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On a fourth order elliptic equation with supercritical exponent

Advances in Difference Equations20142014:319

https://doi.org/10.1186/1687-1847-2014-319

  • Received: 26 August 2014
  • Accepted: 1 December 2014
  • Published:

Abstract

This paper is concerned with the semi-linear elliptic problem involving nearly critical exponent ( P ε ) : Δ 2 u = | u | 8 / ( n 4 ) + ε u in Ω, Δ u = u = 0 on Ω, where Ω is a smooth bounded domain in R n , n 5 , and ε is a positive real parameter. We show that, for ε small, ( P ε ) has no sign-changing solutions with low energy which blow up at exactly three points. Moreover, we prove that ( P ε ) has no bubble-tower sign-changing solutions.

MSC:35J20, 35J60.

Keywords

  • nonlinear problem
  • critical exponent
  • sign-changing solutions
  • bubble-tower solution

1 Introduction and results

We consider the following semi-linear elliptic problem with supercritical nonlinearity:
( P ε ) { Δ 2 u = | u | p 1 + ε u in  Ω , Δ u = u = 0 on  Ω ,

where Ω is a smooth bounded domain in R n , n 5 , ε is a positive real parameter and p + 1 = 2 n n 4 is the critical Sobolev exponent for the embedding of H 2 ( Ω ) H 0 1 ( Ω ) into L p + 1 ( Ω ) .

When the biharmonic operator in ( P ε ) is replaced by the Laplacian operator, there are many works devoted to the study of the counterpart of ( P ε ) ; see for example [16], and the references therein.

When ε < 0 , many works have been devoted to the study of the solutions of ( P ε ) see for example [79]. In the critical case, this problem is not compact, that is, when ε = 0 it corresponds exactly to the limiting case of the Sobolev embedding H 2 ( Ω ) H 0 1 ( Ω ) into L p + 1 ( Ω ) , and thus we lose the compact embedding. In fact, van Der Vorst showed in [10] that ( P 0 ) has no positive solutions if Ω is a starshaped domain. Whereas Ebobisse and Ould Ahmedou proved in [11] that ( P 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 show [12].

In the supercritical case, ε > 0 , the problem ( P ε ) becomes more delicate since we lose the Sobolev embedding which is an important point to overcome. The problem ( P ε ) was studied in [7] where the authors show that there is no one-bubble solution to the problem and there is a one-bubble solution to the slightly subcritical case under some suitable conditions. However, we proved in [13] that ( P ε ) has no sign-changing solutions which blow up exactly at two points. In this work we will show the non-existence of sign-changing solutions of ( P ε ) having three concentration points.

We note that problem ( P ε ) has a variational structure. The related functional is
inf J ( u ) , where  J ( u ) : = Ω | Δ u | 2 ( Ω | u | p + 1 + ε ) 2 / ( p + 1 + ε ) , u H 2 ( Ω ) H 0 1 ( Ω ) , u 0 .
J satisfies the Palais-Smale condition in the subcritical case, while this condition fails in the critical case. Such a failure is due to the functions
δ ( a , λ ) ( x ) = c 0 λ ( n 4 ) / 2 ( 1 + λ 2 | x a | 2 ) ( n 4 ) / 2 , c 0 = ( n ( n 4 ) ( n 2 4 ) ) ( n 4 ) / 8 , λ > 0 , a R n .
(1.1)
c 0 is chosen so that δ ( a , λ ) is the family of solutions of the following problem:
Δ 2 u = u p , u > 0  in  R n .
(1.2)
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 H 0 1 ( Ω ) . It satisfies
Δ 2 P δ ( a , λ ) = Δ 2 δ ( a , λ ) in  Ω , Δ P δ ( a , λ ) = P δ ( a , λ ) = 0 on  Ω .

These functions are almost positive solutions of (1.2).

We denote by G the Green’s function defined by, x Ω ,
Δ 2 G ( x , ) = c n δ x in  Ω , Δ G ( x , ) = G ( x , ) = 0 on  Ω ,
where δ x is the Dirac mass at x and c n = ( n 4 ) ( n 2 ) w n , with w n is the area of the unit sphere of R n . We denote by H the regular part of G, that is,
H ( x 1 , x 2 ) = | x 1 x 2 | 4 n G ( x 1 , x 2 ) for  ( x 1 , x 2 ) Ω 2 .
For x = ( x 1 , x 2 ) Ω 2 Γ , with Γ = { ( y , y ) : y Ω } , we denote by M ( x ) the matrix defined by
M ( x ) = ( m i j ) 1 i , j 2 , where  m i i = H ( x i , x i ) , m 12 = m 21 = G ( x 1 , x 2 ) ,
(1.3)

and let ρ ( x ) be its least eigenvalue.

