Multi-peak solutions for the Henon equation with slightly subcritical growth

Angela Pistoia · Enrico Serra

Multi–peak solutions for the H´ enon equation with slightly subcritical growth? Abstract. We study the Dirichlet problem for the H´enon equation 8 N +2 −ε > 0 > : u=0

in Ω, in Ω, on ∂Ω,

where Ω is the unit ball in IRN , with N ≥ 3, the power α is positive and ε is a small positive parameter. We prove that for every integer k ≥ 1 the above problem has a solution which blows up at k different points of ∂Ω as ε goes to zero. We also show that the ground state solution (which blows up at one point) is unique. Mathematics Subject Classification (2000): 35J60, 35J20, 35J25 Key words: H´enon equation, blowing up solutions, critical exponent, ground state solution.

1. Introduction In a 1973 paper, [11], M. H´enon introduced, as a model in the context of spherically symmetric clusters of stars, the Dirichlet problem  α p  −∆u = |x| u in Ω, (1) u>0 in Ω,   u=0 on ∂Ω, where Ω is the unit ball in IRN , with N ≥ 3, the power α is positive, and p > 1. H´enon studied this equation numerically, for some definite values of α and p, but subsequent researches showed that the above problem exhibits very rich A. Pistoia: Dipartimento di Metodi e Modelli Matematici, Universit` a di Roma “La Sapienza”, e-mail: [email protected] E. Serra: Dipartimento di Matematica, Universit` a di Milano e-mail: [email protected] ?

The first author is supported by the M.I.U.R. National Project “Metodi variazionali e topologici nello studio di fenomeni non lineari” . The second author is supported by the M.I.U.R. National Project “Metodi variazionali ed equazioni differenziali nonlineari”.

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Angela Pistoia, Enrico Serra

features from the functional–analytic point of view. In particular, various questions that arise quite naturally concerning existence, multiplicity and qualitative properties of solutions, have given the H´enon equation the role of a very interesting item in nonlinear analysis and critical point theory. In order to make clear the motivations of this paper, we recall some known properties and results concerning problem (1). The starting remark one can make is that the presence of the weight |x|α modifies, in some sense, the global homogeneity properties of the equation, so that the usual arguments based on the Pohozaev identity yield that N +2 2α problem (1) has no solution if p ≥ N −2 + N −2 . The first existence result is due to Ni, who in [14] proved that for every N +2 2α p ∈ (1, N −2 + N −2 ), problem [11] admits at least one radial solution. It is interesting to notice that since the weight r 7→ rα is increasing, symmetry results such as those of Gidas, Ni and Nirenberg ([9]) do not apply, and therefore nonradial solutions might be present as well. The first to realize this were Smets, Su and Willem, who studied the ground state solutions associated to (1). In [18] they proved, among other results, that for N +2 p ∈ (1, N −2 ) no ground state is radial provided α is large enough. This result shows in particular that in the subcritical range, problem (1) has at least two solutions for α large. In view of the above considerations, existence of nonradial solutions could be expected also for larger values of N +2 p. Indeed, in [17] the author proved that for p = N −2 problem (1) has a nonradial solution for α large. In that paper great use is made of the fact that the weight |x|α prohibits concentration phenomena at zero. This feature was also used in [3], a paper where multiplicity results for some supercritical growths are obtained. These results point out that the presence of the weight acts similarly to that of a hole in the domain Ω. It prevents, as we have observed, concentration phenomena at zero, and therefore enriches the topology of some sublevel sets of the functionals associated to (1). This is the heuristic reason why, apart from the technical estimates, solutions are present also for critical and supercritical growths. At this stage it is therefore tempting to conjecture that the analogy weight– hole reduces the existence results for the H´enon equation to similar ones for problems on annuli. In this paper we will show that this is not at all the case, and that the presence of the weight gives rise to phenomena that are not present for problems on domains with holes. N +2 In order to illustrate this point we set p = N −2 and we consider the “slightly subcritical” problem  α p−ε  −∆u = |x| u u>0   u=0

in Ω, in Ω, on ∂Ω.

(2)

Multi–peak solutions

3

where Ω is the unit ball in IRN centered at the origin, N ≥ 3, α > 0 and ε > 0 is a small parameter. If the weight were not present (α = 0), and the domain Ω were a general open set, results in [4] would ensure that the number of solutions of (2) is bounded independently of ε. In sharp contrast, the main result in this paper will show that problem 2 admits an arbitrarily large number of solutions provided ε is small enough. This is a rather unexpected behavior if one is accustomed to nearly critical problems such as those treated in [4], [13] and [16]. Our main result will actually describe the shape of the solutions as ε tends to zero. In order to state it we introduce some notation. For every δ > 0 and every ξ ∈ IRN we denote Uδ,ξ (x) =

δ

N −2 2

(δ 2 + |x − ξ|2 )

N −2 2

.

N −2

The functions a0 Uδ,ξ , with a0 = [N (N − 2)] 4 , are the only solutions of the equation −∆u = up in IRN . We also denote by P Uδ,ξ the projection onto H01 (Ω) of the function Uδ,ξ , namely ∆P Uδ,ξ = ∆Uδ,ξ in Ω, P Uδ,ξ = 0 on ∂Ω. In Section 1 we will prove the following multiplicity result. Theorem 1. Let N ≥ 3. For every integer k ≥ 1 there exists εk > 0 such that for every ε ∈ (0, εk ) problem (2) has a family of solutions uε blowing up at k different points of the boundary. More precisely there exists δε > 0 and τε > 0 such that, as ε goes to zero, δε → 0, τε → 0 and uε −

k X

a0 P Uδε ,(1−τε )ξi∗ → 0 in H01 (Ω),

i=1

where, for every i = 1, . . . , k, µ ¶ 2π 2π ξi∗ = cos (i − 1), sin (i − 1), 0, . . . , 0 . k k

(3)

The proof of this result makes use of rather heavy estimates (collected in Section 4) in the spirit of [2], [4], or [16], while the technique is inspired by that of [7], [8] and [13]. In the second part of the paper we provide a uniqueness result for the ground states of (2). In [18] the authors proved that R |∇u|2 Ω (4) Sε := inf ¡R ¢ 2 u∈H01 (Ω) α |u|p+1−ε p+1−ε |x| Ω u6=0

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Angela Pistoia, Enrico Serra

is achieved by a positive function uε which is non-radial if ε is small enough. The function uε , suitably scaled, is a ground state solution of (2). N +2 The analysis of the asymptotic behavior as p → N −2 of ground state solutions was carried out in [6], where it is shown that they blow up at a single point of the boundary of Ω. In Section 2 we will prove the following uniqueness result. Theorem 2. Let N ≥ 3. There exists ε0 > 0 such that for every ε ∈ (0, ε0 ) problem (2) has a unique ground state solution uε . We alert the reader from the beginning that the proof of Theorem 2 requires various estimates similar to those used in the proof of Theorem 1, as well as a combination of arguments that can be found in the existing literature. Since technicalities are already rather involved for the multiplicity result, we have decided to be rather sketchy in the proof of Theorem 2, confining ourselves to an outline of the argument and addressing the interested reader to existing works. R Notation. We equip H01 (Ω) with Rthe inner product (u, v) = Ω ∇u∇v dx 2 and corresponding norm = Ω |∇u|2 dx. For r ∈ [1, +∞) and u ∈ R ||u|| r r r L (Ω), we set kukr = Ω |u| dx. In most of the integrals, when no ambiguity is possible, we will omit to write dx, dy and similar expressions.

