Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 77, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SIGN-CHANGING SOLUTIONS OF A FOURTH-ORDER
ELLIPTIC EQUATION WITH SUPERCRITICAL EXPONENT
KAMAL OULD BOUH
Abstract. In this article we study the nonlinear elliptic problem involving
nearly critical exponent
∆2 u = |u|8/(n−4)+ε u
∆u = u = 0
in Ω,
on ∂Ω,
where Ω is a smooth bounded domain in Rn with n ≥ 5, and ε is a positive
parameter. We show that, for ε small, there is no sign-changing solution with
low energy which blow up at exactly two points. Moreover, we prove that this
problem has no bubble-tower sign-changing solutions.
1. Introduction and statement of results
In this article, we consider the semi-linear elliptic problem with supercritical
nonlinearity
∆2 u = |u|p−1+ε u in Ω,
(1.1)
∆u = u = 0 on ∂Ω,
where Ω is a smooth bounded domain in Rn , n ≥ 5, ε is a positive real parameter
2n
is the critical Sobolev exponent for the embedding of H 2 (Ω)∩H01 (Ω)
and p+1 = n−4
p+1
into L (Ω).
Problem (1.1) is related to the limiting problem (when ε = 0) which exhibits
a lack of compactness. In fact, van Der Vorst [25, 26] (see also [19]) showed that
(1.1) with ε = 0 has no positive solutions if Ω is a starshaped domain. Whereas
Ebobisse and Ould Ahmedou [13] proved that (1.1) with ε = 0 has a positive
solution provided that some homology group of Ω is non trivial. This topological
condition is sufficient, but not necessary, as examples of contractible domains Ω
on which a positive solution exists as shown in [14] (see also [15]). Note that
some problems of type (1.1) were studied in case of Riemannian manifolds, see for
example [17] and [20].
In view of this qualitative change of the situation for (1.1) with ε = 0, it is
interesting to study the problem (1.1) with ε < 0 and ε > 0 and to understand
what happens to the solutions of (1.1) (if they exist) as ε → 0.
Observe that, when ε < 0, the existence of solutions of (1.1) has been proved in
[5, 7] (see [3, 4, 9] for the Laplacian case) for each ε ∈ (1 − p, 0). For the positive
2000 Mathematics Subject Classification. 35J20, 35J60.
Key words and phrases. Sign-changing solutions; critical exponent; bubble-tower solution.
c 2014 Texas State University - San Marcos.
Submitted November 15, 2013. Published March 19, 2014.
1
2
K. OULD BOUH
EJDE-2014/77
solutions, Chou and Geng [10] made the first study, when Ω is a convex domain.
They gave the asymptotic behavior of the low energy positive solution. They used
the method of moving planes to show that the blow-up point is away from the
boundary of the domain. The process is standard if the domain is convex. We
note that, for non convex regions, this method still works in the Laplacian case
through the applications of Kelvin transformations, see [16] (since the problem is
invariant under these transformations). However, the Navier boundary conditions
are not invariant under the Kelvin transformation of the biharmonic operator. But
the method of moving planes also works for convex domains, see [10]. To remove
the convexity assumption, Ben Ayed and El Mehdi [5] used another method based
on some ideas introduced by Bahri in [1]. This result is the analogous one to the
one in [24] and [16] where the Laplacian operator was studied.
Concerning the supercritical case, ε > 0, the problem (1.1) becomes more delicate
since we lose the Sobolev embedding which is an important point to overcome. We
recall that, when the biharmonic operator in (1.1) is replaced by the Laplacian
one, there are many works devoted to the study of the positive solutions of the
counterpart of (1.1). It was proved in [6] that (1.1) has no positive solution which
blows up at a single point. This result shows that the situation is different from the
subcritical case. However, Del Pino et al [11] (see also [18]) gave an existence result
for two blow up points, provided that Ω satisfies some geometrical conditions. In
sharp contrast to this, very little study has been made concerning the sign-changing
solutions, see [8].
