Bound states of nonlinear Schrodinger equations with potentials vanishing at infinity

Bound states of Nonlinear Schr¨odinger Equations with Potentials Vanishing at Infinity A. Ambrosetti 1 , A. Malchiodi 1 , D. Ruiz

3

Key words: Nonlinear Schr¨ odinger Equations, Perturbation methods, Concentrations. Abstract We considered the nonlinear Schr¨ odinger equation (1) in the case that V (x) ∼ n+2 (1 + |x|)−α and K is bounded and positive. If 0 ≤ α ≤ 2 and 1 < p < n−2 it is shown that (1) has, for all ε  1, a bound state concentrating at some point x0 , provided x0 is a stable stationary point of the auxiliary potential Q defined in (Q).

1

Introduction

We consider a class of nonlinear Schr¨odinger equations with potentials, like (1)

−ε2 ∆u + V (x)u = K(x)up ,

x ∈ Rn ,

u > 0.

The physically relevant solutions of (1) are those with finite energy, namely belonging to the Sobolev space W 1,2 (Rn ). These solutions are called bound states. A great deal of work has been devoted to find bound states of (1), both in the case when ε is arbitrary, as well as when ε ∼ 0. Regarding the former case, critical point theory has been used to prove the existence of bound states uε of (1) under suitable assumptions on V and K. For example, in [13] it is assumed that K ≡ 1, V > 0 and lim|x|→∞ V (x) = +∞, while in [6] the case in which V ≡ 1, K > 0, lim|x|→∞ K(x) = k0 > 0 and that K decays exponentially to k0 , is handled. For other results, we refer to the book [7] which contains also a broad bibliography. In any case, it is worth pointing out that the common assumptions are: (V0)

(K0)

inf V (x) > 0,

x∈Rn

∃ κ > 0 : 0 < K(x) ≤ κ,

1 S.I.S.S.A.,

∀ x ∈ Rn .

2-4 via Beirut, Trieste 34014, Italy. address: Dept. of Math. Anal., Univ. of Granada, Spain. Part of the work has been carried out during some visits at S.I.S.S.A. 3 Permanent

1

Bound states of (1) when ε  1 are called semiclassical states and are relevant for the links between Classical and Quantum Mechanics. An important feature of semiclassical states uε is that they concentrate as ε → 0. By this we mean that, roughly, out of a neighborhood of a set S, uε tends uniformly to zero as ε → 0. For example, by concentration at a point x0 ∈ Rn we mean that ∀ δ > 0, ∃ ε0 > 0, R > 0 s.t. uε (x) ≤ δ, ∀ |x − x0 | ≥ R, ∀ ε < ε0 . The existence of semiclassical states holds true under much weaker assumptions on V and K (although one always suppose that (V0)-(K0) hold). Let Q denote the function (Q)

Q(x) = [V (x)]θ [K(x)]−2/(p−1) ,

θ=

p+1 n − . p−1 2

If Q is smooth, we say that x0 is an isolated stable stationary point of Q if the LeraySchauder index ind(Q0 , x0 , 0) is different from zero. The index ind(Q0 , x0 , 0) of Q0 with respect to x0 and 0 is defined as limr→0 deg(Q0 , Br (x0 ), 0). Here deg denotes the topological degree and Br (x0 ) is the ball in Rn of radius r, centered at x0 . It is easy to see that local isolated maxima and minima as well as non-degenerate stationary points are stable. The following is a typical result dealing with the existence of semiclassical states, see e.g. [5] and Chapter 8 in the Monograph [3] and the references cited there. Theorem A Suppose that 1 < p < (V0), resp. (K0). There holds

n+2 n−2

and let V , K be smooth and satisfy

(i) if uε concentrates at a point x0 , then Q0 (x0 ) = 0; (ii) conversely, let x0 be an isolated stable stationary point of Q. Then for all ε  1 there exists a semiclassical state of (1) concentrating at x0 . We also mention that solutions concentrating on spheres have been found in the radial case, [4]. In a recent paper [2] (see also some previous partial results in [11, 14]), the new case in which V may decay to zero as |x| → ∞ has been addressed. More precisely, it is assumed that the potentials V, K are smooth and satisfy (V1)

∃ A0 , A1 > 0 :

(K1)

A0 ≤ V (x) ≤ A1 , 1 + |x|α

∃ β, k > 0 :

0 < K(x) ≤

0 ≤ α ≤ 2,

k . 1 + |x|β

Setting (for n ≥ 3) ( σ=

n+2 n−2

1

−

4β α(n−2) ,

otherwise, 2

if 0 < β < α

the following result has been proved, by using Critical Point Theory in weighted Sobolev spaces. Theorem B [2] Let V, K satisfy (V1), resp. (K1), and suppose that 0 ≤ α < 2 and that p satisfies (σ)

σ
n+2 . n−2

Then, for every ε > 0, (1) has a solution uε which is a ground state, namely it is a Mountain Pass solution with minimal energy. Furthermore, as ε → 0, uε concentrates at a global minimum of Q. Let us point out the following facts n+2 and hence the previous result does not (B.1) If α > 0 and β = 0, one has σ = n−2 apply. On the other hand, it is possible to show that, if V (x) ∼ (1 + |x|)−α , K(x) ∼ (1 + |x|)−β and (σ) is violated, then there no ground states at all, see [2, Prop. 15].

(B.2) Under the assumptions of Theorem B, the auxiliary potential Q has indeed a minimum on Rn . The main purpose of the present paper is to show that there exist bound states n+2 of (1) for all p satisfying 1 < p < n−2 , provided ε is sufficiently small. We shall further require that V and K are smooth and satisfy (V2)

∃ V1 > 0 : |V 0 (x)| ≤ V1 ,

∀ x ∈ Rn .

(K2)

∃ κ1 > 0 : |K 0 (x)| ≤ κ1 ,

∀ x ∈ Rn .

Our main result is the following. n+2 (if n ≥ 3, otherwise any p > 1 is allowed) and Theorem 1 Let 1 < p < n−2 suppose that V and K are smooth and satisfy (V1) - (V2) , resp. (K0) - (K2). Moreover, let x0 be an isolated stable stationary point of Q. Then for ε  1, equation (1) has a bound state that concentrates at x0 .

We anticipate that the case in which Q has a compact set of non-isolated critical points can be handled, see Theorems 12 and 13. Of course, if α = 0, namely when V is bounded away from zero, we recover Theorem A. A comparison with Theorem B is also in order. In Theorem 1 • K does not need to decay to zero at infinity; in particular we can deal with the case in which K is constant. • All the range 1 < p <

n+2 n−2

is allowed.

3

• The case α = 2 can be handled: we can deal with potentials V such that A0 ≤ V (x) ≤ A1 . 1 + |x|2 On the other hand, • We prove the existence of solutions only for ε small enough. • We do not find ground states (which may not exist, see remark (B.1) above), but merely bound states. The proof of Theorem 1 relies on arguments different from the ones used in [2]. Precisely, we will use a perturbation method, variational in nature, see [1] and [3, Chapter 2], that we are now going to outline in the next Section. Although the general procedure is similar to the one used in [4, 5], several changes and different estimates are required here, due to the fact that the potential V decays to zero at infinity. The paper consists of 6 more Sections. In Section 2-6 we will carry out the proof of Theorem 1. The final Section 7 contains some further existence results. General Remarks and Notation. • B(y, R) denotes the ball {x ∈ Rn : |x − y| ≤ R}. • If J is a functional, J 0 denotes its gradient. • W 1,2 (Rn ) denotes the usual Sobolev space. • In the sequel we will always take n ≥ 3. The case n = 1, 2 requires minor changes. • Without loss of generality we will assume that Q0 (0) = 0. Hence we will look for solutions concentrating at x0 = 0. • c1 , c2 , . . ., C1 , C2 , . . . denote positive, possibly different, constants. • oε (1) denotes a quantity such that limε→0 oε (1) = 0.

2

Outline of the abstract procedure and other preliminaries

After a change of variables, we are led to study the problem (2)

−∆u + V (εx)u = K(εx)up ,

u ∈ W 1,2 (Rn ),

If u is a solution of (2) then u(x/ε) solves (1).

4

u > 0.

Let us introduce the space E = Eε , Z 1,2 n E = {u ∈ D (R ) :

V (εx)u2 (x)dx < ∞}.

