Interior estimates for some semilinear elliptic problem with critical nonlinearity

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Interior estimates for some semilinear elliptic problem with critical nonlinearity ∗ Pierpaolo Esposito†

Abstract We study compactness properties for solutions of a semilinear elliptic equation with critical nonlinearity. For high dimensions, we are able to show that any solutions sequence with uniformly bounded energy is uniformly bounded in the interior of the domain. In particular, singularly perturbed Neumann equations admit pointwise concentration phenomena only at the boundary. Keywords: compactness, critical exponent, singular perturbations, blow up analysis. AMS subject classification: 35J20, 35J25, 35J60

1

Introduction and statement of the results

The starting point in our investigation has been the study of asymptotic properties for the problem:  p   −∆u + λu = u in Ω, (1) u>0 in Ω,   ∂u on ∂Ω, ∂n = 0

+2 where Ω is a smooth bounded domain in RN , N ≥ 3, p = N N −2 is the critical exponent from the Sobolev viewpoint and λ > 0 is a large parameter. Here, ν(x) is the unit outward normal of Ω at x ∈ ∂Ω.

Under the transformation v(x) = λ

1 − p−1

u(x), d2 =

∗

1 λ,

problem (1) reads equivalently as

The author is supported by M.U.R.S.T., project “Variational methods and nonlinear differential equations”. † Dipartimento di Matematica, Universit` a degli Studi “Roma Tre”, Largo S. Leonardo Murialdo, 1 - 00146 Roma, e-mail: [email protected].

1

a singularly perturbed Neumann problem:  2 p   −d ∆v + v = v in Ω, v>0 in Ω,   ∂v on ∂Ω, ∂n = 0

(2)

N +2 where p = N −2 . For general exponent p > 1, problem (2) is related to the study of stationary solutions for a chemotaxis system (see [14]) proposed by Keller, Segel and Gierer, Meinhardt (see [15]). N +2 Problem (1) for λ large has been widely studied in the subcritical case p < N −2 . The asymptotic behaviour and the construction of blowing up solutions have been considered by several authors. In particular, there exist peak solutions which blow up at many finitely boundary and/or interior points of Ω. +2 The critical case p = N N −2 has different features. Starting from the pioneering works of Adimurthi, Mancini and Yadava [3] (see also [1], [2]), asymptotic analysis (see [10], [12] for low energy solutions) and construction of solutions concentrating at boundary points of Ω have been considered by several authors (see for example [18]). We refer to [17] for an extensive list of references about subcritical and critical case. As far as interior concentration, the situation is quite different since in literature no results are available and it is expected that in general such solutions should not exist. A first partial result in this direction is due to Cao, Noussair and Yan for N ≥ 5 and for isolated blow up points. They show that any concentrating solutions sequence with bounded energy cannot have only interior peaks and so at least one blow up point must lie on ∂Ω. At the same time, Rey in [17] gets the same result for N = 3 by removing any assumption on the nature of interior blow up points. Using some techniques developed by Druet, Hebey and Vaugon in [9] for related problems on Riemannian manifolds, Castorina and Mancini in [6] were able to show, among other things, that the conclusion of previous papers holds without any restriction on the dimension. Namely, for N ≥ 3 at least one blow up point lies on ∂Ω.

However, all these papers do not answer to the full question: are there exist blowing up solutions for (1) with bounded energy which do not remain bounded in the interior of Ω as λ → +∞? For N > 6 the answer is negative since we will show that ALL the blow up points have to lie on ∂Ω:

2

Theorem 1.1. Let N > 6. Let λn → +∞ and un be a solutions sequence of  N +2  N −2  −∆u + λ u = N (N − 2)u in Ω,  n n n n un > 0 in Ω,    ∂un = 0 on ∂Ω, ∂n

(3)

with uniformly bounded energy:

sup

Z

2N

unN −2 < +∞.

n∈N Ω

Then, for any K compact set in Ω there exists CK such that: max un (x) ≤ CK x∈K

for any n ∈ N. Theorem 1.1 is based on a local description of possible compactness loss and does not need any boundary condition. In fact, we realized that Theorem 1.1 is a particular case of a more general interior compactness result, which is still more interesting that our initial question about singularly perturbed Neumann equations and becomes the main content of this paper. There holds: Theorem 1.2. Let N > 6. Let K be a compact set in Ω and Λ > 0. There exists a constant C, depending on K and Λ, such that any solution u of the problem:  N +2   −∆u + u = N (N − 2)u N −2 in Ω, u>0 in Ω,   R 2 2 Ω |∇u| + u ≤ Λ,

satisfies the bound:

max u(x) ≤ C. x∈K

The paper is organized in the following way. In Section 2 we introduce the notion of (geometrical) blow up set, we give a description of this set and we show by a rescaling argument that Theorem 1.1 is a particular case of Theorem 1.2. In Section 3, we provide the proof of Theorem 1.2: based on a technical result contained in [7]-[8] due to Druet, Hebey and Robert (which we report in Appendix for the sake of completeness), for any interior blowing up solutions sequence we are able to prove an upper estimate (in terms of standard bubbles) which contradicts a related local Pohozaev identity.