It is well known that problem (1.1) (with ε < 0) has always a positive least
energy solution uε which is obtained by solving the variational problem
R
|∆u|2
Ω
u ∈ H 2 (Ω) ∩ H01 (Ω), u 6≡ 0.
inf J(u) where J(u) := R
p+1+ε 2/(p+1+ε)
|u|
Ω
Removing the assumption of the positivity of the solutions, the study of the
asymptotic behavior becomes difficult. The main difficulty is that the limit problem,
after a change of variable, which is
∆2 u = |u|p−1 u in Rn ,
(1.2)
has many sign-changing solutions which are unknown. However, an interesting
information about the energy shows that [14, Lemma 2]
Z
|∆w|2 > 2S n/4 ,
(1.3)
Rn
for each sign-changing solution w of (1.2), where S denotes the best minimizers of
the Sobolev inequality on the whole space, that is
S = inf{|∆u|2L2 (Rn ) |u|−2
: ∆u ∈ L2 , u ∈ L2n/(n−4) , u 6≡ 0}.
L2n/(n−4) (Rn )
When we add the positivity assumption, the solutions of (1.2) are the family
δ(a,λ) (x) = c0
(1 +
λ(n−4)/2
,
− a|2 )(n−4)/2
λ2 |x
with λ > 0 and a ∈ Rn .
(n−4)/8
c0 = n(n − 4)(n2 − 4)
,
(1.4)
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SIGN-CHANGING SOLUTIONS
3
The space H 2 (Ω) ∩ H01 (Ω) is equipped with the norm k · k and its corresponding
inner product h·, ·i defined by
Z
Z
1/2
kuk =
, hu, vi =
∆u∆v, u, v ∈ H 2 (Ω) ∩ H01 (Ω).
(1.5)
|∆u|2
Ω
Ω
When we study problem (1.2) in a bounded smooth domain Ω, we need to introduce
the function P δ(a,λ) which is the projection of δ(a,λ) on H01 (Ω). It satisfies
∆2 P δ(a,λ) = ∆2 δ(a,λ)
in Ω;
∆P δ(a,λ) = P δ(a,λ) = 0
on ∂Ω.
These functions are almost positive solutions of (1.1). Our first result deals with
the low energy sign-changing solution of (1.1) with ε > 0. We prove that there
is no solution which blows up at exactly two points. More precisely, we have the
following result.
Theorem 1.1. Let Ω be any smooth bounded domain in Rn with n ≥ 5. There
exists ε0 > 0, such that for each ε ∈ (0, ε0 ), problem (1.1) has no sign-changing
solution uε which satisfies
uε = P δ(aε,1 ,λε,1 ) − P δ(aε,2 ,λε,2 ) + vε ,
(1.6)
with the L∞ -norm of uε at the power ε (|uε |ε∞ ) begin bounded and
aε,i ∈ Ω,
λε,i d(aε,i , ∂Ω) → ∞
hP δ(aε,1 ,λε,1 ) , P δ(aε,2 ,λε,2 ) i → 0
and
for i = 1, 2
kvε k → 0
as ε → 0.
We point out that there are other important phenomena in sign-changing solutions. Indeed, it is possible to find solutions having bubble over bubble (bubbletower solutions). In the case of the Laplacian operator, Pistoia and Weth [21]
constructed a family of sign-changing solutions of (1.1) (ε < 0) with k bubbles,
k ≥ 2, concentrated at the same point. This result gives a new phenomenon compared with the positive case. In their paper, they conjectured that this phenomenon
cannot appear when ε > 0. In [8], we gave an affirmative answer for the conjecture
of Pistoia and Weth. The following result deals with phenomenon of bubble-tower
solutions for the biharmonic problem (1.1) with supercritical exponent.
Theorem 1.2. Let Ω be any smooth bounded domain in Rn with n ≥ 5. There
exists ε0 > 0, such that for each ε ∈ (0, ε0 ), problem (1.1) has no solution uε of the
form
k
X
uε =
(−1)i+1 P δ(aε,i ,λε,i ) + vε ,
(1.7)
i=1
with λε,1 ≤ λε,2 ≤ · · · ≤ λε,k and |uε |ε∞ bounded, where k ≥ 2, aε,i ∈ Ω,
min(λε,i , λε,j )|aε,i − aε,j | is bounded, and vε → 0 in H01 (Ω), λε,i d(aε,i , ∂Ω) → +∞,
hP δ(aε,i ,λε,i ) , P δ(aε,j ,λε,j ) i → 0, for i 6= j, as ε → 0.
