Existence and multiplicity results for the conformal ...

Existence and multiplicity results for the conformal scalar curvature equation ∗ Francesca Gladiali

1

†

Massimo Grossi‡

Introduction and statement of the main results

This paper deals with the problem   −∆u = K(x)up u>0  u ∈ D1,2 (IRn )

in IRn in IRn

(1.1)

p = p − , n ≥ 3,  > 0 and D1,2 (IRn ) is the completion of
1 R C0∞ (IRn ) with respect to the norm kuk1,2 = IRn |∇u|2 2 . The function K ∈ C 1 (IRn ) satisfies the following assumptions: where p =

n+2 n−2 ,

1 ≤ K(x) ≤ C; 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 scalar curvature of the new metric. It is well known that, if  = 0 and K(x) = n(n − 2) problem (1.1) has only the two-parameters family of solutions Uδ,y (x) =

δ (δ 2

n−2 2

+ |x −

y|2 )

n−2 2

,

x ∈ IRn ,

(1.3)

where δ > 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 equations”. † Struttura dipartimentale di Matematica e Fisica, Universit` a di Sassari,via Vienna 2 -07100 Sassari, e-mail [email protected]. ‡ Dipartimento di Matematica, Universit` a di Roma “La Sapienza”, P.le A. Moro 2 - 00185 Roma, e-mail [email protected].

1

In this paper we are interested in single-peak solutions u concentrating at some point, i.e. u (x) = Uδ ,y (x) + φ (x) (1.4) where δ → 0, y → y0 ∈ IRn , φ (x) → 0 in D1,2 (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 of K(x) and the exponent p (see for example [9], [11] and [12] and the reference therein) Let us recall that when K(x) is constant there is no solution to (1.1) (see [3]). The same happens if (∇K(x), x − x0 ) ≥ 0 for some x0 ∈ IRn ([10]). Another nonexistence result can be found in [9] where it was showed that if K(x) grows like or faster than |x|2−(n−2) then equation (1.1) has no positive solutions. Concerning the existence results we mention the one in [12] where the authors showed the existence of solutions of (1.1) that concentrate at a nondegenerate critical point y0 of K(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) is bounded, they had to impose the condition n > 6. The proof of this result uses the finite dimensional reduction method, a tool widely used in this kind of problems. In [11] S. Yan improved this result by removing the condition n > 6 but still assuming (1.5). Note that in [11] the supercritical case  < 0 was also considered. Finally we mention the paper [8] which concerns the problem   −∆u + V (x)u = n(n − 2)up in IRn u>0 in IRn (1.6)  n 1,2 u ∈ D (IR ), 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 suitable critical point of the function K(x). To do this we use some ideas of [5] where the nonlinear Schrodinger equation was considered. This method well suits to our problem and allows us to establish some existence and multiplicity results to (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 . Before stating the main theorem we have to introduce notations and some hypotheses on the function K(x). Let y0 be a fixed critical point of K(x). We make the following assumptions on K(x): there exists functions hi : IRn → IR, Ri : B(0, r) → IR and real constants αi , βi with αi ≥ 1 for i = 1, .., n such that 2

i)

∂K ∂xi (x

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

ii) |Ri (x)| ≤ C|x|βi for x ∈ B(0, r) and βi > αi , iii) hi (tx) = tαi hi (x) for any x ∈ IRn and t > 0, iv) (h1 (x), .., hn (x)) = (0, .., 0) if and only if x = 0. Now let us introduce the following vector field Z  p+1 Ly0 (y) := hi (x + y)U (x)dx IRn

.

(1.7)

i=1,...,n

Note that the definition of L is well posed if αi < n. Then we set Zy0 := {y ∈ IRn |y is a stable zero of Ly0 }.

(1.8)

(see Section 5 for the definition of stable zero) and α = min{αi | i = 1, . . . , n}.

(1.9)

Our first result is the existence of several single-peak solutions concentrating at the 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) and suppose that K(x) satisfies hypotheses i)-iv) in a neighborhood of y0 with 1 ≤ αi ≤ n − 1 and

X Z αj =α

IRn

for i = 1, . . . , n,

hj (x + y˜)xj U p+1 (x)dx < 0, ∀˜ y ∈ Zy0 .

(1.10)

(1.11)

Then there exists 0 such that for any  ∈ (0, 0 ) the number of single-peak solutions of (1.1) blowing up at y0 is greater or equal to #Zy0 . We observe that if y0 is a nondegenerate critical point of K then Zy0 = {0} and (1.11) becomes (1.5) (see Proposition 8.1 and (8.2)). One of the crucial tool of the proof of Theorem 1.1 is splitting of the peak point y of the solution as  1  1 y = y0 +  α+1 y + o  α+1 where y belongs to Zy0 . This decomposition enable us to handle functions K whose gradient at y0 is flat of order α. If all zeroes of Ly0 belong to Zy0 we have that the previous theorem is optimal. Theorem 1.2. Let us suppose that 0 is a regular value for Ly0 and suppose that K(x) satisfies (1.2) and hypotheses i)-iv) in a neighborhood of y0 . If (1.10) and (1.11) hold and #Zy0 < ∞, then the number of single peak solutions of (1.1) blowing up at y0 is equal to #Zy0 . 3

Let us recall that 0 is a regular value for Ly0 if Jac (Ly0 (y)) 6= 0 for any y ∈ IRn such that Ly0 (y) = 0. Since it is known that any single peak solution concentrates at a critical point of K, 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 that the assumptions of the previous theorem are ”almost” sharp. Theorem 1.3. Let y0 ∈ IRn be a critical point of K(x). Assume K(x) satisfies (1.2) and hypotheses i)-iv) in a neighborhood of y0 , with 1 ≤ αi < n − 1 for i = 1, . . . , n. If either Ly0 (y) 6= 0

for any y ∈ IRn ,

(1.12)

or ∃˜ y ∈ IRn such that Ly0 (˜ y ) = 0 and

X Z αj =α

hj (x+ y˜)xj U p+1 (x)dx ≥ 0, (1.13)

IRn

then there is no single-peaked solutions to (1.1) blowing up at y0 . Let us observe that if K(x) is an homogenous polynomial of degree α+1 then R the assumption that y ˜ belongs to #Z becomes ∇K(x + y˜)U p+1 (x)dx = 0 y0 IRn R P p+1 and αj =α IRn hj (x + y˜)xj U (x)dx ≥ 0 becomes (by Eulero’s Theorem for R homogenous functions) IRn K(x + y˜)U p+1 (x)dx ≥ 0. In this setting Chen and Lin 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 and we prove some technical estimates; in Section 3 we perform the finite dimension reduction while in Section 4 we prove a useful estimate on the decay of the solutions. In Section 5 we find some necessary conditions for the existence of solutions and we prove Theorem 1.1. In Section 6 we prove the asymptotic behavior 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 some examples 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 LiapunovSchmidt reduction method. The first step is to scale (1.1). If u is a solution of (1.1) we let 2 uδ,y = δ p −1 u (δx + y) (2.1) with δ > 0 and y ∈ IRn . This function satisfies   p   −∆uδ,y (x) = K(δx + y) uδ,y (x) u > 0   δ,y uδ,y ∈ D1,2 (IRn ). 4

in IRn (2.2)

Of course (1.1) and (2.2) are equivalent. Let us introduce the operator 2n

i∗ : L n+2 (IRn )→D1,2 (IRn ), 2n

the adjoint of the embedding i : D1,2 (IRn ) → L n−2 (IRn ). The operator i∗ is defined as follows: Z ∗ i (u) = v ⇔ (v, φ)1,2 = u(x)φ(x)dx ∀φ ∈ D1,2 (IRn ). IRn

As observed in [7] i∗ is continuous, i.e there exists C > 0 s.t. 2n ki∗ (u)k1,2 ≤ Ckuk n+2

2n

∀u ∈ L n+2 (IRn ).

(2.3) n+2

We let f (s) = (s+ )p where s+ denotes the positive part of s and f0 = (s+ ) n−2 . So we can rewrite problem (2.2) in this way:
u = i∗ K(δx + y)(u+ )p (2.4) where u ∈ D1,2 (IRn ) ∩ Ls (IRn ) for some s > 1. Here we point out that if u solves (2.4) then u > 0 by the Strong Maximum Principle. Now we consider the operator L : D1,2 (IRn ) → D1,2 (IRn ) defined by   4 L(u) = u − i∗ n(n − 2)U n−2 u . It is known that L is self-adjoint and is a compact perturbation of the identity. Moreover Ker L = span {ψ0 , ψ1 , . . . , ψn } , where n − 2 1 − |x|2 n−2 U (x) = n , 2 2 (1 + |x|2 ) 2 ∂U xi i = 1, . . . , n. ψi (x) = (x) = −(n − 2) n , ∂xi (1 + |x|2 ) 2 ψ0 (x) = x · ∇U (x) +

(2.5) (2.6)

Let X = Ls (IRn ) ∩ D1,2 (IRn ) with the norm kukX = max{kuks , kuk1,2 }, for some s > 1. Remark 2.1. Let s > and

n n−2 .