2. A multiplicity result For the proof of Theorem 1 we follow a by now classical reduction argument that was developed in [1], [4], [8] and [13]. First of all we rewrite problem (2) in a different form. We consider the 2N standard immersion operator i : H01 (Ω) → L N −2 (Ω), and we define its 2N adjoint i∗ : L N +2 (Ω) → H01 (Ω), namely Z ∗ i (u) = v ⇐⇒ (v, ϕ) = u(x)ϕ(x) dx ∀ ϕ ∈ H01 (Ω). Ω 2N

It is clear that ki∗ (u)k ≤ ckuk 2N for every u ∈ L N +2 (Ω). N +2 Setting fε (s) = (s+ )p−ε , by means of the operator i∗ , problem (2) can be equivalently written ( u = i∗ [|x|α fε (u)] (1) u ∈ H01 (Ω). We look for a solution of the form uε =

k X i=1

a0 P Uδi ,ξi + φ,

Multi–peak solutions

5

where the concentration parameters verify δ1 = · · · = δk =: δ and the concentration points ξ1 , . . . , ξk are chosen to be µ ¶ 2π 2π ξi := (1 − τ )ξi∗ = (1 − τ ) cos (i − 1), sin (i − 1), 0, . . . , 0 . (2) k k The remainder term φ is symmetric with respect to each of the variables x3 , . . . , xN and is invariant under rotation of 2π k in the space of the first two variables x1 and x2 . In order to define the functional setting we introduce some notation. For i, j = 1, . . . , k we define ψi0 (x) :=

∂Uδi ,ξi N − 2 N −4 |x − ξi |2 − δ 2 δ 2 = a0 , ∂δi 2 (δ 2 + |x − ξi |2 )N/2

x ∈ IRN

and ψij (x) :=

∂Uδi ,ξi ∂ξij

= −a0 (N − 2)δ

N −2 2

xi j − ξi j , (δ 2 + |x − ξi |2 )N/2

x ∈ IRN .

We recall from [5] that the space spanned by ψij , j = 0, 1, . . . , N is the set of the solutions of the linearized problem −∆ψ = p Uδp−1 ψ in IRN . i ,ξi Notice also that ¡ ¢ P ψij (x) = i∗ p Uδp−1 ψij (x), i ,ξi

x ∈ Ω.

(3)

Let He = {φ ∈ H01 (Ω) | φ is symmetric with respect to each of the variables x3 , . . . , xN and is invariant under rotation of 2π k in the space of the first two variables x1 and x2 }. We introduce the spaces n o Kδ,τ = span P ψij | j = 0, 1, . . . , N, i = 1, . . . , k and

n o ⊥ Kδ,τ = φ ∈ He | (φ, P ψij ) = 0, j = 0, 1, . . . , N, i = 1, . . . , k

and the projection operators Πδ,τ (u) =

k X N X (u, P ψij )P ψij

and

⊥ Πδ,τ (u) = u − Πδ,τ (u).

i=1 j=0

Finally we denote, for simplicity, Vδ,τ :=

k X i=1

a0 P U i ,

with Ui := Uδi ,ξi .

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Angela Pistoia, Enrico Serra

Our approach to solve problem (1) will be to find a pair (δ, τ ) and a function ⊥ φ ∈ Kδ,τ such that ⊥ Πδ,τ (Vδ,τ + φ − i∗ [|x|α fε (Vδ,τ + φ)]) = 0

(4)

Πδ,τ (Vδ,τ + φ − i∗ [|x|α fε (Vδ,τ + φ)]) = 0

(5)

and ⊥ First of all we find, for every (δ, τ ) and for small ε, a function φ ∈ Kδ,τ such that (4) is fulfilled. ⊥ ⊥ To this aim we define a linear operator Lδ,τ : Kδ,τ → Kδ,τ by

" Lδ,τ φ = φ −

⊥ ∗ Πδ,τ i

k X

# f00 (a0 Ui )φ

.

i=1

Arguing as in [8] and in [13] one can prove the following result. Proposition 1. There exist δ0 > 0, τ0 > 0, η0 > 0 and C > 0 such that for every δ ∈ (0, δ0 ), τ ∈ (0, τ0 ) with δ/τ ∈ (0, η0 ), the operator Lδ,τ is invertible and ⊥ kLδ,τ φk ≥ Ckφk, ∀ φ ∈ Kδ,τ . We now show that by means of the previous proposition we can solve (4). We assume from now on that δ = O (εγ ) for some γ > 0 to be fixed later. Proposition 2. There exist δ0 > 0, τ0 > 0, η0 > 0 and ε0 > 0 such that for every δ ∈ (0, δ0 ), τ ∈ (0, τ0 ) with δ/τ ∈ (0, η0 ) and ε ∈ (0, ε0 ), there exists ⊥ a unique φεδ,τ ∈ Kδ,τ such that ¡ ¢ ⊥ Πδ,τ Vδ,τ + φεδ,τ − i∗ [|x|α fε (Vδ,τ + φεδ,τ )] = 0.

(6)

Moreover ¸  · ¡ δ ¢ N2+2  O ε| log ε| + τ + τ        h ¡ ¢ ¡ ¢i ε kφδ,τ k = O ε| log ε| + τ + δ 4 log2/3 δ τ τ       h i   O ε| log ε| + τ + ¡ δ ¢N −2 τ

if N ≥ 7, if N = 6,

(7)

if N = 3, 4, 5.