Note that Theorem 1.2 deals with the bubble-tower solutions at one point. However Theorem 1.1 says that there are no solutions which blow up at two points.
Combining the ideas of the proof of Theorems 1.1 and 1.2, we are able to prove the
following result.
Theorem 1.3. Let Ω be any smooth bounded domain in Rn with n ≥ 5. There
exists ε0 > 0, such that for each ε ∈ (0, ε0 ), problem (1.1) has no solution uε of the
4
K. OULD BOUH
EJDE-2014/77
form
uε =
m
X
(−1)i+1 P δ(aε,i ,λε,i ) +
i=1
k
X
(−1)i−m P δ(aε,i ,λε,i ) + vε := u1ε + u2ε + vε ,(1.8)
i=m+1
|uε |ε∞
with
bounded, kvε k → 0, aε,i → a for each i ≤ m, aε,i → b for each i ≥ m + 1
with a 6= b, and, for j = 1, 2, if ujε contains more than one bubble then it satisfies
the assumptions of Theorem 1.2.
The proof of our results will be by contradiction. Thus, throughout this paper
we will assume that there exist solutions (uε ) of (1.1) which satisfy (1.6) or (1.7). In
Section 2, we will obtain some information about such (uε ) which allow us to develop
Sections 3 which deal with some useful estimates to the proof of our Theorems.
Finally, in Section 4, we combine these estimates to obtain a contradiction. Hence
the proof of our results.
2. Preliminary results
In this Section, we assume that there exist solutions (uε ) of (1.1) which satisfy
uε =
k
X
(−1)i+1 P δ(aε,i ,λε,i ) + vε ,
(2.1)
i=1
with |uε |ε∞ bounded, k ≥ 2, aε,i ∈ Ω, and as ε → 0, kvε k → 0, λε,i d(aε,i , ∂Ω) →
+∞, hP δ(aε,i ,λε,i ) , P δ(aε,j ,λε,j ) i → 0 for i 6= j. We will collect some useful information used in the next sections. First, from (2.1), it is easy to see that the following
remark holds.
Remark 2.1. Let (uε ) be a family of sign-changing solutions of (1.1) satisfying
(2.1). Then
(i) Ruε * 0 as ε → R0,
(ii) Ω |uε |p+1+ε = Ω |∆uε |2 = kS n/4 + o(1),
(iii) Mε,+ := maxΩ uε → +∞, Mε,− := − minΩ uε → +∞ as ε → 0.
Secondly, arguing as in [2] and [22], we see that for uε satisfying (2.1), there is
a unique way to choose αi , ai , λi and v such that
uε =
k
X
(−1)i+1 αi P δ(ai ,λi ) + v,
(2.2)
i=1
with
αi ∈ R,
ai ∈ Ω,
v→0
λi ∈ R∗+ ,
2
αi → 1,
λi d(ai , ∂Ω) → +∞,
in H (Ω) ∩
H01 (Ω),
(2.3)
v ∈ E,
where E denotes the subspace of H01 (Ω) defined by
E := w : hw, ϕi = 0, ∀ϕ ∈ span{P δi , ∂P δi /∂λi , ∂P δi /∂aji , i ≤ k; j ≤ n} . (2.4)
Here, aji denotes the j-th component of ai and in the sequel, in order to simplify
the notations, we set
δ(ai ,λi ) = δi , P δ(ai ,λi ) = P δi .
(2.5)
In the following, we always assume that uε (which satisfies (2.1)) is written as in
(2.2) and (2.3) holds.
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SIGN-CHANGING SOLUTIONS
5
Lemma 2.2. Let uε satisfying the assumption of above theorems. Then λi occurring in (2.2) satisfies
λεi → 1
as ε → 0, for each i ≤ k.
Proof. By Remark 2.1, we know that
Z
|uε |p+1+ε = kS n/4 + o(1)
(2.6)
as ε → 0.