Rn

E is a Hilbert space with scalar product and norm given respectively by Z (3) (u, v) = [∇u(x) · ∇v(x) + V (εx)u(x)v(x)] dx, kuk2 = (u, u). Rn

Since the functions in E might not belong to Lp+1 (Rn ), we need to introduce a truncated nonlinearity. Let Fε be defined by setting ( 1 |u|p+1 if |u| < (1 + |εx|)−ϑ (4) Fε (x, u) = p+1 −ϑ(p+1) c¯ (1 + |εx|) if |u| > 2(1 + |εx|)−ϑ , where c¯ and ϑ will be chosen in a suitable way later on. For the moment, we just take ϑ(p + 1) > n + 1, so that Fε (x, u(x)) ∈ L1 (Rn ) for u ∈ E. For u ∈ E we set Z 2 1 K(εx)Fε (x, u(x))dx. (5) Iε (u) = 2 kuk − Rn

Clearly, any critical point u of Iε such that |u(x)| < c¯(1 + |εx|)−ϑ gives rise to a solution of (2). Next, let us denote by U the unique radial positive function satisfying (see [10]) −∆U + U = U p ,

U ∈ W 1,2 (Rn ).

Setting  zε,ξ (x) = σU (λ(x − ξ)),

σ=

V (εξ) K(εξ)

1  p−1

1

, λ = [V (εξ)] 2 ,

one checks that z = zε,ξ satisfies (6)

−∆z + V (εξ)z = K(εξ)z p .

We are now ready to describe the finite dimensional procedure used below. Let us introduce the manifold Zε = Z = {zε,ξ (x) : ξ ∈ R, |εξ| < 1}, and let P = Pξ denote the orthogonal projection onto (Tzε,ξ Z)⊥ , the orthogonal (with respect to the scalar product in 3) to the tangent space to Z at zε,ξ . Critical points of Iε in the form u = zε,ξ + w, with zε,ξ ∈ Z, will be found by using a finite dimensional reduction which takes into account the variational nature of the problem. For a broader discussion of this abstract perturbation method in critical point theory, as well as for several different applications, including nonlinear Schr¨ odinger equations, we refer to the aforementioned monograph [3]. 5

In the present case, we begin by studying, for every ξ ∈ Rn with |εξ| ≤ 1, the auxiliary equation P Iε0 (zε,ξ + w) = 0.

(7)

Roughly, (7) is first transformed into an equivalent fixed point problem like Sε (w) = w. One selects a subset Γε of E, whose functions satisfy appropriate estimates and suitable decays. Finally, one shows that, for ε  1, Sε is a contraction that maps Γε into itself and hence has a unique fixed point wε,ξ satisfying (7). Once (7) is solved, it is a general fact that the manifold {u = zε,ξ + wε,ξ } is a natural constraint for Iε , see [3, Chapter 2]. This means that for finding the critical points of Iε on E, it suffices to find critical points of the reduced (finite dimensional) functional Φε (ξ) = Iε (zε,ξ + wε,ξ ). The existence of solutions of (7) is discussed in Sections 3, 4 and 5. The study of the reduced functional Φε (ξ) is carried out in Section 6 and allows us to conclude the proof of Theorem 1.

3

Solving the auxiliary equation, I

In this section we prove some preliminary results needed to solve the auxiliary equation (7). First of all it is convenient to state the following lemma, which provides an uniform lower bound for V (εx + y). Lemma 2 Let α > 0, suppose that V (x) > a|x|−α for |x| > 1, and let m > 0 be given. Then there exist ε0 > 0 and R > 0 such that V (εx + y) ≥

m , |x|α

∀ |x| ≥ R, ε ≤ ε0 , y ∈ Rn , |y| ≤ 1.

Proof. Let ε0 > 0 and R > 0 be such that  α 2a = m, m R−α < min{V (x) : |x| ≤ 3}, 3ε0 and take ε < ε0 , |x| > R and |y| ≤ 1. Then, for |x| > 2/ε one has εx + y > 1 and hence  α a a |εx| V (εx + y) > > α α |εx + y|α ε |x| |εx| + 1  α a 2 m > = . εα |x|α 3 |x|α On the other hand, if R < |x| < 2/ε we get that |εx + y| ≤ 3 and hence V (εx + y) > This concludes the proof. 6

m m > . α R |x|α

Next we estimate the smallness of Iε0 (zε,ξ ). The proof is similar to that of Lemma 1 in [5] but we carry out the details since we are using a different functional space, with a different norm. Let us point out that we can choose ϑ in the definition of Fε , see (4), in such a way that the functional Iε defined in (5) evaluated on zε,ξ takes the form Z Iε (zε,ξ ) = 21 kzε,ξ k2 − K(εx)|zε,ξ (x)|p+1 dx. Rn

Lemma 3 There exists C > 0 such that kIε0 (zε,ξ )k ≤ Cε provided |εξ| ≤ 1. Proof. Let us evaluate Iε0 (zε,ξ )[v] for an arbitrary v ∈ E. Taking into account that zε,ξ satisfies (6), one finds Z Z p 0 [K(εx) − K(εξ)]zε,ξ vdx [V (εx) − V (εξ)]zε,ξ vdx + |Iε (zε,ξ )[v]| ≤ Rn

Rn

Z ≤

1/2 Z [V (εx) − V (εξ)]2 zε,ξ dx

Rn

Z + Rn

1/2 v 2 zε,ξ dx

Rn

1/2 Z 2 p [K(εx) − K(εξ)] zε,ξ dx

1/2

Rn

p v 2 zε,ξ dx

.

Let us estimate the first integral one the right-hand side. The change of variable ζ = x − ξ yields Z Z (8) [V (εx) − V (εξ)]2 zε,ξ dx = [V (εζ + εξ) − V (εξ)]2 zε,ξ (ζ + ξ)dζ. Rn

Rn

From the definition of zε,ξ it follows that  (9)

zε,ξ (ζ + ξ) = σU (λ(ζ)),

σ=

V (εξ) K(εξ)

1  p−1

, λ2 = V (εξ),

and this makes it clear that zε,ξ (· + ξ) has an uniform exponential decay at infinity, provided |εξ| ≤ 1 (in this region V and K are bounded from above and from below by two fixed positive constants). By the regularity of V , from a Taylor’s formula we obtain |V (εζ + εξ) − V (εξ)|2 ≤ Cε2 , (C > 0), and by the exponential decay of z we infer that Z Z [V (εζ + εξ) − V (εξ)]2 zε,ξ (ζ + ξ)dζ ≤ Cε2 |ζ|2 zε,ξ (ζ + ξ)dζ ≤ Cε2 . Rn

R Regarding the integral Rn v 2 zε,ξ , we claim that there exists c1 > 0, independent of ε, such that zε,ξ (x) ≤ c1 V (εx), provided ε  1 and |εξ| ≤ 1. This is equivalent to show that (10)

∀ ε  1, ζ ∈ Rn , |εξ| ≤ 1.

zε,ξ (ζ + ξ) ≤ c1 V (εζ + εξ), 7

Using Lemma 2 with m = 1, x = ζ and y = εξ, we deduce that V (εζ + εξ) ≥

1 , |ζ|α

∀ ε  1, |ζ| > R, |εξ| ≤ 1.

Since, as remarked before, zε,ξ (· + ξ) has an uniform exponential decay at infinity, taking c sufficiently large, we have that zε,ξ (ζ + ξ) ≤

c ≤ c1 V (εζ + εξ), |ζ|α

∀ ε  1, |ζ| > R, |εξ| ≤ 1.

Taking c1 possibly larger, the preceding inequality holds for |ζ| ≤ R, too. This shows that (10) holds, proving the claim. Hence from (10) we infer that Z Z 2 zε,ξ v dx ≤ c1 V (εx)v 2 dx ≤ c2 kvk2 . Rn

Rn

Similar estimates hold for the terms involving the function K. The proof is thereby concluded. We now start the study of Iε00 (zε,ξ ). Lemma 4 The operator Iε00 (zε,ξ ) : E → E is a compact perturbation of the identity. Proof. The proof follows an argument from [1] but takes into account that the potential V may tend to zero at infinity. One has that Z h i p−1 Iε00 (zε,ξ )[u, v] = ∇u · ∇v + V (εx)uv − pzε,ξ uv dx Z p−1 = (u, v) − p zε,ξ uvdx, where (u, v)R is the scalar product in E. Then, we need to prove that the operator p−1 K(u, v) = zε,ξ uv is a compact operator. Take um * u0 , vm * v0 in E; we claim that K(um , vm ) → K(u0 , v0 ). Clearly, we can restrict ourselves to the case u0 = 0, v0 = 0. Observe that the sequences um , vm must be bounded in E; suppose, for instance, that kum k ≤ 1, kvm k ≤ 1. Given δ > 0, because of the exponential decay of zε,ξ we can take R > 0 such R 2/n (p−1)n/2 that |x−ξ|>R zε,ξ < 2Sδ 2 , where S is the Sobolev constant. Then: Z

p−1 zε,ξ |um vm |dx =

Z ≤ |x−ξ|≤R

Z ≤ |x−ξ|≤R

Z |x−ξ|≤R

p−1 zε,ξ |um vm |dx +

p−1 zε,ξ |um vm |dx +

Z |x−ξ|>R

p−1 zε,ξ |um vm |dx

#2/n

"Z |x−ξ|>R

p−1 zε,ξ |um vm |dx + δ/2.