Finally, the map (δ, τ ) 7→ φεδ,τ is of class C 1 . Proof. By simple manipulations, it is easy to see that φ solves equation (6) ε ⊥ ⊥ if and only if φ is a fixed point of the map Tδ,τ : Kδ,τ → Kδ,τ defined by k k h i X X α ε ⊥ ∗ |x| f (V + φ) − f (a U ) − f00 (a0 Ui )φ . Tδ,τ (φ) = L−1 Π i ε δ,τ 0 0 i δ,τ δ,τ i=1

i=1

Multi–peak solutions

7

ε We prove that Tδ,τ is a contraction. By Proposition 1 we deduce that ε kTδ,τ (φ)k ≤ c k|x|α [fε (Vδ,τ + φ) − fε (Vδ,τ ) − fε0 (Vδ,τ )φ]k

+c k|x|α [fε0 (Vδ,τ ) − f00 (Vδ,τ )] φk +c k[|x|α f00 (Vδ,τ )

2N N +2

2N N +2

f00 (Vδ,τ )] φk 2N N +2

− °" # ° k ° ° X ° 0 ° +c ° f0 (Vδ,τ ) − f00 (a0 Ui ) φ° ° ° i=1

+c k|x|α [fε (Vδ,τ ) − f0 (Vδ,τ )]k

2N N +2

2N N +2

α

+c k|x| f0 (Vδ,τ ) − f0 (Vδ,τ )k 2N N +2 ° ° k ° ° X ° ° +c °f0 (Vδ,τ ) − f0 (a0 Ui )° . ° ° 2N i=1

N +2

Each of the above terms is estimated according to Lemmas 8, 9 and 10. It ε follows that if kφk satisfies (7) then also kTδ,τ (φ)k satisfies (7). Arguing in ε a similar way one proves that, for some L ∈ (0, 1), there results kTδ,τ (φ1 ) − ε Tδ,τ (φ2 )k ≤ Lkφ1 − φ2 k, provided kφ1 k and kφ2 k satisfy (7). That concludes the main part of the proof. In order to check the smoothness of the map (δ, τ ) 7→ φεδ,τ one works as in [4] and in [8]. We now introduce the standard action functional Jε : H01 (Ω) → IR defined by Z Z 1 1 2 Jε (u) = |∇u| dx − |x|α |u|p+1−ε dx (8) 2 p+1−ε Ω

Ω

and the function Jeε defined by ¡ ¢ Jeε (δ, τ ) = Jε Vδ,τ + φεδ,τ .

(9)

It is well known that critical points of Jε are (classical) solutions of problem (2); the role of the function Jeε is described by the following proposition. Proposition 3. The function Vδ,τ + φεδ,τ is a critical point of the functional Jε if and only if the pair (δ, τ ) is a critical point of the function Jeε . Proof. We argue as in the proof of Proposition 1 in [4] and as in the proof of Lemma 3.1 in [8]. The problem is thus reduced to the search for critical points of Jeε . In order to find these, we now compute the first terms of the expansion of Jeε . This is the most technical part of the paper; to simplify the reading we postpone to Section 4 some of the most involved computations.

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Angela Pistoia, Enrico Serra

Proposition 4. There results " # µ ¶N −2 δ p+1 Jeε (δ, τ ) = ka0 c1 + αc2 τ + c3 − c4 ε log δ + c5 ε τ õ ¶ ! N −2 δ + o(τ ) + o + o(ε), τ

(10)

C 1 −uniformly with respect to (δ, τ ). Here γ1 γ1 γ2 , c2 := , c3 := N −1 , N p+1 2 (N − 2)2 γ1 γ1 log a0 γ3 c4 := γ1 , c5 := − + + , 4N (p + 1)2 p+1 p+1 c1 :=

(11) (12)

where the numbers γ1 , γ2 and γ3 are defined in (1), (2) and (3) respectively. Proof. For simplicity we write φ instead of φεδ,τ . The function Jeε (δ, τ ) can be expressed as Z Z 1 1 2 p+1−ε Jeε (δ, τ ) = |∇ (Vδ,τ + φ)| dx − |x|α |Vδ,τ + φ| dx 2 p+1−ε Ω Ω Z Z 1 1 2 p+1 |∇Vδ,τ | dx − |x|α Vδ,τ dx = 2 p+1 Ω Ω   Z Z 1 1 p+1 p+1 −ε  |x|α Vδ,τ dx − |x|α Vδ,τ log Vδ,τ dx (p + 1)2 p+1 Ω

Ω

+I1 + I2 + I3 , where

Z Z 1 1 p+1 α p+1−ε I1 = − |x| Vδ,τ dx + |x|α Vδ,τ dx p+1−ε p+1 Ω Ω   Z Z 1 1 p+1 p+1 +ε |x|α Vδ,τ dx − |x|α Vδ,τ log Vδ,τ dx , (p + 1)2 p+1 Ω

I2 = − 1 I3 = 2

1 p+1−ε

Z

h i p+1−ε p+1−ε p−ε |x|α |Vδ,τ + φ| − Vδ,τ − (p + 1 − ε)Vδ,τ φ dx,

Ω

Z

Ω

Z

2

|∇φ| dx + Ω

Z |x|α fε (Vδ,τ )φ dx

∇Vδ,τ ∇φdx − Ω

Ω

The first terms in the expansion of Jeε (δ, τ ) are estimated by means of Lemmas 6 and 7.

Multi–peak solutions

9

As far as I1 , I2 and I3 are concerned, we see that working as in Lemma 3.2 of [7], there results I1 = o(ε), while one can easily check that ¡ ¢ I2 = O kφk2 . The treatment of I3 is more involved. We write it as Z Z 1 2 I3 = |∇φ| dx + |x|α [fε (Vδ,τ ) − f0 (Vδ,τ )] φ dx 2 Ω Ω Z Z " [|x|α f0 (Vδ,τ ) − f0 (Vδ,τ )] φ dx +

+ Ω

f0 (Vδ,τ ) −

k X

# f0 (a0 Ui ) φ dx

i=1

Ω

and we estimate each term with the aid of Lemmas 8, 9 and 10. Keeping in mind also (7) we conclude that I1 , I2 and I3 are lower order terms. Therefore expansion (10) follows C 0 −uniformly with respect to (δ, τ ). In a similar way we prove that the expansion holds also in a C 1 sense (see also [4] and [8] for similar computations). To complete the proof of the main result we assume from now on that

and we define

N −1

δ = dε N −2

(13)

³ ´ Jε∗ (d, t) := Jeε δ(d, ε), τ (t, ε) .

(14)

τ = tε

and

The link between Jε∗ and Jeε is given by the following proposition. In its statement the constants ci are those defined in (11) and (12). Proposition 5. The pair (δ, τ ) is a critical point of Jeε if and only if (d, t) is a critical point of Jε∗ . Moreover letting F (d, t) := αc2 t + c3 we have that Jε∗ (d, t)

=

kap+1 0

µ ¶N −2 d − c4 log d, t

· ¸ N −1 c1 + εF (d, t) − c4 ε log ε + c5 ε + o(ε) N −2

(15)

(16)

C 1 −uniformly with respect to (d, t) in compact subsets of IR+ × IR+ . Proof. It follows by (13) and Proposition 4. Proposition 6. The function F has a unique critical point which is a nondegenerate minimum.

