 − 2α8 S5 + O ε5/2 + 2 + 5/2 .
=− 2
2λ
∂a
λ
λ
(3.21)
On the other hand, one can easily verify that
∂2 Pδ 1
O
,
=
∂λ2 λ2
∂2 Pδ = O λ2 .
∂ai ∂a j (3.22)
Now, we take the following change of variables:
α = α0 + β,
a = x0 + ξ,
1
λ1/2
=
⎛
⎞
√
c2 ⎝
1
+ ρ⎠ ε.
c1
H x0 ,x0
(3.23)
Then, using estimates (3.18)–(3.22), Lemma 2.12, Proposition 2.4 and the fact that x0 is
a nondegenerate critical point of ϕ, the system (E1 ) becomes
β = O ε| log ε| + |β|2
ρ = O ε| log ε| + |β|2 + |ξ |2 + ρ2
(E2 )
ξ = O |β|2 + |ξ |2 + ε1/2 .
Thus Brower’s fixed point theorem shows that the system (E2 ) has a solution (βε ,ρε ,ξε )
for ε small enough such that
βε = O ε| log ε| ,
ρε = O ε| logε| ,
ξε = O ε1/2 .
(3.24)
Khalil El Mehdi 19
By construction, the corresponding uε is a critical point of Jε that is wε =Jε (uε )(5−ε/2)/(8−ε) uε
satisfies
8−ε
Δ2 wε = wε wε
in Ω, wε = Δwε = 0 on ∂Ω
(3.25)
with |wε− |L10 (Ω) very small, where wε− = max(0, −wε ).
As in [7, Proposition 4.1], we prove that wε− = 0. Thus, since wε is a non-negative
function which satisfies (3.25), the strong maximum principle ensures that wε > 0 on Ω
and then wε is a solution of (Pε ), which blows up at x0 as ε goes to zero. This ends the
proof of Theorem 1.2.
References
[1] F. V. Atkinson and L. A. Peletier, Elliptic equations with nearly critical growth, Journal of Differential Equations 70 (1987), no. 3, 349–365.
[2] A. Bahri, Critical Points at Infinity in Some Variational Problems, Pitman Research Notes in Mathematics Series, vol. 182, Longman Scientific & Technical, Harlow, 1989.
[3] A. Bahri, Y. Li, and O. Rey, On a variational problem with lack of compactness: the topological
effect of the critical points at infinity, Calculus of Variations and Partial Differential Equations 3
(1995), no. 1, 67–93.
[4] M. Ben Ayed and K. El Mehdi, Existence of conformal metrics on spheres with prescribed Paneitz
curvature, Manuscripta Mathematica 114 (2004), no. 2, 211–228.
, On a biharmonic equation involving nearly critical exponent, to appear in Nonlinear
[5]
Differential Equations and Applications, 2006.
, The Paneitz curvature problem on lower dimensional spheres, to appear in Annals of
[6]
Global Analysis and Geometry.
[7] M. Ben Ayed, K. El Mehdi, and M. Hammami, Some existence results for a Paneitz type problem via the theory of critical points at infinity, Journal de Mathématiques Pures et Appliquées.
Neuvième Série 84 (2005), no. 2, 247–278.
[8] M. Ben Ayed and M. Hammami, On a fourth order elliptic equation with critical nonlinearity in
dimension six, to appear in Nonlinear Analysis TMA.
[9] H. Brezis and L. A. Peletier, Asymptotics for elliptic equations involving critical growth, Partial Differential Equations and the Calculus of Variations, Vol. 1, Progr. Nonlinear Differential Equations Appl., vol. 1, Birkhäuser Boston, Massachusetts, 1989, pp. 149–192.
[10] S.-Y. A. Chang, On a fourth-order partial differential equation in conformal geometry, Harmonic
Analysis and Partial Differential Equations (Chicago, Ill, 1996) (M. Christ, C. Kenig, and C.
Sadorsky, eds.), Chicago Lectures in Math., University of Chicago Press, Illinois, 1999, pp. 127–
150, essays in honor of Alberto P. Calderon.