8

(p−1)n/2 zε,ξ dx

kum kL2∗ kvm kL2∗

Since V (εx) ≥ γ > 0 for any x, |x − ξ| ≤ R, we have that um , vm belong to W 1,2 (B(ξ, R)). Moreover, um * 0 in W 1,2 (B(ξ, R)). By the compact embedding, um → 0 in L2 (B(0, R)); the same holds for vm . So, we just need to take m large enough so that Z p−1 n max{zε,ξ (x) : x∈R } |um vm |dx ≤ δ/2. B(ξ,R)

This completes the proof. Lemma 5 ∃ C 0 > 0 such that if ε is small enough then P Iε00 (zε,ξ ) is uniformly invertible ∀ ξ ∈ Rn , with |εξ| ≤ 1, and there results k[P Iε00 (zε,ξ )]−1 k ≤ C 0 . Proof. We only need to verify that ∃ θ > 0 such that the interval (−θ, θ) does not contain any eigenvalue of P Iε00 (zε,ξ ), provided ε is small enough and |εξ| ≤ 1. By direct computation, using (6), we have Z
Iε00 (zε,ξ )[z, z] = |∇z|2 + V (εx)z 2 − pK(εx)z p+1 = Z

[V (εx) − V (εξ)]z 2 + (1 − p)K(εξ)

Z

z p+1 + p

Z

Rn

[K(εξ) − K(εx)]z p+1

Rn

Z = oε (1) + (1 − p)K(εξ)

z p+1 < −ckzk2 .

Rn

Next, let X = hzε,ξ ,

∂zε,ξ ∂zε,ξ ,... i. ∂ξ1 ∂ξn

Let us remark that X ⊂ W 1,2 (Rn ). We will show that the following inequality holds Iε00 (zε,ξ )[v, v] ≥ ckvk2 ,

∀ v ∈ X ⊥.

Let us fix v ∈ X ⊥ , and suppose that kvk = 1. We will need the following technical result Claim 1: There exists R ∈ (ε−1/4 , ε−1/2 ) such that Z [|∇v|2 + v 2 ]dx < C1 ε1/2 kvk2 = C1 ε1/2 . R<|x−ξ|
Take into account that if |x − ξ| < ε−1/2 + 1, then |x| < |ξ| + ε−1/2 + 1. Thus, for ε small we have |εx| < 2. Define 0 < m < min{V (x) : |x| ≤ 2}, m < 1. Then there holds Z Z V (εx) 2 v ]dx [|∇v|2 + v 2 ]dx ≤ [|∇v|2 + m |x−ξ|<ε−1/2 +1 |x−ξ|<ε−1/2 +1 9

1 m

<

Z

[|∇v|2 + V (εx)v 2 ]dx ≤

1 . m

[|∇v|2 + v 2 ]dx ≤

1 m

|x−ξ|<ε−1/2 +1

Note that the sum ε

−1 4


−1 2

Z R<|x−ξ|
R∈N −1/2

has more than ε 2 summands (for ε small). Then, it is always possible to choose R ∈ N, R ∈ (ε−1/4 , ε−1/2 ) so that claim 1 holds. c For any R > 0, define BR = B(ξ, R), BR = Rn \ BR , CR = BR+1 \ BR . Let us now fix R as in the previous claim, and choose χR : Rn → R to be a C ∞ c function such that χR = 1 in BR , χR = 0 in BR+1 , |∇χR | < 2. We decompose v as v = v1 + v2 , where v1 = χR v, v2 = (1 − χR )v. First of all, we estimate the norms of v1 and v2 Z
kv1 k2 = |χR ∇v + v∇χR |2 + V (εx)χ2R v 2 dx Rn

Z

2

Z

2

[v |∇χR | + 2vχR ∇v · ∇χR ]dx +

=

[χ2R |∇v|2 + V (εx)χ2R v 2 ]dx

BR+1

CR

= O(ε1/2 ) +

Z

[|∇v|2 + V (εx)v 2 ]dx.

BR+1

In the same way, we can show that kv2 k2 = O(ε1/2 ) +

Z

[|∇v|2 + V (εx)v 2 ]dx.

c BR

Therefore, we conclude that kv1 k2 + kv2 k2 = 1 + O(ε1/2 ).

(11) We also get

1 = kvk2 = kv1 k2 + kv2 k2 + 2(v1 , v2 ) = 1 + O(ε1/2 ) + 2(v1 , v2 ), which implies (v1 , v2 ) = O(ε1/2 ). After these preliminaries, we decompose Iε00 (zε,ξ )[v, v] in the following way (12) Iε00 (zε,ξ )[v1 + v2 , v1 + v2 ] = Iε00 (zε,ξ )[v1 , v1 ] + Iε00 (zε,ξ )[v2 , v2 ] + 2Iε00 (zε,ξ )[v1 , v2 ]. As for the last term, one has that (13)

|Iε00 (zε,ξ )[v1 , v2 ]| = oε (1).

10

Actually, one finds |Iε00 (zε,ξ )[v1 , v2 ]|

Z p−1 = [∇v1 ∇v2 + V (εx)v1 v2 − pK(εx)zε,ξ v1 v2 ]dx

Z ≤ |(v1 , v2 )| + C



CR

p−1 2 pzε,ξ v dx

≤ C1 ε1/2 ,

(C1 > 0).

p−1 Furthermore, we apply Lemma 2 to obtain that V (εx) − pK(εx)zε,ξ (x) ≥ 21 V (εx) for any x with |x − ξ| > R. Using this inequality, we estimate the second term in the right-hand side of (12) in the following way Z h i p−1 2 00 Iε (zε,ξ )[v2 , v2 ] = |∇v2 |2 + [V (εx) − pK(εx)zε,ξ ]v2 dx c BR

  1 |∇v2 |2 + V (εx)v22 dx. c 2 BR

Z ≥ This implies

Iε00 (zε,ξ )[v2 , v2 ] ≥

(14)

1 kv2 k2 . 2

We now focus our attention to the first term on the right-hand side of (12). Observe that, since v1 has compact support, it belongs to W 1,2 (Rn ). Actually, we have Z Z Z 1 1 2 2 V (εx)v12 dx ≤ + C2 ε1/2 , v1 dx = v1 dx ≤ m m BR+1 BR+1 where 0 < m < min{V (x) : |x| ≤ 2}, and C2 > 0. As we mentioned above, we are concerned with the estimate of Z h i p−1 2 Iε00 (zε,ξ )[v1 , v1 ] = |∇v1 |2 + [V (εx) − pK(εx)zε,ξ ]v1 dx Rn

Z

h

= BR+1

Z +

i p−1 2 |∇v1 |2 + [V (εξ) − pK(εξ)zε,ξ ]v1 dx

[V (εx) − V (εξ)]v12 dx + p

Z

BR+1

BR+1

p−1 2 [K(εξ) − K(εx)]zε,ξ v1 dx.