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Angela Pistoia, Enrico Serra

Proof. Since ∂F dN −2 = αc2 − (N − 2)c3 N −1 ∂t t

∂F dN −3 c4 = (N − 2)c3 N −2 − , ∂d t d

and

one sees immediately that F has unique critical point (d0 , t0 ). A trivial 2 2 4 4 computation shows that ∂∂tF2 (d0 , t0 ) = (N −1)c , ∂∂dF2 (d0 , t0 ) = (N −2)c and t2 d2 ∂2F ∂d∂t (d0 , t0 )

0

4 = − (Nd−2)c , so that 0 t0

det ∇2 F (d0 , t0 ) =

0

(N − 2)c24 > 0. d20 t20

This proves our claim. End of the proof of Theorem 1. By Proposition 6 it follows that there exists a critical point (d0 , t0 ) of F which is stable under C 1 perturbations. Proposition 5 then yields that for ε small enough there exists a critical point (dε , tε ) of Jε∗ such that dε → d0 and tε → t0 , as ε goes to zero. If δε and τε are defined as in (13), Proposition 5 tell us that (δε , τε ) is a critical point of Jeε . Finally the proof is completed by the application of Proposition 3.

3. Uniqueness of the ground state solution In [6] the authors studied the profile of the ground state solutions uε (functions where Sε in (4) is attained) as ε → 0, and they proved the following result. Theorem 3 ([6]). If ε is small enough there exist aε > 0, δε > 0, ξε ∈ Ω and φε ∈ H01 (Ω) such that uε = aε P Uδε ,ξε + φε ,

(1)

where, as ε → 0, aε → a0 ,

ξε → ξ0 ∈ ∂Ω,

dist(ξε , ∂Ω) →0 δε

and

kφε k → 0.

(2)

In this section we give an outline of the proof of Theorem 2, namely, the ground state solution is unique for ε small. As we pointed out in the Introduction, we decided to omit most of the details since they can be reconstructed from existing results in the literature, and they consist in nothing really different from the estimates of Section 4. Thus, we confine ourselves to illustrate the sequence of the arguments needed in the proof, and we give the corresponding references when we suppress the computations. First of all we point out a symmetry result for ground state solutions.

Multi–peak solutions

11

Theorem 4. If ε is small enough, uε is axially symmetric with respect to the axis passing trough the origin and its unique maximum point xε . Proof. By [12] we obtain that the solution uε has index one. In Theorem 1.2 of [6] it was proved that uε has a unique maximum point xε provided ε is small enough. The symmetry property follows from Theorem 3.1 of [15]. By Theorem 3 and Theorem 4 we can assume that there exists τε > 0 such that uε = a0 P Uδε ,ξε + φε , (3) where, as ε → 0, ξε = (1 − τε , 0, . . . , 0),

τε → 0 and

δε →0 τε

(4)

and φε is even with respect to x3 , . . . , xN .

(5)

We will prove that problem (2) has a unique solution which satisfies (3), (4) and (5). The idea is similar to the one considered in [4] and [10]. ⊥ First of all we point out the following result. In its statement, Kδ,τ is the set defined in Section 2. Proposition 7. For ν > 0, let ½ ⊥ Mν = (δ, τ, φ) ∈ (0, ∞) × (0, 1) × Kδ,τ | δ < ν,

δ < ν, kφk < ν τ

¾

and define Jˆε : Mν → IR by Jˆε (δ, τ, φ) := Jε (a0 P Uδ,τ + φ) ,

(6)

where Jε is the functional in (8). Then the function uε = a0 P Uδε ,τε + φε with (3), (4) and (5), is a critical point of Jε if and only if (δε , τε , φε ) ∈ Mν for some ν > 0, and is a critical point of Jˆε . Proof. We argue as in [1] and [16]. Secondly we have: Proposition 8. Let (δε , τε , φε ) ∈ Mν , for some ν > 0, be a critical point of Jˆε . Then φε = φεδ,τ and estimates (7) hold. Proof. If (δε , τε , φε ) is a critical point of Jˆε , then equation (6) is satisfied and by Proposition 2 the claim follows. Finally we have:

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Angela Pistoia, Enrico Serra

Proposition 9. Let (δε , τε , φε ) ∈ Mν , for some ν > 0, be a critical point of Jˆε . Then, as ε goes to zero, tε :=

τε → t0 > 0 ε

and

dε :=

δε ε

N −1 N −2

→ d0 > 0.

(7)

Proof. By Proposition 4 we deduce that ¢ ∂ Jˆε ¡ δ N −3 ε δ, τ, φεδ,τ = c3 (N − 2) N −2 − c4 + l.o.t. ∂δ τ δ and ¢ δ N −2 ∂ Jˆε ¡ δ, τ, φεδ,τ = αc2 − c3 (N − 2) N −1 + l.o.t. ∂τ τ C 0¡−uniformly with respect to (δ, τ ). Therefore t = O(ε) and δ N −2 = ¢ N −1 O t . Then the proof can be easily completed. Proposition 10. (δ, τ, φ) ∈ Mν is a critical point of Jˆε (see (6)) if and only if (dε , tε ) ∈ IR+ × IR+ (see (7)) is a critical point of Jε∗ (see (14)). Moreover Jε∗ (d, t) = c1 + εF (d, t) − c4 ε log ε + o(ε)

(8)

C 2 −uniformly with respect to (d, t) in compact subsets of IR+ × IR+ . ¡ ¢N −2 Here F (d, t) := αc2 t + c3 dt − c4 log d (see (15). Proof. We use Proposition 5: estimate (8) is equivalent to estimate (16) with k = 1. It remains only to prove that estimate (8) holds in a C 2 sense. To this aim we have to compute the first and second derivates of Jε∗ argue as in [10]. Proposition 11. If ε is small enough problem (2) has a unique solution uε which satisfies (3), (4) and (5). Proof. By Proposition 6 we deduce that the function F has a unique critical point which is nondegenerate. By estimate (8) we see that the functions Jε∗ has a unique critical point, provided ε is small enough. By Proposition 10 it follows also that the function Jˆε introduced in (6) has a unique critical point, provided ε is small enough. The application of Proposition 7 concludes the proof. Proof of Theorem 2. The claim follows by Theorem 3, Theorem 4 and Proposition 11.

