Gladiali F., Grossi M. - Existence and multiplicity results for the conformal scalar curvature equation(36).pdf

7/25/2019 Gladiali F., Grossi M. - Existence and multiplicity results for the conformal scalar curvature equation(36).pdf

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Existence and multiplicity results for the

conformal scalar curvature equation

Francesca Gladiali Massimo Grossi

1 Introduction and statement of the main re-

sultsThis paper deals with the problem

u= K(x)up in IRn

u >0 in IRn

u D1,2(IRn)(1.1)

where p = n+2n2 , p = p , n 3, > 0 and D1,2(IRn) is the completion of

C0 (IRn) with respect to the norm u1,2 =

IRn |u|

2 12 .

The function K C1(IRn) satisfies the following assumptions:

1

C K(x) C; |K(x)| C, (1.2)

for some positive constant C.The main motivation in studying (1.1) arises from the problem of finding

a metric conformal to the standard one of IRn such that K(x) is the scalarcurvature of the new metric.

It is well known that, if = 0 and K(x) =n(n 2) problem (1.1) has onlythe two-parameters family of solutions

U,y(x) =

n22

(2 + |x y|2)n22

, x IRn, (1.3)

where >0 and y IRn. Hereafter we denote by U(x) = U1,0(x).

Supported by M.I.U.R., project Variational methods and nonlinear differential equa-tions.

Struttura dipartimentale di Matematica e Fisica, Universita di Sassari,via Vienna 2 -07100

Sassari, e-mail [email protected] di Matematica, Universita di Roma La Sapienza, P.le A. Moro 2 - 00185

Roma, e-mail [email protected].

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In this paper we are interested in single-peak solutions u concentrating atsome point, i.e.

u(x) = U,y(x) + (x) (1.4)

where 0, y y0 IRn, (x) 0 inD

1,2(IRn).Problem (1.1) has been studied by many authors. Indeed there exist many

existence or nonexistence results of solutions of (1.1) depending on the shape ofK(x) and the exponent p (see for example [9], [11] and [12] and the referencetherein)

Let us recall that whenK(x) is constant there is no solution to (1.1) (see [3]).The same happens if (K(x), x x0) 0 for some x0 IR

n ([10]). Anothernonexistence result can be found in [9] where it was showed that ifK(x) growslike or faster than |x|2(n2) then equation (1.1) has no positive solutions.

Concerning the existence results we mention the one in [12] where the authorsshowed the existence of solutions of (1.1) that concentrate at a nondegenerate

critical point y0 ofK(x) satisfying the crucial condition

K(y0)< 0. (1.5)

Here the authors looked for solutions with a suitable decay at infinity, i.e.solutions that belong to a suitable Sobolev space. In their setting, if K(x) isbounded, they had to impose the condition n > 6. The proof of this resultuses the finite dimensional reduction method, a tool widely used in this kind ofproblems.

In [11] S. Yan improved this result by removing the condition n > 6 butstill assuming (1.5). Note that in [11] the supercritical case < 0 was alsoconsidered.

Finally we mention the paper [8] which concerns the problem

u + V(x)u= n(n 2)up in IRn

u >0 in IRn

u D1,2(IRn),(1.6)

where the authors obtain existence and multiplicity results assuming that V(x)has stable (not necessarily nondegenerate) critical points.

In this paper we focus on the multiplicity of single-peak solutions at a suita-ble critical point of the function K(x). To do this we use some ideas of [5] wherethe nonlinear Schrodinger equation was considered. This method well suits toour problem and allows us to establish some existence and multiplicity resultsto (1.1).

One of the main results will be the existence, under suitable assumptions on

K(x), of several single-peak solutions concentrating at the same point y0. Beforestating the main theorem we have to introduce notations and some hypotheseson the functionK(x).

Let y0 be a fixed critical point ofK(x). We make the following assumptionson K(x): there exists functions hi : IR

n IR, Ri : B(0, r) IR and realconstantsi, i with i 1 for i = 1, . . ,nsuch that

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i) Kxi (x + y0) = hi(x) + Ri(x) in B (0, r),

ii) |Ri(x)| C|x|i forx B(0, r) and i > i,

iii) hi(tx) = tihi(x) for any x IR

n and t >0,

iv) (h1(x), . . ,hn(x)) = (0,.., 0) if and only ifx = 0.

Now let us introduce the following vector field

Ly0(y) :=

IRn

hi(x + y)Up+1(x)dx

i=1,...,n

. (1.7)

Note that the definition ofL is well posed ifi < n. Then we set

Zy0 :={y IRn|y is a stable zero ofLy0}. (1.8)

(see Section 5 for the definition of stable zero) and

= min{i | i= 1, . . . , n}. (1.9)

Our first result is the existence of several single-peak solutions concentrating atthe point y0. This is linked to the number of points of the set Zy0 .

Theorem 1.1. Let K(x) satisfy (1.2). Let y0 be a critical point of K(x) andsuppose thatK(x) satisfies hypotheses i)-iv) in a neighborhood ofy0 with

1 i n 1 for i= 1, . . . , n , (1.10)

and

j=IRn

hj(x + y)xjUp+1(x)dx


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Let us recall that 0 is a regular value for Ly0 if Jac(Ly0(y)) = 0 for anyy IRn such thatLy0(y) = 0.

Since it is known that any single peak solution concentrates at a criticalpoint ofK, the previous result classifies the number of single peak solution to(1.1) (in the assumption that 0 is a regular value for Ly0).

We end with a nonexistence result of single peak solutions. It shows thatthe assumptions of the previous theorem are almost sharp.

Theorem 1.3. Lety0 IRn be a critical point ofK(x). AssumeK(x)satisfies

(1.2) and hypotheses i)-iv) in a neighborhood of y0, with 1 i < n1 fori= 1, . . . , n. If either

Ly0(y)= 0 for anyy IRn, (1.12)

or

y IRnsuch thatLy0(y) = 0andj=

IRn

hj(x+ y)xjUp+1(x)dx0, (1.13)

then there is no single-peaked solutions to (1.1) blowing up aty0.

Let us observe that ifK(x) is an homogenous polynomial of degree +1 thenthe assumption that y belongs to #Zy0 becomes

IRn

K(x + y)Up+1(x)dx= 0and

j=

IRn

hj(x+ y)xjUp+1(x)dx 0 becomes (by Euleros Theorem for

homogenous functions)IRn

K(x+ y)Up+1(x)dx 0. In this setting Chen andLin proved that there is no single peak solution to (1.1)(see [1]).