0

We now use the boundedness of V and K 0 , see (V2) and (K2), to conclude |V (εx) − V (εξ)| ≤ M ε|x − ξ| and |K(εx) − K(εξ)| ≤ M ε|x − ξ|, for some M > 0. Since x ∈ BR+1 and R < ε−1/2 , we obtain Z h i p−1 2 (15) Iε00 (zε,ξ )[v1 , v1 ] − |∇v1 |2 + [V (εξ) − pK(εξ)zε,ξ ]v1 dx ≤ M ε1/2 . Rn

We let

Z (u, v)ξ =

[∇u · ∇v + V (εξ)uv] dx, 11

R denote a scalar product in W 1,2 (Rn ) equivalent to the standard one Rn (∇u · ∇v + uv) (uniformly for |εξ| ≤ 1), and we let kuk2ξ = (u, u)ξ be the associated norm, also equivalent to the standard one. We now write v1 = φ + w, where φ ∈ X and w⊥ξ X, where ⊥ξ stands for orthogonality in the (·, ·)ξ sense. From [12] we know that Z h i p−1 (16) |∇w|2 + (V (εξ) − pK(εξ)zε,ξ )w2 dx ≥ c1 kwk2ξ . Rn

Roughly speaking, our aim is to show that φ is small compared to w, so v1 turns out to be close to w. After that, we estimate Iε00 (zε,ξ )[v1 , v1 ] taking into account (15) and (16). Claim 2: As ε → 0, there holds kφkξ = oε (1). Since φ=

(v1 , zε,ξ )ξ zε,ξ kzε,ξ k−2 ξ

−2 ∂zε,ξ ∂zε,ξ ∂zε,ξ

, )ξ + (v1 , ∂ξi ∂ξi ∂ξi ξ ∂z

in order to prove this claim, it suffices to evaluate (v1 , zε,ξ )ξ and (v1 , ∂ξε,ξ )ξ . We i first show that |(v1 , zε,ξ ) − (v1 , zε,ξ )ξ | ≤ C1 ε1/2 , for some C1 > 0. Indeed Z |(v1 , z) − (v1 , z)ξ | = [V (εx) − V (εξ)]v1 zε,ξ dx BR+1 ! 21

Z ≤

[V (εx) − V

(εξ)]2 v12 dx

BR+1

≤ C1 ε

! 12

Z

2 zε,ξ dx

BR+1

! 21

Z

1/2

! 21

Z

v12 dx

BR+1

2 zε,ξ dx

≤ C2 ε1/2 .

BR+1

In the last formula we have used the inequality |V (εx) − V (εξ)| ≤ M ε1/2 whenever x ∈ BR+1 . Since v = v1 + v2 and v⊥X with respect to the scalar product in E, we have that |(v1 , zε,ξ )| = |(v2 , zε,ξ )|. Then we find Z [|∇v2 · ∇zε,ξ | + V (εx)|v2 |zε,ξ ] dx |(v1 , zε,ξ )| = |(v2 , zε,ξ )| ≤ c BR

!1/2

Z ≤

!1/2

Z

2

2

|∇v2 | dx

|∇zε,ξ | dx

c BR

c BR

!1/2

Z

2

Z

V (εx)|v2 | dx

+C2 c BR

c BR

12

!1/2 2 zε,ξ dx

.

1

Recall that R > ε− 4 . Then, because of the exponential decay of zε,ξ and its derivatives, taking into account that kv2 k2 ≤ 1 + O(ε1/2 ), see (11), we find that |(v1 , zε,ξ )| = oε (1). We point out that this convergence to zero is uniform in v. Then, we conclude that |(v1 , zε,ξ )ξ | ≤ |(v1 , zε,ξ )ξ − (v1 , zε,ξ )| + |(v1 , zε,ξ )| = oε (1). ∂z )|ξ = oε (1) for any i = 1 . . . n. In the same way we can prove that |(v1 , ∂ξε,ξ i Claim 2 is thereby proved. Taking into account (15), there holds Z h i p−1 2 Iε00 (zε,ξ )[v1 , v1 ] = |∇v1 |2 + [V (εξ) − pK(εξ)zε,ξ ]v1 dx + oε (1). BR+1

Next, using Claim 2 we get Z i h p−1 2 |∇v1 |2 + [V (εξ) − pK(εξ)zε,ξ ]v1 dx BR+1 Z i h p−1 |∇w|2 + [V (εξ) − pK(εξ)zε,ξ ]w2 dx + oε (1), = BR+1

and hence Iε00 (zε,ξ )[v1 , v1 ] =

Z

h

BR+1

i p−1 |∇w|2 + [V (εξ) − pK(εξ)zε,ξ ]w2 dx + oε (1).

This equality and (16) imply Iε00 (zε,ξ )[v1 , v1 ] ≥ c2 kwk2ξ . Then, using Claim 2 we deduce (17)

Iε00 (zε,ξ )[v1 , v1 ] ≥ c3 kv1 k2ξ + oε (1) ≥ c4 kv1 k2 + oε (1).

Finally, (17), (13) and (14) imply Iε00 (zε,ξ )[v, v] ≥ c5 kv1 k2 + c6 kv2 k2 + oε (1). Since kv1 k2 +kv2 k2 = 1+O(ε1/2 ) = kvk+O(ε1/2 ), we get Iε00 (zε,ξ )[v, v] ≥ c7 kvk2 , and this concludes the proof. We are now ready to transform the auxiliary equation (7) into a fixed point problem. Specifically, solving the equation P Iε0 (zε,ξ + w) = 0 is clearly equivalent to finding fixed points of the map Sε , defined by setting (18)

−1

Sε (w) = w − [P Iε00 (zε,ξ )]

(P Iε0 (zε,ξ + w)) .

Note that Sε is well defined in virtue of Lemma 5. Fixed points of Sε will be found in a suitable subset of E consisting of functions satisfying an appropriate decay estimate, by means of Bessel functions. In the next section, we are going to discuss some preliminary material on this topic. 13

4

Linear equation and decay estimate

Motivated by some comparison arguments we need in the sequel (see in particular the proof of Lemma 9) we are interested in the behavior of the radial solutions of the linear problem  m n  −∆u + |x|α u = f (|x|) x ∈ R , |x| > R, (LR ) u(x) = 1 |x| = R,  u(x) → 0 |x| → ∞. where R > 0, m > 0, α ∈ (0, 2] and f : (R, +∞) → R is positive and has a certain decay (the exact hypotheses on f will be described later on). If u(x) = u(|x|) is a radial solution of (LR ) then the function u(r) is a solution of the problem  0  −u00 (r) − (n − 1) u r(r) + m u(r) r α = f (r) r > R (19) u(R) = 1  u(r) → 0 r → +∞ Since this is a linear problem, we will be interested in the solutions of the homogeneous equation, namely

(20)

−u00 (r) − (n − 1)

u0 (r) u(r) + m α = 0. r r n−1

By making the change of variables v(r) = u(r)r 2 , from equations (19), (20) we obtain respectively  n−1 (n−1)(n−3) 00  r>R v(r) + rmα v(r) = f (r)r 2  −v (r) + 2 4r n−1 2 (21) v(R) = R   v(r)r− n−1 2 →0 r → +∞

(22)

−v 00 (r) +

(n − 1)(n − 3) v(r) v(r) + m α = 0. 2 4r r

As mentioned before, we will first solve the homogeneous problem. Afterwards we will use this information to study the solutions of (21), see Lemma 6 later on. Equations (20), (22) admit a two-dimensional vector space of solutions. We will denote by ψ1 and ψ2 two generators the space of solution of (22). In so n−1 n−1 doing, u1 (r) = r− 2 ψ1 (r) and u2 (r) = r− 2 ψ2 (r) span the space of solutions of (20). We will choose u1 , u2 so that both of them are positive and u1 (r) → 0, u2 (r) → +∞ as r → +∞. Since equation (22) has the form v 00 (r) = q(r)v(r), there exists a constant d0 6= 0 such that the Wronskian ψ1 (r)ψ20 (r) − ψ10 (r)ψ2 (r) = d0 . We now show that d0 is a positive constant. Since u2 /u1 = ψ2 /ψ1 is positive and tends to infinity as

14

0

r → +∞, its derivative must be positive at a certain point. But (ψ2 /ψ1 ) = d0 /ψ12 , and this implies d0 > 0. The algebraic expressions defining ψ1 and ψ2 depend upon the coefficient α, and this forces us to distinguish two cases, α < 2 and α = 2. Case 1: α < 2. In this case, the functions ψ1 , ψ2 can be written by means of Bessel functions. Precisely, let B`I (r), respectively B`K (r), denote the modified Bessel function of the first kind, respectively of the second kind. Recall B`I and B`K are both positive solutions of the modified Bessel’s equation r2 y 00 + ry 0 − (r2 + `2 )y = 0. Furthermore, B`I is increasing and tends to infinity, while B`K is decreasing and tends to zero. By direct computation one finds that  √  √ 2 m 2−α n−2 K r 2 , `= ; ψ1 (r) = r · B` 2−α 2−α  √  √ 2 m 2−α n−2 ψ2 (r) = r · B`I r 2 , `= , 2−α 2−α From the asymptotics of the Bessel functions, there holds √ 2 m

ψ1 (r) ∼ rα/4 e− 2−α r

2−α 2

√ 2 m

ψ2 (r) ∼ rα/4 e 2−α r

;

2−α 2

Having ψ1 and ψ2 , we then obtain (23)

u1 (r) = r−

n−1 2

u2 (r) = r−

ψ1 (r);

n−1 2

ψ2 (r),

the solutions of (20). Observe that u1 (r) → 0, u2 (r) → ∞ for r → +∞. In particular, from the asymptotic behavior of Bessel functions, see [9, Sections 5.7 and 5.16.4], one has that α

u1 (r) ∼ r 4 −

(24)

n−1 2

√ 2 m

e− 2−α r

2−α 2

.