Multi–peak solutions

13

4. Estimates In this section we establish the technical estimates that were used in Section 2. First of all we recall a classical result which will be useful throughout the section and whose proof is given in [16]. In its statement we denote by H the regular part of the Green function of the laplacian on the unit ball Ω, namely 1 H(x, y) = . N −2 ||x|y − x/|x|| Proposition 12. Let ξ ∈ Ω and τ = dist(ξ, ∂Ω). For every δ > 0, define ϕδ,ξ = Uδ,ξ − P Uδ,ξ . Then 0 ≤ ϕδ,ξ (x) ≤ Uδ,ξ (x), and ϕδ,ξ (x) = δ

N −2 2

x∈Ω

H(x, ξ) + Rδ,ξ (x),

x ∈ Ω,

with Rδ,ξ verifying the uniform estimate à N +2 ! δ 2 Rδ,ξ = O . τN Moreover kϕδ,ξ k

2N N −2

õ ¶ N −2 ! 2 δ =O and τ

ÃÃ kϕδ,ξ kL∞ (B(ξ,τ )) = O

For future reference we define the constants Z 1 γ1 := dy, (1 + |y|2 )N N IR Z 1 γ2 := N +2 dy, (1 + |y|2 ) 2 N IR Z 1 1 γ3 := log N −2 dy. 2 N 2) 2 (1 + |y| ) (1 + |y| N

N −2

δ 2 τ N −2

!! .

(1) (2) (3)

IR

We begin with the estimate of the principal part in the expansion of Jeε (δ, τ ), needed in the proof of Proposition 4. In particular, the next Lemma deals with the first term of the principal part. Lemma 1. There reslts " õ ¶ !# µ ¶N −2 Z N −2 1 1 p+1 1 δ δ 2 |∇Vδ,τ | dx = ka0 γ1 − N −2 γ2 +o . 2 2 2 τ τ Ω

14

Angela Pistoia, Enrico Serra

Proof. For the sake of simplicity we assume k = 2. We write 1 2

Z 2

|∇Vδ,τ | dx = Ω

=

Z

2 ¯ X ¯2 ¯∇ P Uδ,ξi ¯ dx i=1

Ω

2 Z X

1 2 a 2 0 i=1

1 2 a 2 0

Z

2

|∇P Uδ,ξi | dx + a20

Ω

Ω

2 Z X

1 = ap+1 2 0 i=1

∇P Uδ,ξ1 ∇P Uδ,ξ2 dx

p P Uδ,ξi Uδ,ξ dx i

+

Z

ap+1 0

Ω

p P Uδ,ξ1 Uδ,ξ dx. 2

(4)

Ω

By estimate (5.36) in [1] we see that Z p P Uδ,ξ1 Uδ,ξ dx 2

! õ ¶ N −2 δ . =O τ

(5)

Ω

For every given i, we split Z p P Uδ,ξi Uδ,ξ dx = i Ω

Z

Z ··· +

B(ξi ,τ )

...

(6)

Ω\B(ξi ,τ )

and we evaluate the integrals separately. By (3.21) in [16] we obtain õ ¶ ! Z N δ p P Uδ,ξi Uδ,ξi dx = O . τ

(7)

Ω\B(ξi ,τ )

Using Proposition 12 we can write Z p P Uδ,ξi Uδ,ξ dx i B(ξi ,τ )

Z

h

=

Uδ,ξi − δ

N −2 2

i p H(x, ξi ) + Rδ,ξi Uδ,ξ dx i

B(ξi ,τ )

Z

Z p+1 Uδ,ξ dx − δ i

=

Z

N −2 2

B(ξi ,τ )

p H(x, ξi )Uδ,ξ dx + i B(ξi ,τ )

p Rδ,ξi Uδ,ξ dx. i B(ξi ,τ )

(8) By the estimates of page 30 in [16] we see that Z

Z p+1 Uδ,ξ dx i

B(ξi ,τ )

= IRN

1 dy + O (1 + |y|2 )N

õ ¶ ! N δ , τ

(9)

Multi–peak solutions

15

while by Proposition 12 we obtain Z p Rδ,ξi Uδ,ξ dx i

õ ¶ ! N δ =O . τ

(10)

B(ξi ,τ )

Finally by Lemma 2 below we have Z Z N −2 p N −2 δ 2 H(x, ξi )Uδ,ξ dx = δ i B(ξi ,τ )

µ ¶N −2 δ = τ

H(δy + ξ, ξi )

B(0,τ /δ)

Z τ N −2 H(δy + ξi , ξi )

B(0,τ /δ)

1 (1 + |y|2 ) 

N +2 2

1 (1 + |y|2 )

N +2 2

dy

dy

 µ ¶N −2 Z δ 1  1 . = N +2 dy + o(1) 2) 2 τ 2N −2 (1 + |y| N

(11)

IR

The claim then follows by (4)–(11). Lemma 2. There results 0 ≤ τ N −2 H(δy + ξi , ξi ) ≤ 1

∀ y ∈ B(0, τ /δ)

and τ N −2 H(δy + ξi , ξi ) → 2−(N −2)

as

τ, δ → 0

and

δ → 0. τ

Proof. Since ¡ ¢− N2−2 H(x, y) = |x|2 |y|2 + 1 − 2(x, y) ,

x, y ∈ Ω,

the definition (2) of the points ξi , through an easy computation shows that τ N −2 H(δy + ξi , ξi ) "µ ¶ #− N2−2 µ ¶ 2 δ δ 2 2 ∗ 2 = |y| (1 − τ ) − 2 (y, ξi )(1 − τ )(2 − τ ) + (2 − τ ) . τ τ Therefore it is clear that τ N −2 H(δy + ξi , ξi ) → 2−(N −2) as τ, δ, τδ → 0. Moreover µ ¶2 µ ¶ δ δ |y|2 (1 − τ )2 − 2 (y, ξi∗ )(1 − τ )(2 − τ ) + (2 − τ )2 τ τ µ ¶2 µ ¶ δ δ 2 2 ≥ |y| (1 − τ ) − 2 |y|(1 − τ )(2 − τ ) + (2 − τ )2 τ τ ·µ ¶ ¸2 δ = |y|(1 − τ ) − (2 − τ ) ≥ 1, τ

16

since

Angela Pistoia, Enrico Serra δ τ |y|

≤ 1 and τ ∈ (0, 1). Then also the first part of the claim follows.