3

2

The blow up set

Let un be a solutions sequence of  N +2  N −2   −∆un + µn un = N (N − 2)un un > 0  R   sup 2 2 n∈N Ω (|∇un | + un ) < +∞,

in Ω, in Ω,

(4)

where Ω is a domain in RN , N ≥ 3, and 0 ≤ µn → µ ∈ [0, +∞]. We define the (geometrical) blow up set of un in Ω as S = {x ∈ Ω : ∃ xn → x s.t. lim sup un (xn ) = +∞}, n→+∞

0 (Ω \ S). and, by definition of S, clearly un is uniformly bounded in Cloc Further, define the set Z 2N unN −2 ≥ c ∀ r > 0}. Σc = {x ∈ Ω : lim sup n→+∞

Br (x)

2N

Let SN be the best constant related to the immersion of H01 (Ω) into L N −2 (Ω): R 2 Ω |∇u| . SN = inf R 2N N −2 u∈H01 (Ω)\{0} ( N −2 ) N Ω |u|

(5)

By means of an iterative Moser-type scheme, we can describe the set S in the following way:

Proposition 2.1. There exists c = cN such that it holds S = Σc . In particular, S is a finite 0 (Ω \ S) (up to a subsequence). set and, if µ = +∞, we have that un → 0 in Cloc Proof First of all, we show the following implication: Z

2N N −2

un

B2r (x)

N SN )2, q ≥2 ⇒ ≤( qN (N − 2)

N q N −2

Z

un

Br (x)

! N −2 N

8 ≤ SN r 2

Z

B2r (x)

uqn

for any r < 12 dist (x, RN \ Ω). Let ϕ ∈ C0∞ (B2r (x)) be so that 0 ≤ ϕ ≤ 1, ϕ ≡ 1 in Br (x) and k∇ϕk∞ ≤ 2r . Multiplying (4) for ϕ2 uq−1 and integrating by parts, by (5) and H¨ older’s n inequality we get that: Z Z Z 4 q 2 q 2 q−1 ϕ un = N (N − 2) unN −2 (ϕun2 )2 ∇un ∇(ϕ un ) + µn Ω

Ω

Ω

≤

N (N − 2) SN 4

Z

B2r (x)

2N N −2

un

!2 Z N

Ω

q

|∇(ϕun2 )|2 .

On the other hand, we can write: Z Z Z 2 q−1 2 q−2 2 ∇un ∇(ϕ un ) = (q − 1) ϕ un |∇un | + 2 ϕuq−1 n ∇ϕ∇un Ω Ω Ω Z Z Z q 2 q−2 2 2 |∇u | − = |∇(ϕun2 )|2 + ϕ2 uq−2 |∇ϕ|2 uqn n n q Ω 2 q Ω Ω Z Z q 2 2 2 2 q |∇(ϕun2 )| − |∇ϕ| un . ≥ q Ω q Ω Combining these two estimates, we get that 2 q

Z

!2 Z Z N 2N q N (N − 2) |∇(ϕun2 )|2 unN −2 |∇ϕ|2 uqn + SN Ω B2r (x) Ω Z Z q 8 1 uq + |∇(ϕun2 )|2 qr2 B2r (x) n q Ω

2 q

q 2

|∇(ϕun )|2 ≤

Ω

≤

Z

2N

in view of Z

R

N q N −2

un

Br (x)

Since

N N −2 q

N

SN N −2 2 ≤ ( qN (N B2r (x) un −2) ) . Therefore, by (5) we obtain that

! N −2 N

≤

Z

q 2

(ϕun )

2N N −2

Ω

 N −2 N

≤

1 SN

Z

q

|∇(ϕun2 )|2 ≤

Ω

8 SN r 2

Z

B2r (x)

uqn .

> q, we can iterate the procedure starting from q = 2 up to get a-priori bounds 2N

in Lp -norms around x for any p > 2 provided the L N −2 -norm around x is sufficiently small. N +2 N Namely, we find 0 < δ < 1, p > N −2 2 and c > 0, depending only on N , such that, if 2N R N −2 ≤ c, then B2r (x) un !2 Z Z p p un ≤ C(N, r) u2n , Bδr (x)

B2r (x)

for some constant C(N, r) depending only on N and r. Let u1n be the solution of (

N +2

−∆u1n = N (N − 2)unN −2 u1n = 0

in Bδr (x), on ∂Bδr (x),

and u2n be an harmonic function such that u2n = un on ∂Bδr (x). Since ! Z N +2

kN (N − 2)unN −2 kLs (Bδr (x)) = O (

for some s >

N 2,

1

B2r (x)

u2n ) 2

by elliptic regularity theory (cfr. [11]) we get that ! Z ku1n kC 0 (Bδr (x)) = O ( 5

1

B2r (x)

u2n ) 2

2N

provided

R

N −2 ≤ c. By the representation formula for harmonic function, we get that B2r (x) un ! Z

ku2n kC 0 (Bδr/2 (x)) = O

un

.

∂Bδr (x)

Since by the maximum principle 0 < un ≤ u1n + u2n , we get that kun kC 0 (Bδr/2 (x)) ≤ C

Z (

B2r (x)

1 u2n ) 2

+

Z

∂Bδr (x)

un

!