The paper is organized as follows. In Section 2 we report some known facts andwe prove some technical estimates; in Section 3 we perform the finite dimensionreduction while in Section 4 we prove a useful estimate on the decay of thesolutions. In Section 5 we find some necessary conditions for the existence ofsolutions and we prove Theorem 1.1. In Section 6 we prove the asymptoticbehavior of the maximum points of solutions of (1.1) and prove Theorem 1.2,while in Section 7 we prove Theorem 1.3. Finally in Section 8 we give someexamples where Theorem 1.1-1.3 apply.

2 Preliminaries and notations

We rewrite problem (1.1) in an equivalent way in order to use the Liapunov-Schmidt reduction method. The first step is to scale (1.1). Ifu is a solution of(1.1) we let

u,y= 2p1 u(x + y) (2.1)

with >0 and y IRn. This function satisfies

u,y(x) = K(x + y)

u,y(x)p

in IRn

u,y>0u,y D

1,2(IRn).

(2.2)

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Of course (1.1) and (2.2) are equivalent.

Let us introduce the operator

i :L 2nn+2 (IRn)D1,2(IRn),

the adjoint of the embedding i : D1,2(IRn) L 2nn2 (IRn). The operator i is

defined as follows:

i(u) = v (v, )1,2 =

IRn

u(x)(x)dx D1,2(IRn).

As observed in [7] i is continuous, i.e there exists C >0 s.t.

i(u)1,2 Cu 2nn+2

u L 2nn+2 (IRn). (2.3)

We letf(s) = (s+)p wheres+ denotes the positive part ofs andf0 = (s

+)n+2n2 .

So we can rewrite problem (2.2) in this way:

u= i

K(x + y)(u+)p

(2.4)

where u D1,2(IRn) Ls(IRn) for some s > 1. Here we point out that if usolves (2.4) thenu >0 by the Strong Maximum Principle. Now we consider the

operator L : D1,2(IRn) D1,2(IRn) defined by

L(u) = u i

n(n 2)U 4n2 u

.

It is known that L is self-adjoint and is a compact perturbation of the identity.MoreoverKerL= span{0, 1, . . . , n} ,

where

0(x) = x U(x) +n 2

2 U(x) =

n 2

2

1 |x|2

(1 + |x|2)n2

, (2.5)

i(x) = U

xi(x) = (n 2)

xi

(1 + |x|2)n2

, i= 1, . . . , n . (2.6)

Let X = Ls(IRn) D1,2(IRn) with the norm uX = max{us, u1,2}, forsome s >1.

Remark 2.1. Lets > nn2 . Ifu L

2n

n+2 (IRn

)L ns

n+2s (IRn

)theni(u) Ls(IRn

)and

i(u)s C(n, s)u nsn+2s

. (2.7)

We quote the following result:

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Lemma 2.2. Let s > nn2 . If u X then L(u) X. Moreover L|X : XXis a continuous function , i.e. there existsC >0 s.t.

L|XX CuX u X (2.8)

Finally KerL|X =KerL.

Proof See Lemmas 2.10 and 2.11 in [7].

Let W = (KerL) Ls(IRn) = {u X : (u, )1,2 = 0, KerL}. We now

consider the projection : XW. Recall the following result,

Lemma 2.3. Let s > nn+2 andL : W W defined byL(u) =

L|X(u)

.

ThenL is invertible andL1 is a continuous operator, i.e. there exists C >0s.t.

L1(u)X CuX u W. (2.9)

Proof See Lemma 2.13 in [7].Now we recall the following useful estimates

Lemma 2.4. There exist 0 > 0 and C > 0 such that for each (0, 0) and1, 2 D1,2(IR

n) ands >1 we have

Up Up + (log U) Up 2nn+2

C 2 (2.10)

Up Up 2nn+2

C , Up Up nsn+2s

C (2.11)

pUp1 pUp1n

2C (2.12)(U+ 1)+p (U+ 2)+pp (U+ 2)+p1(1 2) C|1 2|p (2.13)

(U+ 1)+

p1

(U+ 2)+

p1

C|1 2|p1 (2.14)

Proof See [7] Lemma 2.20 and Remark 2.21, 2.22.

Lemma 2.5. LetK(x) satisfy (1.2) ands > nn1 . Theny IRn it holds

(K(x + y) K(y)) Up 2nn+2

C (2.15)

(K(x + y) K(y)) Up nsn+2s

C (2.16)

(K(x + y) K(y)) Up1n2

C (2.17)

Proof Let us prove (2.15). For some t = t(x, y) (0, 1) we have

(K(x + y) K(y)) Up 2nn+2

= IRn (K(x + y) K(y)) U

n+2n2 (x)

2nn+2

dx

n+22n

=

IRn

10

K(tx + y) x dt Un+2n2 (x)

2nn+2

dx

n+22n

C

IRn

|x|U

n+2n2 (x) 2nn+2

dx

n+22n

C .

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where in the last line we used that K C. Arguing in the same way weget

(K(x + y) K(y)) Up nsn+2s

=

IRn

(K(x + y) K(y)) Un+2n2 (x) nsn+2s dxn+2sns

IRn

|x|U

n+2n2 (x) ns

n+2s

dx

n+2sns

C ,

and

(K(x + y) K(y)) Up1(x)n2

= IRn(K(x + y) K(y)) U 4n2 (x)

n2

dx2n

C .

3 The finite dimension reduction

We are looking for solutions to (2.2) of the type

u,y=U+ ,y and

,y 0 D

1,2(IRn) as 0, and 0.

This turns the problem (2.2) into a finite dimensional one.The first step is to show the existence of the function ,y.

Proposition 3.1. Let (0, 1)be fixed and nn2 < s < 2nn2 . Then there exist

0 > 0 and0 > 0 such that (0, 0) (0, 0) andy IRn, ! ,y W

such that

,yX ( + )

(3.1)

{U+ ,y i

K(x + y)

(U+ ,y)+

= 0 (3.2)

(see Section 2 for the definition ofW).

Proof With no loss of generality we can assume that K(y) = n(n 2). Set= ,y. We first observe that solves (3.2) if and only if is a fixed point forthe operator T,y: W Wdefined by

T,y() = L1

i

K(x + y)

(U+ )+p

K(y)Up K(y)pUp1

.

(3.3)

Step A We begin by proving that inequality X ( + )

implies T,y()X ( + )

.

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Using (3.3) and (2.9) we have

T,y()X Ci

K(x + y)

(U+ )

+p K(y)U

p

K(y)pUp1

X

C

i

K(x + y)

(U+ )+p

Up pUp1

X

+ i [K(x + y) (Up Up)] X

+ i [(K(x + y) K(y)) Up] X

+ i

K(x + y)pU

p1 pUp1

X

+ i

(K(x + y) K(y))pUp1

X

= C(A1+ A2+ A3+ A4+ A5) . (3.4)

By the boundedness ofK(x) inL, we get

i

K(x + y)

(U+ )+

p Up pU

p1

1,2 ( by (2.3))

C

(U+ )

+p U

p

pU

p1

2nn+2 ( by (2.13))

Cp 2nn+2

CpX (3.5)by interpolation since

2n

n + 2p (s,

2n

n 2) if is sufficiently small

.