We now turn to the estimate of the second term of the principal part of Jeε (δ, τ ), which is more complicate. Lemma 3. There results Z 1 p+1 |x|α Vδ,τ dx p+1 Ω " õ ¶ ! µ ¶N −2 # N −2 p+1 ka0 p+1 δ δ = γ1 − αγ1 τ − N −2 γ2 + o (τ ) + o . p+1 2 τ τ Proof. For the sake of simplicity we again assume k = 2. First of all we write Z Z p+1 p+1 α p+1 |x| Vδ,τ dx = a0 |x|α (P Uδ,ξ1 + P Uδ,ξ2 ) dx Ω

=

Z ap+1 0

Ω

h i p+1 p+1 p+1 |x| (P Uδ,ξ1 + P Uδ,ξ2 ) − Uδ,ξ − U δ,ξ2 dx 1 α

Ω

Z + ap+1 0

h i p+1 p+1 |x|α Uδ,ξ + U dx. δ,ξ 1 2

(12)

Ω

The second integral is the object of Lemma 4. We split the first integral as Z Ω

h i p+1 p+1 p+1 |x|α (P Uδ,ξ1 + P Uδ,ξ2 ) − Uδ,ξ − U δ,ξ2 dx 1 Z

=

Z ... +

B(ξ1 ,τ )

B(ξ2 ,τ )

Z ... +

...

(13)

Ω\(B(ξ1 ,τ )∪B(ξ2 ,τ ))

By (7) we immediately have Z

h i p+1 p+1 p+1 |x|α (P Uδ,ξ1 + P Uδ,ξ2 ) − Uδ,ξ − U δ,ξ2 dx 1

Ω\(B(ξ1 ,τ )∪B(ξ2 ,τ ))

Z ≤c

õ ¶ ! N h i δ p+1 p+1 . Uδ,ξ1 + Uδ,ξ2 dx = O τ

(14)

Ω\(B(ξ1 ,τ )∪B(ξ2 ,τ ))

We now estimate the integral over B(ξ1 , τ ), the other one being analogous.

Multi–peak solutions

Z

17

h i p+1 p+1 p+1 |x|α (P Uδ,ξ1 + P Uδ,ξ2 ) − Uδ,ξ − U dx δ,ξ 1 2

B(ξ1 ,τ )

Z p

|x|α [Uδ,ξ1 + θ (P Uδ,ξ1 − Uδ,ξ1 + P Uδ,ξ2 )] ·

= (p + 1) B(ξ1 ,τ )

· (P Uδ,ξ1 − Uδ,ξ1 dx + P Uδ,ξ2 )

Z

p+1 |x|α Uδ,ξ dx 2

+ B(ξ1 ,τ )

Z p |x|α Uδ,ξ (P Uδ,ξ1 − Uδ,ξ1 ) dx + I1 + I2 + I3 , 1

= (p + 1)

(15)

B(ξ1 ,τ )

where I1 , I2 and I3 are defined and estimated right below. By Proposition 12 we easily deduce Z n o p p |x|α [Uδ,ξ1 + θ (P Uδ,ξ1 − Uδ,ξ1 + P Uδ,ξ2 )] − Uδ,ξ · I1 := (p + 1) 1 B(ξ1 ,τ )

· (P Uδ,ξ1 − Uδ,ξ1 ) dx

Z

p−1

θ|x|α [Uδ,ξ1 + ηθ (P Uδ,ξ1 − Uδ,ξ1 + P Uδ,ξ2 )]

= p(p + 1) B(ξ1 ,τ )

(P Uδ,ξ1 − Uδ,ξ1 + P Uδ,ξ2 ) (P Uδ,ξ1 − Uδ,ξ1 ) dx Z p−1 2 − Uδ,ξ1 ) dx + c Uδ,ξ |P Uδ,ξ1 − Uδ,ξ1 | Uδ,ξ2 dx 1

Z

p−1 Uδ,ξ (P Uδ,ξ1 1

≤c B(ξ1 ,τ )

B(ξ1 ,τ )

Z

+c

Z

|P Uδ,ξ1 − Uδ,ξ1 |

p+1

dx + c

B(ξ1 ,τ )

p |P Uδ,ξ1 − Uδ,ξ1 | Uδ,ξ dx 2

B(ξ1 ,τ )

õ ¶ ! N −2 δ =o , τ

(16)

Z I2 := (p + 1) Z

p

|x|α [Uδ,ξ1 + θ (P Uδ,ξ1 − Uδ,ξ1 + P Uδ,ξ2 )] P Uδ,ξ2 dx

B(ξ1 ,τ )

Z

p Uδ,ξ Uδ,ξ2 dx + c 1

≤c B(ξ1 ,τ )

Z +c B(ξ1 ,τ )

(P Uδ,ξ1 − Uδ,ξ1 )p Uδ,ξ2 dx

B(ξ1 ,τ )

õ ¶ ! N −2 δ p+1 Uδ,ξ2 dx = o , τ

(17)

18

Angela Pistoia, Enrico Serra

Z α

I3 :=

|x|

p+1 Uδ,ξ dx 2

õ ¶ ! N −2 δ =o . τ

(18)

B(ξ1 ,τ )

Finally, the remaining term in (15) is estimated in Lemma 5. The complete estimate follows then by (12)–(18), Lemma 4 and Lemma 5. Lemma 4. We have Z α

p+1 Uδ,ξ dx i

α

p+1 Uδ,ξ dx i

|x|

õ ¶ ! N δ = γ1 − αγ1 τ + o(τ ) + O . τ

Ω

Proof. Since Z

Z |x|

Ω

Z p+1 Uδ,ξ dx i

=

p+1 (|x|α − 1) Uδ,ξ dx i

+

Ω

(19)

Ω

and by [16] we have Z p+1 Uδ,ξ dx i

õ ¶ ! N δ , = γ1 + O τ

(20)

Ω

we only have to estimate the second term in 19. We split it as Z

Z

Z α

(|x| −

p+1 1) Uδ,ξ dx i

Ω

...

··· +

= B(ξi ,τ /2)

(21)

Ω\B(ξi ,τ /2)

and by (2.68) in [1] we see that Z

õ ¶ ! N δ p+1 (|x|α − 1) Uδ,ξ dx = O . i τ

Ω\B(ξi ,τ /2)

Since by the mean value theorem we have |δy + (1 − τ )ξi∗ |α − 1 = αδ(ξi∗ , y) − ατ + R(δ, τ, y), where R satisfies the uniform estimate ¡ ¢ |R(δ, τ, y)| ≤ c δ 2 |y|2 + δτ |y| + τ 2 ∀ y ∈ B(0, τ /2δ),

(22)

Multi–peak solutions

19

using the definition of the points ξi , we conclude that Z p+1 (|x|α − 1) Uδ,ξ dx i B(ξi ,τ /2)

Z

(|δy + (1 − τ )ξi∗ |α − 1)

=

1 dy (1 + |y|2 )N

B(0,τ /2δ)

Z

(αδ(ξi∗ , y) − ατ + R(δ, τ, y))

=

1 dy (1 + |y|2 )N

B(0,τ /2δ)

Z

= −ατ

1 dy + o(τ ), (1 + |y|2 )N

(23)

IRN

The claim follows by (19)–(23).