(6)

for some C > 0. 2N R By (5) we get that supn∈N Ω unN −2 < +∞ and therefore, ΣcN is a finite set. Moreover, up to a subsequence we can assume that un ⇀ u weakly in H 1 (Ω) and un → u in L2 (Ω), in view of R the compact embedding of H 1 (Ω) into L2 (Ω). Since µn Ω u2n is uniformly bounded, we have that u = 0 if µ = +∞. By the compactness of the embedding of H 1 (B2r (x)) into L2 (B2r (x)) and of H 1 (Bδr (x)) into L1 (∂Bδr (x)) in the sense of traces, in view of (6) we get that S = ΣcN is a finite set and, if 0 (Ω \ S) (up to a subsequence). µ = +∞, un → 0 in Cloc Remark 2.2. Blowing up the sequence un around a point x ∈ S, by means of the same techniques which we will exploit strongly in Appendix, it is easy to show that the value N N SN SN 2 2 c = cN in Proposition 2.1 must satisfy c ≥ ( N (N −2) ) . Hence, S = Σc with c = ( N (N −2) ) . We are now in position to deduce Theorem 1.1 by Theorem 1.2. Proof (of Theorem 1.1) Multiplying (3) for un and integrating by parts, we get that: Z Z 2N (7) (|∇un |2 + λn u2n ) ≤ Λ := N (N − 2) sup unN −2 < +∞. n∈N Ω

Ω

We can define the blow up set S of the sequence un . By the validity of Theorem 1.2, we 0 (Ω). deduce that S has to be an empty set and therefore, un is uniformly bounded in Cloc Otherwise, if S 6= ∅, up to a subsequence, we can assume that there exists x0 ∈ S such that maxx∈Br (x0 ) un (x) → +∞ as n → +∞, for any r > 0. By Proposition 2.1, we know that S is a finite set. Let 0 < r < dist (x0 , S \ {x0 }) and xn be such that un (xn ) = maxx∈Br (x0 ) un (x) → 0 (Ω\S), x → x as n → +∞. +∞ as n → +∞. Clearly, since un is uniformly bounded in Cloc n 0 N −2

2

Introduce εn = un (xn )− N −2 → 0 and define Un (y) = εn 2 un (εn y+xn ) for y ∈ Bn := B 2εr (0). n We have that ( N +2 −∆Un + µn Un = N (N − 2)UnN −2 in Bn , 0 < Un (y) ≤ Un (0) = 1, where µn = λn ε2n . Assume that µn = λn ε2n → µ ∈ [0, +∞]. Since Un is uniformly bounded 1 (RN ) and in C 0 (RN ), if µ = +∞, Proposition 2.1 implies that U → 0 in C 0 (RN ) in Hloc n loc loc 6

(up to a subsequence) contradicting Un (0) = 1. 2 (RN ) So, µ < +∞. By standard elliptic estimates (cfr. [11]), we have that Un → U in Cloc where U ∈ H 1 (RN ) is a solution of ( N +2 −∆U + µU = N (N − 2)U N −2 in RN , (8) 0 < U (y) ≤ U (0) = 1 (in view of (7)). By a Pohozaev identity on RN (see [16]), we must have that µn = λn ε2n → µ = 0. − N −2

Now, we do the following rescaling. Let vn (x) = λn 4 un ( √xλ +xn ) be defined for x ∈ B1 (0). n The function vn satisfies:  N +2  N −2  in B1 (0),  −∆vn + vn = N (N − 2)vn vn > 0 in B1 (0),    R 2 2 B1 (0) (|∇vn | + vn ) ≤ Λ, since

Z

B1 (0)

2

(|∇vn | +

vn2 )

=

Z

2

(|∇un | +

B √1 (xn )

λn u2n )

≤

λn

Z

Ω

(|∇un |2 + λn u2n ) ≤ Λ.

By Theorem 1.2 we get that there exists C > 0 such that max vn (x) ≤ C.

x∈B 1 (0) 2

So, we reach a contradiction since we have already shown that − N 4−2

vn (0) = λn

un (xn ) = (

1 N −2 ) 4 → +∞ λn ε2n

as n → +∞. The proof is now complete.

3

Non existence of interior blow up points

The proof of Theorem 1.2 is based on a contradiction argument. In view of Proposition 2.1, let us assume the existence of a solutions sequence un of the following problem:  N +2 N −2  −∆u + u = N (N − 2)u  n n n  un > 0  2N  R  supn∈N B1 (0) unN −2 < +∞, 7

in B1 (0), in B1 (0),

which blows up in B1 (0) only at 0: maxx∈B1 (0) un (x) → +∞ as n → +∞ and un is uniformly 0 (B (0) \ {0}). bounded in Cloc 1 By means of Propositions 4.1, 4.2 and by elliptic regularity theory (cfr. [11]), up to a subsequence, we will assume throughout this section the existence of sequences x1n , . . . , xkn → 0, ε1n , . . . , εkn → 0 and x1 , . . . , xk ∈ RN such that for any i = 1, . . . , k: Uni (y) = (εin )

N −2 2

1

un (εin y + xin ) →

(1 + |y − xi un → u0 in Cloc (B1 (0) \ {0}) as n → +∞,

|2 )

N −2 2

2 (RN \ Si ) as n → +∞, (9) in Cloc

0

(10)

N −2 2

un (x) ≤ C for any n ∈ N , |x| < 1,   N −2 lim lim sup maxn dk (x) 2 un (x) − u0 (x) = 0,

dk (x)

(11) (12)

R→+∞ n→+∞ x∈BR

for some constant C > 0 and for some smooth solution u0 ≥ 0 of the equation: N +2