In the same way, using (2.7) and (2.13) we have

i

K(x + y)

(U+ )+p

Up pUp1

s

C

(U+ )+p Up pUp1 ns

n+2s

Cp nsn+2s

CpX . (3.6)

Hence from (3.5) and (3.6) we can write

A1 = i

K(x + y)

(U+ )+p Up pUp1

X CpX . (3.7)

Using (2.3), (2.7) and (2.11) we get

A2 = i (K(x + y) (Up Up)) X

CUp UpX C . (3.8)

Using (2.3) and (2.15) we easily have

i [(K(x + y) K(y)) Up] 1,2

C (K(x + y) K(y)) Up 2nn+2

C (3.9)

while from (2.7) and (2.16) it follows that

i ((K(x + y) K(y)) Up) s

C (K(x + y) K(y)) Up nsn+2s

C . (3.10)

From (3.9) and (3.10)

A3 = i ((K(x + y) K(y)) Up) X C . (3.11)

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Using (2.3), (2.12) and interpolating we have

i

K(x + y)pUp1 pUp1

1,2

CK(x + y)pU

p1 pUp1

2nn+2

CpUp1 pUp1n

2 2n

n2C X . (3.12)

In the same way from (2.7), (2.12) we get that

i

K(x + y)pU

p1 pUp1

s C X . (3.13)

Estimates (3.12) and (3.13) imply

A4 = i

K(x + y)pU

p1 pUp1

X C X . (3.14)

From (2.3) and (2.17) we have

i

(K(x + y) K(y)) Up1

1,2

C (K(x + y) K(y)) Up1 2nn+2

C (K(x + y) K(y)) Up1n2

2nn2

C X (3.15)

while from (2.7) and (2.17) it follows that

i

(K(x + y) K(y)) Up1

s

C (K(x + y) K(y)) Up1 nsn+2s

C (K(x + y) K(y)) Up1n2

s C X . (3.16)

From (3.15) and (3.16) then

A5 = i

(K(x + y) K(y)) Up1

X C X . (3.17)

Putting together (3.7), (3.8), (3.11), (3.14) and (3.17) we finally get

T,y()X C(pX + + + X+ X)

C( + )

( + )(p1) + ( + )

1+ +

( + ) (3.18)

ifand are small enough.Step B Here we want to show that

T,y :{ X :X ( + )} { X: X ( + )}

is a contraction map.

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Arguing as in the previous step we have

T

,y(1) T

,y(2)X Ci(K(x + y)

(U+ 1)

+p (U+ 2)+p K(y)Up1 (1 2) X

Ci K(x + y) (U+ 1)+p (U+ 2)+p

p

(U+ 2)+p1

(1 2)

X(3.19)

+ Ci

K(x + y)

p

(U+ 2)+p1 pUp1 (1 2) X

+ Ci

K(x + y)pU

p1 pUp1

(1 2)

X

+ Ci

(K(x + y) K(y)) Up1 (1 2)

X

.

Estimating each term in (3.20) as done in the previous step we get

T,y(1) T,y(2)X

C

1 2pX + 2

p1X 1 2X+ 1 2X+ 1 2X

(choosing and small enough )

C01 2X (3.20)

for some constant C0 < 1. This show thatT,y is a contraction mapping from

{ X: X ( + )} into itself and the claim of the proposition follows.

4 Some useful estimates.

In Section 3 we proved the existence of,y D1,2(IRn) Ls(IRn) with nn2 0, 0 > 0, 0 > 0 suchthat for any (0, 0), (0, 0) andy G

|u,y(x)| C U(x) in IRn (4.15)

u,y

xi(x)

C|U(x)| inIRn. (4.16)

Proof The proof will be divided into three steps.Step A Here we will prove thatR > 0 there exist C(R) > 0, 0 >0, 0 >0such that for any y G, (0, 0) and (0, 0) holds

|u,y(x)| C(R) in B (0, R). (4.17)

From (4.2) we have that u,y satisfies the equation

u,y=K(x + y)

u,y+ n+2n2

ni=0

ci(,,y)i in IRn. (4.18)

By Lemma 4.1 we have

ni=0

ci(,,y)iL(IRn)

=o(1)

where o(1) 0 as and go to 0. Our assumptions imply that

K(x + y)

u,y+ 4n2

Ln2(B(0,4R))

C.

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By Lemma 6 of [6], (see also [8]) we derive that

u,yLq(B(0,2R)) Cu

,yL 2nn2 (B(0,4R))

where q= ( 2n

n2 )2

2 . This implies that

K(x + y)

u,y+ 4n2

Lq

n24(n2) (B(0,2R))

C

withq n24(n2) > n2 . So, by elliptic regularity (see [4] or [6]), we can derive

supB(0,R)

|u,y| C

B(0,2R)

1 +

u,y2

dx C.

Step B In this step we will prove that there existR >0, C >0, 0 > 0, 0 > 0

such thaty G, (0, 0), (0, 0) holds

|u,y(x)| C U(x) x IRn \ B(0, R). (4.19)

Let u,y(x) L 2nn2 (IRn) the Kelvin transform ofu,y i.e.

u,y(x) = 1

|x|n2u,y

x

|x|2

x IRn \ {0}.

We want to prove that |u,y(x)| C x IRn \ {0} such that |x| R. It is

standard to see that u,y(x) satisfies

u,y(x) = K

x

|x|2+ y

1

|x|(n2)(

u,y

+)

4n2u,y(x)

ni=0

1

|x|n+2ci(,,y)i

x

|x|2

in IRn \ {0}. (4.20)

We observe that 1|x|n+2 i

x|x|2

L (B(0, 4R)) so that the last term in

(4.20) converges to zero asand go to zero. Moreover using that u,y U(x)

in L 2nn2 (IRn) then

K

x

|x|2+ y

1

|x|(n2)

u,y 4n2Ln2(B(0,4R))

C.

Proceeding as in the previous step we get

u

,yLq(B(0,2R)) Cu

,yL 2nn2 (B(0,4R)) C

where q=( 2nn2 )

2

2 . This implies that

K

x

|x|2+ y

1

|x|(n2)

u,y 4n2

Ln+12 (B(0,2R))

Cu,yLq(B(0,2R)).

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By elliptic regularity it follows

u,y(x)L(B(0,R)) C.