[11] K.-S. Chou and D. Geng, Asymptotics of positive solutions for a biharmonic equation involving
critical exponent, Differential and Integral Equations. An International Journal for Theory &Applications 13 (2000), no. 7-9, 921–940.
[12] Z. Djadli, E. Hebey, and M. Ledoux, Paneitz-type operators and applications, Duke Mathematical
Journal 104 (2000), no. 1, 129–169.
[13] Z. Djadli, A. Malchiodi, and M. O. Ahmedou, Prescribing a fourth order conformal invariant on
the standard sphere. I. A perturbation result, Communications in Contemporary Mathematics 4
(2002), no. 3, 375–408.
, Prescribing a fourth order conformal invariant on the standard sphere. II. Blow up analysis
[14]
and applications, Annali della Scuola Normale Superiore di Pisa. Classe di Scienze. Serie V 1
(2002), no. 2, 387–434.
20
Single blow-up solutions for a biharmonic equation
[15] F. Ebobisse and M. O. Ahmedou, On a nonlinear fourth-order elliptic equation involving the critical Sobolev exponent, Nonlinear Analysis. Theory, Methods & Applications. An International
Multidisciplinary Journal. Series A: Theory and Methods 52 (2003), no. 5, 1535–1552.
[16] D. E. Edmunds, D. Fortunato, and E. Jannelli, Critical exponents, critical dimensions and the
biharmonic operator, Archive for Rational Mechanics and Analysis 112 (1990), no. 3, 269–289.
[17] V. Felli, Existence of conformal metrics on Sn with prescribed fourth-order invariant, Advances in
Differential Equations 7 (2002), no. 1, 47–76.
[18] J. Garcı́a Azorero and I. Peral Alonso, On limits of solutions of elliptic problems with nearly critical
exponent, Communications in Partial Differential Equations 17 (1992), no. 11-12, 2113–2126.
[19] F. Gazzola, H.-C. Grunau, and M. Squassina, Existence and nonexistence results for critical growth
biharmonic elliptic equations, Calculus of Variations and Partial Differential Equations 18 (2003),
no. 2, 117–143.
[20] D. Geng, On blow-up of positive solutions for a biharmonic equation involving nearly critical exponent, Communications in Partial Differential Equations 24 (1999), no. 11-12, 2333–2370.
[21] Z.-C. Han, Asymptotic approach to singular solutions for nonlinear elliptic equations involving
critical Sobolev exponent, Annales de l’Institut Henri Poincaré. Analyse Non Linéaire 8 (1991),
no. 2, 159–174.
[22] R. Lewandowski, Little holes and convergence of solutions of −Δu = u(n+2)/(n−2) , Nonlinear Analysis. Theory, Methods & Applications. An International Multidisciplinary Journal. Series A: Theory and Methods 14 (1990), no. 10, 873–888.
[23] C.-S. Lin, A classification of solutions of a conformally invariant fourth order equation in Rn , Commentarii Mathematici Helvetici 73 (1998), no. 2, 206–231.
[24] O. Rey, Proof of two conjectures of H. Brézis and L. A. Peletier, Manuscripta Mathematica 65
(1989), no. 1, 19–37.
, The role of the Green’s function in a nonlinear elliptic equation involving the critical
[25]
Sobolev exponent, Journal of Functional Analysis 89 (1990), no. 1, 1–52.
, Blow-up points of solutions to elliptic equations with limiting nonlinearity, Differential
[26]
and Integral Equations. An International Journal for Theory and Applications 4 (1991), no. 6,
1155–1167.
, The topological impact of critical points at infinity in a variational problem with lack of
[27]
compactness: the dimension 3, Advances in Differential Equations 4 (1999), no. 4, 581–616.
[28] R. C. A. M. Van der Vorst, Fourth-order elliptic equations with critical growth, Comptes Rendus
de l’Académie des Sciences. Série I. Mathématique 320 (1995), no. 3, 295–299.
Khalil El Mehdi: Faculté des Sciences et Techniques, Université de Nouakchott, BP 5026,
Nouakchott, Mauritania
E-mail address: khalil@univ-nkc.mr
Current address: The Abdus Salam ICTP, 34014 Trieste, Italy