−∆u0 + u0 = N (N − 2)u0N −2 in B1 (0),  n where dk (x) = min |x − xin | : i = 1, . . . , k , BR = {|x| < 1 : |x − xin | ≥ Rεin ∀ i = 1, . . . , k} and |xjn − xin | xjn − xin : j < i s.t. = O(1)}. Si = {yj = lim n→+∞ εin εin i

n Let now x be so that |x − xin | = Rεin for some i = 1, . . . , k. We have that y = x−x satisfies: εin |y| = R and |y − yj | ≥ R − |yj | ≥ 1, for R large. Hence, by (9) we get that for any R > 0 large and C > 1 there exists N0 such that for any n ≥ N0 and |x − xin | = Rεin

un (x) ≤ CUεin ,xin +εin xi (x), where Uε,y (x) = ε−

N −2 2

max

N −2

U(

ε 2 x−y )= N −2 . ε (ε2 + |x − y|2 ) 2

|x |

i i=1,...,k )|x − xin | for |x − xin | = Rεin , we obtain that for Since |x − (xin + εin xi )| ≥ (1 − R any R > 0 large there exists N0 such that

un (x) ≤ 2Uεin ,xin (x)

(13)

for any n ≥ N0 and |x − xin | = Rεin . The aim will be to estimate from above un (x) in terms of the standard bubbles Uεin ,xin (x), S i = 1, . . . , k, in Bδ (0) \ ki=1 BRεin (xin ), 0 < δ < 1. By performing some simple asymptotic analysis we get the following result (see also the techniques developed by Schoen in [19] and exploited in [12], [13]): 8

Lemma 3.1. Let α ∈ (0, N 2−2 ). There exist R > 0, 0 < δ < 1 and N0 ∈ N such that un (x) ≤

k  X

N −2 (εin ) 2 −α |x

−

xin |2−N +α

+ Mn |x −

xin |−α

i=1



,

for any n ≥ N0 and |x| ≤ δ with |x − xin | ≥ Rεin , i = 1, . . . , k, where Mn = 2δ α sup|x|=δ un (x). 4

Proof Let us introduce the operator Ln = −∆ + 1 − N (N − 2)unN −2 . Since un is a positive solution of Ln un = 0 in Bδ (0), we have that Ln satisfies the minimum principle in Bδ (0) for any 0 < δ < 1 (see [11]). Since u0 is a smooth function, by (12) we have that there exist 1 R > 2 α , 0 < δ < 1 and N0 ∈ N such that dk (x)

N −2 2

un (x) ≤ (

α(N − 2 − α) N −2 ) 4 kN (N − 2)

(14)

for any n ≥ N0 and x ∈ Bδ (0): |x − xin | ≥ Rεin , i = 1, . . . , k. P Define now a comparison function ϕn in the form ϕn = ki=1 ϕin , where ϕin (x) = (εin )

N −2 −α 2

|x − xin |2−N +α + Mn |x − xin |−α ,

and compute Ln on ϕn − un : Ln (ϕn − un ) =

k X

Ln ϕin

=

i=1

k  X

α(N − 2 − α)|x −

xin |−2

4 N −2

+ 1 − N (N − 2)un

i=1



ϕin .

Let x ∈ Bδ (0) be such that |x − xin | ≥ Rεin , i = 1, . . . , k. There exists j ∈ {1, . . . , k} so that |x − xjn | = min{|x − xin | : i = . . . , k}. Since |x − xjn | ≤ |x − xin |, we have that ϕjn (x) ≥ ϕin (x) for any i = 1, . . . , k and therefore, Ln (ϕn − un )(x) =

k  X i=1

≥ α(N − 2 − α)|x −

 4 α(N − 2 − α)|x − xin |−2 + 1 − N (N − 2)un (x) N −2 ϕin (x)

xjn |−2 ϕjn (x)

− N (N − 2)un (x)

4 N −2

k X

ϕin (x)

i=1

  4   N −2 j N 2−2 un (x) |x − xjn |−2 ϕjn (x) ≥ 0 ≥ α(N − 2 − α) − kN (N − 2) |x − xn |

in view of (14), for any n ≥ N0 and x ∈ Bδ (0): |x − xin | ≥ Rεin , i = 1, . . . , k. In view of the validity of (13) on ∂BRεin (xin ), we can always assume that R and N0 are such that un (x) ≤ 2(εin )

N −2 2

|x − xin |2−N

for any n ≥ N0 and |x − xin | = Rεin . Therefore, we have that un (x) ≤ 2R−α (εin )

N −2 −α 2

|x − xin |2−N +α ≤ (εin ) 9

N −2 −α 2

|x − xin |2−N +α ≤ ϕin (x) ≤ ϕn (x)