2n

ns

If u ∈ L n+2 (IRn )∩L n+2s (IRn ) then i∗ (u) ∈ Ls (IRn )

ns . ki∗ (u)ks ≤ C(n, s)kuk n+2s

We quote the following result:

5

(2.7)

n Lemma 2.2. Let s > n−2 . If u ∈ X then L(u) ∈ X. Moreover L|X : X → X is a continuous function , i.e. there exists C > 0 s.t.

kL|X kX ≤ CkukX

∀u ∈ X

(2.8)

Finally Ker L|X = Ker L. Proof

See Lemmas 2.10 and 2.11 in [7]. ⊥

Let W = (Ker L) ∩ Ls (IRn ) = {u ∈ X : (u, φ)1,2 = 0, ∀φ ∈ Ker L}. We now consider the projection Π : X → W . Recall the following result,   n ˜ : W → W defined by L(u) ˜ Lemma 2.3. Let s > n+2 and L = Π L|X (u) . ˜ is invertible and L ˜ −1 is a continuous operator, i.e. there exists C > 0 Then L s.t. ˜ −1 (u)kX ≤ CkukX ∀u ∈ W. kL (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 ) and ∀φ1 , φ2 ∈ D1,2 (IRn ) and ∀s > 1 we have 2 2n ≤ C kU p − U p +  (log U ) U p k n+2

kU

p

p

2n ≤ C , kU − U k n+2

p −1

p

(2.10)

p

ns − U k n+2s ≤ C

(2.11)

p−1

kp U − pU k n2 ≤ C (2.12)       p p p −1 (φ1 − φ2 ) ≤ C|φ1 − φ2 |p (2.13) (U + φ1 )+ − (U + φ2 )+ − p (U + φ2 )+  p −1  p −1 − (U + φ2 )+ (2.14) (U + φ1 )+ ≤ C|φ1 − φ2 |p −1 Proof

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

Lemma 2.5. Let K(x) satisfy (1.2) and s >

n n−1 . p

Then ∀y ∈ IRn it holds

2n ≤ Cδ k (K(δx + y) − K(y)) U k n+2

p

ns k (K(δx + y) − K(y)) U k n+2s ≤ Cδ

k (K(δx + y) − K(y)) U Proof

p−1

k n2 ≤ Cδ

Let us prove (2.15). For some t = t(x, y) ∈ (0, 1) we have 2n k (K(δx + y) − K(y)) U p k n+2 Z  n+2 2n n+2 2n n+2 n−2 dx = (x) (K(δx + y) − K(y)) U

IRn

(Z = δ IR

Z n

1

∇K(δtx + y) · x dt U

0

Z ≤ Cδ

h

|x|U

n+2 n−2

n+2 n−2

) n+2 2n n+2 2n (x) dx

 n+2 2n i n+2 2n (x) dx ≤ Cδ.

IRn

6

(2.15) (2.16) (2.17)

where in the last line we used that k∇Kk∞ ≤ C. Arguing in the same way we get ns k (K(δx + y) − K(y)) U p k n+2s Z  n+2s ns n+2s ns n+2 n−2 + y) − K(y)) U = (x) dx (K(δx

IRn

Z ≤ δ



 n+2s ns  n+2s ns n+2 |x|U n−2 (x) ≤ Cδ, dx

IRn

and k (K(δx + y) − K(y)) U p−1 (x)k n2 Z n2  n2 4 ≤ Cδ. = (K(δx + y) − K(y)) U n−2 (x) dx IRn

3

The finite dimension reduction

We are looking for solutions to (2.2) of the type uδ,y = U + φδ,y and φδ,y → 0 ∈ D1,2 (IRn ) as  → 0, and δ → 0. This turns the problem (2.2) into a finite dimensional one. The first step is to show the existence of the function φδ,y . n 2n Proposition 3.1. Let γ ∈ (0, 1) be fixed and n−2 < s < n−2 . Then there exist 0 > 0 and δ0 > 0 such that ∀ ∈ (0, 0 ) ∀δ ∈ (0, δ0 ) and ∀y ∈ IRn , ∃ ! φδ,y ∈ W such that γ

kφδ,y kX ≤ ( + δ) Π{U +

φδ,y

∗

−i

(3.1)

K(δx + y) (U +

φδ,y )+




=0

(3.2)

(see Section 2 for the definition of W ). Proof With no loss of generality we can assume that K(y) = n(n − 2). Set φ = φδ,y . We first observe that φ solves (3.2) if and only if φ is a fixed point for  the operator Tδ,y : W → W defined by  

  ˜ −1 Π i∗ K(δx + y) (U + φ)+ p − K(y)U p − K(y)pU p−1 φ Tδ,y (φ) = L . (3.3) γ  Step A We begin by proving that inequality kφkX ≤ ( + δ) implies kTδ,y (φ)kX ≤ γ ( + δ) .

7

Using (3.3) and (2.9) we have 
p   kTδ,y (φ)kX ≤ Cki∗ K(δx + y) (U + φ)+ − K(y)U p − K(y)pU p−1 φ kX
  
p ≤ C ki∗ K(δx + y) (U + φ)+ − U p − p U p −1 φ kX + ki∗ [K(δx + y) (U p − U p )] kX + ki∗ [(K(δx + y) − K(y)) U p ] kX 
 + ki∗ K(δx + y) p U p −1 φ − pU p−1 φ kX   + ki∗ (K(δx + y) − K(y)) pU p−1 φ kX = C (A1 + A2 + A3 + A4 + A5 ) .

(3.4)

By the boundedness of K(x) in L∞ , we get
p

− U p − p U p −1 φ k1,2 ( by (2.3)) ki∗ K(δx + y) (U + φ)+
p
2n ( by (2.13)) ≤ Ck (U + φ)+ − U p − p U p −1 φ k n+2 p 2n ≤ Ckφk ≤ Ckφp k n+2 X  by interpolation since

(3.5)  2n 2n p ∈ (s, ) if  is sufficiently small . n+2 n−2

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

ki∗ K(δx + y) (U + φ)+ − U p − p U p −1 φ ks
p
ns ≤ Ck (U + φ)+ − U p − p U p −1 φ k n+2s ns ≤ Ckφp k n+2s ≤ CkφkpX .

(3.6)

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

A1 = ki∗ K(δx + y) (U + φ)+ − U p − p U p −1 φ kX ≤ CkφkpX . (3.7) Using (2.3), (2.7) and (2.11) we get A2 = ki∗ (K(δx + y) (U p − U p )) kX ≤ CkU p − U p kX ≤ C.

(3.8)

Using (2.3) and (2.15) we easily have ki∗ [(K(δx + y) − K(y)) U p ] k1,2 2n ≤ Cδ ≤ Ck (K(δx + y) − K(y)) U p k n+2

(3.9)

while from (2.7) and (2.16) it follows that ki∗ ((K(δx + y) − K(y)) U p ) ks ns ≤ Cδ. ≤ Ck (K(δx + y) − K(y)) U p k n+2s

(3.10)

From (3.9) and (3.10) A3 = ki∗ ((K(δx + y) − K(y)) U p ) kX ≤ Cδ. 8

(3.11)

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

ki∗ K(δx + y) p U p −1 φ − pU p−1 φ k1,2
2n ≤ CkK(δx + y) p U p −1 φ − pU p−1 φ k n+2 2n ≤ CkφkX . ≤ Ckp U p −1 − pU p−1 k n2 kφk n−2

(3.12)

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

ki∗ K(δx + y) p U p −1 φ − pU p−1 φ ks ≤ CkφkX .

(3.13)

Estimates (3.12) and (3.13) imply A4 = ki∗ K(δx + y) p U p −1 φ − pU p−1 φ





kX ≤ CkφkX .

(3.14)

From (2.3) and (2.17) we have
ki∗ (K(δx + y) − K(y)) U p−1 φ k1,2 ≤

2n Ck (K(δx + y) − K(y)) U p−1 φk n+2

≤

2n ≤ CδkφkX Ck (K(δx + y) − K(y)) U p−1 k n2 kφk n−2

(3.15)

while from (2.7) and (2.17) it follows that
ki∗ (K(δx + y) − K(y)) U p−1 φ ks ≤

ns Ck (K(δx + y) − K(y)) U p−1 φk n+2s

≤

Ck (K(δx + y) − K(y)) U p−1 k n2 kφks ≤ CδkφkX .

(3.16)

From (3.15) and (3.16) then
A5 = ki∗ (K(δx + y) − K(y)) U p−1 φ kX ≤ CδkφkX .

(3.17)

Putting together (3.7), (3.8), (3.11), (3.14) and (3.17) we finally get  kTδ,y (φ)kX

≤ ≤

C (kφkpX +  + δ + kφkX + δkφkX )   γ γ(p −1) 1−γ C ( + δ) ( + δ)  + ( + δ) ++δ ≤ ( + δ)

γ

(3.18)

if  and δ are small enough. Step B Here we want to show that γ

γ

 Tδ,y : {φ ∈ X : kφkX ≤ ( + δ) } → {φ ∈ X : kφkX ≤ ( + δ) }

is a contraction map.