The space H 2 ( Ω ) H 0 1 ( Ω ) is equipped with the norm and its corresponding inner product , defined by
u = ( Ω | Δ u | 2 ) 1 / 2 and u , v = Ω Δ u Δ v , u , v H 2 ( Ω ) H 0 1 ( Ω ) .
(1.4)

Now, we are able to state our result.

Theorem 1.1 Let Ω be any smooth bounded domain in R n , n 6 . If 0 is a regular value of ρ ( x ) , then there exists ε 0 > 0 , such that, for each ε ( 0 , ε 0 ) , problem ( P ε ) has no sign-changing solutions u ε which satisfy
u ε = P δ ( a ε , 1 , λ ε , 1 ) P δ ( a ε , 2 , λ ε , 2 ) + P δ ( a ε , 3 , λ ε , 3 ) + v ε ,
(1.5)
with | u ε | ε is bounded and
{ a ε , i Ω , λ ε , i d ( a ε , i , Ω ) for  i = 1 , 2 , 3 , P δ ( a ε , i , λ ε , i ) , P δ ( a ε , j , λ ε , j ) 0 for  i j  and  v ε 0  as  ε 0 .

The second result deals with the phenomenon of bubble-tower solutions for the biharmonic problem ( P ε ) with supercritical exponent. We will give a generalization of the result found in [13]. More precisely, we have the following.

Theorem 1.2 Let Ω be any smooth bounded domain in R n , n 5 . There exists ε 0 > 0 , such that, for each ε ( 0 , ε 0 ) , problem ( P ε ) has no solutions u ε of the form
u ε = i = 1 k γ i P δ ( a ε , i , λ ε , i ) + v ε , with  λ ε , 1 λ ε , 2 λ ε , k  and  | u ε | ε  is bounded ,
(1.6)

where k 2 , γ i { 1 , 1 } , a ε , i Ω , for each i j , λ ε , i | a ε , i a ε , j | is bounded and as ε 0 , v ε 0 , λ ε , i d ( a ε , i , Ω ) + , P δ ( a ε , i , λ ε , i ) , P δ ( a ε , j , λ ε , j ) 0 for i j , and if l { k 1 , k } , λ ε , l | a ε , l a ε , l + 1 | 0 , where l = min { q : γ q = = γ k } .

The proof of our results will be by contradiction. Thus, throughout this paper we will assume that there exist solutions ( u ε ) of ( P ε ) which satisfy (1.5) or (1.6). In Section 2, we will obtain some information as regards such ( u ε ) which allows us to develop Section 3 which deals 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 follows.

2 Preliminary results

In this section, we assume that there exist solutions ( u ε ) of ( P ε ) which satisfy
u ε = i = 1 k γ i P δ ( a ε , i , λ ε , i ) + v ε ,
(2.1)
with | u ε | ε is bounded, k 2 , a ε , i Ω , and as ε 0 , v ε 0 , λ ε , i d ( a ε , i , Ω ) + , P δ ( a ε , i , λ ε , i ) , P δ ( a ε , j , λ ε , j ) 0 for i j . Arguing as in [14] and [15], we see that for u ε satisfying (2.1), there is a unique way to choose α i , a i , λ i , and v such that
u ε = i = 1 k γ i α i P δ ( a i , λ i ) + v ,
(2.2)
with  { α i R , α i 1 , a i Ω , λ i R + , λ i d ( a i , Ω ) + , v 0 in  H 2 ( Ω ) H 0 1 ( Ω ) , v E ,
(2.3)
where E denotes the subspace of H 0 1 ( Ω ) defined by
E : = { w : w , φ = 0 , φ Span { P δ i , P δ i / λ i , P δ i / a i j , i k ; j n } } .
(2.4)

Here, a i j denotes the j th component of a i and in the sequel, in order to simplify the notations, we set δ ( a i , λ i ) = δ i and P δ ( a i , λ i ) = P δ i . We always assume that u ε (which satisfies (2.1)) is written as in (2.2) and (2.3) holds. From (2.1), it is easy to see that the following remark holds.

Lemma 2.1 [13]

Let u ε satisfying the assumption of the theorems. λ i occurring in (2.2) satisfies
λ i ε 1 as  ε 0  for each  i k .
(2.5)

Remark 2.2 [2, 16]

We recall the following estimate:
δ i ε ( x ) c 0 ε λ i ε ( n 4 ) / 2 = O ( ε log ( 1 + λ i 2 | x a i | 2 ) ) in  Ω .
(2.6)

3 Some useful estimates

As usual in this type of problems, we first deal with the v-part of u ε , 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 following estimate:
v c ε + c { i 1 ( λ i d i ) n 4 + i j ε i j ( log ε i j 1 ) ( n 4 ) / n if  n < 12 , i 1 ( λ i d i ) ( n + 4 ) / 2 ε ( n 4 ) + i j ε i j ( n + 4 ) / 2 ( n 4 ) ( log ε i j 1 ) ( n + 4 ) / 2 n if  n 12 ,
where d i : = d ( a i , Ω ) for i k and for i j , ε i j is defined by
ε i j = ( λ i λ j + λ j λ i + λ i λ j | a i a j | 2 ) ( 4 n ) / 2 .
(3.1)

Proof The proof is the same as that of Lemma 3.1 of [13], so we omit it. □

Now, we state the crucial points in the proof of our theorems.