Case 2: α = 2. In this case, the functions ψ1 , ψ2 are given by √

ψ1 (r) = r

1−

(n−2)2 +4m 2

√

;

ψ2 (r) = r

1+

(n−2)2 +4m 2

.

Observe that here we still obtain the asymptotics ψ1 (r) → 0, ψ2 (r) → ∞ for r → +∞. 15

The functions u1 and u2 are given by √

(25)

u1 (r) = r

2−n−

(n−2)2 +4m 2

√

, u2 (r) = r

2−n+

(n−2)2 +4m 2

.

Clearly, we still have that u1 (r) → 0, u2 (r) → +∞ as r → ∞. We can now state and prove the main result of the section. Lemma 6 Let u1 , u2 be defined as above, and ϕ be a solution of problem (LR ), where f : (R, +∞) → R is a positive continuous function satisfying the integrability condition Z +∞ (26) rn−1 f (r)u2 (r) dr < +∞. R

Then there exists γ(R) > 0 such that ϕ(r) ≤ γ(R)u1 (r) for all r ∈ (R, +∞). n−1

Proof. We argue by passing to problem (21), so we write v(r) = r 2 ϕ(r), n−1 fe(r) = r 2 f (r). In terms of ψ2 and fe, condition (26) can be rewritten as +∞

Z (27)

fe(r)ψ2 (r) dr < +∞. R

The variation of parameters yields that v(r) is of the form   Z r Z r 1 v(r) = ψ1 (r) ψ2 (s)fe(s) ds − ψ2 (r) ψ1 (s)fe(s) ds +a1 ψ1 (r)+a2 ψ2 (r), d0 R R where d0 = ψ1 ψ20 − ψ10 ψ2 > 0 is the constant given by the Wronskian, and a1 , a2 are suitable constants. Observe that   Z −1 r ψ1 (s)fe(s) ds + a2 ψ2 (r)r−(n−1)/2 . 0 = lim r−(n−1)/2 v(r) = r→+∞ d0 R Recall that in both cases 1 and 2 we have u2 (r) = ψ2 (r)r−(n−1)/2 → +∞ as R +∞ r → +∞. This implies that a2 must coincide with − d10 R ψ1 (s)fe(s) ds. The constant a1 can be found by taking into account the boundary condition v(R) = n−1 R 2 . We now compute the limit of the quotient v(r)/ψ1 (r). v(r) 1 lim = lim r→∞ ψ1 (r) r→+∞ d0 1 d0

Z

+∞

R

Z

r

R

ψ2 (r) ψ2 (s)fe(s) ds + ψ1 (r)

1 ψ2 (r) ψ2 (s)fe(s) ds + a1 + lim r→+∞ d0 ψ1 (r)

16

+∞

Z

 e ψ1 (s)f (s) ds + a1 =

r

Z

+∞

ψ1 (s)fe(s) ds. r

Let us now compute the last limit. In order to do that, we first note that ψ1 /ψ2 is a positive decreasing function ((ψ1 /ψ2 )0 = −d0 /ψ22 < 0). Thus, we can write ψ2 (r) r→+∞ ψ1 (r)

+∞

Z

0 ≤ lim

ψ2 (r) r→+∞ ψ1 (r)

+∞

Z

ψ1 (s)fe(s) ds = lim r

ψ2 (r) ψ1 (r) r→+∞ ψ1 (r) ψ2 (r)

Z

lim

+∞

r +∞

Z

ψ2 (s)fe(s) ds = 0.

ψ2 (s)fe(s) ds = lim

r→+∞

r

ψ1 (s) ψ2 (s)fe(s) ds ≤ ψ2 (s)

r

Hence, we obtain that v(r) 1 = r→∞ ψ1 (r) d0

Z

+∞

lim

ψ2 (s)fe(s) ds + a1 . R

To complete the proof it suffices to note that

5

v(r) ψ1 (r)

=

ϕ(r) u1 (r) .

Solving the auxiliary equation, II

Using the analysis carried out in the previous section, we can now choose the set where we shall find the fixed points of the map Sε defined in (18). As before, it is always understood that |εξ| ≤ 1. Let us introduce the set Wε (R) of the functions w ∈ E such that ( √ γ(R) ε u1 (|x|), if |x| ≥ R, (28) |w(x + ξ)| ≤ √ ε, if |x| ≤ R, where u1 (r) is defined in (23) (respectively in (25)) if 0 ≤ α < 2 (respectively if α = 2), and γ(R) is the constant found in Lemma 6. Next, for |εξ| ≤ 1 we set Γε (R) = {w ∈ E : kwk ≤ c0 ε,

w ∈ Wε (R) ∩ (Tzε,ξ )⊥ },

where c0 is a fixed positive constant that will be chosen later on (see (31)). Clearly, we can choose c¯ and ϑ in the equation (4) defining Fε , in such way that |zε,ξ (x) + w(x)| < c¯(1 + |εx|)−ϑ for any w ∈ Γε (R). Thus we have Z 1 K(εx)|zε,ξ + w|p+1 dx, Iε (zε,ξ + w) = 21 kzε,ξ + wk2 − p + 1 Rn and any critical point u = zε,ξ + w of Iε , with w ∈ Γε (R), gives rise to a solution of (2). By Lemma 2, given any m > 0, we can find R1 > 0 such that (29)

V (εx + εξ) ≥

m , |x|α

∀ |x| ≥ R1 .

The choice of m depends upon the fact that α < 2 or α = 2. In the former case, we can take, say, m = 2. In the latter we have to choose m sufficiently large (see 17

below). Furthermore, let z0 (x) = zε,ξ (x + ξ). Let us point out that z0 depends on ε, ξ, but has a uniform decay at infinity because |εξ| ≤ 1. Hence there exists R2 > 0 such that pz0p−1 (x) ≤

(30)

1 , |x|α

∀ |x| ≥ R2 .

We set ρ = max{R1 , R2 }. We point out that this choice is independent of ε. Proposition 7 Sε (Γε (ρ)) ⊂ Γε (ρ), and is a contraction provided ε is sufficiently small. Proposition 7 is an immediate consequence of the following two lemmas. Lemma 8 For c0 large enough and for ε sufficiently small one has kSε (w)k ≤ c0 ε, for all w ∈ Γε (ρ), and Sε is a contraction in Γε (ρ). Lemma 9 For all ε sufficiently small, one has that Sε (Γε (ρ)) ⊂ Wε (ρ), for every w ∈ Γε (ρ). Proof of Lemma 8. Let C be given by Lemma 3. Observe that by Lemma 5, k[P Iε00 (zε,ξ )]−1 k ≤ C 0 for some C 0 > 0. Choose c0 = 2C 0 C,

(31)

in the definition of Γε . We first compute Sε0 (w) for some w ∈ Γε . There holds Sε0 (w)[v] = v − [P Iε00 (zε,ξ )]−1 (P Iε00 (zε,ξ + w)[v]). We apply P Iε00 (zε,ξ ), and obtain (32)

kP Iε00 (zε,ξ ) [Sε0 (w)[v] ] k = kP Iε00 (zε,ξ )[v] − P Iε00 (zε,ξ + w)[v])k.