9

Arguing as in the previous step we have   kTδ,y (φ1 ) − Tδ,y (φ2 )kX


p 
p − K(y)U p−1 (φ1 − φ2 ) kX − (U + φ2 )+ ≤ Cki∗ (K(δx + y) (U + φ1 )+
p
p  − (U + φ2 )+ ≤ C i∗ K(δx + y) (U + φ1 )+ 
p −1 (3.19) − p (U + φ2 )+ (φ1 − φ2 ) X    
p −1 + Cki∗ K(δx + y) p (U + φ2 )+ − p U p −1 (φ1 − φ2 ) kX

+ Cki∗ K(δx + y) p U p −1 − pU p−1 (φ1 − φ2 ) kX
+ C ki∗ (K(δx + y) − K(y)) U p−1 (φ1 − φ2 ) kX . Estimating each term in (3.20) as done in the previous step we get   kTδ,y (φ1 ) − Tδ,y (φ2 )kX n o ≤ C kφ1 − φ2 kpX + kφ2 kpX −1 kφ1 − φ2 kX + kφ1 − φ2 kX + δkφ1 − φ2 kX

(choosing  and δ small enough ) ≤ C0 kφ1 − φ2 kX

(3.20)

 for some constant C0 < 1. This show that Tδ,y is a contraction mapping from γ {φ ∈ X : kφkX ≤ ( + δ) } into itself and the claim of the proposition follows.

4

Some useful estimates.

n In Section 3 we proved the existence of φδ,y ∈ D1,2 (IRn ) ∩ Ls (IRn ) with n−2 < γ 2n   s < n−2 such that kφδ,y kX ≤ ( + δ) and U + φδ,y is a solution of the equation 
p
Π U + φδ,y − i∗ K(δx + y) (U + φδ,y )+ = 0. (4.1)

From (4.1) we deduce the existence of real numbers ci (, δ, y) such that uδ,y = U + φδ,y is a solution of uδ,y − i∗ K(δx + y) (uδ,y )+


p


=

n X

ci (, δ, y)ψi

(4.2)

i=0

where the functions ψi are defined in (2.5) and (2.6). In order to find solutions of (2.4) we need to find , δ , y such that ci (, δ , y ) = 0 for i = 0, . . . , n. We start by showing the following Lemma 4.1. Let k , δk , and yk ∈ IRn be sequences such that lim k = 0, lim δk = 0 and lim yk = y. Let uδkk ,yk = U + φδkk ,yk be a solution to (4.2) such that uδkk ,yk → U in D1,2 (IRn ). Then ci (k , δk , yk ) = o(1) where o(1) goes to zero as k → ∞. 10

∀i = 1, . . . , n

(4.3)

Proof Set φk = φδkk ,yk and pk = pk . The proof will be divided into two steps. Step A We first show that c0 (k , δk , yk ) = o(1). Taking the scalar product of (4.2) and ψ0 we have     pk   + U + φk − i∗ K(δk x + yk ) (U + φk ) , ψ0 1,2

n X

=

ci (k , δk , yk ) (ψi , ψ0 )1,2 = c0 (k , δk , yk ) kψ0 k21,2 .

(4.4)

i=0

Now we estimate the left hand side of (4.4). Z Z Z   pk + ∇U · ∇ψ0 dx + ∇φk · ∇ψ0 dx − K(δk x + yk ) (U + φk ) ψ0 dx IRn IRn IRn Z Z n+2 4 U n−2 ψ0 φk dx = K(y) U n−2 ψ0 dx + K(y) IRn IRn Z   pk + − K(δk x + yk ) (U + φk ) ψ0 dx IRn Z h  pk i + =− K(δk x + yk ) (U + φk ) − U pk − pk U pk −1 φk ψ0 dx n Z IR + K(δk x + yk ) [−U pk + U p − k ln U U p ] ψ0 dx IRn Z   + K(δk x + yk ) −pk U pk −1 + pU p−1 φk ψ0 dx n ZIR Z + [K(y) − K(δk x + yk )] U p ψ0 dx + k K(y) U p ln U ψ0 dx IRn IRn Z +k [−K(y) + K(δk x + yk )] U p ln U ψ0 dx IRn Z + [K(y) − K(δk x + yk )] pU p−1 φk ψ0 dx. (4.5) IRn

We now estimate each term in (4.5). Using (2.13) and interpolating we have Z h  pk i + pk pk −1 K(δ x + y ) (U + φ ) − U − p U φ ψ dx k k k k k 0 n IR Z p−k 2n 2n kφ k n+2 ≤ C |φk |p−k |ψ0 |dx ≤ Ckψ0 k n−2 k IRn

p−k 2n kφk k = o(1) ≤ Ckψ0 k n−2 X

since

2n n+2 (p

(4.6)

2n − k ) ∈ (s, n−2 ) for  small enough. Using (2.10) we get

Z

IRn

K(δk x + yk ) [−U pk + U p − k ln U U p ] ψ0 dx

2 2n kψ0 k 2n ≤ C ≤ CkU pk − U p + k ln U U p k n+2 k n−2

11

(4.7)

From (2.12) we have Z   pk −1 p−1 K(δ x + y ) −p U + pU φ ψ dx k k k k 0 n IR

2n kψ0 k 2n ≤ Ck kφk kX . ≤ Ckpk U pk −1 − pU p−1 k n2 kφk k n−2 n−2

(4.8)

Reasoning as in the proof of (2.15) one can see that Z p n [K(y) − K(δk x + yk )] U ψ0 dx IR

≤

2n kψ0 k 2n ≤ C (δk + |yk − y|) k [K(y) − K(δk x + yk )] U p k n+2 (4.9) n−2

and Z k

IRn

[K(y) − K(δk x + yk )] U p ln U ψ0 dx

2n kψ0 k 2n ≤ Ck . ≤ Ck k ln U U p k n+2 n−2

(4.10)

In the same way, from (2.17), we have Z p−1 [K(y) − K(δ x + y )] U φ ψ dx k k k 0 IRn

≤

2n kψ0 k 2n k [K(y) − K(δk x + yk )] U p−1 )k n2 kφk k n−2 n−2

≤ C (δk + |y − yk |) kφk kX .

(4.11)

Finally Z k K(y)

IRn

U ln U ψ0 dx ≤ Ck . p

(4.12)

Using estimates (4.6)-(4.12) we get c0 (k , δk yk ) kψ0 k21,2 ≤ o(1).

(4.13)

which gives the claim. Step B Here we will show that ci (k , δk , yk ) = o(1) for i = 1, . . . , n. Taking the scalar product of (4.2) and ψi for i = 1, . . . , n we have     pk   + U + φk − i∗ K(δk x + yk ) (U + φk ) , ψi 1,2

=

n X

cj (k , δk , yk ) (ψj , ψi )1,2 = ci (k , δk , yk ) kψi k21,2

j=0

12

(4.14)

for i = 1, . . . , n. As before we expand the left hand side of (4.14), Z Z Z   pk + ∇U · ∇ψi dx + ∇φk · ∇ψi dx − K(δk x + yk ) (U + φk ) ψi dx IRn IRn IRn Z h  pk i + = − K(δk x + yk ) (U + φk ) − U pk − pk U pk −1 φk ψi dx n ZIR + K(δk x + yk ) [−U pk + U p − k U p ln U ] ψi dx IRn Z   + K(δk x + yk ) −pk U pk −1 + pU p−1 φk ψi dx n ZIR + [K(y) − K(δk x + yk )] U p ψi dx IRn Z +k [−K(y) + K(δk x + yk )] U p ln U ψi dx IRn Z + [K(y) − K(δk x + yk )] U p−1 φk ψi dx. IRn

Proceeding in the same way as in the previous step we get the claim. Lemma 4.2. Let uδ,y ∈ X be a solution of (4.2) such that uδ,y → U in D1,2 (IRn ). For each compact set G ⊂ IRn there exist C > 0, δ0 > 0, 0 > 0 such that for any  ∈ (0, 0 ), δ ∈ (0, δ0 ) and y ∈ G |uδ,y (x)| ≤ CU (x) in IRn  ∂uδ,y n ∂xi (x) ≤ C|∇U (x)| in IR .

(4.15) (4.16)

Proof The proof will be divided into three steps. Step A Here we will prove that ∀R > 0 there exist C(R) > 0, δ0 > 0, 0 > 0 such that for any y ∈ G,  ∈ (0, 0 ) and δ ∈ (0, δ0 ) holds |uδ,y (x)| ≤ C(R)

in B(0, R).

(4.17)

From (4.2) we have that uδ,y satisfies the equation −∆uδ,y = K(δx + y)



uδ,y

n+2 −
+  n−2

−

n X

ci (, δ, y)∆ψi in IRn .

i=0

By Lemma 4.1 we have n X ci (, δ, y)∆ψi i=0

= o(1)

L∞ (IRn )

where o(1) → 0 as  and δ go to 0. Our assumptions imply that kK(δx + y)



uδ,y

4 −
+  n−2

13

kL n2 (B(0,4R)) ≤ C.