Then (4.15) follows by (4.17) and (4.19).Step C This step is devoted to show (4.16). By Greens representation formulawe have

u,y

xj(x) = cn

xj

IRn

K(z + y)

u,y(z)p

n

i=0 ci(,,y)i(z)

|x z|n2 dz

= cn(2 n)

IRn

K(z + y)

u,y(z)p

n

i=0 ci(,,y)i(z)

|x z|n (xj zj)dz

wherecn= 1

n(2n)nandnis the area of the unit sphere in IR

n. Using estimate

(4.15) we have

u,yxj

(x) C

IRn

1

(1+|z|2)

n22 p

+n

i=0 |ci(,,y)|pUp1(z)|i(z)|

|x z|n1 dz

C

IRn

1

(1+|z|2)

n22 p

+

1(1+|z|2)

n2+2

1 + |z| + |z|2

|x z|n1 dz. (4.21)

Now we let IRn = A1 A2 where A1 =

z IRn : |x z|> |x|2

and A2 =

z IRn : |x z| |x|2

. We can now split (4.21) into two integrals and esti-

mate these two integrals

A1

1

(1+|z|2)

n22 p

+

1(1+|z|2)

n2+2

1 + |z| + |z|2

|x z|n1 dz

C|x|1nA1

1

(1 + |z|2)

n22 p

+

1

(1 + |z|2)

n2+2

1 + |z| + |z|2

dz

C|x|1n (4.22)

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and

A2

1(1+|z|2)n22 p

+

1(1+|z|2)n2+2

1 + |z| + |z|2

|x z|n1 dz

C

A2

|x z|1n

|z|n2+(n2) + |z|n4(1 + |z| + |z|2)

dz.

C

|x|n2+(n2) + |x|n4

1 + |x| + |x|2

|xz||x|2

|x z|1ndz

C

|x|n2+(n2) + |x|n4

1 + |x| + |x|2 |x|2

0

1nn1d

C

|x|n1+(n2) + |x|n3

1 + |x| + |x|2

. (4.23)

Using (4.22) and (4.23) we get estimate (4.16). Note that the same can be donefor the second derivatives.

Lemma 4.3. Let u,y X be a solution of (4.2) such that u,y U in

D1,2(IRn). Then

u,yp u,yp + ln u,y u,yp 2nn+2 C 2. (4.24)

Proof By mean value theorem we getx IRn

u,y(x)

p

u,y(x)p

+

ln u,y(x)

u,y(x)p

= 2

2

ln u,y(x)

2 u,y(x)

pxfor some x [0, 1]. Using estimate (4.15) it follows that

ln u,y(x)2 u,y(x)px Lt(IRn)t > nn+2 . The claim is proved.

Lemma 4.4. Let u,y be a solution of (4.2) such that u,y U in D

1,2(IRn)andK(x) be a bounded function, then

IRnu,y(x)

u,y

xi(x)dx= 0 (4.25)

IRn

xi

K(x + y)

u,y(x)

p+1dx= 0 (4.26)

Proof From the Divergence Theorem and estimate (4.16) we get

B(0,R)

u,y(x) u,y

xi

(x)dx

=

B(0,R)

xi

u,y(x)22

dx=

B(0,R)

u,y(x)22

id

C

1

Rn1

2Rn1 C

1

Rn1.

16


17/36

Then by the Lebesgue theorem we have

IRn

u,y(x) u

,y

xi(x)dx= 0

The estimate (4.26) follows in the same way.

Lemma 4.5. Letu,y be a solution of (4.2) such thatu,y U inD

1,2(IRn)as 0, and letK(x) be a bounded function, then

IRndiv

xu,y(x)2 dx= 0 (4.27)

IRndiv

K(x + y)x

u,y(x)p+1

dx= 0 (4.28)

IRn

divK(x + y)x u,y

(x)p+1 ln u,y

(x) dx= 0 (4.29)Proof The proof follows exactly as in Lemma 4.4.

5 A lower bound on the number of solutions

In this section we want to prove Theorem 1.1. A crucial assumption is to writey as

y= y0+ y

where y0 is a critical point of K and y IRn will be chosen later. We start

proving two lemmas.

Lemma 5.1. Let y0 be a critical point of K(x). Assume K(x) satisfies hy-potheses (1.2) and i)-iv) in a neighborhood of y0 with 1 i < n for anyi= 1, . . . , n. Assumeu,y is a solution to (4.2) such thatu

,yU inD

1,2(IRn)as 0 and 0. Then, for small enough andy= y0+ y, it holds

u,y i

K(x + y)f(u,y)

,u,y

xi

1,2

= i+1

p + 1

IRn

hi(x + y)Up+1(x)dx

+ o

i+1

. (5.1)

Proof We haveu,y i

K(x + y)f(u

,y)

,u,y

xi

1,2

=

IRn

u,y(x) u,y

xi(x)dx

IRn

K(x + y)

u,y(x)+p u,y

xi(x)dx.

17


18/36

Then by (4.25) and (4.26) we have

u,y i

K(x + y)f(u,y)

, u,y

xi

1,2

= 1

p+ 1

IRn

K(x + y)

xi

u,y(x)

+p+1dx

=

p+ 1

IRn

K

xi(x + y)

u,y(x)

+p+1dx. (5.2)

Now we want to estimate the integral in (5.2) using the properties ofK. SoIRn

K

xi(x + y)

u,y(x)

+p+1dx

= |x+y|>r

K

xi(x + y)u,y(x)+p+1 dx

+

|x+y|r

K

xi(x + y)

u,y(x)

+p+1dx= I1+I2. (5.3)

Using (4.15) and that K Cthen

|I1| C

+r

1

1 + 2

n22 (

2nn2 )

n1d C n. (5.4)

Now we lety= y0+ y (5.5)

with y IRn. Then using properties i)-iv) ofKwe have

I2 =

|x+y|r

(hi(x + y) + Ri(x + y))

u,y(x)+p+1

dx= A2+ B2

(5.6)and|x+y|r

hi(x + y)

u,y(x)p+1

dx= i|x+y|r

hi(x + y)

u,y(x)+p+1

dx

= iIRn

hi(x + y)Up+1(x)dx + o(i) (5.7)

since|hi(x + y)| C(|x|i + 1) and i< n, while

|B2|= |x+y|r

Ri(x + y)

u,y(x)+p+1

dx

Ci|x+y|r

|x + y|i

u,y(x)+p+1

dx

Ci + Cn. (5.8)

18


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Then by (5.6), (5.7) and (5.8) we have

I2 = iIRn

hi(x + y) (U(x))p+1 dx + o(i). (5.9)

Finally we getu,y i

K(x + y)f(u,y)

,u,y

xi

1,2

= i+1

p + 1

IRn

hi(x + y) (U(x))p+1

dx

+ o(i+1). (5.10)

which gives the claim.