In the next Lemma we estimate the above quantity. Lemma 10 ∃ C1 > 0 and δ > 0 such that for all w ∈ Γε there holds kP Iε00 (zε,ξ ) − P Iε00 (zε,ξ + w)k ≤ C1 kwkδ , provided ε small enough. Proof. Let w1 , w2 ∈ E. By direct computation, since |K| ≤ κ, we find Z p−1 p−1 00 00 [(zε,ξ + w) − zε,ξ ]w1 w2 dx |Iε (zε,ξ )[w1 , w2 ] − Iε (zε,ξ + w)[w1 , w2 ]| ≤ κ Rn

Z ≤ C1

p−1

(|w| + |w|

Z )w1 w2 dx ≤ C2

Rn


p−1 n/2

|w| + |w|

2/n dx

Rn

×kw1 kL2∗ kw2 kL2∗ . We now estimate the term Rn |w|(p−1)n/2 dx. Let q > 1 and q 0 = q/(q − 1). Then we obtain Z 1/q Z 1/q0 Z (p−1)n/2 q(p−1)n/4 q 0 (p−1)n/4 |w| ≤ |w| |w| . R

Rn

Rn

Rn

18

Let us now fix q so that τ = q(p − 1)n/4 > 2. Since w ∈ Γε , the expression R 1/q0 q 0 (p−1)n/4 |w| is finite. From Lemma 2 we deduce that there exists C3 > 0 n R so that |w(x)|τ −2 < C3 V (εx) (recall again that w ∈ Γε ). Therefore we have Z

q(p−1)n/4

1/q

Z

τ −2

1/q

|w| |w|

≤ C4

|w|

2

Rn

Z ≤ C5 The estimate of the term

R

1/q |w| V (εx) ≤ C6 kwk1/q . 2

|w|n/2 can be carried out in the same way.

Proof of Lemma 8 completed. Using Lemma 10 and (32), we get kP Iε00 (zε,ξ ) [Sε0 (w)[v] ] k ≤ C1 kwkδ kvk. Then, for any w1 , w2 ∈ Γε , we have k[P Iε00 (zε,ξ )]−1 k kP Iε00 (zε,ξ )(Sε (w1 ) − Sε (w2 ))k Z 1 ≤ C0 kP Iε00 (z) (Sε0 (w2 + s(w1 − w2 ))[w1 − w2 ]) kds.

kSε (w1 ) − Sε (w2 )k

≤

0

Then we obtain kSε (w1 ) − Sε (w2 )k ≤ C

00

δ

 max kw2 + s(w1 − w2 )k

kw1 − w2 k.

s∈[0,1]

for some C 00 , δ > 0. Since both w1 and w2 belong to Γε (ρ), we easily find that (33)

kSε (w1 ) − Sε (w2 )k = oε (1)kw1 − w2 k.

Equation (33) yields the contraction property for Sε . Next, we show that kSε (w)k ≤ c0 ε for any w ∈ Γε . Using (33) with w1 = w and w2 = 0, we obtain kSε (w) − Sε (0)k = oε (1)kwk. On the other hand, by using Lemma 3 and Lemma 5 we obtain kSε (0)k = k[P Iε00 (zε,ξ )]−1 (P Iε0 (zε,ξ ))k ≤ C 0 kP Iε0 (zε,ξ )k ≤ C 0 Cε. Hence, we finally deduce kSε (w)k

≤ kSε (w) − Sε (0)k + kSε (0)k ≤ oε (1)kwk + C 0 Cε = oε (1)kwk +

Since w ∈ Γε (ρ), then kwk ≤ c0 ε, and hence we get kSε (w)k ≤ oε (1)c0 ε + 19

1 2

c0 ε ≤ c0 ε,

1 2

c0 ε.

provided ε is sufficiently small. This concludes the proof. Proof of Lemma 9. First, let us introduce some notation. We set w e = Sε (w), p−1 L[v] = −∆v + V (εx)v − pK(εx)zε,ξ v, z˙ε,ξ = Dξ zε,ξ , η = kz˙ε,ξ k−2 (Iε00 (zε,ξ )[w e − w] + Iε0 (zε,ξ + w), z˙ε,ξ ), g(v)

p p−1 = K(εx)[(zε,ξ + v)p − zε,ξ − pzε,ξ v] + η [−∆z˙ε,ξ + V (εx)z˙ε,ξ ] h i p − −∆zε,ξ + V (εx)zε,ξ − K(εx)zε,ξ .

Here, for brevity, the symbol Dξ zε,ξ stands for a linear combination of the derivatives Dξ1 zε,ξ , . . . , Dξn zε,ξ (related to the projection of the equation w ˜ = Sε (w) onto (Tz Z)⊥ ). With all this notation, using integration by parts and the definitions of Iε0 (z + w), Iε00 (z), one finds that the function w e satisfies L[w] e = g(w). Moreover, if we set z0 (x) = zε,ξ (x + ξ) (as at the beginning of this section), w0 (x) = w(x + ξ);

w e0 (x) = w(x e + ξ);

L0 = −∆ + V (εx + εξ) − pK(εξ + εx)z0p−1 . and g1 (v) = K(εx + εξ)[(z0 + v)p − z0p − pz0p−1 v], g2 = η [−∆z˙0 + V (εx + εξ)z˙0 ] , g3 = −∆z0 + V (εx + εξ)z0 − K(εx + εξ)z0p . we can write that w e0 satisfies (34)

L0 [w e0 ] = g0 (w0 ) := g1 (w0 ) + g2 − g3

Below, for q > n, we will need the following estimates √ (35) kg0 (w0 )kLq/2 (B2ρ ) ≤ oε (1) ε; (36)

|g0 (w0 )(x)| ≤ oε (1)

√

ε u2∧p 1 ,

∀ |x| ≥ ρ.

The proof of (35) and (36) are postponed to the end of this section. Now we will prove separately that for ε  1, √ if |x| ≤ ρ, (37) |w e0 (x)| ≤ ε, and (38)

|w e0 (x)| ≤ γ(ρ)

√

ε u1 (|x|), 20

if |x| ≥ ρ.

Concerning the former estimate, we apply Theorem 8.24 of [8] to (34) to infer that (39)

kw e0 kL∞ (Bρ ) ≤ c1 kw e0 kL2 (B2ρ ) + c2 kg0 (w0 )kLq/2 (B2ρ ) ,

(q > n).

Using Lemma 8 and recalling that kzε,ξ k ≤ const., we get (40)

kw e0 kL2 (B2ρ ) ≤ c3 ε.

√ Inserting (35) and (40) into (39), we find that kw e0 kL∞ (Bρ ) ≤ c4 ε + c2 oε (1) ε and (37) follows, provided that ε is sufficiently small. Let us now prove (38). It is convenient to consider first the case α < 2. From (34) and taking into account (37), it follows that w e0 verifies the equation (41)

|x| > ρ,

L0 [w e0 ] = g0 (w0 ),

together with the boundary condition (42)

|w e0 (ρ)| ≤

√

ε.

Let ϕ denote the solution of the linear problem ( √ −∆ϕ + |x|1α ϕ = ε u2∧p 1 , |x| > ρ; (43) √ |x| = ρ. ϕ(x) = ε, √ Obviously, ϕ = εϕ, where ϕ is the solution of (LR ) with R = ρ and f (r) = u2∧p 1 (r). Recalling the discussion carried out in Section 4, in particular (24), we infer that f (r) = u2∧p 1 (r) satisfies the integrability condition (26). Hence Lemma 6 applies yielding √ (r > ρ). (44) ϕ(r) ≤ γ(ρ) ε u1 (r), Using (29) (with m = 2) and (30) we infer that V (εx + εξ) − pK(εx + εξ)z0p−1 (x) ≥ From (36), we clearly deduce that √ g0 (w0 ) ≤ ε u2∧p 1

1 , |x|α

(|x| > ρ).

for |x| > ρ.

This allows us to compare (41) - (42) with (43) yielding √ |w e0 (r)| ≤ ϕ(r) = ε ϕ(r), (r > ρ). √ Finally, using (44), we get |w e0 (r)| ≤ γ(ρ) ε u1 (r), for r > ρ, proving (38) in the case α < 2. To complete the proof, we have to handle the case α = 2. We will indicate the changes required in the above arguments. First, let us point out that the solution 21

u1 of (LR ) has a polynomial decay that depends on m, see (25). Therefore, if we take m sufficiently large, the function f (r) = u2∧p 1 (r) still satisfies the integrability condition (26). Substituting the comparison problem (43) with ( √ 2∧p −∆ϕ + (m−1) ε u1 , |x| > ρ, |x|α ϕ = √ ϕ(x) = ε, |x| = ρ, and noticing that V (εx + εξ) − pK(εx + εξ)z0p−1 (x) ≥ m−1 |x|α , for |x| > ρ, we can repeat the preceding arguments to find that (38) holds. The proof of Lemma 9 is thereby completed. It remains to carry out the proofs of (35) and (36). Proof of (35). We will estimate separately g1 (w0 ), g2 and g3 . Since w ∈ Wε (ρ) and u1 is decreasing, we infer that w0 can be estimated as follows ( √ |w0 (x)| ≤ ε if |x| < ρ, √ |w0 (x)| ≤ γ(ρ) ε u1 (ρ) if ρ < |x| < 2 ρ. Moreover, one has |g1 (w0 )| ≤ c1 |w0 (x)|2∧p .