(4.18)

By Lemma 6 of [6], (see also [8]) we derive that kuδ,y kLq (B(0,2R)) ≤ Ckuδ,y k

2n

L n−2 (B(0,4R))

where q =

2n 2 ( n−2 ) . 2

This implies that

kK(δx + y) n−2 > with q 4−(n−2)



uδ,y

4 −
+  n−2

k L

q

n−2 4−(n−2)

≤C (B(0,2R))

n 2.

So, by elliptic regularity (see [4] or [6]), we can derive Z 
2  dx ≤ C. sup |uδ,y | ≤ C 1 + uδ,y

B(0,R)

B(0,2R)

Step B In this step we will prove that there exist R > 0, C > 0, δ0 > 0, 0 > 0 such that ∀y ∈ G, δ ∈ (0, δ0 ),  ∈ (0, 0 ) holds |uδ,y (x)| ≤ CU (x) ∀x ∈ IRn \ B(0, R).

(4.19)

2n

Let u ˆδ,y (x) ∈ L n−2 (IRn ) the Kelvin transform of uδ,y i.e.   1 x   u ˆδ,y (x) = u ∀x ∈ IRn \ {0}. |x|n−2 δ,y |x|2 We want to prove that |ˆ uδ,y (x)| ≤ C ∀x ∈ IRn \ {0} such that |x| ≤ R. It is  standard to see that u ˆδ,y (x) satisfies  
+ 4 1 x −∆ˆ uδ,y (x) = K δ 2 + y ( u ˆδ,y ) n−2 − u ˆδ,y (x) (n−2) |x| |x|   n X 1 x − c (, δ, y)∆ψi in IRn \ {0}. (4.20) n+2 i 2 |x| |x| i=0 We observe that

1 |x|n+2

∆ψi



x |x|2



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

(4.20) converges to zero as δ and  go to zero. Moreover using that u ˆδ,y → U (x) 2n

in L n−2 (IRn ) then   4
n−2 1 −  K δ x + y u ˆ ≤ C. n |x|2 |x|(n−2) δ,y L 2 (B(0,4R)) Proceeding as in the previous step we get kˆ uδ,y kLq (B(0,2R)) ≤ Ckˆ uδ,y k 2

2n

L n−2 (B(0,4R))

≤C

( 2n ) where q = n−2 . This implies that 2   4
n−2 x 1 −  kK δ 2 + y u ˆ k n+1 ≤ Ckˆ uδ,y kLq (B(0,2R)) . δ,y L 2 (B(0,2R)) |x| |x|(n−2) 14

By elliptic regularity it follows kˆ uδ,y (x)kL∞ (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 Green’s representation formula we have   p P n  Z K(δz + y) u (z) − i=0 ci (, δ, y)∆ψi (z) ∂uδ,y δ,y ∂ dz (x) = cn ∂xj ∂xj IRn |x − z|n−2   p P n Z K(δz + y) uδ,y (z) − i=0 ci (, δ, y)∆ψi (z) = cn (2 − n) (xj − zj )dz |x − z|n IRn 1 and ωn is the area of the unit sphere in IRn . Using estimate where cn = n(2−n)ω n (4.15) we have

∂uδ,y ∂xj

 (x) ≤ C

Z

≤C

 n−2 2 p

IRn

 Z

1 (1+|z|2 )

1 (1+|z|2 )

 n−2 2 p

+

Pn

i=0

|ci (, δ, y)| pU p−1 (z) |ψi (z)|

|x − z|n−1   n2 +2
1 + (1+|z| 1 + |z| + |z|2 2)

dz

dz. (4.21) |x − z|n−1 n o Now we let IRn = A1 ∪ A2 where A1 = z ∈ IRn : |x − z| > |x| and A2 = 2 o n |x| n z ∈ IR : |x − z| ≤ 2 . We can now split (4.21) into two integrals and estimate these two integrals IRn

 Z A1

1 (1+|z|2 )

 n−2 2 p

+



1 (1+|z|2 )

 n2 +2

1 + |z| + |z|2




dz |x − z|n−1  n−2  n2 +2  Z  2 p
1 1 1−n ≤ C|x| + 1 + |z| + |z|2 dz 2) 2) (1 + |z| (1 + |z| A1 ≤ C|x|1−n

(4.22)

15

and  Z

1 (1+|z|2 )

 n−2 2 p

+



1 (1+|z|2 )

 n2 +2

1 + |z| + |z|2




dz |x − z|n−1   ≤C |x − z|1−n |z|−n−2+(n−2) + |z|−n−4 (1 + |z| + |z|2 ) dz. A2 Z 
 ≤ C |x|−n−2+(n−2) + |x|−n−4 1 + |x| + |x|2 |x − z|1−n dz

A2

Z

|x−z|≤

Z


≤ C |x|−n−2+(n−2) + |x|−n−4 1 + |x| + |x|2 0 
 −n−1+(n−2) −n−3 2 ≤ C |x| + |x| 1 + |x| + |x| .

|x| 2

|x| 2

ρ1−n ρn−1 dρ (4.23)

Using (4.22) and (4.23) we get estimate (4.16). Note that the same can be done for the second derivatives. Lemma 4.3. Let uδ,y ∈ X be a solution of (4.2) such that uδ,y → U in D1,2 (IRn ). Then
p
p

p 2 2n ≤ C . k uδ,y (4.24) − uδ,y +  ln uδ,y uδ,y k n+2 Proof

By mean value theorem we get ∀x ∈ IRn


p
p

p
2 
p−θx  2 uδ,y (x) − uδ,y (x) +  ln uδ,y (x) uδ,y (x) = ln uδ,y (x) uδ,y (x) 2 for some θx ∈ [0, 1]. Using estimate (4.15) it follows that
2 
p−θx  ln uδ,y (x) uδ,y (x) ∈ Lt (IRn ) t>

n n+2 .

The claim is proved.

Lemma 4.4. Let uδ,y be a solution of (4.2) such that uδ,y → U in D1,2 (IRn ) and K(x) be a bounded function, then Z ∂uδ,y ∇uδ,y (x) · ∇ (x)dx = 0 (4.25) n ∂xi ZIR
p+1  ∂  K(δx + y) uδ,y (x) dx = 0 (4.26) IRn ∂xi Proof

From the Divergence Theorem and estimate (4.16) we get Z ∂uδ,y ∇uδ,y (x) · ∇ (x)dx ∂xi B(0,R)  !  Z Z ∇u (x) 2 ∇u (x) 2 ∂ δ,y δ,y = dx = νi dσ 2 2 B(0,R) ∂xi ∂B(0,R)  2 1 1 ≤ C Rn−1 ≤ C n−1 . Rn−1 R 16

Then by the Lebesgue theorem we have Z ∂uδ,y ∇uδ,y (x) · ∇ (x)dx = 0 ∂xi IRn The estimate (4.26) follows in the same way. Lemma 4.5. Let uδ,y be a solution of (4.2) such that uδ,y → U in D1,2 (IRn ) as  → 0, and let K(x) be a bounded function, then Z  2  (4.27) div x ∇uδ,y (x) dx = 0 n ZIR 
p+1  div K(δx + y)x uδ,y (x) dx = 0 (4.28) IRn Z  
p+1 div K(δx + y)x uδ,y (x) ln uδ,y (x) dx = 0 (4.29) IRn

Proof

5

The proof follows exactly as in Lemma 4.4.

A lower bound on the number of solutions

In this section we want to prove Theorem 1.1. A crucial assumption is to write y as y = y0 + δ 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 hypotheses (1.2) and i)-iv) in a neighborhood of y0 with 1 ≤ αi < n for any i = 1, . . . , n. Assume uδ,y is a solution to (4.2) such that uδ,y → U in D1,2 (IRn ) as  → 0 and δ → 0. Then, for  small enough and y = y0 + δ y˜, it holds   Z
∂uδ,y δ αi +1  ∗  uδ,y − i K(δx + y)f (uδ,y ) , hi (x + y˜)U p+1 (x)dx = ∂xi 1,2 p + 1 IRn
+ o δ αi +1 . (5.1) Proof

We have  
∂uδ,y uδ,y − i∗ K(δx + y)f (uδ,y ) , ∂xi 1,2 Z Z  
+ p ∂uδ,y ∂uδ,y = ∇uδ,y (x) · ∇ (x)dx − K(δx + y) uδ,y (x) (x)dx. ∂xi ∂xi IRn IRn

17

Then by (4.25) and (4.26) we have  
∂uδ,y uδ,y − i∗ K(δx + y)f (uδ,y ) , ∂xi 1,2   Z
+ p +1 1 ∂  = − K(δx + y) uδ,y (x) dx p + 1 IRn ∂xi Z 
+ p +1 ∂K δ (δx + y) uδ,y (x) = dx. p + 1 IRn ∂xi

(5.2)

Now we want to estimate the integral in (5.2) using the properties of K. So Z 
+ p +1 ∂K (δx + y) uδ,y (x) dx IRn ∂xi Z 
+ p +1 ∂K (δx + y) uδ,y (x) dx = |δx+y|>r ∂xi Z 
+ p +1 ∂K (δx + y) uδ,y (x) dx = I1 +I2 . (5.3) + |δx+y|≤r ∂xi Using (4.15) and that k∇Kk∞ ≤ C then Z

+∞

|I1 | ≤ C r δ



1 1 + ρ2

2n  n−2 2 ( n−2 )

ρn−1 dρ ≤ Cδ n .