(2.7)
Ω
Furthermore,
Z
Z
Z
X
|uε |p+1+ε =
−∆uε uε =
|uε |p−1+ε uε
(−1)i+1 αi P δi +O(kvk2 ). (2.8)
Ω
Ω
Ω
Observe that
Z
X
|uε |p−1+ε uε
(−1)i+1 αi P δi
Ω
XZ
X p+ε+1 Z
p+ε+1
P δip+ε P δj
P δi
+O
=
αi
Ω
+O
XZ
j6=i
αi P δip+ε |v|
+
XZ
Ω
(2.9)
Ω
X
αi P δi |v|p+ε :=
Ai ,
Ω
where
Ai :=
αip+ε+1
Z
P δip+ε+1
+O
XZ
Ω
P δip+ε P δj
Ω
j6=i
Z
+
P δip+ε |v|
Z
+
Ω
P δi |v|p+ε .
Ω
Easy computations show that
Z
ε(n−4)/2
P δip+1+ε = λi
S n/4 + o(1) ,
Ω
Z
ε(n−4)/2
P δip+ε |v| = λi
O(|v|Lp+1 ),
Ω
Z
ε(n−4)/2
P δi |v|p+ε = λi
O(|v|p+ε
Lp+1 ).
Ω
Recall that for i 6= j (see [1])
Z
Z
δip δj =
Rn
Rn
n/(n−4)
δjp δi = cεij + O εij
log ε−1
,
ij
where c is a positive constant and, for i 6= j, εij is defined by
(4−n)/2
λ
λj
i
+
+ λi λj |ai − aj |2
.
εij =
λj
λi
Hence, we obtain
Z
Z
ε(n−4)/2
ε(n−4)/2
P δip+ε P δj = O λi
P δip P δj = λi
O(εij ),
Ω
(2.10)
for i 6= j.
Ω
Thus
ε(n−4)/2
Ai = αip+ε+1 λi
S n/4 + o(1) .
Therefore (2.8), (2.9), and (2.11) provide us with
Z
X
ε(n−4)/2
|uε |p+1+ε =
αip+1+ε λi
S n/4 + o(1) + o(1).
(2.11)
(2.12)
Ω
Combining (2.7), (2.12), and the fact that αi satisfies (2.3), the lemma follows. 6
K. OULD BOUH
EJDE-2014/77
Remark 2.3 ([6, 24] ). We recall the estimate
ε(n−4)/2
δiε (x) − cε0 λi
= O ε log(1 + λ2i |x − ai |2 )
in Ω,
(2.13)
which will be very useful in the next section.
3. Some useful estimates
As usual in this type of problems, we first deal with the v-part of uε , in order to
show that it is negligible with respect to the concentration phenomenon.
Lemma 3.1. The function v defined in (2.2), satisfies the estimate
kvk
P
 i
≤ cε + c P

i
(n−4)/n
εij (log ε−1
ij )
P
(n+4)/2(n−4)
1
(n+4)/2n
+ i6=j εij
(log ε−1
ij )
(λi di )(n+4)/2−ε(n−4)
1
(λi di )n−4
+
P
i6=j
if n < 12,
if n ≥ 12,
where εij is defined in (2.10) and di := d(ai , ∂Ω) for i ≤ k.
P
Proof. Since uε = (−1)i+1 αi P δi + v is a solution of (1.1) and v ∈ E (see (2.4)),
we obtain
Z
Z
−∆uε v = kvk2 =
|uε |p−1+ε uε v
Ω
Ω
Z X
X
=
|
(−1)i+1 αi P δi |p−1+ε ( (−1)i+1 αi P δi )v
Ω
Z X
+p
|
(−1)i+1 αi P δi |p−1+ε v 2 + o(kvk2 ).
Ω
Hence, we have
Q(v, v) = f (v) + o(kvk2 ),
(3.1)
where
Z X
Q(v, v) =kvk2 − p
|
(−1)i+1 αi P δi |p−1+ε v 2 ,
Z X Ω
X
f (v) =
|
(−1)i+1 αi P δi |p−1+ε ( (−1)i+1 αi P δi )v.
Ω
Using Remark 2.3 and according to [1], it is easy to see that
k Z
X
2
Q(v, v) = kvk − p
(P δi )p−1+ε v 2 + o(kvk2 )
i=1
Ω
is positive definite; that is, there exists c > 0 independent of ε, satisfying Q(v, v) ≥
ckvk2 , for each v ∈ E. Then, from (3.1) we get
kvk2 = O(kf (v)k).
Now, using Lemma 2.2, we obtain
Z
X
i+1
f (v) =
(−1)
(αi P δi )p+ε v
Ω
XZ
X Z p−1
(δi δj )p/2 |v| +
δi δj |v|(if n < 12) .