Proposition 3.2 Assume that n 5 and let α i , a i and λ i be the variables defined in (2.2) with k = 3 and γ 1 = γ 2 = γ 3 . We have
| α i c 1 n 4 2 H ( a i , a i ) λ i n 4 + j i ( 1 ) i + j α j c 1 ( λ i ε i j λ i + n 4 2 H ( a i , a j ) ( λ i λ j ) ( n 4 ) / 2 ) + α i n 4 2 c 2 ε | c ε 2 + c { k 1 ( λ k d k ) n 2 + j i ( ε i j n n 4 log ε i j 1 + ε i j 2 ( log ε 12 1 ) 2 ( n 4 ) n ) if  n 6 , k 1 ( λ k d k ) 2 + j i ε i j 2 ( log ε 12 1 ) 2 / 5 if  n = 5 ,
(3.2)

where i , j { 1 , 2 , 3 } with i j and c 1 , c 2 are positive constants.

Proof Let
c 1 = c 0 2 n n 4 R n d x ( 1 + | x | 2 ) ( n + 4 ) / 2
and
c 2 = n 4 2 c 0 2 n n 4 R n log ( 1 + | x | 2 ) | x | 2 1 ( 1 + | x | 2 ) n + 1 d x .
It suffices to prove the proposition for i = 1 . Multiplying ( P ε ) by λ 1 P δ 1 / λ 1 and integrating on Ω, we obtain
α 1 Ω δ 1 p λ 1 P δ 1 λ 1 α 2 Ω δ 2 p λ 1 P δ 1 λ 1 + α 3 Ω δ 3 p λ 1 P δ 1 λ 1 = Ω | u ε | p 1 + ε u ε λ 1 P δ 1 λ 1 .
(3.3)
Using [17], we derive
Ω δ 1 p λ 1 P δ 1 λ 1 = n 4 2 c 1 H ( a 1 , a 1 ) λ 1 n 4 + O ( log ( λ 1 d 1 ) ( λ 1 d 1 ) n 1 ) ,
(3.4)
Ω δ j p λ 1 P δ 1 λ 1 = c 1 ( λ 1 ε 1 j λ 1 + n 4 2 H ( a 1 , a j ) ( λ 1 λ j ) ( n 4 ) / 2 ) + R j ,
(3.5)
where j = 2 , 3 and R j satisfies
R j = O ( k = 1 , j log ( λ k d k ) ( λ k d k ) n 1 + ε 1 j n n 4 log ε 1 j 1 ) .
(3.6)
For the other term of (3.3), we have
Ω | u ε | p 1 + ε u ε λ 1 P δ 1 λ 1 = Ω | α 1 P δ 1 α 2 P δ 2 + α 3 P δ 3 | p 1 + ε ( α 1 P δ 1 α 2 P δ 2 + α 3 P δ 3 ) λ 1 P δ 1 λ 1 + ( p + ε ) Ω | α 1 P δ 1 α 2 P δ 2 + α 3 P δ 3 | p 1 + ε v λ 1 P δ 1 λ 1 + O ( v 2 + i j ε i j n n 4 log ε i j 1 ) .
(3.7)
Concerning the last integral, it can be written as
Ω | α 1 P δ 1 α 2 P δ 2 + α 3 P δ 3 | p 1 + ε v λ 1 P δ 1 λ 1 = Ω ( α 1 P δ 1 ) p 1 + ε v λ 1 P δ 1 λ 1 + O ( Ω A j P δ j p 1 P δ 1 | v | + A j P δ 1 p 1 P δ 2 | v | ) ,
(3.8)

where A j = { x : 2 α j P δ j α 1 P δ 1 } for j = 2 , 3 .