(5.4)

Now we let y = y0 + δ y˜

(5.5)

n

with y˜ ∈ IR . Then using properties i)-iv) of K we have Z 
+ p +1 I2 = (hi (δx + δ y˜) + Ri (δx + δ y˜)) uδ,y (x) dx = A2 + B2 |δx+y|≤r

(5.6) and Z


p +1 uδ,y (x)

αi

Z

hi (δx + δ y˜) dx = δ hi (x + y˜) |δx+y|≤r |δx+y|≤r Z = δ αi hi (x + y˜)U p+1 (x)dx + o(δ αi )




+ p +1 uδ,y (x) dx (5.7)

IRn

since |hi (x + y˜)| ≤ C (|x|αi + 1) and αi < n, while Z 
+ p +1 |B2 | = Ri (δx + δ y˜) uδ,y (x) dx |δx+y|≤r Z 
+ p +1 βi ≤ Cδ |x + y˜|βi uδ,y (x) dx |δx+y|≤r

≤ Cδ βi + Cδ n .

(5.8)

18

Then by (5.6), (5.7) and (5.8) we have Z p+1 I2 = δ αi hi (x + y˜) (U (x)) dx + o(δ αi ).

(5.9)

IRn

Finally we get   Z
∂uδ,y δ αi +1 p+1  ∗  uδ,y − i K(δx + y)f (uδ,y ) , = hi (x + y˜) (U (x)) dx ∂xi 1,2 p + 1 IRn + o(δ αi +1 ).

(5.10)

which gives the claim. Lemma 5.2. Let y0 be a critical point of K(x). Assume K(x) satisfies hypotheses (1.2) and i)-iv) in a neighborhood of y0 with 1 ≤ αi < n − 1 for any i = 1, . . . , n. Assume uδ,y is a solution to (4.2) such that uδ,y → U in D1,2 (IRn ) as  → 0 and δ → 0. Then, for  small enough and y = y0 + δ y˜, it holds  
n−2   ∗   uδ,y − i K(δx + y)f (uδ,y ) , x · ∇uδ,y + uδ,y 2 1,2 Z 2 (n − 2) p+1 K(y0 ) (U (x)) dx + o() = n 4n IR Z n X
δ αi +1 p+1 + hi (x + y˜)xi (U (x)) dx + o δ αi +1 (5.11) p + 1 IRn i=1 Proof We have  
n−2  uδ,y (5.12) uδ,y − i∗ K(δx + y)f (uδ,y ) , x · ∇uδ,y + 2 1,2   Z n−2    uδ,y (x) dx = ∇uδ,y (x) · ∇ x · ∇uδ,y (x) + 2 IRn   Z 
+  p n−2    − K(δx + y) uδ,y (x) x · ∇uδ,y (x) + uδ,y (x) dx 2 IRn   Z Z
1 n−2    2 = div x|∇uδ,y (x)| dx − K(δx + y) x · ∇uδ,y (x) + uδ,y (x) 2 IRn 2 IRn h      i
+ p
+ p
+ p × uδ,y (x) − uδ,y (x) +  uδ,y (x) ln uδ,y (x) dx

19

− − + + + −

Z 
+ p+1 n−2 K(δx + y) uδ,y (x) dx 2 IRn   Z 
+ p+1 1  div K(δx + y)x uδ,y (x) dx p + 1 IRn Z 
+ p+1 1 div (K(δx + y)x) uδ,y (x) dx p + 1 IRn Z 
+ p+1 n−2  K(δx + y) uδ,y (x) ln uδ,y (x)dx n 2    Z IR 
+ p+1  1 div K(δx + y)x uδ,y (x) dx ln uδ,y (x) − p + 1 IRn p+1   Z 
+ p+1  1 div (K(δx + y)x) uδ,y (x) dx. ln uδ,y (x) − p + 1 IRn p+1

Using (4.27), (4.28) and (4.29) we get  
n−2  uδ,y = uδ,y − i∗ K(δx + y)f (uδ,y ) , x · ∇uδ,y + 2 1,2   Z n−2   − K(δx + y) x · ∇uδ,y (x) + uδ,y (x) × 2 IRn h     i
+ p
+ p
+  p × uδ,y (x) − uδ,y (x) +  uδ,y (x) ln uδ,y (x) dx Z 
+ p+1 δ ∇K(δx + y) · x uδ,y (x) + dx p + 1 IRn Z 
+ p+1 n  u (x) dx +  K(δx + y) δ,y (p + 1)2 IRn   Z 
+ p+1 δ 1   −  ∇K(δx + y) · x uδ,y (x) ) ln uδ,y (x) − dx p + 1 IRn p+1 = I1 + I2 + I3 + I4 . (5.13) We can estimate the first integral as follows Z  
+  p  
+  p
+  p I1 ≤ C − uδ,y (x) +  uδ,y (x) ln uδ,y (x) × uδ,y (x) IRn n−2   uδ,y (x) dx (5.14) × x · ∇uδ,y (x) + 2  
+ p  
+ p
+ p n−2   2n 2n kx · ∇u uδ,y k n−2 ≤ k uδ,y − uδ,y +  uδ,y ln uδ,y k n+2 δ,y + 2 ≤ C2 .

20

In order to estimate I2 we split the integral I2 as follows Z 
+ p+1 δ I2 = ∇K(δx + y) · x uδ,y (x) dx p + 1 IRn Z 
+ p+1 δ ∇K(δx + y) · x uδ,y (x) dx = p + 1 |δx+y|≥r Z 
+ p+1 δ ∇K(δx + y) · x uδ,y (x) dx + p + 1 |δx+y|≤r δ (I21 + I22 ) . = p+1

(5.15)

From (4.15) and the boundedness of k∇Kk∞ we have +∞

Z |I21 | ≤ C

ρ r δ

1 n−1 ρ dρ ≤ Cδ n−1 . ρ2n

(5.16)

To estimate the term I22 we use our crucial assumption y = δ y˜ + y0 . Then Z ∇K(δx + y) · x

I22 = = + =

|δx+y|≤r Z n X αi

δ

uδ,y (x)

hi (x + y˜)xi




+ p+1

uδ,y (x)

dx


+ p+1

dx

|δx+y|≤r

i=1 n Z X

Ri (δx + δ y˜)xi

i=1

|δx+y|≤r

n X

αi

δ



Z




+ p+1 uδ,y (x) dx

hi (x + y˜)xi U p+1 (x)dx + o(δ αi ) + A22

(5.17)

IRn

i=1

since αi < n − 1 and |hi (x + y˜)| ≤ C(|x|αi + 1), where n X A22 ≤ C

Z |δx+y|≤r

i=1

≤ Cδ βi

Z


p+1 Ri (δx + δ y˜)xi uδ,y (x) + dx x + y˜ βi |x| (U (x))p+1 dx

|δx+y|≤r

≤ Cδ βi + Cδ n−1 .

(5.18)

Putting together (5.15)-(5.18) we get I2 =

Z n X δ αi +1 i=1

p+1

p+1

hi (x + y˜)xi (U (x))

IRn

21


dx + o δ αi +1 + O (δ n ) .

(5.19)

Concerning I3 we have Z 
+ p+1 n  I3 =  K(δx + y) u (x) dx δ,y (p + 1)2 IRn Z n K(y)U p+1 (x)dx + o() = (p + 1)2 IRn and since y = y0 + δ y˜ I3 = 

n K(y0 ) (p + 1)2

Z (U (x))

p+1

dx + o().

(5.20)

IRn

Finally Z

p+1

|I4 | ≤ Cδ

|x| (U (x)) IRn

ln U (x) dx ≤ Cδ.