(45)

From these estimates we readily deduce kg1 (w0 )kLq/2 (B2ρ ) ≤ c3

√
2∧p ε .

Since 2 ∧ p > 1 we get (46)

kg1 (w0 )kLq/2 (B2ρ ) ≤ oε (1)

√

ε,

where oε (1) → 0 as ε → 0.

Let us now turn to the estimate if g2 . From the definition of zε,ξ (see Section 2), its exponential decay, and from the boundedness of V 0 and K 0 one finds kz˙ε,ξ k ≤ C;

k − ∆z˙0 + V (εx + εξ)z˙0 kLq/2 (B2ρ ) ≤ C.

Moreover, from Lemma 3 and Lemma 10 one finds Z 1 kIε0 (zε,ξ + w)k ≤ kIε0 (zε,ξ )k + kIε00 (zε,ξ + sw)[w]kds 0

Z 1 + kIε00 (zε,ξ )[w]k + k(Iε00 (zε,ξ + sw) − Iε00 (zε,ξ ))[w]kds 0 √ ≤ C1 ε + C2 ε + C3 εδ ε ≤ oε (1) ε.

≤

kIε0 (zε,ξ )k

In addition, since kwk e ≤ c0 ε, see Lemma 8, and w ∈ Γε (ρ), we easily infer that kIε00 (zε,ξ )[w e − w]k ≤ C4 ε. 22

√ Therefore we find that |η| ≤ oε (1) ε, and (47)

kg2 kLq/2 (B2ρ ) = oε (1)

√

ε.

We finally turn to g3 . From the fact that −∆z0 + V (εξ)z0 = K(εξ)z0p , we get g3 = [V (εx + εξ) − V (εξ)] z0 (x) + [K(εξ) − K(εx)]z0p (x). Using the assumption (V2), we deduce that |g3 | ≤ V1 ε |x| |z0 (x)| + K0 ε |x| |z0 (x)|p . Since z0 has an exponential decay, it follows that √ (48) kg3 kLq/2 (B2ρ ) = oε (1) ε. Putting together (46), (47) and (48) we find that (35) holds. Proof of (36). Using (45) and the fact w ∈ Wε (ρ) we deduce  2∧p √ √ |g1 (w0 )(x)| ≤ c1 · γ(ρ) ε u1 (x) = oε (1) ε (u1 (x))2∧p ,

(|x| > ρ).

Furthermore, since z0 (and its derivatives, even multiplied by polynomials in x) decays faster than u2∧p 1 , repeating the arguments carried out above, we get that √ (|x| > ρ), i = 1, 2. |gi | = oε (1) ε u2∧p 1 , This completes the proof of (36). As a consequence of the above arguments we obtain the existence of a function w satisfying Sε (w) = w. We summarize the existence result, collecting some properties about the dependence on ξ in the next proposition. Proposition 11 Under the assumptions of Theorem 1 there exists a unique w ∈ Γ(ρ) satisfying Sε (w) = w. Moreover w is differentiable with respect to ξ and there exist C1 > 0 and δ > 0 such that k∇ξ wk ≤ C1 εδ .

kwk ≤ C1 ε;

Proof. The estimate on the norm of w has already been shown. We turn now to the dependence of w on ξ. The equation Sε (w) = w is equivalent to H(ξ, w, µ) = 0, where µ ∈ Rn , and where H : Rn × E × Rn → E × Rn is given by ! Pn ∂z Iε0 (zε,ξ + w) − i=1 µi ∂ξε,ξ i H(ξ, w, µ) = . (w, ∂ξ1 zε,ξ ), . . . , (w, ∂ξn zε,ξ ) Let us fix some ξ ∈ Rn with |εξ| ≤ 1. Then we know that for ε sufficiently small there exists a (locally unique) solution of H(ξ, w, µ) = 0, which coincides with the one found by Proposition 7. We notice that the function H is of class C 1 in ξ, w and µ. Moreover there holds ! Pn ∂z ∂H Iε00 (zε,ξ + w)[v] − i=1 νi ∂ξε,ξ i [v, ν] = . ∂(w, µ) (v, ∂ξ1 zε,ξ ), . . . , (v, ∂ξn zε,ξ ) 23

∂H Using Lemma 5, Lemma 10, and arguing as in [1], one can prove that ∂(w is ξ ,µ) uniformly invertible for |εξ| ≤ 1. As a byproduct of this fact one obtains an estimate in norm for µ similar to that for w. Then, by the local uniqueness of the function w, applying the implicit function theorem we obtain

∂H

k(w, µ)k ≤ c1 (ξ, µ, w)

, ∂ξ

where c1 is independent of ξ for |εξ| ≤ 1. Without loss of generality we can consider the derivative with respect to ξ1 , which gives ! Pn ∂ 2 zε,ξ ∂H Iε00 (z + w)[∂ξ1 zε,ξ ] − i=1 µi ∂ξi ∂ξ 1 = . ∂ξ1 (w, ∂ξ22 zε,ξ ), . . . , (w, ∂ξ21 ξn zε,ξ ) 1

Since (w, µ) is bounded by c2 ε (see also the estimate of η before (47)), we immediately find that

∂H 00

∂ξ1 ≤ c3 (ε + kIε (zε,ξ + w)[∂ξ1 z]k) . By Lemma 10 we can write kIε00 (zε,ξ + w)[∂ξ1 zε,ξ ]k ≤ kIε00 (zε,ξ )[∂ξ1 zε,ξ ]k + c4 εδ , where δ is some fixed positive constant. Therefore at this point it is sufficient to estimate the norm kIε00 (zε,ξ )[∂ξ1 zε,ξ ]k. Arguing as in [5], formula (6), one finds ∂ξ1 zε,ξ = −∂x1 zε,ξ + O(ε)

in E.

It follows that kIε00 (zε,ξ )[∂ξ1 zε,ξ ]k ≤ c5 ε + kIε00 (zε,ξ )[∂x1 zε,ξ ]k. Given any function v ∈ E, since ∂x1 zε,ξ belongs to the kernel of the linearization of (6), we have Z 00 Iε (zε,ξ )[∂x1 zε,ξ , v] = [V (εx) − V (εξ)](∂x1 zε,ξ )v. Rn

Reasoning as for the proof of Lemma 3 one finds Z ≤ c6 εkvk. [V (εx) − V (εξ)](∂ z)v x1 Rn

Therefore, by the arbitrarity of v, from the last formulas we deduce k∂ξ1 wk ≤ c7 εδ for some δ > 0. This concludes the proof.

24

6

Proof of Theorem 1

In this final section we complete the proof of Theorem 1 by showing that the reduced functional Φε has a critical point ξε /ε, with ξε ∼ 0, provided ε is small enough. According to the discussion carried out in Section 2, this implies that (2) has a solution uε , and therefore uε (x/ε) solves (1). Moreover, since uε ∼ zε,ξ , then ε uε (x/ε) ∼ zε,ξε ( x−ξ ε ) and hence such a solution concentrates at x0 = 0. First of all, let us expand the reduced functional Φε (ξ) = Iε (zξ + wε,ξ ) in the variable ξ. To simplify notation, we will write z instead of zε,ξ and w instead of wε,ξ . As in [5, Subsection 4.2], we find 2

= C1 [V (εξ)]θ [K(εξ)]− p−1 + Iε0 (z)[w] + Λε (ξ) + Ψε (ξ) = C1 Q(εξ) + Iε0 (z)[w] + Λε (ξ) + Ψε (ξ),

Φε (ξ)

where θ = where

p+1 p−1

Λε (ξ) =

n 2,

− 1 2

Z

C1 is a positive constant depending only on n and p, and

[V (εx) − V (εξ)]z 2 −

Rn

1 Ψε (ξ) = kwk2 − 2

Z

1 p+1

Z

[K(εx) − K(εξ)]z p+1 ,

Rn

  K(εx) |z + w|p+1 − z p+1 − (p + 1)z p w .