+O
i6=j
Ω
i6=j
Ω
(3.2)
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SIGN-CHANGING SOLUTIONS
7
Using Remark 2.3 and the fact that v ∈ E, we obtain
Z
P δip+ε v
Ω
Z
Z
= | δip+ε v| + O
δip−1+ε θi |v|
Z
Z
≤ cε log(1 + λ2i |x − ai |2 )δip |v| + c|θi |L∞ δip−1+ε |v|
1
1
≤ ckvk ε +
(if n < 12) +
(if n ≥ 12) ,
n+4
n−4
(λi di )
(λi di ) 2 +ε(n−4)
where θi := θai ,λi := δi − P δi .
For the other integrals of (3.2), we use Holder’s inequality and we obtain for
i 6= j
Z
Z
(n+4)/2n
p/2
(δi δj ) |v| ≤ ckvk
(δi δj )n/(n−4)
Ω
≤
Ω
(n+4)/2(n−4)
(n+4)/2n
ckvkεij
(log ε−1
ij )
and if n < 12, we have p − 1 = 8/(n − 4) > 1; therefore
Z
Z
(n−4)/n
(n−4)/n
δip−1 δj |v| ≤ ckvk
(δi δj )n/(n−4)
≤ ckvkεij (log ε−1
.
ij )
Ω
(3.3)
Ω
Combining (3.2)–(3.3), the proof follows.
Now, we need to introduce some notations before to state the crucial point in the
proof of our Theorems. We denote by G the Green’s function defined by : ∀x ∈ Ω
∆2 G(x, .) = cn δx
in Ω,
∆G(x, .) = G(x, .) = 0
on ∂Ω,
where δx is the Dirac mass at x and cn = (n − 4)(n − 2)ωn , with ωn is the area of
the unit sphere of Rn . We denote by H the regular part of G, that is,
H(x1 , x2 ) = |x1 − x2 |4−n − G(x1 , x2 )
for (x1 , x2 ) ∈ Ω2 \ Γ
with Γ = {(y, y) : y ∈ Ω}.
Proposition 3.2. Assume that n ≥ 5 and let αi , ai and λi be the variables defined
in (2.2) with k = 2. We have
∂ε
n − 4 H(a1 , a2 ) n−4
n − 4 H(ai , ai )
12
αi c1
−
α
c
λ
+
+ αi
c2 ε
j
1
i
n−4
(n−4)/2
2
∂λi
2 (λ1 λ2 )
2
λi
(P
n
n−4
−1
−1 2(n−4)
1
2
n
if n ≥ 6),
2
k=1,2 (λk dk )n−2 + ε12 log ε12 + ε12 (log ε12 )
≤ cε + c P
−1 2/5
1
2
if n = 5,
k=1,2 (λk dk )2 + ε12 (log ε12 )
(3.4)
where i, j ∈ {1, 2} with i 6= j and c1 , c2 are positive constants.
Proof. Let
Z
2n
n−4
c1 = c0
Rn
dx
,
(1 + |x|2 )(n+4)/2
2n
n − 4 n−4
c2 =
c0
2
Z
Rn
log(1+|x|2 )
|x|2 − 1
dx.
(1 + |x|2 )n+1
8
K. OULD BOUH
EJDE-2014/77
It suffices to prove the proposition for i = 1. Multiplying (1.1) by λ1 ∂P δ1 /∂λ1 and
integrating on Ω, we obtain
Z
Z
Z
∂P δ1
∂P δ1
∂P δ2
α1
δ1p λ1
− α2
=
|uε |p−1+ε uε λ1
.
(3.5)
δ2p λ1
∂λ
∂λ
∂λ1
1
2
Ω
Ω
Ω
Using [1], we derive
Z
log(λ d ) ∂P δ1
n − 4 H(a1 , a1 )
1 1
,
δ1p λ1
=
c1
+
O
n−4
∂λ1
2
(λ1 d1 )n−1
λ1
Ω
Z
∂ε
∂P δ1
n − 4 H(a1 , a2 ) 12
δ2p λ1
+ R,
= c1 λ 1
+
∂λ1
∂λ1
2 (λ1 λ2 )(n−4)/2
Ω
where R satisfies
R=O
X log(λ d )
n
k k
n−4
−1
+
ε
log
ε
.