(5.21)

Putting together (5.13), (5.15), (5.19), (5.20) and (5.21) we get (5.11). Definition 5.3. Let L ∈ C (IRn , IRn ) be a vector field. We say that y˜ is a stable zero for L if a) L(˜ y ) = 0, b) y˜ is isolated, c) if Ln is a sequence of vector fields such that kLn − LkC(B(˜y,r)) → 0 for some r > 0 then there exists y˜n ∈ B(˜ y , r) such that Ln (˜ yn ) = 0 and y˜n → y˜. Note that if there exists a neighborhood B(˜ y , r) of y˜ such that deg (L, B(˜ y , r), 0) 6= 0

(5.22)

then y˜ is a stable zero for L. Proof of Theorem 1.1 We want to prove that any value y˜0 ∈ Zy0 gives rise to a single peak solution of (1.1) which blows up at y0 . We note that we have already shown, in Section 3 that if  and δ are sufficiently small for each y ∈ IRn there exists a function φδ,y ∈ W such that U + φδ,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 δα+1 = d1  and y = y0 + δ y˜ , then the function uδ ,y = U + φδ ,y satisfies equation (4.2) with coefficients ci (, δ , y ) = 0 for all i = 0, . . . , n. To this end we let    
+  p  1 n−2   ∗   0 G (d, y˜) = uδ,y − i K(δx + y) uδ,y , x · ∇uδ,y + uδ,y  2 1,2        p
∂u  1 + δ,y Gi (d, y˜) = αi +1 uδ,y − i∗ K(δx + y) uδ,y , . δ ∂xi 1,2

22

∂u

Taking the scalar product of equation (4.2) and the function ∂xδ,y for i = i 1  α+1 1, . . . , n or x · ∇uδ,y + n−2 u , and letting δ =  and y = y + δ y ˜ , we have  0   δ,y 2 d G0 (d, y˜)

=

Gi (d, y˜)

=


1 c0 (, d, y˜) kψ0 k21,2 + o(1) 
1 ci (, d, y˜) kψi k21,2 + o(1) δ αi +1

Finding a solution of (2.2) means finding d and y˜ such that G0 (d , y˜ )

=

0

Gi (d , y˜ )

=

0

for i = 1, . . . , n and for each  in (0, 0 ) for some 0 > 0. Using (5.1) and (5.11) we have that   Z 1 (n − 2)2 0 p+1 G (d, y˜) = K(y0 ) U (x)dx + ψ(˜ y ) + o(1) (5.23) d d 4n IRn Z 1 Gi (d, y˜) = hi (x + y˜)U p+1 (x)dx + o(1) p + 1 IRn R P 1 where ψ(˜ y ) = p+1 ˜)xj U p+1 (x)dx. We observe that the αj =α IRn hj (x + y leading term of Gi (d, y˜) is independent of d. So, since y˜0 is a stable zero to Ly0 , there exists y˜ → y˜0 independent of d 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˜ ) =

1 (Ad + ψ(˜ y )) + o(1), d

R 2 y) p+1 where A = (n−2) (x)dx. Hence there exists d close to − ψ(˜ 4n K(y0 ) IRn U A 0 such that G (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 y˜01 and y˜02 be two different points of Zy0 , and let
α+1 di and y˜i be the parameter associated with this two points. Let δi = d1i ,  i i i   y = y0 + δ y˜ and uδ1 ,y1 uδ2 ,y2 be the two solutions of (2.2) generated by the     parameters. For i = 1, 2 we let  
− 2 x − yi ui (x) = δi p−−1 uδi ,yi δi the corresponding solutions of (1.1). Then δ1 δ2

2
p−−1

2
p−−1

u1 (y1 ) = uδ1 ,y1 (0) → U (0) u2 (y1 ) = uδ2 ,y2 ( 23

δ1 1 y˜ − y˜2 ) → U (c˜ y01 − y˜02 ) δ2 

δ1 2 δ →0 

where c = lim

= lim

→0



d2 d1

1  α+1

=



d20 d10

1  α+1

. We have the following alternative:

1)c = 1. Then u1 (y1 ) U (0) = 6= 1 →0 u2 (y1 ) U (˜ y01 − y˜02 ) lim

and the claim follows. 2) c 6= 1. There is no loss of generality in assuming c < 1. In this case we have 2  1 − p−−1 uδ1 ,y1 (0) δ u (y 1 )   = lim lim 1 1 δ1 1  →0 δ2 →0 u2 (y ) u 2 2 ( 2 y˜ − y˜2 ) δ ,y δ

2 2   p−1   p−1
n−2 1 U (0) 1 = 1 + |c˜ y01 − y˜02 |2 2 > 1. = 1 2 c U (c˜ y0 − y˜0 ) c

This proves that u1 6= u2 .

6

A useful estimate

In this section we want to prove that if u is a solution of (1.1) that blows-up and concentrate at y0 and if y stands for its peak then y0 − y 1

→ y˜

 α+1 where y˜ satisfies Ly0 (˜ 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 that goes to zero and un = un the corresponding single-peaked solutions of (1.1). If yn = yn denotes the peak of un and if yn → y0 as  → 0, then there exists a positive constant C > 0 such that y − y 0 n (6.1) ≤C 1 nα+1 where α is defined in (1.9) and depends only on the shape of K(x) in a neighborhood of y0 . Proof We suppose by contradiction that there exists a sequence n such that y − y 0 n (6.2) → ∞. 1 nα+1  1  p−2 −1  1  n α+1 α+1 We let vn (x) = n un n x + yn . Then vn satisfies the equation 1

−∆vn (x) = K(nα+1 x + yn ) (vn (x)) 24

p−n

in IRn .

(6.3)

Here we note that vn satisfies the hypotheses of Lemma 4.3 and hence it verifies vn ≤ CU. Moreover vn * v weakly in D1,2 (IRn ) where v is a solution of −∆v = K(y0 )v p and hence v =

1 K(y0 )

n−2 4

U (x). Multiplying (6.3) by 

Z 0

∇vn · ∇

=

in IRn

IRn

∂vn ∂xi



Z

∂vn ∂xi

and integrating we have

1

K(nα+1 x + yn ) (vn (x))

dx − n

p−n

IR Z 1 1 ∂ p− +1 = − K(nα+1 x + yn ) (vn (x)) n dx p − n + 1 IRn ∂xi ( using vn ≤ CU ) 1 Z 1 ∂K α+1 nα+1 p− +1 = (n x + yn ) (vn (x)) n dx. p − n + 1 IRn ∂xi

∂vn dx ∂xi

(6.4) (6.5)

Arguing as in the proof of the Lemma 5.1, using properties i)-iv) of K(x) we find Z 1 ∂K α+1 p− +1 0 = (n x + yn ) (vn (x)) n dx n ∂x i ZIR 1 ∂K α+1 p− +1 = (n x + yn ) (vn (x)) n dx 1 α+1 ∂x i | x+yn |>r Z n 1 p− +1 hi (nα+1 x + yn − y0 ) (vn (x)) n dx + 1 α+1 x+yn |≤r | Z n 1 p− +1 + Ri (nα+1 x + yn − y0 ) (vn (x)) n dx 1 |nα+1 x+yn |≤r

= I1 + I2 + I3 .

(6.6)

As in Lemma 5.1 we get 

1 α+1

|I1 | ≤ C n

n .

(6.7)

Moreover using (6.2) we have (up to a subsequence)   1 Z α+1  y − y n n 0  (vn (x))p−n +1 dx I2 = |yn − y0 |αi hi  x+ 1 α+1 |y − y | |y − y | n 0 n 0 |n x+yn |≤r Z p+1 = |yn − y0 |αi hi (z) (U (x)) dx + o(|yn − y0 |αi ) (6.8) IRn

where z ∈ ∂B(0, 1) is given by (up to a subsequence) z = lim

n→∞

yn − y0 . |yn − y0 |

25

Reasoning as before we get  1 n |I3 | ≤ C|yn − y0 |βi + C nα+1

(6.9)

n

and nα+1 = o (|yn − y0 |αi ) by (6.2). We have shown so far that Z p+1 0 = |yn − y0 |αi hi (z) (U (x)) dx + o (|yn − y0 |αi ) IRn

and this implies hi (z) = 0 which is not possible from the properties iv) on K. 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 ≤ αi < n − 1 for all i = 1, . . . , n. Let n be a sequence that goes to zero and un = un the corresponding single-peaked solutions of (1.1). If yn = yn denotes the peak of un and if yn → y0 as  → 0, then up to a subsequence, we have  1  1 yn = y0 − nα+1 y˜ + o nα+1 , (6.10) Ly0 (˜ y) = 0 and X αj

1 p + 1 =α

Z

hj (x + y˜)xj U p+1 (x)dx = −

IRn

(6.11)

n−2 K(y0 ) 2

Z

U p+1 (x)dx (6.12)

IRn

where α is defined in (1.9) and depends only on the shape of K(x) in a neighborhood of y0 . Proof of Proposition 6.2 sition 6.1 we let

Let us prove (6.10) and (6.11). As in proof of Propo

vn (x) =

 p−2 −1

1

nα+1

n

 1  un nα+1 x + yn .

Since un is a single peaked solution of (1.1) then vn → U (x) in D1,2 (IRn ). n Multiplying (2.2) by ∂v ∂xi we have (see also (5.2)) 1

nα+1 0= p − n + 1

Z IRn

1 ∂K α+1 p− +1 (n x + yn ) (vn (x)) n dx ∂xi

1 α+1

n (I1,n + I2,n + I3,n ) p − n + 1

=

(6.13)

where I1,n =

 1 αi Z = nα+1

1

|nα+1 x+yn −y0 |≤r

αi α+1

Z

n

hi (x + y˜) (U (x))

hi

p+1

x+

yn − y0 1 α+1

! (vn (x))

p−n +1

dx

n

αi

dx + o(nα+1 )

IRn

26

(6.14)

and up to a subsequence y˜ = lim

yn − y0 1

n→∞

.