Lemma 5. We have Z |x|

α

p Uδ,ξ i

(P Uδ,ξi

õ ¶ ! µ ¶N −2 N −2 δ δ − Uδ,ξi ) dx = − N −2 γ2 +o . 2 τ τ 1

B(ξi ,τ )

Proof. By Proposition 12 we can write Z p |x|α Uδ,ξ (P Uδ,ξi − Uδ,ξi ) dx i B(ξi ,τ )

Z

i h N −2 p |x|α Uδ,ξ −δ 2 H(x, ξi )dx + Rδ,ξi (x) dx i

= B(ξi ,τ )

Z

= −δ N −2

|δy + ξi |α H(δy + ξi , ξi )

B(0,τ /δ)

Z

+δ

N −2 2

|δy + ξi |α Rδ,ξi (δy + ξi ) B(0,τ /δ)

1 (1 + |y|2 )

N +2 2

1 (1 + |y|2 )

N +2 2

dy

dy.

(24)

Therefore, still by Proposition 12 we have Z δ

N −2 2

õ ¶ ! N δ |δy + ξi | Rδ,ξi (δy + ξi ) dy = O . (25) N +2 τ (1 + |y|2 ) 2 α

B(0,τ /δ)

1

20

Angela Pistoia, Enrico Serra

As far as the other term in 24 is concerned, by Lemma 2 and by the dominated convergence theorem, we deduce that Z 1 N −2 δ |δy + ξi |α H(δy + ξi , ξi ) N +2 dy (1 + |y|2 ) 2 B(0,τ /δ)

µ ¶N −2 δ = τ

Z |δy + ξi |α τ N −2 H(δy + ξi , ξi )

B(0,τ /δ)

1 (1 + |y|2 )

N +2 2

dy

  µ ¶N −2 Z δ 1  1 , = N +2 dy + o(1) 2) 2 τ 2N −2 (1 + |y| N

(26)

IR

because |δy + ξi | ≤ 1 and |δy + ξi | → 1 as δ, τ go to zero. The claim follows by (24)–(26). We are now ready to describe the first two terms of the principal part of Jeε (δ, τ ). Lemma 6. There results Z Z 1 1 2 p+1 |∇Vδ,τ | dx − |x|α Vδ,τ dx 2 p+1 Ω Ω " ! õ ¶ µ ¶N −2 # N −2 γ1 1 δ δ p+1 γ1 = ka0 +α τ + N −1 γ2 + o (τ ) + o . N p+1 2 τ τ Proof. The the single terms are evaluated in Lemma 1 and Lemma 3. The next result describes the behavior of the last term in the principal part of Jeε (δ, τ ). Lemma 7. There results   Z Z 1 1 p+1 p+1 ε |x|α Vδ,τ dx − |x|α Vδ,τ log Vδ,τ dx (p + 1)2 p+1 Ω Ω ¶ µ γ1 log a0 γ3 (N − 2)γ1 γ1 − − + log δ + o(1) . = kεap+1 0 (p + 1)2 p+1 p+1 2(p + 1) Proof. We argue exactly as in Lemma 3.2 of [7]. We now turn to the quantities that appear in the proof of Proposition 2. Lemma 8. Assume that δ = O(εγ ) for some γ > 0. Then k|x|α [fε (Vδ,τ ) − f0 (Vδ,τ )]k and

2N N +2

= O (ε| log ε|)

k|x|α [fε0 (Vδ,τ ) − f00 (Vδ,τ )]k N = O (ε| log ε|) . 2

Multi–peak solutions

21

Proof. We prove only the first estimate since for the second one the argument is the same. First of all, by the mean value theorem, for every x ∈ Ω there exists θ = θ(x) ∈ [0, 1] such that p−θε

|fε (Vδ,τ ) − f0 (Vδ,τ )| = ε |log(Vδ,τ (x))| |Vδ,τ (x)|

.

Therefore we have Z 2N ||x|α [fε (Vδ,τ ) − f0 (Vδ,τ )]| N +2 dx Ω

Z 2N

≤ ε N +2

2N

2N

|log(Vδ,τ (x))| N +2 |Vδ,τ (x)| N +2

(p−θε)

.

(27)

Ω

For the sake of simplicity, we let k = 2, Ui = Uδ,ξi and Bi = B(ξi , τ ), for i = 1, 2. We estimate Z 2N 2N (p−θε) |log(a0 P U1 + a0 P U2 )| N +2 |a0 P U1 + a0 P U2 | N +2 Ω

Z

=

Z ··· +

B1

Z ··· +

B2

...,

(28)

Ω\{B1 ∪B2 }

for some r > 0 small enough. Define g(y) = + a0 δ

a0 (1 + |y|2 )

N −2 2

N −2 2

− a0 δ N −2 H(δy + ξ1 , ξ1 )

Rδ,ξ1 (δy + ξ1 ) + a0 δ

N −2 2

P U2 (δy + ξ1 )

Then, setting x = δy + ξ1 , we have, using Proposition 12 and familiar estimates, Z

Z ··· = δ

N

B1

¯ ³ N −2 ¯ ¯ N2N ´¯ N2N ¯ ¯ +2 ¯ N2−2 ¯ +2 (p−θε) g(y)¯ dy ¯log δ − 2 g(y) ¯ ¯δ

B(0,r/δ)

¯ N2N Z ¯ ¯ N −2 ¯ +2 −2 2N NN −θε ¯ ¯ N +2 δ log δ + log g(y) |g(y)| N +2 (p−θε) dy − ¯ ¯ 2 B1

≤ c| log δ|.

(29)

The integral over B2 is treated in the same way. Finally, since in Ω \ {B1 ∪ B2 } ! Ã N −2 N −2 δ 2 2a0 N −2 δ 2 + ≤ N −2 δ 2 , 0 ≤ a 0 P U1 + a 0 P U2 ≤ a 0 |x − ξ1 |N −2 |x − ξ2 |N −2 r

22

Angela Pistoia, Enrico Serra

we easily see that Z 2N 2N (p−θε) |log(a0 P U1 + a0 P U2 )| N +2 |a0 P U1 + a0 P U2 | N +2 = o(1). Ω\{B1 ∪B2 }

Putting together the previous estimates completes the proof. Lemma 9. There results k|x|α f0 (Vδ,τ ) − f0 (Vδ,τ )k

2N N +2

= O (τ + δ)

and k|x|α f00 (Vδ,τ ) − f00 (Vδ,τ )k N = O (τ + δ) . 2

Proof. As above, we prove only the first estimate. First of all, we have Z

2N

2N

||x|α − 1| N +2 |f0 (Vδ,τ )| N +2 ≤ c

k Z X

2N

2N

N −2 Uδ,ξ (1 − |x|α ) N +2 . i

(30)

i=1 Ω

Ω

For every i = i, . . . , k, the change of variables x = ξi + δy shows that Z Z 2N 2N 2N 2N 1 N −2 N −2 α N +2 Uδ,ξi (1 − |x| ) dx = a0 (1 − |δy + ξi |α ) N +2 dy. 2 N (1 + |y| ) Ω

Ω−ξi δ

(31) Now, by the choice of ξi (i.e. ξi = (1−τ )ξi∗ ) and by the mean value theorem, there exists θ = θ(δ, τ, x) ∈ [0, 1] such that |δy + ξi |α = |δy − τ ξi∗ + ξ ∗ |α = 1 + α|θδy − θτ ξi∗ + ξ ∗ |α−2 (θδy − θτ ξi∗ + ξ ∗ , δy − τ ξi∗ ).