12
12
(λk dk )n−1
(3.6)
k=1,2
For the other term of (3.5), we have
Z
∂P δ1
|uε |p−1+ε uε λ1
∂λ1
Ω
Z
∂P δ1
=
|α1 P δ1 − α2 P δ2 |p−1+ε (α1 P δ1 − α2 P δ2 )λ1
(3.7)
∂λ1
Ω
Z
n
∂P δ1
n−4
+ O kvk2 + ε12
+ (p + ε)
|α1 P δ1 − α2 P δ2 |p−1+ε vλ1
log ε−1
12 .
∂λ1
Ω
The above integral can be written as
Z
∂P δ1
|α1 P δ1 − α2 P δ2 |p−1+ε vλ1
∂λ1
Ω
Z
Z
Z
(3.8)
∂P δ1
+O
P δ2p−1 P δ1 |v| +
= (α1 P δ1 )p−1+ε vλ1
P δ1p−1 P δ2 |v| ,
∂λ1
Ω\A
Ω
A
where A = {x : 2α2 P δ2 ≤ α1 P δ1 }. Observe that, for n ≥ 12, we have p − 1 =
8/(n − 4) ≤ 1, thus
Z
Z
Z
n+4
p−1
p−1
P δ2 P δ1 |v| +
P δ1 P δ2 |v| ≤ c
|v|(δ1 δ2 ) 2(n−4)
Ω\A
A
Ω
(n+4)/2(n−4)
≤ ckvkε12
(n+4)/2n
(log ε−1
.
12 )
For n < 12, we have
Z
Z
p−1
(n−4)/n
P δ2 P δ1 |v| +
P δ1p−1 P δ2 |v| ≤ cε12 (log ε−1
kvk.
12 )
Ω\A
(3.9)
A
For the other integral in (3.8), using [1], [24] and Remark 2.3, we obtain
Z
∂P δ1
P δ1p−1+ε vλ1
∂λ1
Ω
h
i
1
log(λ1 d1 )
= O kvk ε +
(if n 6= 12) +
(if n = 12)
.
4
inf(n−4,(n+4)/2)
(λ1 d1 )
(λ1 d1 )
It remains to estimate the second integral of (3.7). We have
Z
∂P δ1
|α1 P δ1 − α2 P δ2 |p−1+ε (α1 P δ1 − α2 P δ2 )λ1
∂λ1
Ω
EJDE-2014/77
SIGN-CHANGING SOLUTIONS
9
Z
=
Z
∂P δ1
∂P δ1
(α1 P δ1 )p+ε λ1
− (α2 P δ2 )p+ε λ1
∂λ1
∂λ1
Ω
Ω
Z
n
∂P
δ
1
n−4
− (p + ε)
α2 P δ2 (α1 P δ1 )p−1+ε λ1
+ O ε12
log ε−1
12 .
∂λ1
Ω
Now, using Remark 2.3 and [1], we have
Z
H(a1 , a1 ) n − 4
∂P δ1
c2 ε + 2c1
=
P δ1p+ε λ1
∂λ1
2
λn−4
Ω
1
1
log(λ1 d1 )
2
+
(if n = 5) ,
+O ε +
n−1
2
(λ1 d1 )
(λ1 d1 )
Z
∂ε
∂P δ1
n − 4 H(a1 , a2 ) 12
P δ2p+ε λ1
+ R2 ,
= c1 λ1
+
∂λ1
∂λ1
2 (λ1 λ2 )(n−4)/2
Ω
Z
∂ε
n − 4 H(a1 , a2 ) ∂P δ1
12
p
P δ2 P δ1p−1+ε λ1
= c1 λ1
+
+ R1 ,
∂λ1
∂λ1
2 (λ1 λ2 )(n−4)/2
Ω
where for i = 1, 2,
n
n−4
log(λi di )
n−4
−1
n
+
ε
(log
ε
)
+
Ri = O εε12 (log ε−1
)
if
n
≥
8
12
12
12
(λi di )n
n−4
ε (log ε−1 ) n
12
12
if n < 8
+
n−4
(λi di )
(3.10)
Therefore, combining (3.5)–(3.10), and Lemma 3.1, the proof of Proposition 3.2
follows.