Observe that, for n 12 , we have p 1 = 8 / ( n 4 ) 1 , thus
Ω A j P δ j p 1 P δ 1 | v | + A j P δ 1 p 1 P δ j | v | c Ω | v | ( δ 1 δ j ) n + 4 2 ( n 4 ) c v ε 1 j ( n + 4 ) / 2 ( n 4 ) ( log ε 1 j 1 ) ( n + 4 ) / 2 n .
(3.9)
But for n < 12 , we have
Ω A j P δ j p 1 P δ 1 | v | + A P δ 1 p 1 P δ j | v | c ε 1 j ( log ε 1 j 1 ) ( n 4 ) / n v .
(3.10)
For the other integral in (3.8), using [16, 17], and Remark 2.2, we get
Ω P δ 1 p 1 + ε v λ 1 P δ 1 λ 1 = O ( v [ ε + ( 1 ( λ 1 d 1 ) inf ( n 4 , ( n + 4 ) / 2 ) ( if  n 12 ) + log ( λ 1 d 1 ) ( λ 1 d 1 ) 4 ( if  n = 12 ) ) ] ) .
(3.11)
It remains to estimate the second integral of (3.7). We have
Ω | α 1 P δ 1 α 2 P δ 2 + α 3 P δ 3 | p 1 + ε ( α 1 P δ 1 α 2 P δ 2 + α 3 P δ 3 ) λ 1 P δ 1 λ 1 = Ω ( α 1 P δ 1 ) p + ε λ 1 P δ 1 λ 1 Ω ( α 2 P δ 2 ) p + ε λ 1 P δ 1 λ 1 + Ω ( α 3 P δ 3 ) p + ε λ 1 P δ 1 λ 1 ( p + ε ) ( Ω α 2 P δ 2 ( α 1 P δ 1 ) p 1 + ε λ 1 P δ 1 λ 1 Ω α 3 P δ 3 ( α 1 P δ 1 ) p 1 + ε λ 1 P δ 1 λ 1 ) + O ( ε 1 j n n 4 log ε 1 j 1 ) .
(3.12)
Now, using Remark 2.2 and [17], we have
Ω P δ 1 p + ε λ 1 P δ 1 λ 1 = n 4 2 ( c 2 ε + 2 c 1 H ( a 1 , a 1 ) λ 1 n 4 ) Ω P δ 1 p + ε λ 1 P δ 1 λ 1 = + O ( ε 2 + log ( λ 1 d 1 ) ( λ 1 d 1 ) n 1 + 1 ( λ 1 d 1 ) 2 ( if  n = 5 ) ) ,
(3.13)
Ω P δ j p + ε λ 1 P δ 1 λ 1 = c 1 ( λ 1 ε 1 j λ 1 + n 4 2 H ( a 1 , a j ) ( λ 1 λ j ) ( n 4 ) / 2 ) + T j ,
(3.14)
p Ω P δ j P δ 1 p 1 + ε λ 1 P δ 1 λ 1 = c 1 ( λ 1 ε 1 j λ 1 + n 4 2 H ( a 1 , a j ) ( λ 1 λ j ) ( n 4 ) / 2 ) + T j ,
(3.15)
where for i = 2 , 3 ,
T i = O ( ε ε 1 j ( log ε 1 j 1 ) n 4 n ) + ( ε 1 j n n 4 ( log ε 1 j 1 ) + log ( λ i d i ) ( λ i d i ) n ( if  n 8 ) ) + ( ε 1 j ( log ε 1 j 1 ) n 4 n ( λ i d i ) n 4 ( if  n < 8 ) ) .

Therefore, combining (3.3)-(3.15), and Lemma 3.1, the proof of Proposition 3.2 follows. □

Proposition 3.3 Let n 6 . We have the following estimate:
α i 1 λ i n 3 H ( a i , a i ) a i 2 λ i j i ( 1 ) i + j α j ( ε i j a i 1 ( λ i λ j ) ( n 4 ) / 2 H a i ( a i , a j ) ) = O ( k 1 ( λ k d k ) n 2 + j i ε i j n n 4 log ε i j 1 + ε i j 2 ( log ε 1 j 1 ) 2 ( n 4 ) n + ε 2 + ε ( λ i d i ) n 3 ) ,

where i , j { 1 , 2 , 3 } and j i .

Proof The proof is similar to the proof of Proposition 3.2. But there exist some integrals which have different estimates. We will focus in those integrals. In fact, (3.3), (3.7)-(3.12) are also true if we change λ 1 P δ 1 / λ 1 by ( 1 / λ 1 ) P δ 1 / a 1 . It remains to deal with the other equations. Following [17], we get
Ω δ 1 p 1 λ 1 P δ 1 a 1 = 1 2 c 1 λ 1 n 3 H ( a 1 , a 1 ) a 1 + O ( 1 ( λ 1 d 1 ) n 1 ) ,
(3.16)
Ω δ j p 1 λ 1 P δ 1 a 1 = c 1 λ 1 ( ε 1 j a 1 1 ( λ 1 λ j ) ( n 4 ) / 2 H a 1 ( a 1 , a j ) ) Ω δ j p 1 λ 1 P δ 1 a 1 = + O ( k = 1 , j 1 ( λ k d k ) n 1 + λ j | a 1 a j | ε 1 j ( n 1 ) / ( n 4 ) ) ,
(3.17)
Ω P δ 1 p + ε 1 λ 1 P δ 1 a 1 = c 0 ε λ 1 ε ( n 4 ) / 2 c 1 λ 1 n 3 H ( a 1 , a 1 ) a 1 + O ( 1 ( λ 1 d 1 ) n 2 + ε ( λ 1 d 1 ) n 3 ) ,
(3.18)
Ω P δ j p + ε 1 λ 1 P δ 1 λ 1 = c 0 ε λ j ε ( n 4 ) / 2 ( P δ j , 1 λ 1 P δ 1 a 1 ) + O ( ε ε 1 j ( log ε 1 j 1 ) ( n 4 ) / n ) + T j ,
(3.19)
Ω P δ j 1 λ 1 ( P δ 1 p + ε ) a 1 = c 0 ε λ 1 ε ( n 4 ) / 2 ( P δ j , 1 λ 1 P δ 1 a 1 ) + O ( ε ε 1 j ( log ε 1 j 1 ) ( n 4 ) / n ) + T j .
(3.20)