(6.15)

nα+1 Arguing as in the proof of Lemma 5.1 we get βi

n

I2,n = O(nα+1 ) + O(nα+1 ) and

(6.16)

n

I3,n = O(nα+1 ).

(6.17)

Hence from (6.13)-(6.17) we get Z hi (x + y˜)U p+1 (x)dx = 0 IRn

and then Ly0 (˜ y ) = 0. Let us prove (6.12). Taking the scalar product of (4.2) with x · ∇vn + n−2 2 vn we get from (5.11)

0=

αj +1 Z n X nα+1

j=1

p+1

hj (x + y˜)xj (U (x))

p+1

n−2 K(y0 )n 2

dx +

IRn

+o(n )

Z (U (x))

p+1

dx

IRn

(6.18)

and this proves the claim. Proof of Theorem 1.3

7

It follows from Proposition 6.2.

An exact multiplicity result

Proof of Theorem 1.2 By contradiction let us suppose that #{ single peak solutions of (1.1) concentrating at y0 } > #Zy0 . Since #Zy0 < ∞ from Proposition 6.2 there exist y˜ ∈ Zy0 , a sequence n → 0 and two distinct solutions u1n and u2n of (1.1) such that if yn1 and yn2 denote their peaks we have lim

n→∞

yn1 − y0 1 α+1

n

= lim

n→∞

yn2 − y0 1

= y˜.

nα+1 2

1

Letting δn = nα+1 the functions vni (x) = δnp−n −1 uin (δn x + y0 ) satisfy the equation
p−n −∆vni (x) = K(δn x + y0 ) vni (x) in IRn (7.1) and vni (x) → U (x − y˜) uniformly on IRn for i = 1, 2. By Lemma 4.2 we have that 0 < vni (x) ≤ CU (x − y˜) in IRn . Since u1n (x) 6= u2n (x) we can consider the function v 1 (x) − vn2 (x) in IRn . (7.2) vn (x) = 1 n kvn (x) − vn2 (x)k1,2 27

The function vn satisfies −∆vn (x) = K(δn x + y0 )cn (x)vn (x) where Z

1



cn (x) = (p − n )

in IRn

(7.3)

p−1−n tvn1 (x) + (1 − t)vn2 (x) dt.

0

Here we observe that from (4.15) we have 0 ≤ cn (x) ≤

C

2,

(1 + |x − y˜|2 )

∀x ∈ IRn .

(7.4)

Up to a subsequence we can assume that vn * v weakly in D1,2 (IRn ) and almost everywhere in IRn where v satisfies −∆v(x) = K(y0 )pU p−1 (x − y˜)v(x)

in IRn .

(7.5)

Then there exist real numbers ai such that v(x) =

n X

ai ψi (x − y˜)

(7.6)

i=0

with ψi as defined in (2.5) and (2.6). Moreover reasoning exactly as in the proof of Lemma 4.2 one can see that |vn (x)| ≤

C 1 + |x − y˜|n−2

in IRn .

(7.7)

∂v i

Now we multiply the equation (7.1) for ∂xnj and we integrating over IRn . Using (4.26) we get Z
p−n +1 1 ∂K 0= δn (δn x + y0 ) vni (x) dx. (7.8) n p − n + 1 ∂x j IR Subtracting equation (7.8) evaluated for i = 1, 2 we have Z h
p−n +1
p−n +1 i ∂K (δn x + y0 ) vn1 (x) 0 = − vn2 (x) dx IRn ∂xj Z ∂K (δn x + y0 )˜ cn (x)vn (x)dx 0 = n ∂x j IR where Z c˜n (x) = (p + 1 − n )

1

 1 p−n tvn (x) + (1 − t)vn2 (x) dt

(7.9)

0

and by (4.15) 0 < c˜n (x) ≤ CU p (x − y˜). 28

(7.10)

Using the properties i)-iv) of K(x) in a neighborhood of y0 we get Z 0

= |δn x+y0 |>r

∂K (δn x + y0 )˜ cn (x)vn (x)dx ∂xj

Z +

hj (δn x)˜ cn (x)vn (x)dx |δn x+y0 |≤r

Z +

Rj (δn x)˜ cn (x)vn (x)dx = I1 + I2 + I3 .

(7.11)

|δn x+y0 |≤r

By (7.7), (7.10) and (1.2) we get I1 ≤ Cδ n . Moreover by (7.10) and αj < n, we have Z I2 = δnαj hj (x)˜ cn (x)vn (x)dx |δn x+y0 |≤r Z = δnαj hj (x)U p (x − y˜)v(x)dx + o (δnαj )

(7.12)

(7.13)

IRn

and I3 ≤ Cδnβj

Z

|x|βj c˜n (x)vn (x)dx ≤ Cδnβj + Cδnn .

(7.14)

|δn x+y0 |≤r

From (7.11)-(7.14) we have Z 0= hj (x)U p (x − y˜)v(x)dx (7.15) n IR   Z n−2 = a0 hj (x)U p (x − y˜) (x − y˜) · ∇U (x − y˜) + U (x − y˜) dx 2 IRn Z n X ∂U + ai hj (x)U p (x − y˜) (x − y˜)dx n ∂x i IR i=1

29

for j = 1, . . . , n. On the other hand we have that   Z n−2 hj (x)U p (x − y˜) (x − y˜) · ∇U (x − y˜) + U (x − y˜) dx 2 IRn (since y˜ is a zero of Ly0 ) n Z X ∂U = (x − y˜)dx hj (x)U p (x − y˜)(xi − (˜ y )i ) n ∂xi i=1 IR Z X n 1 ∂hj (x) =− (xi − y˜i )U p+1 (x − y˜)dx p + 1 IRn i=1 ∂xi

(7.16)

(using the Eulero Theorem for homogenous functions) Z Z αj y˜i ∂hj (x) p+1 p+1 U (x − y˜)dx hj (x)U (x − y˜)dx + =− p + 1 IRn p + 1 IRn ∂xi Z Z y˜i ∂hj (x) p+1 ∂U = U (x − y˜)dx = −˜ yi hj (x)U p (x − y˜) (x − y˜)dx. n p + 1 IRn ∂xi ∂x i IR Hence (7.15) becomes 0=

n X i=1

Z (ai − a0 y˜i )

hj (x)U p (x − y˜)

IRn

∂U (x − y˜)dx ∂xi

(7.17)

Recalling the definition of Ly0 we have that  Z ∂U Jac (Ly0 (˜ (x − y˜)dx y )) = (p + 1) hj (x)U p (x − y˜) ∂xi IRn i,j=1,...,n Since 0 is a regular value for Ly0 , then Ly0 (˜ y ) = 0 implies that the linear system (7.17) admits only the solution ai = a0 y˜i

for i = 1, . . . , n.

(7.18)

Next aim is to show that a0 = 0. To do this let us multiply equation (7.1) for x · ∇vni and integrating over IRn we get Z 
p−n +1 n n−2 − K(δn x + y0 ) vni (x) dx 0= 2 p − n + 1 IRn Z
p−n +1 δn − ∇K(δn x + y0 ) · x vni (x) dx. (7.19) p − n + 1 IRn Subtracting (7.19) evaluated for i = 1, 2, we get  Z n−2 n 0= − K(δn x + y0 )˜ cn (x)vn (x)dx 2 p − n + 1 IRn Z δn − ∇K(δn x + y0 ) · x˜ cn (x)vn (x)dx = I1 + I2 p − n + 1 IRn 30

where c˜n are as defined in (7.11). Let us estimate I1 using (7.7) and (7.10)  Z n−2 n I1 = − K(δn x + y0 )˜ cn (x)vn (x)dx 2 p − n + 1 IRn   Z (n − 2)2 = − n + o(n ) K(y0 ) U p (x − y˜)v(x)dx n 4n IR = o(n ) R because a straightforward computation gives that IRn U p (x − y˜)v(x)dx = 0. Concerning I2 we get Z δn ∇K(δn x + y0 ) · x˜ cn (x)vn (x)dx I2 = p − n + 1 IRn "Z ∂K δn (δn x + y0 )xj c˜n (x)vn (x)dx = p − n + 1 |δn x+y0 |>r ∂xj Z + hj (δn x)xj c˜n (x)vn (x)dx |δn x+y0 |≤r

#

Z +

Rj (δn x)xj c˜n (x)vn (x)dx = J1 + J2 + J3 . |δn x+y0 |≤r

Using (7.7) and (7.10) one gets J1 = O δ n−1




and J3 = O δ βj + δ n−1




Finally δn p − n + 1 X δ α+1

Z hj (δn x)xj c˜n (x)vn (x)dx

J2 = =

|δn x+y0 |≤r

n

αj =α

p − n + 1

Z IRn

hj (x)xj U p (x − y˜)v(x)dx + o δnα+1

since αj < n − 1. Since  = δnα+1 we derive that (7.20) becomes X Z hj (x)xj U p (x − y˜)v(x)dx = 0 αj =α