for n ≥ N0 and |x − xin | = Rεin for some i = 1, . . . , k. Since k

un (x) ≤

X 1 M ≤ M |x − xin |−α ≤ ϕn (x) n n 2δ α i=1

for |x| = δ and n large, by the minimum principle for Ln we get the desired estimate in the region x ∈ Bδ (0) with |x − xin | ≥ Rεin ∀ i = 1, . . . , k. We have to combine the estimate contained in Lemma 3.1 with the following Pohozaev-type inequality proved in [6] (for the Pohozaev identity, refer to [16]): Lemma 3.2. There exists C > 0, depending only on the dimension N , such that for any |x| < 1 and 0 < δ < 1−|x| 4   Z Z 2N 1 2 N −2 2 u + un . (15) un ≤ C 2 n Bδ (x) B4δ (x)\Bδ (x) δ Let N > 6 and fix 0 < α
1 is a constant. By (16), we obtain that for x ∈ B4Drn (xkn )\BDrn (xkn ) there hold: ( |x − xin | ≥ |xin − xkn | − |x − xkn | ≥ |xin − xkn | − 4Drn ≥ rn if i = 1, . . . , s (17) |x − xin | ≥ |x − xkn | − |xkn − xin | ≥ Drn − (D − 1)rn = rn if i = s + 1, . . . , k. We apply now (15) on BDrn (xkn ). Let r = 21 min{|yj | : yj ∈ Sk , yj 6= 0}. Since rn >> εkn for n large in view of α < N 3−4 , we have that Z Z Z Z 2 2 k 2 k 2 k 2 un ≥ un = (εn ) (Un ) ≥ (εn ) (Unk )2 . BDrn (xkn )

Since Sk ∩

{ 2r

Brεk (xkn )

Br (0)

n

≤ |y| ≤ r} = ∅, by (9) we get that Z Z (Unk )2 → Br (0)\B r (0)

Br (0)\B r (0)

2

2

Br (0)\B r (0) 2

1 >0 (1 + |y − xk |2 )N −2

and hence, Z

BDrn (xkn )

u2n

1 ≥ (εkn )2 2

Z

Br (0)\B r (0) 2

10

1 (1 + |y − xk |2 )N −2

1 δ > rn ≥ Rεkn ≥ Rεin for any i = 1, . . . , k. for n large. Let us remark that for n large 8D Therefore, we can use Lemma 3.1 and (17) to provide that

un (x) ≤

k  X

(εin )

N −2 −α 2

|x − xin |2−N +α + Mn |x − xin |−α

i=1

≤ k(εkn )

N −2 −α 2

rn2−N +α + kMn rn−α ,

 (18)

for any n ≥ N0 and x ∈ B4Drn (xkn ) \ BDrn (xkn ). Hence, by (18) we get that   Z 2N 1 N −2 2 u + un 2 2 n B4Drn (xkn )\BDrn (xkn ) D rn   k 2N N −α N2N εkn N −α N2N εn N −2−2α N −2 −2 2 N −2−2α −2 +( ) + Mn r n ≤C ( ) + Mn rn rn rn  k    εn N −2−2α ′ 2 N −2−2α = C ′ (εkn )N −4−3α + Mn2 (εkn )2+α + Mn r n ≤C ( ) rn since α
0 in B1 (0),  2N  R  supn∈N B1 (0) unN −2 < +∞.

We assume that 0 is the only blow up point of un in B1 (0):

max un (x) → +∞ as n → +∞,

x∈B1 (0)

0 un → u0 in Cloc (B1 (0) \ {0})

(up to a subsequence), where u0 ≥ 0 is a smooth solution of the limit equation: N +2

−∆u0 + u0 = N (N − 2)u0N −2 in B1 (0).

First of all, the following classical result holds (see for example [7]-[8] and [19]): 11

(19) (20)

Proposition 4.1. Up to a subsequence of un , there exist s ∈ N and sequences x1n , . . . , xsn → 0 such that for any i, j = 1, . . . , s, i 6= j, we have: εin + εjn

2

εin := un (xin )− N −2 → 0 , un (xin ) =

max

x∈Bεi (xin )

|xin − xjn |

un (x)

→ 0 as n → +∞,

(21)

for n large,

(22)

n

Uni (y) ds (x)

N −2 (εin ) 2 un (εin y

=

N −2 2

+ xin ) →

1 (1 +

|y|2 )

N −2 2

2 (RN ) as n → +∞, in Cloc

un (x) ≤ C for any n ∈ N , |x| < 1,

(23) (24)

 where dj (x) = min |x − xin | : i = 1, . . . , j , 1 ≤ j ≤ s. In particular, there holds Z Z 2N 2N N SN u0N −2 , )2 + unN −2 ≥ s( lim inf n→+∞ B (0) N (N − 2) B1 (0) 1

(25)

where SN is the Sobolev constant. Proof Let x1n be the maximum point of un in B1 (0). By (19)-(20) we deduce that x1n → 0 and 2 1 N −2 ε1n = un (x1n )− N −2 → 0. Define Un1 (y) = (ε1n ) 2 un (ε1n y + x1n ) for y ∈ Bn := B(ε1n )−1 (− xε1n ). n We have that ( N +2 −∆Un1 + (ε1n )2 Un1 = N (N − 2)(Un1 ) N −2 in Bn , 0 < Un1 (y) ≤ Un1 (0) = 1. By standard elliptic estimates (cfr. [11]) and a diagonal process, (up to a subsequence) we 2 (RN ) where U is a solution of have that Un1 → U in Cloc (

N +2

−∆U = N (N − 2)U N −2 0 < U (y) ≤ U (0) = 1.

in RN ,

(26)

By the classification in [4], problem (26) admits only the solution U (y) = Hence, we get that Un1 (y) →

1 (1 + |y|2 ) 1

(1 + |y|2 )

N −2 2

N −2 2

.