The proof of Proposition 3.3 is thereby completed. □

4 Proof of the theorems

Proof of Theorem 1.1

Arguing by contradiction, let us assume that problem ( P ε ) has solutions ( u ε ) as stated in Theorem 1.1. Recall that u ε is written as
u ε = α ε , 1 P δ ( a ε , 1 , λ ε , 1 ) α ε , 2 P δ ( a ε , 2 , λ ε , 2 ) + α ε , 3 P δ ( a ε , 3 , λ ε , 3 ) + v ε ,
with v ε orthogonal to each P δ ( a i , λ i ) and their derivatives with respect to λ i and ( a i ) k , where ( a i ) k denotes the k th component of a i (see (2.2) and (2.3)). For simplicity, we will write α i : = α ε , i , λ i : = λ ε , i , and a i : = a ε , i . From Proposition 3.2, for each i = 1 , 2 , 3 , with γ 1 = γ 3 = 1 , γ 2 = 1 . We have
( E i ) c 1 n 4 2 H ( a i , a i ) λ i n 4 + γ i c 1 j i γ j ( λ i ε i j λ i + n 4 2 H ( a i , a j ) ( λ i λ j ) ( n 4 ) / 2 ) + n 4 2 c 2 ε ( E i ) = o ( ε + j = 1 3 1 ( λ j d j ) n 4 + r j ε r j ) .
Furthermore, an easy computation shows that
λ i ε i j λ i = n 4 2 ε i j ( 1 2 λ j λ i ε i j 2 / n 4 ) for  i , j = 1 , 2 , 3 , j i ,
(4.1)
λ i ε i j λ i 2 λ j ε i j λ j n 4 2 ε i j for  λ i λ j .
(4.2)
On the other hand, following the proof of Proposition 3.3, we have, for each i = 1 , 2 , 3 ,
( F i ) 1 λ i n 3 H ( a i , a i ) a i j i 2 ( 1 ) j + i λ i ( ε j i a i H ( a j , a i ) a i 1 ( λ j λ i ) ( n 4 ) / 2 ) ( F i ) = o ( j 1 ( λ j d j ) n 3 + r j ε r j n 3 n 4 + ε n 3 n 4 ) .
(4.3)
We distinguish many cases depending on the set
Ϝ : = { ( i , j ) : i j  and  min ( λ i , λ j ) | a i a j |  is bounded }

and we will prove that all these cases cannot occur.

We remark that if ( i , j ) Ϝ we derive λ i / λ j 0  or  and d i / d j = 1 + o ( 1 ) as ε 0 .

Furthermore, the behavior of ε i j depends on the set Ϝ. In fact we have, assuming that λ i λ j ,
c ( λ i λ j ) ( n 4 ) / 2 ε i j ( λ i λ j ) ( n 4 ) / 2 if  ( i , j ) Ϝ ,
(4.4)
ε i j = 1 ( λ i λ j | a i a j | 2 ) ( n 4 ) / 2 + o ( ε i j ) if  ( i , j ) Ϝ .
(4.5)

First we start by proving the following crucial lemmas.

Remark 4.1 Ordering the λ i ’s: λ i 1 λ i 2 λ i 3 , adding ( E i 1 ) + 2 ( E i 2 ) + 4 ( E i 3 ) , and using (4.2), it is easy to derive a contradiction if we have ε 13 = o ( ( λ i d i ) 4 n + ε r j + ε ) .

Lemma 4.2 Let n 4 . Then there exists a positive constant c ̲ 0 > 0 such that
( i ) c ̲ 0 1 d 1 d 3 c ̲ 0 ; ( ii ) c ̲ 0 1 λ 1 λ 3 c ̲ 0 ; ( iii ) c ̲ 0 1 | a 1 a 3 | d i c ̲ 0 1 for  i = 1 , 3 .

Proof The proof will be by contradiction.