4. Proof of main theorems
Proof of Theorem 1.1. Arguing by contradiction, let us suppose that the problem
(1.1) has a solution uε as stated in Theorem 1.1. This solution has to satisfy (2.2)
and from Proposition 3.2, we have
∂ε
n − 4 H(a1 , a2 ) n − 4
n − 4 H(ai , ai )
12
−
c
+
c2 ε
c1
λ
+
1
i
2
∂λi
2 (λ1 λ2 )(n−4)/2
2
λn−4
i
(4.1)
X
1
=o ε+
+
ε
,
for
i
=
1,
2.
12
(λk dk )n−4
k=1,2
Furthermore, an easy computation shows that
λi
∂ε12
n−4
λj 2/n−4 =−
ε12 1 − 2 ε12
,
∂λi
2
λi
for i, j = 1, 2; j 6= i.
(4.2)
Without loss of generality, we can assume that λ2 ≥ λ1 . We distinguish two cases
and in each one, we will find a contradiction which implies our theorem.
Case 1.
λ1 λ2 |a1 −a2 |2
λ2 /λ1
→ +∞. In this case, it is easy to obtain
ε12 =
1
+ o(ε12 ),
(λ1 λ2 |a1 − a2 |2 )(n−4)/2
(4.3)
which implies that
λi
∂ε12
n−4
1
=−
+ o(ε12 )
∂λi
2 (λ1 λ2 |a1 − a2 |2 )(n−4)/2
for i = 1, 2.
(4.4)
10
K. OULD BOUH
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Then from (4.1) and (4.4), we obtain
c1 H(a1 , a1 ) H(a2 , a2 ) 1
c1
+
+
−
H(a
,
a
)
+ c2 ε
1
2
2
(λ1 λ2 )(n−4)/2 |a1 − a2 |n−4
λn−4
λn−4
1
2
X
1
=o ε+
+
ε
.
12
(λk dk )n−4
k=1,2
Using the fact that
1
− H(a1 , a2 ) > 0,
|a1 − a2 |n−4
H(a , a )
G(a1 , a2 ) 1 2
=O
+
,
(λ1 λ2 )(n−4)/2
(λ1 λ2 )(n−4)/2
G(a1 , a2 ) :=
ε12
we derive a contradiction in this case.
2
1 −a2 |
Case 2. λ1 λ2λ|a
→ c ≥ 0. In this case, we remark that λ2 /λ1 → +∞ (since
2 /λ1
ε12 → 0). Multiplying (4.1) by 2 for i = 2 and adding to (4.1) for i = 1, we obtain:
H(a , a )
H(a2 , a2 ) 2c1 ∂ε12
∂ε12 1 1
c1
+
2
−
λ
+
2λ
1
2
n−4
∂λ1
∂λ2
λn−4
λn−4
1
2
3H(a1 , a2 )
+ 3c2 ε
−
(4.5)
(λ1 λ2 )(n−4)/2
X
1
=o ε+
+ ε12 .
n−4
(λk dk )
k=1,2
Now, using (4.2) and the fact that λ2 ≥ λ1 , an easy computation shows that
− λ1
∂ε12
n−4
∂ε12
ε12 .
− 2λ2
≥
∂λ1
∂λ2
4
Furthermore, since H(a1 , a2 ) ≤ cd4−n
and λ2 /λ1 → ∞, we obtain
1
H(a1 , a2 )
1
.
=
o
(λ1 d1 )n−4
(λ1 λ2 )(n−4)/2
(4.6)
(4.7)
Then we derive a contradiction from (4.5), (4.6) and (4.7). Our proof is thereby
complete.
Proof of Theorem 1.2. Arguing by contradiction, let us assume that problem (1.1)
has solutions (uε ) as stated in Theorem 1.2. From Section 2, these solutions have
to satisfy (2.2) and (2.3). As in the proof of Proposition 3.2, we have for each
i = 1, . . . , k
∂ε
X
n − 4 H(ai , ai )
n − 4 H(ai , aj ) n − 4
ij
j+1
+
c2 ε
c1
+
c
(−1)
λ
+
1
i
2
∂λi
2 (λi λj )(n−4)/2
2
λn−4
i
j6=i
k
X
=o ε+
j=1
X 1
+
εrj .