Lemma 5.2. Let y0 be a critical point of K(x). Assume K(x) satisfies hy-potheses (1.2) and i)-iv) in a neighborhood ofy0 with1 i < n 1 for any

i= 1, . . . , n. Assumeu,y is a solution to (4.2) such thatu

,yU inD

1,2

(IRn

)as 0 and 0. Then, for small enough andy= y0+ y, it holds

u,y i

K(x + y)f(u,y)

, x u,y+n 2

2 u,y

1,2

= (n 2)2

4n K(y0)

IRn

(U(x))p+1

dx + o()

+

ni=1

i+1

p + 1

IRn

hi(x + y)xi(U(x))p+1

dx + o

i+1

(5.11)

Proof We have

u,y i

K(x + y)f(u,y)

, x u,y+ n 22 u

,y1,2

(5.12)

=

IRn

u,y(x)

x u,y(x) +

n 2

2 u,y(x)

dx

IRn

K(x + y)

u,y(x)+p

x u,y(x) +n 2

2 u,y(x)

dx

= 1

2

IRn

div

x|u,y(x)|2

dx

IRn

K(x + y)

x u,y(x) +

n 2

2 u,y(x)

u,y(x)+p

u,y(x)+p

+

u,y(x)+p

ln u,y(x)

dx

19


20/36

n 2

2 IRn K(x + y)u,y(x)

+

p+1

dx

1p + 1

IRn

div

K(x + y)x

u,y(x)+p+1

dx

+ 1

p + 1

IRn

div(K(x + y)x)

u,y(x)+p+1

dx

+ n 2

2

IRn

K(x + y)

u,y(x)+p+1

ln u,y(x)dx

+

p + 1

IRn

div

K(x + y)x

u,y(x)

+p+1ln u,y(x)

1

p + 1

dx

p + 1

IRn

div(K(x + y)x)

u,y(x)+p+1

ln u,y(x) 1

p + 1

dx.

Using (4.27), (4.28) and (4.29) we getu,y i

K(x + y)f(u,y)

, x u,y+n 2

2 u,y

1,2

=

IRn

K(x + y)

x u,y(x) +

n 2

2 u,y(x)

u,y(x)+p

u,y(x)+p

+

u,y(x)+p

ln u,y(x)

dx

+

p + 1

IRn

K(x + y) x

u,y(x)+p+1

dx

+ n

(p + 1)2

IRn

K(x + y)

u,y(x)+p+1

dx

p + 1IRn

K(x + y) x

u,y(x)+)p+1

ln u,y(x) 1p + 1

dx

= I1+ I2+ I3+ I4. (5.13)

We can estimate the first integral as follows

I1 CIRn

u,y(x)+p u,y(x)+p + u,y(x)+p ln u,y(x)

x u,y(x) + n 22 u,y(x) dx (5.14)

u,y+p

u,y+p

+

u,y+p

ln u,y 2nn+2

x u,y+n 2

2 u,y 2n

n2

C2.

20


21/36

In order to estimate I2 we split the integral I2 as follows

I2 = p + 1

IRn

K(x + y) x

u,y(x)+p+1 dx

=

p + 1

|x+y|r

K(x + y) x

u,y(x)+p+1

dx

+

p + 1

|x+y|r

K(x + y) x

u,y(x)+p+1

dx

=

p + 1(I21+ I22) . (5.15)

From (4.15) and the boundedness ofK we have

|I21| C +

r

1

2nn1d C n1. (5.16)

To estimate the term I22 we use our crucial assumption

y= y+ y0.

Then

I22 =

|x+y|r

K(x + y) x

u,y(x)+p+1

dx

=n

i=1

i|x+y|r

hi(x + y)xi

u,y(x)

+p+1dx

+

ni=1

|x+y|r

Ri(x + y)xi

u,y(x)+p+1

dx

=n

i=1

iIRn

hi(x + y)xiUp+1(x)dx + o(i) + A22 (5.17)

since i < n 1 and |hi(x + y)| C(|x|i + 1), where

A22 C ni=1

|x+y|r

Ri(x + y)xi u,y(x)+p+1 dxC i

|x+y|r

x + yi |x| (U(x))

p+1dx

C i

+ Cn1

. (5.18)

Putting together (5.15)-(5.18) we get

I2 =ni=1

i+1

p + 1

IRn

hi(x + y)xi(U(x))p+1

dx + o

i+1

+ O (n) . (5.19)

21


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ConcerningI3 we have

I3 = n(p + 1)2

IRn

K(x + y)

u,y(x)+p+1 dx

= n

(p + 1)2

IRn

K(y)Up+1(x)dx + o()

and since y = y0+ y

I3 = n

(p + 1)2K(y0)

IRn

(U(x))p+1

dx + o(). (5.20)

Finally

|I4| C

IRn

|x| (U(x))p+1

ln U(x)

dx C. (5.21)

Putting together (5.13), (5.15), (5.19), (5.20) and (5.21) we get (5.11).

Definition 5.3. LetL C(IRn, IRn)be a vector field. We say thaty is a stablezero forL if

a) L(y) = 0,

b) y is isolated,

c) if Ln is a sequence of vector fields such that Ln LC(B(y,r)) 0 forsome r > 0 then there exists yn B(y, r) such that Ln(yn) = 0 andyn y.

Note that if there exists a neighborhood B (y, r) of y such that

deg(L, B(y, r), 0)= 0 (5.22)

then y is a stable zero for L.

Proof of Theorem 1.1 We want to prove that any value y0 Zy0 gives rise to asingle peak solution of (1.1) which blows up at y0. We note that we have alreadyshown, in Section 3 that if and are sufficiently small for each y IRn thereexists a function ,y W such that U+

,y is a solution of (3.2). Here we

want to show that there exist d such that 1c < d < c and y such that letting

+1 = 1d

and y = y0 + y, then the function u,y

= U+,y satisfiesequation (4.2) with coefficients ci(, , y) = 0 for all i = 0, . . . , n. To this endwe let

G0(d,y) = 1

u,y i

K(x + y)

u,y+p

, x u,y+ n 22 u,y1,2

Gi(d,y) = 1

i+1

u,y i

K(x + y)

u,y+p

,u,y

xi

1,2

.