Proof of (i). Assume that d 1 / d 3 0 . In this case, we have
| a 1 a 3 | c d 3 and ε 13 = 1 ( λ 1 λ 3 | a 1 a 3 | 2 ) ( n 4 ) / 2 + o ( ε 13 ) ,
(4.6)

which implies that ε 13 = o ( ( λ 1 d 1 ) 4 n + ( λ 3 d 3 ) 4 n ) . Using Remark 4.1, we derive a contradiction. In the same way, we prove that d 3 / d 1 0 . Hence the proof of Claim (i) is completed.

Proof of (ii). Assume that λ 1 / λ 3 0 . By Claim (i), we have ( λ 3 d 3 ) 1 = o ( ( λ 1 d 1 ) 1 ) . Four cases may occur.

Case 1. λ 2 / λ 3 0 or { ( 1 , 2 ) , ( 2 , 3 ) } Ϝ = ϕ . Using (4.5), ( E 2 ) implies that
H ( a 2 , a 2 ) λ 2 n 4 + ε 12 + ε 23 + ε = o ( 1 ( λ 1 d 1 ) n 4 + ε 13 ) .

By Claim (i) and ( E 3 ) , we obtain ε 13 = o ( ( λ 1 d 1 ) 4 n ) . By Remark 4.1, this case cannot occur.

Case 2. λ 2 / λ 3 0 , { ( 1 , 2 ) , ( 2 , 3 ) } Ϝ ϕ , and λ 2 / λ 1 + . In this case, it is easy to obtain ε 13 = o ( ε 12 + ε 23 ) . Using Remark 4.1, we derive a contradiction.

Case 3. λ 2 / λ 3 0 , ( 2 , 3 ) Ϝ , ( 1 , 2 ) Ϝ , and λ 2 / λ 1 + . In this case, we see that λ 2 | a 2 a 3 | is bounded and λ 2 | a 1 a 2 | + . Hence, we derive that λ 2 | a 1 a 3 | + , which implies that λ k | a 1 a 3 | + for k = 1 , 3 . Thus
ε 13 = 1 + o ( 1 ) ( λ 1 λ 3 | a 1 a 3 | 2 ) ( n 4 ) / 2 = ( λ 2 λ 3 ) ( n 4 ) / 2 1 + o ( 1 ) ( λ 1 λ 2 | a 1 a 3 | 2 ) ( n 4 ) / 2 = o ( ε 23 ) .

Then by Remark 4.1, we get a contradiction.

Case 4. λ 2 / λ 3 0 , ( 1 , 2 ) Ϝ , and λ 2 / λ 1 + . In this case, it is easy to get ε 23 = o ( ε 12 ) .

Using the formula [ ( E 1 ) + ( E 2 ) ( E 3 ) ] , we deduce that ε = o ( ε 12 + ε 13 ) , which implies that ε 13 = o ( ε 12 ) . Hence by Remark 4.1, we derive a contradiction and Claim (ii) is thereby completed.

Proof of (iii). Without loss of generality, we can assume that d 1 d 3 . First, as in the proof of Claim (i), we get | a 1 a 3 | c 0 d 1 . Now assume that | a 1 a 3 | / d 1 0 , which implies
H ( a i , a i ) λ i n 4 = o ( ε 13 ) for  i = 1 , 3 .

Two cases may occur.

Case 1. λ 1 λ 2 or { ( 1 , 2 ) , ( 2 , 3 ) } Ϝ = ϕ . Using ( E 2 ) , we obtain
H ( a 2 , a 2 ) λ 2 n 4 = o ( ε 13 ) , ε i 2 = o ( ε 13 ) for  i = 1 , 3 and ε = o ( ε 13 ) ,

and we derive a contradiction from ( E 1 ) .

Case 2. λ 2 λ 1 and { ( 1 , 2 ) , ( 2 , 3 ) } Ϝ ϕ . Let k { 1 , 3 } such that ( 2 , k ) Ϝ . Using Claim (ii) and the fact that λ 2 λ 1 , we derive that ε 2 k c ( λ 2 / λ k ) ( n 4 ) / 2 , which implies that d 2 d k , λ 2 / λ k 0 , and λ 2 | a 2 a k | is bounded. Using (4.3) for i = k , we get
λ 2 | a 2 a k | ε 2 k n n 4 + λ 1 λ 3 λ k | a 1 a 3 | ε 13 n n 4 = o ( 1 ( λ 2 d 2 ) n 3 + r j ε r j n 3 n 4 + ε n 3 n 4 ) .
(4.7)
Since λ 2 | a 2 a k | is bounded and ε 13 ( λ 1 λ 3 | a 2 a k | 2 ) ( 4 n ) / 2 , we derive that
ε 13 n 3 n 4 = o ( 1 ( λ 2 d 2 ) n 3 + ε 12 n 3 n 4 + ε 23 n 3 n 4 + ε n 3 n 4 ) ,
which implies that
ε 13 = o ( 1 ( λ 2 d 2 ) n 4 + ε 12 + ε 23 + ε ) .
(4.8)

By Remark 4.1, we get a contradiction. □

Lemma 4.3 There exists a positive constant c ̲ 0 such that
( i ) c ̲ 0 λ 1 λ 2 ; ( ii ) d i c ̲ 0 for  i = 1 , 3 .