Rn

First of all, we notice that, from Lemma 3 and Proposition 11, there holds Iε0 (z)[w] ≤ Cεkwk ≤ C2 ε2 . Moreover, from a Taylor’s expansion of V and K one finds V (εx) − V (εξ) = εV 0 (εξ)(x − ξ) + O(ε2 |x − ξ|2 ); K(εx) − K(εξ) = εK 0 (εξ)(x − ξ) + O(ε2 |x − ξ|2 ). 2 Therefore, from elementary estimates involving the evenness of zε,ξ (x − ξ) and the 0 oddness of V (εξ)x, one finds

|Λε (ξ)| = o(ε). Furthermore, arguing as in the proof of Lemma 10, one finds |Ψε (ξ)| = o(kwk) = o(ε) Hence it follows that (49)

Φε (ξ) = C1 Q(εξ) + o(ε).

25

Let us remark that (49) would suffice to prove Theorem 1 in the case that x0 is an e ε (ξ) = Φε (ξ / ε), one finds isolated local minimum or maximum. Actually, setting Φ e ε (ξ) = C1 Q(ξ) + o(ε). Φ e ε (ξ) possesses a critical point ξε ∼ 0 and hence From this one readily deduces that Φ ε Φε has a critical point ξε /ε with ξε ∼ 0, yielding a solution uε (x/ε) ∼ zε,ξε ( x−ξ ε ) concentrating at x0 = 0. In order to handle the more general case in which x0 is an isolated stable critical point of Q, we need to estimate the derivatives of Φε with respect to ξ. We will write z 0 for ∂ξ z and w0 for ∂ξ w. One has ∂ξ (Iε0 (z)[w]) = Iε00 (z)[z 0 , w] + Iε0 (z)[w0 ]. As for Lemma 3, one can prove that kIε00 (z)[z 0 ]k = oε (1), and hence from Proposition 11 and the fact that kIε0 (z)k ≤ Cε (see Lemma 3) we obtain |∂ξ (Iε0 (z)[w])| = εoε (1),

as ε → 0.

Regarding the function Λε , using a change of variables we can write Z Z 1 1 [V (εx + εξ) − V (εξ)]z02 − [K(εx + εξ) − K(εξ)]z0p+1 , Λε (ξ) = 2 Rn p + 1 Rn where z0 , as before, stands for the function z0 (x) = σU (λx), with λ2 = V (εξ) and   1 V (εξ) p−1 σ = K(εξ) . Hence, there holds Λ0ε (ξ)

=

ε 2

Z

Z [V 0 (εξ + εx) − V 0 (εξ)]z02 + [V (εx + εξ) − V (εξ)]z0 z00 n n R R Z ε 0 0 [K (εξ + εx) − K (εξ)]z0p+1 − p + 1 Rn Z + [K(εx + εξ) − K(εξ)]z0p z00 . Rn

By using the Dominated Convergence Theorem, we obtain |Λ0ε (ξ)| = o(ε). We have also Ψ0ε (ξ) = (w, w0 ) + (p + 1)

Z K(εx)G(z, w)dx, Rn

where G(z, w) = |z + w|p−1 (z + w)(z 0 + w0 ) − z p z 0 − pz p−1 z 0 w − z p w0 . From (K2) we infer |Ψ0ε (ξ)|

Z

0

≤ kwkkw k + C7

G(z, w)dx, Rn

26

and hence, using Proposition 11 we find Z 0 (50) |Ψε (ξ)| ≤ C7 G(z, w)dx + o(ε). Rn

Next, let us write G(z, w) = G1 (z, w)z 0 + G2 (z, w)w0 with G1 (z, w) = |z + w|p−1 (z + w) − z p − pz p−1 w, G2 (z, w) = |z + w|p−1 (z + w) − z p , R R and let us estimate separately Rn G1 (z, w)z 0 dx and Rn G2 (z, w)w0 dx. As for the former, since G1 (z, w) = |z + w|p−1 (z + w) − z p − pz p−1 w = O(|w|2∧p) ), we infer Z Z wz 0 p 0 1∧(p−1) p V (εx)dx. G1 (z, w)z dx ≤ kwk∞ V (εx) Rn Rn Using the H¨ older inequality we get Z

0

G1 (z, w)z dx ≤ Rn

kwk1∧(p−1) kwk ∞

Z Rn

1 2 z 02 dx . V (εx)

Taking into account the exponential decay of z 0 as well as the fact that kwk∞ = O(ε1/2 ) and kwk ≤ ε (see the definition of Γε ), we infer Z (51) G1 (z, w)z 0 dx = o(ε). Rn

In a quite similar way, using once more that kw0 k = oε (1), we find Z (52) G2 (z, w)w0 dx ≤ O(ε)kw0 k = o(ε). Rn

In conclusion, from (50), (51) and (52), we deduce : |Ψ0ε (ξ)| = o(ε). From these estimates it follows that (53)

Φ0ε (ξ) = C1 ε Q0 (εξ) + o(ε),

|εξ| ≤ 1.

e ε (ξ) = Φε (ξ / ε). Then (53) implies that, for ε  1, As before, we set Φ e 0ε , 0, 0) = ind(Q0 , 0, 0) 6= 0. ind(Φ e ε has a critical point ξε ∼ 0. As a consequence, the reduced functional Φε Hence Φ possesses a critical point ξε /ε with ξε ∼ 0, and the conclusion follows.

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7

Further Results

In this final section we discuss some extensions of Theorem 1. Let X0 be a compact set of critical points of Q. We say that X0 is a stable critical set of Q if the topological degree deg(Q0 , X0,δ , 0) 6= 0, ∀ δ > 0 small, where X0,δ is a δ-neighborhood of X0 . Then the same arguments carried out in the previous sections lead to show: n+2 and suppose that V and K are smooth and satisfy Theorem 12 Let 1 < p < n−2 (V1) - (V2) , resp. (K0) - (K2). Moreover, let X0 be a compact, stable critical set of Q. Then (1) has, for ε  1, a bound state that concentrates at some point of X0 .

Of course, if X0 = {x0 }, Theorem 12 is nothing but Theorem 1. However, in the more general case covered by Theorem 12, we cannot establish a priori at which point of X0 the concentration occurs. In some circumstances, one can also find a multiplicity result. Let Σ be a smooth compact manifold of critical points of Q. We say that Σ is non-degenerate, if every x ∈ Σ is a non-degenerate critical point of Q| Σ⊥ . Combining the arguments used in [5] and those carried over in the present paper, one can prove the existence of multiple solutions of (1), concentrating at points of Σ. n+2 and suppose that V and K are smooth and satisfy Theorem 13 Let 1 < p < n−2 (V1) - (V2) , resp. (K0) - (K2). Moreover, let Σ be either a non-degenerate compact manifold of critical points of Q, or a compact set of minima/maxima of Q. Then for ε > 0 small, (1) has at least l(Σ), resp. cat(Σ, Σδ ), solutions concentrating near points of Σ.

Above, l(Σ) is the cup long of Σ, defined by ˇ ∗ (Σ) \ 1, α1 ∪ . . . ∪ αk 6= 0}. l(Σ) = 1 + sup{k ∈ N : ∃ α1 , . . . , αk ∈ H ˇ ∗ (Σ) is the Alexander cohomology If no such class exists, we set l(Σ) = 1. Here H of Σ with real coefficients and ∪ denotes the cup product. Moreover, cat(Σ, Σδ ) denotes the Lusternik-Schnirelman category of Σ with respect Sk to the δ-neighborhood Σδ of Σ, namely the least number k such that Σ ⊆ 1 Ti , with Ti closed and contractible in Σδ . In general, one has that l(Σ) ≤ cat(Σ, Σδ ).

Acknowledgements The first two authors have been supported by M.U.R.S.T within the PRIN 2004 ”Variational methods and nonlinear differential equations”. The third author has been partially supported by S.I.S.S.A., by the Ministry of Science and Technology (Spain), under grant number BFM2002-02649 and by J. Andalucia (FQM 116). He would also like to thank S.I.S.S.A. for the warm hospitality.

28

Added in proofs: work

After this paper was completed, we became aware of the

P. Souplet and Qi S. Zhang: Stability for semilinear parabolic equations with decaying potentials in Rn and dynamical approach to the existence of ground states, Ann. I. H. Poincar´e - AN, 19-5 (2002), 683-703, where the Schrodinger equation with a decaying potential V is studied. However, they do not deal with semiclassical states and do not study spikes. Moreover, they only consider radial potentials V satisfying (V1) with 0 ≤ α < 2(n−1)(p−1) , wich is p+3 strictly smaller than 2.

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