2 (RN ). in Cloc

Moreover, by taking the liminf as n → +∞ and the limit as δ → 0, R → +∞ in the following inequality chain: Z Z Z Z Z 2N 2N 2N 2N N −2 N −2 N −2 1 N2N ≥ un = (Un ) −2 + un + unN −2 , un B1 (0)

BRε1 (x1n ) n

B1 (0)\Bδ (0)

BR (0)

12

B1 (0)\Bδ (0)

which is true for δ, R > 0 and n ≥ N = N (δ, R), we deduce that Z Z Z Z 2N 2N 2N N SN dy N −2 2 + lim inf unN −2 ≥ + u ) = ( u0N −2 . 0 2 N n→+∞ B (0) N (N − 2) B1 (0) RN (1 + |y| ) B1 (0) 1 If (24) holds true for x1n , we take s = 1 and the proof is complete since (21)-(23) are already verified. Otherwise, by induction assume that there exist j sequences x1n , . . . , xjn → 0 as n → +∞ satisfying (21)-(23). If (24) holds true, we have done. Otherwise, there exists a sequence yn ∈ B1 (0) such that dj (yn )

N −2 2

un (yn ) = max dj (x)

N −2 2

x∈B1 (0)

un (x) → +∞ as n → +∞.

(27) 2

Since un → u0 away from 0, by (27) we have that yn → 0 and, for µn = un (yn )− N −2 , property (27) reads equivalently as: µn → 0 as n → +∞. (28) dj (yn ) N −2

Since (21)-(23) imply dj (x) 2 un (x) ≤ 1 for |x−xin | ≤ Rεin , i = 1, . . . , j, and n ≥ N = N (R), we deduce that yn must be outside this region, i.e. εin → 0 as n → +∞ , ∀ i = 1, . . . , j. |yn − xin |

(29)

N −2

Introduce the function Un (y) = µn 2 un (µn y + yn ) for y ∈ Bn , where the balls Bn = {y ∈ d (y ) RN : |y| ≤ j2µnn } expand to RN as n → +∞ in view of (28). Moreover, since 1 |µn y + yn − xin | ≥ |yn − xin | − µn |y| ≥ |yn − xin | 2 for any y ∈ Bn and i = 1, . . . , j, we have the following estimate: N −2

N −2 dj (µn y + yn ) 2 un (µn y + yn ) N −2 dj (yn ) ) 2 Un (y) = ( ≤2 2 N −2 dj (µn y + yn ) dj (yn ) 2 un (yn )

for any y ∈ Bn , in view of the maximality property in the definition (27) of yn . Hence, by standard elliptic estimates and a diagonal process, (up to a subsequence) we get that Un → U 2 (RN ), where U is a solution of in Cloc (

N +2

−∆U = N (N − 2)U N −2 in RN 0 < U (y) ≤ 2

N −2 2

, U (0) = 1.

By the classification in [4], the function U satisfies: U (y) =

µ

N −2 2

(µ2 + |y − y0 |2 ) 13

N −2 2

,

for some µ > 0 and y0 ∈ RN . Since U (y) has a non degenerate maximum point at y = y0 , for n large Un (y) has a unique maximum point zn close to y0 . Hence, for the original sequence, the point xj+1 := µn zn + yn is a maximum point of un on B2µn (xj+1 n n ) for n large. Since N −2

N −2

2

j+1 − = un (xj+1 2 2 ≥ Un (zn ) = µn 2 un (xj+1 n ) ≥ Un (0) = 1, we get that, for εn n ) N −2 , there holds: 2 εj+1 1 n → U (y0 )− N −2 = µ ∈ [ , 1]. µn 2

By (28)-(29) we get that i |xj+1 − xin | ≥ |yn − xin | − |yn − xj+1 n n | = |yn − xn | − µn |zn | >>

(

2µn ≥ εj+1 n i εn

for any i = 1, . . . , j and hence (21)-(22) hold. Moreover, there holds: (εj+1 n )

N −2 2

j+1 un (εj+1 n y + xn ) = (

N −2 N −2 εj+1 εj+1 1 n n ) 2 Un ( y + zn ) → µ 2 U (µy + y0 ) = N −2 µn µn (1 + |y|2 ) 2

2 (RN ) and, so (23) holds. So, the inductive step holds for j + 1. By (21) there holds in Cloc

Z

B1 (0)

2N

unN −2 ≥

j+1 Z X i=1

2N

BRεi (xin ) n

unN −2 +

Z

2N

unN −2 =

B1 (0)\Bδ (0)

j+1 Z X i=1

BR (0)

2N

(Uni ) N −2 +

for δ, R > 0 and n ≥ N = N (δ, R), and hence, by (23) we deduce that Z Z Z 2N 2N dy N −2 N −2 ≥ (j + 1) un u + lim inf 0 2 N n→+∞ B (0) RN (1 + |y| ) B (0) 1 Z 1 2N N SN u0N −2 . = (j + 1)( )2 + N (N − 2) B1 (0)

Z

2N

unN −2

B1 (0)\Bδ (0)

(30)

2N R Since supn∈N B1 (0) unN −2 < +∞, by (30) the induction process has to stop after s steps giving that also (24) holds for the sequences x1n , . . . , xsn . Further, there holds (25) by means of the validity of (30) for s = j + 1.