22


23/36

Taking the scalar product of equation (4.2) and the function u,yxi

for i =

1, . . . , nor x u,y

+ n22

u

,y, and letting+1

= 1

dand y

= y

0+

y, we have

G0(d,y) = 1

c0(,d,y) 0

21,2+ o(1)

Gi(d,y) =

1

i+1

ci(,d,y) i

21,2+ o(1)

Finding a solution of (2.2) means finding d and y such that

G0(d,y) = 0

Gi(d,y) = 0

fori = 1, . . . , nand for each in (0, 0) for some 0 > 0. Using (5.1) and (5.11)we have that

G0(d,y) = 1

d

d (n 2)

2

4n K(y0)

IRn

Up+1(x)dx + (y)

+ o(1) (5.23)

Gi(d,y) = 1

p + 1

IRn

hi(x + y)Up+1(x)dx + o(1)

where (y) = 1p+1

j=

IRn

hj(x + y)xjUp+1(x)dx. We observe that the

leading term ofGi(d,y) is independent ofd. So, since y0 is a stable zero toLy0 ,there exists y y0 independent ofd such that

Gi(d,y) = 0 for any d IR.

Inserting y into (5.23) we derive that G0(d,y) becomes

G0(d,y) = 1d(Ad + (y)) + o(1),

where A = (n2)2

4n K(y0)IRn

Up+1(x)dx. Hence there exists d close to (y)A

such that G0(d,y) = 0 which proves the existence of the solution.It remains to prove that two different stable zeros of Ly0 give rise to two

different solutions of (1.1). So let y10 and y20 be two different points ofZy0 , and let

di and yi be the parameter associated with this two points. Let

i+1

= 1di,

yi = y0+ iy

i and u

1 ,y

1

u2 ,y2be the two solutions of (2.2) generated by the

parameters. For i = 1, 2 we let

ui(x) = i

2p1 ui,yi

x yi

i

the corresponding solutions of (1.1). Then

1 2p1 u1(y

1 ) = u

1 ,y

1

(0) U(0)

2 2p1 u2(y

1 ) = u

2 ,y

2

(12

y1 y2 ) U(cy

10 y

20)

23


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wherec= lim0

12

= lim0

d2d1

1+1

= d20d10

1+1

. We have the following alternative:

1)c= 1. Then

lim0

u1(y1 )

u2(y1 )

= U(0)

U(y10 y20)

= 1

and the claim follows.2) c = 1. There is no loss of generality in assuming c < 1. In this case we

have

lim0

u1(y1 )

u2(y1 )

= lim0

12

2p1 u1 ,y1

(0)

u2 ,y2(12

y1 y2 )

=

1

c

2p1 U(0)

U(cy10 y20)

=

1

c

2p1

1 + |cy10 y

20|2

n22 >1.

This proves that u1 =u2.

6 A useful estimate

In this section we want to prove that if u is a solution of (1.1) that blows-upand concentrate at y0 and ify stands for its peak then

y0 y

1+1

y

where y satisfiesLy0(y) = 0. We start proving the following:

Proposition 6.1. Let y0 be a critical point of K(x). Assume K(x) satisfies

(1.2) and hypotheses i)-iv) in a neighborhood of y0. Let n be a sequence thatgoes to zero andun= un the corresponding single-peaked solutions of (1.1). Ifyn = yn denotes the peak of un and if yn y0 as 0, then there exists apositive constant C >0 such thaty0 yn

1

+1n

C (6.1)where is defined in (1.9) and depends only on the shape ofK(x) in a neigh-borhood ofy0.

Proof We suppose by contradiction that there exists a sequencen such that

y0 yn

1+1

n . (6.2)

We letvn(x) =

1+1n

2pn1

un

1+1n x + yn

. Thenvnsatisfies the equation

vn(x) = K(1

+1n x + yn) (vn(x))

pn in IRn. (6.3)

24


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Here we note that vn satisfies the hypotheses of Lemma 4.3 and hence it verifies

vn C U.

Moreover vn v weakly in D1,2(IRn) where v is a solution of

v= K(y0)vp in IRn

and hencev = 1K(y0)

n24

U(x). Multiplying (6.3) by vnxi and integrating we have

0 =

IRn

vn

vn

xi

dx

IRn

K(1

+1n x + yn) (vn(x))

pn vn

xidx

= 1

p n+ 1

IRn

K(1

+1n x + yn)

xi(vn(x))

pn+1 dx

( using vn C U) (6.4)

=

1+1n

p n+ 1

IRn

K

xi(

1+1n x + yn) (vn(x))

pn+1 dx. (6.5)

Arguing as in the proof of the Lemma 5.1, using properties i)-iv) of K(x) wefind

0 =

IRn

K

xi(

1+1n x + yn) (vn(x))

pn+1 dx

=

|

1+1n x+yn|>r

K

xi(

1+1n x + yn) (vn(x))

pn+1 dx

+

|

1+1n x+yn|r

hi(1

+1n x + yn y0) (vn(x))

pn+1 dx

+|

1+1n x+yn|r

Ri(1+1

n x + yn y0) (vn(x))pn+1 dx

= I1+ I2+ I3. (6.6)

As in Lemma 5.1 we get

|I1| C

1+1n

n. (6.7)

Moreover using (6.2) we have (up to a subsequence)

I2 = |yn y0|i

|

1+1n x+yn|r

hi

1+1n

|yn y0|x +

yn y0|yn y0|

(vn(x))

pn+1 dx

= |yn y0|ihi(z)IRn

(U(x))p+1 dx + o(|yn y0|i) (6.8)

where z B(0, 1) is given by (up to a subsequence)

z= limn

yn y0|yn y0|

.

25


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Reasoning as before we get

|I3| C|yn y0|i + C

1

+1nn

(6.9)

and n

+1n =o (|yn y0|i) by (6.2). We have shown so far that

0 = |yn y0|ihi(z)

IRn

(U(x))p+1

dx + o (|yn y0|i)

and this implies hi(z) = 0 which is not possible from the properties iv) onK.

Proposition 6.2. Let y0 be a critical point of K(x). Assume K(x) satisfies(1.2) and hypotheses i)-iv) in a neighborhood of y0 with 1 i < n 1 forall i = 1, . . . , n. Let n be a sequence that goes to zero and un = un thecorresponding single-peaked solutions of (1.1). If yn = yn denotes the peak of

un and ifyn y0 as 0, then up to a subsequence, we have

yn= y0 1

+1n y+ o

1+1n

, (6.10)

Ly0(y) = 0 (6.11)

andj=

1

p + 1

IRn

hj(x + y)xjUp+1(x)dx=

n 2

2 K(y0)

IRn

Up+1(x)dx (6.12)

where is defined in (1.9) and depends only on the shape ofK(x) in a neigh-borhood ofy0.

Proof of Proposition 6.2 Let us prove (6.10) and (6.11). As in proof of Propo-sition 6.1 we let

vn(x) =

1+1n

2pn1

un

1+1n x + yn

.

Since un is a single peaked solution of (1.1) then vn U(x) in D1,2(IRn).