IRn

31




(7.20)

and then (recalling that ai = a0 y˜i ) X Z hj (x)xj U p (x − y˜)v(x)dx αj =α

= a0

IRn

X Z αj =α

! ∂U n−2 xi (x − y˜) + U (x − y˜) dx ∂xi 2 i=1

n X

p

hj (x)xj U (x − y˜)

IRn

Z n a0 X X ∂ =− (hj (x)xi xj ) U p+1 (x − y˜)dx p + 1 i=1 α =α IRn ∂xi j Z (n − 2)a0 X hj (x)xj U p+1 (x − y˜)dx + n 2 I R αj =α   n X Z X ∂hj (x) a0 i hj (x)xj + hj (x)δj xi + =− xi xj U p+1 (x − y˜)dx p + 1 i=1 α =α IRn ∂xi j Z (n − 2)a0 X hj (x)xj U p+1 (x − y˜)dx + n 2 I R α =α j

(using the Eulero theorem for homogenous functions)  Z X n−2 n+1 α = a0 − − hj (x)xj U p+1 (x − y˜)dx. 2 p+1 p+1 IRn α =α j

Since



n−2 n+1 α − − 2 p+1 p+1

 =−

α+1 < 0, p+1

using (1.11) we derive that (7.20) becomes Z  α n−2 n+1 0 = a0 − − hj (x)xj U p+1 (x − y˜)dx 6= 0 n 2 p+1 p+1 IR a contradiction if a0 6= 0. Since a0 = 0, from (7.18) we get ai = 0

for i = 1, . . . , n.

Then, by (7.6), v(x) ≡ 0. In order to end the proof we have to show that v(x) ≡ 0 is impossible. Multiplying (7.3) by vn and integrating we have Z Z ∇vn 2 = 1 = K(δx + y0 )cn (x)vn2 (x)dx n n IR IR ! Z Z ≤

cn (x)vn2 (x)dx +

C

cn (x)vn2 (x)dx

|x|≤R

(7.21)

|x|>R

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

cn (x)vn2 (x)dx ≤

Z

! n2 n 2

cn (x) dx

vn (x) |x|>R

|x|>R

32

! n−2 n

Z

2n n−2

dx

≤

1 2C

if R is big enough. Moreover since vn * 0 in D1,2 (IRn ) and vn → 0 uniformly on compact set of IRn then Z 1 cn (x)vn2 (x)dx ≤ 4C |x|≤R if  is small enough. Then a contradiction arise from (7.21) and we get the claim.

8

Examples and final remarks

In this section we consider some particular cases of function K(x) where our results apply. We always assume that y0 = 0. Let us start by considering the case of the nondegenerate critical point. Proposition 8.1. Let us assume that 0 is a nondegenerate critical point of K satisfying ∆K(0) < 0. Then there exists exactly one solution to (1.1) blowing up at 0. Proof

By Taylor’s formula we have that K(x) = K(0) +

n 1 X ∂2K (0)xj xk + R(x) 2 ∂xj ∂xk

(8.1)

j,k=1

in a neighborhood of 0. So we have that n

hi (x) =

1 X ∂2K (0)xj 2 j=1 ∂xj ∂xi

Hence we derive that Z L0 (y) = n

=

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

IRn 2

1X ∂ K (0) 2 j=1 ∂xj ∂xi

Z

(xi + yi ) U p+1 (x)dx

IRn

Z n X ∂2K = (0)yi U p+1 (x)dx. n ∂x ∂x j i I R j=1 Since the matrix

∂2K ∂xj ∂xi (0)

is invertible we get that the linear system L0 (y) = 0

admits only the solution y = 0. So Z0 = {0}.

(8.2)

Since αi = 1 for any i = 1, .., n, by Theorem 1.1 and Theorem 1.2 we get the existence and uniqueness of the solution blowing up at zero. 33

Remark 8.2. The same computations of the previous proposition allow us to handle the following case, K(x) = 1 +

n X

s

cj xj j

j=1

where cj ∈ IR and sj ≥ 2 are even positive integers. We again have that Z0 = {0} and under the assumption Z X s cj sj xj j U p+1 (x)dx < 0 IRn

αj =α

we get the existence and uniqueness of the solution blowing up at zero. Proposition 8.3. Let us assume that K(x) = 1 +

n X

s

cj xj j

j=1

where cj ∈ IR and sj are positive integers. Then if at least one of the integers sj is odd there is no solution to (1.1) blowing up at 0. Proof Since (1.13) is verified by Theorem (1.3) the claim follows. In the final proposition we exhibit a function K(x) such that (1.1) admits exactly two single-peak solutions concentrating at the origin. Proposition 8.4. Let us assume that K(x) : IR3 → IR is given by K(x) = K(x1 , x2 , x3 ) = 1 + x31 − x1 x22 − x23

for x ∈ B(0, 1).

Then there exists exactly two single-peak solutions concentrating at the origin. y) = 0 Proof Computing the functions hi (x) for i = 1, .., 3 we derive that L0 (˜ at the points s R ! x2 U p+1 (x)dx IR Rn 1 y˜1 = 0, − 2 ,0 U p+1 (x)dx IRn and

s R y˜2 =

0,

! x21 U p+1 (x)dx 2 R ,0 . U p+1 (x)dx IRn IRn

Note that in this case α1 = α2 = 2 and α = α3 = 1. Since Z X Z p+1 hj (x + y˜)xj U (x) = −2 x23 U p+1 (x)dx < 0, αj =α

IRn

IRn

Theorem 1.11 applies and then (1.1) admits at least two single peak solutions concentrating at the origin. In order to show that we have exactly two solutions 34

we apply Theorem 1.2. This leads to check that y˜1 and y˜2 are regular values for L0 . Indeed det JacL0 (˜ y1 ) = det JacL0 (˜ y2 ) = Z  Z 2 −16 x21 U p+1 (x)dx U p+1 (x)dx 6= 0, IRn

(8.3)

IRn

and this gives the claim. Remark 8.5. The same computation of the previous proposition can be performed in any dimension n ≥ 3. With a little of work it is also possible construct examples of function K(x) such that (1.1) admits exactly m solutions, for any integer m ≥ 1.

References [1] Chen, Chiun-Chuan; Lin, Chang-Shou, Estimate of the conformal scalar curvature equation via the method of moving planes. II, J. Differential Geom.49 (1998), no. 1, 115–178. [2] Floer, Andreas; Weinstein, Alan, Nonspreading wave packets for the cubic Schr¨ odinger equation with a bounded potential. J. Funct. Anal.69 (1986), no. 3, 397–408. [3] Gidas, Basilis; Spruck, Joel, Global and local behavior of positive solutions of nonlinear elliptic equations, Comm. Pure Appl. Math,34 (1981), no. 4, 525–598. 261–280. [4] Gilbarg, David; Trudinger, Neil S., Elliptic Partial Differential Equations of Second Order, Springer-Verlag, Berlin, (1983). [5] Grossi, Massimo, On the number of single-peak solutions of the nonlinear Schr¨ odinger equation, Ann. Inst. H. Poincar´e Anal. Non Lin´eaire,19 (2002), no. 3, 261–280. [6] Han, Zheng-Chao , Asymptotic approach to singular solutions for nonlinear elliptic equations involving critical Sobolev exponent,Ann. Inst. H. Poincar´e Anal. Non Lin´eaire 8 (1991), no. 2, 159–174. [7] Micheletti, Anna Maria; Pistoia, Angela, Existence of blowingup solutions for a slightly subcritical or a slightly supercritical non-linear elliptic equation on Rn ,Nonlinear Anal. 52 (2003), no. 1, 173–195. [8] Micheletti, Anna Maria; Pistoia, Angela; Visetti, Daniela, On the number of blowing-up solutions to a nonlinear elliptic equation with critical growth, Rocky Mountain J. Math. 37 (2007), no. 1, 291–325.

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[9] Ni, Wei Ming, On the elliptic equation ∆u + K(x)u(n+2)/(n−2) = 0, its generalizations, and applications in geometry, Indiana Univ. Math. J. 31 (1982), no. 4, 493–529. [10] Pohozaev, Stanislav, On the eigenfunctions of the equation ∆u + λf (u) = 0,Dokl. Akad. Nauk SSSR, 165, (1965), 36–39. [11] Yan, Shusen, High-energy solutions for a nonlinear elliptic problem with slightly supercritical exponent, Nonlinear Anal. 38 (1999), no. 4, Ser. A: Theory Methods, 527–546. [12] Wang, Xuefeng; Wei, Juncheng, On the equation ∆u + 2 K(x)u(n+2)/(n−2)± = 0 in Rn , Rend. Circ. Mat. Palermo 44 (1995), no. 3, 365–400.

36