Now, in the spirit of the blow up techniques developed by Druet, Hebey and Robert in [7]-[8], we can perform a finer analysis. Proposition 4.2 below has already been showed in [7] for compact Riemannian manifolds. A careful reader could observe that, in fact, their analysis is quite local and does not use any geometric features and/or compactness of the underlying space, and therefore it extends directly to our situation. For the sake of completeness, we re-write their proof in the simpler context of an Euclidean domain: k Proposition 4.2. Up to a subsequence of un , there exist k ∈ N, k ≥ s, sequences xs+1 n , . . . , xn → k N 0, εs+1 n , . . . , εn → 0 and points xs+1 , . . . , xk ∈ R such that for any i = s+1, . . . , k, 1 ≤ j < i,

14

we have: εjn |xin − xjn |

→0,

Uni (y) = (εin )

N −2 2

εjn → 0 as n → +∞, εin

(31) 1

un (εin y + xin ) → 

lim lim sup max dk (x)

n R→+∞ n→+∞ x∈BR

N −2 2

|2 )

N −2 2

(1 + |y − xi  un (x) − u0 (x) = 0,

2 (RN \ Si ) as n → +∞, (32) in Cloc

(33)

 n = where x1n , . . . , xsn are given in Proposition 4.1, dk (x) = min |x − xin | : i = 1, . . . , k , BR {|x| < 1 : |x − xin | ≥ Rεin ∀ i = 1, . . . , k} and xjn − xin |xjn − xin | : j < i s.t. = O(1)} n→+∞ εin εin

Si = {yj = lim

is a non empty set. In particular, there holds Z Z 2N 2N N SN N −2 2 ≥ k( u0N −2 . ) + lim inf un n→+∞ B (0) N (N − 2) B1 (0) 1

(34)

Proof If (33) holds already true for x1n , . . . , xsn , we take k = s and the proof is complete. 2N R 2N P R Moreover, since S s B i (xi ) unN −2 = si=1 BR (0) (Uni ) N −2 for R > 0 and n ≥ N = N (R), i=1

by (23) we deduce that

n

Rεn

Z

lim inf

2N

Ss

n→+∞

i=1

BRεi (xin )

unN −2 ≥ s

BR (0)

n

for any R > 0. In particular, we get that Z Z 2N N −2 ≥s lim inf un n→+∞

BR (0)

B1 (0)

Z

dy + (1 + |y|2 )N

dy (1 + |y|2 )N

Z

B1 (0)\Bδ (0)

2N

u0N −2

for any δ, R > 0, and hence (34) holds. j s+1 k Otherwise, by induction assume that there exist j−s sequences xs+1 n , . . . , xn → 0, εn , . . . , εn → 0 as n → +∞ such that (31)-(32) hold and

lim inf

n→+∞

Z

Sj

i=1

2N N −2

BRεi (xin )

un

n

≥

j Z X i=1

BR (0)

dy , (1 + |y − xi |2 )N

(35)

where we take xi = 0 if i = 1, . . . , s. If (33) holds true, we take k = j and, as before, (35) implies the validity of (34). n such that Otherwise, (up to a subsequence) there exist Rn → +∞ and yn ∈ BR n   N −2 N −2 N −2 dj (yn ) 2 |un (yn ) − u0 (yn )| = max dj (x) 2 |un (x) − u0 (x)| ≥ (4δ0 ) 2 (36) n x∈BRn

15

n = {|x| < 1 : |x − xi | ≥ R εi ∀ i = 1, . . . , j}. Since u → u away for some δ0 > 0, where BR n n n 0 n n N −2 N −2 from 0, by (36) we have that yn → 0 and dj (yn ) 2 un (yn ) ≥ (2δ0 ) 2 for n large. Hence, n so that there exists a sequence xj+1 ∈ BR n n   N −2 N −2 N −2 j+1 2 u (x 2 u (x) 2 . ) = max ≥ (2δ ) d (x) dj (xj+1 ) n n n 0 j n n x∈BRn

2

− For εj+1 = un (xj+1 n n ) N −2 there holds:

|xj+1 − xin | n εj+1 n

εin

≥ 2δ0 ,

|xj+1 − xin | n

→ 0 as n → +∞

(37)

for any i = 1, . . . , j. In particular, εj+1 → 0 as n → +∞. Moreover, observing that property n s+1 (24) is still true if we add the points xn , . . . , xjn : dj (x) we have that

i |xj+1 n −xn | εj+1 n

N −2 2

un (x) ≤ C

|x| < 1 , n ∈ N,

(38)

2

≤ C N −2 for some i = 1, . . . , j (up to a subsequence) and so, Sj+1 6= ∅

and (31) is satisfied. Up to a subsequence, assume that the limits yi = limn→+∞ i

j+1

xin −xj+1 n εj+1 n

n | exist for any i = 1, . . . , j so that |xn −x = O(1). Introduce the set Sj+1 = {yi : i ≤ j} and εj+1 n remark that by (37) we have that Sj+1 ∩ Bδ0 (0) = ∅. N −2 j+1 j+1 Consider the function Un (y) = (εj+1 n ) 2 un (εn y + xn ) for y ∈ Bn , where the balls 1 } expand to RN as n → +∞. We want to show that Un is Bn = {y ∈ RN : |y| ≤ j+1

εn

1 0 (RN \ S uniformly bounded in Cloc j+1 ). Let y ∈ BR (0) be so that |y − yi | ≥ R for any yi ∈ Sj+1 . Since for any i = 1, . . . , j:  xin −xj+1 εj+1 n n j+1  εj+1 i |) ≥ (|y − y | − |y − n i i j+1 x − x 2R if yi ∈ Sj+1 n n j+1 i j+1 εn | ≥ |εj+1 y+x −x | = ε |y− j+1 j+1 i n n n n xn −xn  εj+1 εj+1 | − |y|) ≥ εn otherwise n n (| j+1 2R