(λj dj )n−4
r6=j
(4.8)
Observe that, if j < i, we have λj |ai − aj | is bounded (by the assumption) which
implies that
|ai − aj | = o(dj ),
di /dj = 1 + o(1) ∀i, j,
εij ≥ c(λj /λi )(n−4)/2 ∀j < i,
(4.9)
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SIGN-CHANGING SOLUTIONS
11
where c is a positive constant. Using (4.9), easy computations show that
ε(i−1)j + εi(j+1) = o(εij ) ∀i < j,
(4.10)
1
H(ai , aj )
) if (i, j) 6= (1, 1).
= o(
(λ1 d1 )n−4
(λi λj )(n−4)/2
Thus, using (4.10), (4.8) can be written as
c1
X n − 4 H(a1 , a1 )
n−4
1
∂ε12
+
εrj , (4.11)
+
c2 ε = o ε +
− c1 λ 1
n−4
n−4
2
∂λ1
2
(λ1 d1 )
λ1
r6=j
X ∂ε(k−1)k
n−4
1
−c1
+
εrj , (4.12)
+
c2 ε = o ε +
∂λk
2
(λ1 d1 )n−4
r6=j
and for 1 < i < k,
X ∂ε(i−1)i
∂εi(i+1)
1
n−4
− c1 λi
− c1 λ i
+
c2 ε = o ε +
+
εrj . (4.13)
∂λi
∂λi
2
(λ1 d1 )n−4
r6=j
Using (4.2) and (4.12), we derive that
X X 1
1
ε=o
+
ε
+
εrj .
,
ε
=
o
ij
(k−1)k
(λ1 d1 )n−4
(λ1 d1 )n−4
r6=j
(4.14)
r6=j
Now, using (4.14) and (4.12) with k − 1 instead of k, we derive the estimate of
ε(k−2)(k−1) and by induction we get
X 1
+
εrj
for i = 2, . . . , k.
(4.15)
ε(i−1)i = o
(λ1 d1 )n−4
r6=j
Finally, using (4.10), (4.14), (4.15) and (4.11), we obtain
H(a1 , a1 )
1
),
= o(
n−4
(λ1 d1 )n−4
λ1
which gives a contradiction. Hence, our theorem is proved.
Proof of Theorem 1.3. Arguing by contradiction, let us assume that problem (1.1)
has solutions (uε ) as stated in Theorem 1.3. From Section 2, these solutions have
to satisfy (2.2) and (2.3). Without loss of generality, in the sequel, we will assume
that λ1 d1 ≤ λm+1 dm+1 . As in the proof of Theorem 1.2, (4.8) is satisfied for each
i = 1, . . . , k. Furthermore, (4.10) holds if i, j ≤ m or i, j > m (in the last case, we
require that (i, j) 6= (m + 1, m + 1)).
Observe that since a 6= b, it is easy to obtain that |ai − aj | ≥ c > 0 for each
i ≤ m and j ≥ m + 1. Hence for i ≤ m and j ≥ m + 1 we have
∂εij
n−4
1
λr
=−
+ o(εij ), for r = i, j,
(4.16)
∂λr
2 (λi λj |ai − aj |2 )(n−4)/2
H(ai , aj )
1
εij +
= o ε1(m+1) +
for (i, j) 6= (1, m + 1). (4.17)
n−4
(n−4)/2
(λ1 d1 )
(λi λj )
Now using (4.10), (4.16) and (4.17), we derive that (4.13) holds for each i 6∈ {1, m +
1}. However, since the first bubble in the second bubble tower u2ε has negative sign,
for i = 1, m + 1, we have
X ∂εi(i+1)
H(ai , ai )
2c1
1
c1
−
λi
+ c1 ε1(m+1) + c2 ε = o ε +
+
εrj .
n−4
n−4
n−4
∂λi
(λ1 d1 )
λi
r6=j
12
K. OULD BOUH
EJDE-2014/77
Finally, arguing as in Theorem 1.2, we derive a contradiction. Hence our result is
proved.
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Kamal Ould Bouh
Department of Mathematics, Taibah University, P.O. Box: 30002, Almadinah Almunawwarah, Saudi Arabia
E-mail address: hbouh@taibahu.edu.sa, kamal bouh@yahoo.fr