Multiplying (2.2) by vnxi we have (see also (5.2))

0 =

1+1n

p n+ 1

IRn

K

xi(

1+1n x + yn) (vn(x))

pn+1 dx

=

1+1n

p n+ 1(I1,n+ I2,n+ I3,n) (6.13)

where

I1,n=

1+1n

i |

1+1n x+yny0|r

hi

x +

yn y0

1

+1n

(vn(x))

pn+1 dx

= i+1n

IRn

hi(x + y) (U(x))p+1

dx + o(i+1n ) (6.14)

26


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and up to a subsequence

y = limn

yn y0

1+1

n

. (6.15)

Arguing as in the proof of Lemma 5.1 we get

I2,n= O(i+1n ) + O(

n+1n ) (6.16)

andI3,n= O(

n+1n ). (6.17)

Hence from (6.13)-(6.17) we getIRn

hi(x + y)Up+1(x)dx= 0

and then Ly0(y) = 0.Let us prove (6.12). Taking the scalar product of (4.2) with x vn +

n22

vnwe get from (5.11)

0 =n

j=1

j+1

+1n

p + 1

IRn

hj(x + y)xj(U(x))p+1

dx +n 2

2 K(y0)n

IRn

(U(x))p+1

dx

+o(n) (6.18)

and this proves the claim.

Proof of Theorem 1.3 It follows from Proposition 6.2.

7 An exact multiplicity result

Proof of Theorem 1.2 By contradiction let us suppose that #{ single peaksolutions of (1.1) concentrating at y0}> #Zy0 . Since #Zy0


28/36

The function vn satisfies

vn(x) = K(nx + y0)cn(x)vn(x) in IRn (7.3)

where

cn(x) = (p n)

10

tv1n(x) + (1 t)v

2n(x)p1n

dt.

Here we observe that from (4.15) we have

0 cn(x) C

(1 + |x y|2)2 , x IR

n. (7.4)

Up to a subsequence we can assume that vn vweakly in D1,2(IRn) and almost

everywhere in IRn where v satisfies

v(x) = K(y0)pUp1(x y)v(x) in IRn. (7.5)

Then there exist real numbers ai such that

v(x) =n

i=0

aii(x y) (7.6)

withi as defined in (2.5) and (2.6). Moreover reasoning exactly as in the proofof Lemma 4.2 one can see that

|vn(x)| C

1 + |x y|n2 in IRn. (7.7)

Now we multiply the equation (7.1) for vin

xjand we integrating over IRn. Using

(4.26) we get

0 = 1

p n+ 1n

IRn

K

xj(nx + y0)

vin(x)pn+1

dx. (7.8)

Subtracting equation (7.8) evaluated for i = 1, 2 we have

0 =

IRn

K

xj(nx + y0)

v1n(x)pn+1

v2n(x)pn+1

dx

0 =

IRn

K

xj(nx + y0)cn(x)vn(x)dx

where

cn(x) = (p + 1 n)

10

tv1n(x) + (1 t)v

2n(x)pn

dt (7.9)

and by (4.15)0< cn(x) C U

p(x y). (7.10)

28


29/36

Using the properties i)-iv) ofK(x) in a neighborhood ofy0 we get

0 =

|nx+y0|>r

K

xj(nx + y0)cn(x)vn(x)dx

+

|nx+y0|r

hj(nx)cn(x)vn(x)dx

+

|nx+y0|r

Rj(nx)cn(x)vn(x)dx= I1+ I2+ I3. (7.11)

By (7.7), (7.10) and (1.2) we get I1 C n. (7.12)

Moreover by (7.10) and j < n, we have

I2 = jn

|nx+y0|r

hj(x)cn(x)vn(x)dx

= jn

IRn

hj(x)Up(x y)v(x)dx + o (jn ) (7.13)

and

I3 C jn|nx+y0|r

|x|j cn(x)vn(x)dx C jn + C

nn . (7.14)

From (7.11)-(7.14) we have

0 =IRn

hj(x)Up(x y)v(x)dx (7.15)

=a0

IRn

hj(x)Up(x y)

(x y) U(x y) +

n 2

2 U(x y)

dx

+ni=1

ai

IRn

hj(x)Up(x y)

U

xi(x y)dx

29


30/36

for j = 1, . . . , n. On the other hand we have that

IRn

hj(x)Up(x y)

(x y) U(x y) + n 2

2 U(x y)

dx

(since y is a zero ofLy0) (7.16)

=ni=1

IRn

hj(x)Up(x y)(xi (y)i)

U

xi(x y)dx

= 1

p + 1

IRn

ni=1

hj(x)

xi(xiyi)U

p+1(x y)dx

(using the Eulero Theorem for homogenous functions)

= j

p + 1

IRn

hj(x)Up+1(x y)dx +

yip + 1

IRn

hj(x)

xiUp+1(x y)dx

= yip + 1IRn

hj(x)xi

Up+1(x y)dx= yiIRn

hj(x)Up(x y) Uxi

(x y)dx.

Hence (7.15) becomes

0 =n

i=1

(ai a0yi)

IRn

hj(x)Up(x y)

U

xi(x y)dx (7.17)

Recalling the definition ofLy0 we have that

Jac(Ly0(y)) = (p + 1)

IRn

hj(x)Up(x y)

U

xi(x y)dx

i,j=1,...,n

Since 0 is a regular value for Ly0 , thenLy0(y) = 0 implies that the linear system(7.17) admits only the solution

ai= a0yi for i = 1, . . . , n . (7.18)

Next aim is to show that a0 = 0. To do this let us multiply equation (7.1) forx vin and integrating over IR

n we get

0 =

n 2

2

n

p n+ 1

IRn

K(nx + y0)

vin(x)pn+1

dx

n

p n+ 1

IRn

K(nx + y0) x

vin(x)pn+1

dx. (7.19)

Subtracting (7.19) evaluated for i = 1, 2, we get

0 =

n 2

2

n

p n+ 1

IRn

K(nx + y0)cn(x)vn(x)dx

n

p n+ 1

IRn

K(nx + y0) xcn(x)vn(x)dx= I1+ I2

30


31/36

where cn are as defined in (7.11). Let us estimate I1 using (7.7) and (7.10)

I1 =

n 22

np n+ 1

IRn

K(nx + y0)cn(x)vn(x)dx

=

(n 2)2

4n n+ o(n)

K(y0)

IRn

Up(x y)v(x)dx

=o(n)

because a straightforward computation gives thatIRn

Up(x y)v(x)dx= 0.ConcerningI2 we get

I2 = n

p n+ 1

IRn

K(nx + y0) xcn(x)vn(x)dx

= n

p n+ 1

|nx+y0|>r

K

xj(

nx + y

0)x

jcn

(x)vn

(x)dx

+

|nx+y0|r

hj(nx)xj cn(x)vn(x)dx

+

|nx+y0|r

Rj(nx)xj cn(x)vn(x)dx

= J1+ J2+ J3.