εn

j+1 for n ≥ N = N (R), we get that dj (εj+1 n y + xn ) ≥

Un (y) = (εj+1 n )

N −2 2

≤ C(2R)

j+1 un (εj+1 n y + xn ) ≤ (2R)

N −2 2

εj+1 n 2R

and by (38) we get that

j+1 dj (εj+1 n y + xn )

N −2 2

j+1 un (εj+1 n y + xn )

N −2 2

for n ≥ N = N (R). By standard elliptic estimates and a diagonal process, (up to a subse2 (RN \ S quence) we get that Un → U in Cloc j+1 ), where U is a nonnegative solution of ( N +2 −∆U = N (N − 2)U N −2 in RN \ Sj+1 , U (0) = 1, since Bδ0 (0) ⊂ RN \ Sj+1 . By the result of Caffarelli, Gidas and Spruck [4], we have that: U (y) =

µ

N −2 2

(µ2 + |y − xi |2 ) 16

N −2 2

,

for some µ > 0 and xi ∈ RN . Since (µεj+1 n )

N −2 2

j+1 un (µεj+1 n y + xn ) = µ

N −2 2

N −2 2

Un (µy) → µ

U (µy) =

1 (1 + |y − µ−1 xi |2 )

N −2 2

,

j+1 −1 we can replace εj+1 with µεj+1 with Sj+1 with n n , xi with µ xi and Sj+1 with respect to εn j+1 respect to µεn to obtain the validity of (32). Now, we want to show the validity of (35) with j replaced by j + 1. Let I = {i = 1, . . . , j : yi ∈ Sj+1 } and let C > 0 be such that j+1 |xin − xj+1 for any i ∈ I. Clearly, we have that n | ≤ Cεn [ Bδεj+1 (xin ) ⊂ B(C+δ)εj+1 (xj+1 n ) n n i∈I

for any δ > 0, and by (31) we also deduce that [ BRεin (xin ) ⊂ B2Cεj+1 (xj+1 n ) n i∈I

for n ≥ N = N (R). Similarly, there holds   [  BRεin (xin ) ∩ BRεj+1 (xj+1 n )=∅ n i=1,...,j, i∈I /

for n ≥ N = N (R). Hence, we have that for R ≥ 2C and 0 < δ < C j+1 [

BRεin (xin ) = BRεj+1 (xj+1 n )⊕ n

i=1

\

!

[

Bδεj+1 (xin ) n

[

Bδεj+1 (xin ) n

i∈I

BRεj+1 (xj+1 n )\ n

⊃

BRεin (xin )

i=1,...,j, i∈I /

BRεj+1 (xj+1 n ) n

=

[

!

i∈I

[

⊕

Bδεj+1 (xin ) ⊕ n

i∈I

j [

⊕

[

BRεin (xin )

i=1,...,j, i∈I /

BRεin (xin )

i=1

for n ≥ N = N (δ, R) in view of (31), where ⊕ stands for the union of disjoint sets. Hence, we can write Z Z Z 2N 2N 2N N −2 N −2 u ≥ + u unN −2 n n S S S j+1 i=1

=

BRεi (xin )

B

n

Z

BR (0)\

S

j+1 xin −xn ) i∈I Bδ ( j+1 εn

j+1 j+1 (xn )\ Rεn

(Unj+1 )

2N N −2

i∈I

+

Z

B

j i=1

i j+1 (xn ) δεn

Sj

i=1

BRεi (xin ) n

2N N −2

BRεi (xin )

un

n

for n ≥ N = N (δ, R). Passing to the liminf as n → +∞, by the validity of (35), (32) for j + 1 j+1 i n and xn −x → yi ∈ Sj+1 as n → +∞ for any i ∈ I, we get that j+1 εn

lim inf

n→+∞

Z

S j+1 i=1

2N N −2

BRεi (xin ) n

un

≥

Z

BR (0)\

S

i∈I

Bδ (yi )

j Z X dy dy + 2 N 2 N (1 + |y − xj+1 | ) BR (0) (1 + |y − xi | )

17

i=1

for any R large and δ small. Letting δ → 0+ , we obtain that lim inf

n→+∞

Z

2N N −2

S j+1

i i=1 BRεin (xn )

un

≥

j+1 Z X i=1

BR (0)

dy . (1 + |y − xi |2 )N

(39)

2N R So, the inductive step holds for j + 1. Since supn∈N B1 (0) unN −2 < +∞, by (39) the induction process has to stop after k − s steps giving that also (33) holds for the sequences k x1n , . . . , xsn , xs+1 n , . . . , xn . Moreover, arguing as before, (39) for k = j + 1 implies the validity of

lim inf

n→+∞

Z

B1 (0)

2N N −2

un

≥

k Z X i=1

BR (0)

dy + (1 + |y − xi |2 )N

Z

B1 (0)\Bδ (0)

2N

u0N −2

for any δ, R > 0, and hence, taking δ → 0 and R → +∞, Z Z 2N 2N N SN N −2 2 ≥ k( u0N −2 . lim inf un ) + n→+∞ B (0) N (N − 2) B1 (0) 1 Hence, (34) holds and the proof is complete. Acknowledgements: I would like to thank Prof. Giovanni Mancini and Dr. Daniele Castorina which introduced me to the study of this topics. I’m indebted to them for many stimulating discussions and useful comments during the completion of this work.

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