Proof Without loss of generality, we can assume that d 1 d 3 .

Proof of (i). Assume that λ 2 / λ 1 0 . First we claim that d 1 / d 2 0 . In fact, arguing by contradiction we assume that d 1 / d 2 0 , we get d 1 0 , | a 1 a 2 | c d 2 , and | a 2 a 3 | c d 2 . Hence, { ( 1 , 2 ) , ( 2 , 3 ) } Ϝ = ϕ . From ( E 2 ) , we obtain
H ( a 2 , a 2 ) λ 2 n 4 + ε 12 + ε 23 + ε = o ( 1 ( λ 1 d 1 ) n 4 + 1 ( λ 3 d 3 ) n 4 + ε 13 ) .
(4.9)
Let ν i be the outward normal vector at a i . Since d 1 , d 3 , and | a 1 a 3 | are of the same order, we have (see [18] and [19])
1 λ 1 n 3 H ( a 1 , a 1 ) ν 1 c ( λ 1 d 1 ) n 3 and G ( a 1 , a 3 ) ν 1 0 .
(4.10)

Using ( F 1 ) , we get 1 / ( λ 1 d 1 ) n 3 = o ( ε 13 ( n 3 ) / ( n 4 ) ) , which implies that 1 / ( λ 1 d 1 ) n 4 = o ( ε 13 ) . From ( E 1 ) , we derive a contradiction. Hence our claim is proved.

Thus there exists a positive constant c so that d 1 c d 2 . Now, since we have assumed that λ 2 / λ 1 0 , Lemma 4.2 implies that ε 13 = o ( ( λ 2 d 2 ) 4 n ) . Finally, using Remark 4.1, we get a contradiction and the proof of Claim (i) follows.

Proof of (ii). Assume that d 1 0 . Note that Claim (i) and ( E 2 ) imply that (4.9) holds.

Now, following the proof of (i), we obtain a contradiction. □

We turn now to the proof of Theorem 1.1. By the previous lemmas, we know that λ 1 and λ 3 are of the same order, | a 1 a 3 | c and λ 2 c λ i , for i = 1 , 3 where c is a positive constant.

Hence, ( E 2 ) implies that (4.9) holds. Furthermore, for i = 1 , 3 ( E i ) implies that
H ( a i , a i ) λ i n 4 G ( a 1 , a 3 ) ( λ 1 λ 3 ) n 4 = o ( 1 ( λ 1 d 1 ) n 4 + 1 ( λ 3 d 3 ) n 4 + ε 13 ) .
(4.11)

We denote by r ( x ) the eigenvector associated to ρ ( x ) whose norm is 1. We point out that we can choose r ( x ) so that all their components are positive (see [18] and [19]).

Let Λ i = λ i ( 4 n ) / 2 , Λ = ( Λ 1 , Λ 3 ) , and x = ( a 1 , a 3 ) . From (4.11), we have
M ( x ) Λ t Λ = o ( 1 ) .
(4.12)
The scalar product of (4.12) by r ( x ) gives
ρ ( x ) r ( x ) Λ t Λ = o ( 1 ) .
(4.13)
Since the components of r ( x ) are positive and λ 1 , λ 3 are of the same order, there exists a positive constant c, such that r ( x ) Λ t Λ c > 0 . Hence, we get
ρ ( x ) = o ( 1 ) .
(4.14)
We deduce from (4.3) and (4.11) that
M x i ( x ) Λ t Λ = o ( 1 ) .
(4.15)
Observe that Λ may be written in the form
Λ = β r ( x ) + r ¯ ( x ) , with  r ( x ) r ¯ ( x ) = 0 , r ¯ = o ( β )  and  β Λ .
(4.16)
Using (4.15), we get
M x i ( x ) t r ( x ) + M x i ( x ) r ¯ ( x ) Λ = o ( 1 ) .
(4.17)

Since d i c 0 for i = 1 , 3 and | a 1 a 3 | c 0 , the matrix M x i ( x ) is bounded.

Furthermore, we have r ¯ = o ( Λ ) , which implies that
M x i ( x ) t r ( x ) = o ( 1 ) .
(4.18)
Let us consider the equality
M ( x ) t r ( x ) = ρ ( x ) t r ( x )
and derivative it with respect to x i ; we obtain
M x i ( x ) t r ( x ) + M ( x ) t r x i ( x ) = ρ x i ( x ) t r ( x ) + ρ ( x ) t r x i ( x ) .
The scalar product with r ( x ) gives
r ( x ) M x i ( x ) t r ( x ) = ρ x i ( x ) .
(4.19)
Using (4.18), we obtain
ρ x i ( x ) = o ( 1 ) .
(4.20)

Hence, we derive a contradiction from (4.14), (4.20), and the fact that 0 is a regular value of ρ. Thus the proof of our theorem follows.