Using (7.7) and (7.10) one gets

J1 = O

n1

and J3 = O

j + n1

Finally

J2 = np n+ 1

|nx+y0|r

hj(nx)xj cn(x)vn(x)dx

=j=

+1np n+ 1

IRn

hj(x)xjUp(x y)v(x)dx + o

+1n

since j < n 1. Since = +1n we derive that (7.20) becomes

j=

IRn

hj(x)xjUp(x y)v(x)dx= 0 (7.20)

31


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and then (recalling that ai= a0yi)

j=

IRn

hj(x)xjUp(x y)v(x)dx

=a0j=

IRn

hj(x)xjUp(x y)

ni=1

xiU

xi(x y) +

n 2

2 U(x y)

dx

= a0

p + 1

ni=1

j=

IRn

xi(hj(x)xixj) U

p+1(x y)dx

+(n 2)a0

2

j=

IRn

hj(x)xjUp+1(x y)dx

= a0

p + 1

n

i=1j=IR

nhj(x)xj+ hj(x)ijxi+ hj(x)

xi

xixjUp+1(x y)dx+

(n 2)a02

j=

IRn

hj(x)xjUp+1(x y)dx

(using the Eulero theorem for homogenous functions)

=a0

n 2

2

n + 1

p + 1

p + 1

IRn

j=

hj(x)xjUp+1(x y)dx.

Since n 2

2

n + 1

p + 1

p + 1

=

+ 1

p + 1 R

cn(x)v2n(x)dx (7.21)

for any R >0. By (7.3) we have

|x|>R

cn(x)v2n(x)dx

|x|>R

cn(x)n2 dx

2n

|x|>R

vn(x) 2nn2 dx

n2n

1

2C

32


33/36

ifR is big enough. Moreover since vn 0 in D1,2(IRn) and vn 0 uniformly

on compact set of IRn then|x|R

cn(x)v2n(x)dx

1

4C

if is small enough. Then a contradiction arise from (7.21) and we get theclaim.

8 Examples and final remarks

In this section we consider some particular cases of function K(x) where ourresults apply. We always assume that y0 = 0. Let us start by considering thecase of the nondegenerate critical point.

Proposition 8.1. Let us assume that0 is a nondegenerate critical point of Ksatisfying

K(0)< 0.

Then there exists exactly one solution to (1.1) blowing up at0.

Proof By Taylors formula we have that

K(x) = K(0) +1

2

nj,k=1

2K

xjxk(0)xjxk+ R(x) (8.1)

in a neighborhood of 0. So we have that

hi(x) =1

2

n

j=12K

xjxi(0)xj

Hence we derive that

L0(y) =

IRn

hj(x + y)Up+1(x)dx

=1

2

nj=1

2K

xjxi(0)

IRn

(xi+ yi) Up+1(x)dx

=n

j=1

2K

xjxi(0)yi

IRn

Up+1(x)dx.

Since the matrix 2K

xjxi(0) is invertible we get that the linear system

L0(y) = 0

admits only the solution y = 0. So

Z0 = {0}. (8.2)

Since i = 1 for any i = 1, . . ,n, by Theorem 1.1 and Theorem 1.2 we get theexistence and uniqueness of the solution blowing up at zero.

33


34/36

Remark 8.2. The same computations of the previous proposition allow us tohandle the following case,

K(x) = 1 +n

j=1

cjxsjj

where cj IR and sj 2 are even positive integers. We again have thatZ0 = {0} and under the assumption

j=

cjsj

IRn

xsjj U

p+1(x)dx


35/36

we apply Theorem 1.2. This leads to check that y1 and y2 are regular values forL0. Indeed

det JacL0(y1) = det JacL0(y2) =

16

IRn

x21Up+1(x)dx

IRn

Up+1(x)dx

2= 0, (8.3)

and this gives the claim.

Remark 8.5. The same computation of the previous proposition can be per-formed in any dimensionn 3. With a little of work it is also possible constructexamples of functionK(x) such that (1.1) admits exactly m solutions, for anyintegerm 1.

References[1] Chen, Chiun-Chuan; Lin, Chang-Shou, Estimate of the conformal

scalar curvature equation via the method of moving planes. II, J. Differen-tial Geom.49 (1998), no. 1, 115178.

[2] Floer, Andreas; Weinstein, Alan, Nonspreading wave packets for thecubic Schrodinger equation with a bounded potential. J. Funct. Anal.69(1986), no. 3, 397408.

[3] Gidas, Basilis; Spruck, Joel, Global and local behavior of positive so-lutions of nonlinear elliptic equations, Comm. Pure Appl. Math,34 (1981),no. 4, 525598. 261280.

[4] Gilbarg, David; Trudinger, Neil S., Elliptic Partial Differential Equa-tions of Second Order, Springer-Verlag, Berlin, (1983).

[5] Grossi, Massimo, On the number of single-peak solutions of the nonlin-ear Schrodinger equation, Ann. Inst. H. Poincare Anal. Non Lineaire,19(2002), no. 3, 261280.

[6] Han, Zheng-Chao , Asymptotic approach to singular solutions for non-linear elliptic equations involving critical Sobolev exponent,Ann. Inst. H.Poincare Anal. Non Lineaire8 (1991), no. 2, 159174.

[7] Micheletti, Anna Maria; Pistoia, Angela, Existence of blowing-up solutions for a slightly subcritical or a slightly supercritical non-linearelliptic equation on Rn,Nonlinear Anal. 52 (2003), no. 1, 173195.

[8] Micheletti, Anna Maria; Pistoia, Angela; Visetti, Daniela, Onthe number of blowing-up solutions to a nonlinear elliptic equation withcritical growth, Rocky Mountain J. Math. 37 (2007), no. 1, 291325.

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[9] Ni, Wei Ming, On the elliptic equation u+ K(x)u(n+2)/(n2) = 0, itsgeneralizations, and applications in geometry, Indiana Univ. Math. J. 31

(1982), no. 4, 493529.

[10] Pohozaev, Stanislav, On the eigenfunctions of the equation u +f(u) = 0,Dokl. Akad. Nauk SSSR, 165, (1965), 3639.

[11] Yan, Shusen, High-energy solutions for a nonlinear elliptic problem withslightly supercritical exponent, Nonlinear Anal. 38 (1999), no. 4, Ser. A:Theory Methods, 527546.

[12] Wang, Xuefeng; Wei, Juncheng, On the equation u +

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