(32)

Moreover, it is easy to check that 0 ≤ 1 − |δy + ξi |α ≤ α(δ|y| + τ ).

(33)

Therefore, by (30), (31) and (33), we immediately deduce that Z 2N 2N ||x|α − 1| N +2 |f0 (Vδ,τ )| N +2 Ω



Z 2N

≤ c δ N +2

2N N +2

2N |y| dy + τ N +2 (1 + |y|2 )N

IRN

and the proof is complete.

Z IRN

 1 dy  (1 + |y|2 )N

(34)

Multi–peak solutions

23

Lemma 10. There results

° ° k ° ° X ° ° f0 (a0 Ui )° °f0 (Vδ,τ ) − ° ° i=1

and

2N N +2

 ·¡ ¢ N +2 ¸   O τδ 2        h ¡ ¢ ¡ ¢i = O δ 4 log2/3 δ τ τ         h¡ δ ¢N −2 i  O τ

if N ≥ 7, if N = 6, if N = 3, 4, 5.

°" # ° "µ ¶ N −2 # k ° ° 2 X δ ° ° 0 0 f0 (a0 Ui ) φ° = O . ° f0 (Vδ,τ ) − °N ° τ i=1

2

Proof. As usual, we prove only the first estimate, and we assume that k = 2. We set Ui := Uδ,ξi , i = 1, 2 and we split Z

2N

|f0 (P U1 + P U2 ) − f0 (U1 ) − f0 (U2 )| N +2 Ω

Z

Z

Z

··· +

= B(ξ1 ,τ )

··· +

B(ξ2 ,τ )

(35)

Ω\(B(ξ1 ,τ )∪B(ξ2 ,τ ))

First of all, it is easy to see that Z · · · ≤ cδ N Ω\(B(ξ1 ,τ )∪B(ξ2 ,τ ))

...

Z

Ω\(B(ξ1 ,τ )∪B(ξ2 ,τ ))

1 dx ≤ cδ N . |x − ξi |2N

(36)

We turn to the estimate of the integral over B(ξ1 , τ ); the other one is treated in the same way. We have Z 2N |f0 (P U1 + P U2 ) − f0 (U1 ) − f0 (U2 )| N +2 B(ξ1 ,τ )

Z

2N

|f0 (P U1 + P U2 ) − f0 (U1 )| N +2

≤c B(ξ1 ,τ )

Z

+c

2N

|f0 (U2 )| N +2 .

(37)

B(ξ1 ,τ )

We immediately observe that Z 2N |f0 (U2 )| N +2 ≤ cδ N B(ξ1 ,τ )

Z B(ξ1 ,τ )

1 dx ≤ cδ N . |x − ξ2 |2N

(38)

24

Angela Pistoia, Enrico Serra

By the mean value theorem, we deduce that for some θ = θ(x) ∈ [0, 1] Z 2N |f0 (P U1 + P U2 ) − f0 (U1 )| N +2 B(ξ1 ,τ )

Z

2N

|f00 (U1 + θ(P U1 − U1 + P U2 ))(P U1 − U1 + P U2 )| N +2

= B(ξ1 ,τ )

Z ≤c B(ξ1 ,τ )

Z +c B(ξ1 ,τ )

Z

+c

¯ ¯ N2N ¯ N4−2 ¯ +2 ¯U ¯ (P U − U ) +c 1 1 ¯ ¯ 1 ¯ N2N ¯ ¯ +2 ¯ N4−2 ¯ ¯U +c ¯ 2 (P U1 − U1 )¯

Z

2N

|P U1 − U1 | N −2

B(ξ1 ,τ )

Z

¯ 2N ¯ ¯ N4−2 ¯ N −2 ¯ ¯U ¯ 1 U2 ¯

B(ξ1 ,τ )

¯ ¯ N2N 4 ¯ ¯ +2 +c ¯(P U1 − U1 ) N −2 U2 ¯

B(ξ1 ,τ )

Z

2N

|U2 | N −2

B(ξ1 ,τ )

=: I1 + I2 + I3 + I4 + I5 + I6 .

(39)

We now deal with the six integrals Ii . By Proposition 12 we have µ ¶N δ ≤c . τ

(40)

1 ≤ cδ N . |x − ξ2 |2N

(41)

2N N −2 2N N −2

I2 ≤ kϕδ,ξ1 k It is also easy to see that Z I6 ≤ cδ

N B(ξ1 ,τ )

By the H¨older inequality, (40) and (41), we obtain I5 ≤ cδ

N (N −2)/(N +2)

and also I3 ≤ cδ 4N/(N +2)

µ ¶4N/(N +2) µ ¶N δ δ ≤c τ τ

µ ¶N (N −2)/(N +2) µ ¶N δ δ ≤c . τ τ

Finally, we recall estimate (B.23) in [16]:  ³ 4N N (N −6) ´  O δ N +2 τ N +2       Z  ¡ 8N ¢ (N +2)(N −2) U1 = O δ 3 log(τ /δ)     B(ξ1 ,τ )   (N −2)  cδ N N +2

(42)

(43)

if N ≥ 7, if N = 6, if N = 3, 4, 5.

(44)

Multi–peak solutions

25

Therefore, we have 2N N +2 L∞ (B(ξ1 ,r))

Z

8N (N +2)(N −2)

I4 ≤ kU2 k

U1

Z ≤ cδ

N (N −2)/(N +2)

B(ξ1 ,τ )

8N

U1(N +2)(N −2)

B(ξ1 ,τ )

(45) and, using also Proposition 12, I1 ≤ kU1 −P U1 k

2N N +2 L∞ (B(ξ

Z 1 ,r))

B(ξ1 ,τ )

8N (N +2)(N −2)

U1

≤c

δ τ

N (N −2) N +2

Z

2N (N −2) N +2

8N

U1(N +2)(N −2) .

B(ξ1 ,τ )

(46) Collecting all the estimates, (35)–(38), (40)–(46) and (44), the claim follows.

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Angela Pistoia, Enrico Serra: Multi–peak solutions

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