Proof of Theorem 1.2

Arguing by contradiction, let us assume that problem ( P ε ) has solutions ( u ε ) 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 ,
( E i ) c 1 n 4 2 H ( a i , a i ) λ i n 4 + γ i c 1 j i γ j ( λ i ε i j λ i + n 4 2 H ( a i , a j ) ( λ i λ j ) ( n 4 ) / 2 ) + n 4 2 c 2 ε ( E i ) = o ( ε + j = 1 k 1 ( λ j d j ) n 4 + r j ε r j ) .
Observe that, if j < i , we have λ j | a i a j | is bounded (by the assumption) which implies that
| a i a j | = o ( d j ) , d i / d j = 1 + o ( 1 ) , i , j and ε i j c ( λ j / λ i ) ( n 4 ) / 2 , j < i ,
(4.21)
where c is a positive constant. Using (4.21), easy computations show that
ε ( i 1 ) j + ε i ( j + 1 ) = o ( ε i j ) , i < j , H ( a i , a j ) ( λ i λ j ) ( n 4 ) / 2 = o ( 1 ( λ 1 d 1 ) n 4 ) if  ( i , j ) ( 1 , 1 ) .
(4.22)
Thus, using (4.22), ( E i ) can be written as
( E 1 ) c 1 n 4 2 H ( a 1 , a 1 ) λ 1 n 4 + c 1 γ 1 γ 2 λ 1 ε 12 λ 1 + n 4 2 c 2 ε = o ( ε + 1 ( λ 1 d 1 ) n 4 + r j ε r j ) , ( E k ) c 1 γ k 1 γ k λ k ε ( k 1 ) k λ k + n 4 2 c 2 ε = o ( ε + 1 ( λ 1 d 1 ) n 4 + r j ε r j ) ,
and for 1 < i < k ,
( E i ) c 1 γ i 1 γ i λ i ε ( i 1 ) i λ i + c 1 γ i γ i + 1 λ i ε i ( i + 1 ) λ i + n 4 2 c 2 ε = o ( ε + 1 ( λ 1 d 1 ) n 4 + r j ε r j ) .

The proof will depend on the value of l which is defined in the theorem.

Case 1. l = k . From the definition of l we get γ k 1 γ k = 1 . Now using (4.1) and ( E k ) , we derive that
ε = o ( 1 ( λ 1 d 1 ) n 4 + r j ε i j ) and ε ( k 1 ) k = o ( 1 ( λ 1 d 1 ) n 4 + r j ε r j ) .
(4.23)
Now, using (4.23) and ( E k 1 ) , we derive the estimate of ε ( k 2 ) ( k 1 ) and by induction we get
ε ( i 1 ) i = o ( 1 ( λ 1 d 1 ) n 4 + r j ε r j ) for each  i = 2 , , k .
(4.24)
Finally, using (4.22), (4.23), (4.24), and ( E 1 ) we obtain
H ( a 1 , a 1 ) λ 1 n 4 = o ( 1 ( λ 1 d 1 ) n 4 ) ,

which gives a contradiction.

Case 2. l = k 1 . Using (4.1), an easy computation implies that
λ k 1 ε ( k 1 ) k λ k 1 λ k ε ( k 1 ) k λ k c ε ( k 1 ) k .
(4.25)
Then from ( E k 1 ) , ( E k ) , (4.1), (4.25), and the fact that γ k 1 γ k = 1 and γ k 2 γ k 1 = 1 (since l = k 1 ), we obtain
c ε ( k 1 ) k + ε ( k 2 ) ( k 1 ) = o ( ε + 1 ( λ 1 d 1 ) n 4 + r j ε r j ) .
(4.26)

Now using ( E k ) and (4.26) we get (4.23) and as before, (4.24) is satisfied. Hence we also derive a contradiction from ( E 1 ) .

Case 3. l { k , k 1 } . Recall that in this case we have assumed that λ l | a l a l + 1 | 0 . This implies that
λ l ε l ( l + 1 ) λ l = ( ( n 4 ) / 2 ) ε l ( l + 1 ) ( 1 + o ( 1 ) ) .
(4.27)

Hence, using ( E l ) , the definition of l and (4.1) we obtain the first part of (4.23). The second part follows from ( E k ) and the first one. Finally, as before we derive a contradiction from  ( E 1 ) .

Hence, our theorem is proved.

Declarations

Acknowledgements

The author gratefully acknowledges the Deanship of Scientific Research at Taibah University on material and moral support, in particular by financing this research project.

Authors’ Affiliations

(1)
Department of Mathematics, Taibah University, P.O. Box 30002, Almadinah Almunawwarah, Kingdom of Saudi Arabia

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© Ould Bouh; licensee Springer. 2014

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

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