Multiplicity and concentration behavior of positive solutions ... - PPGME

Multiplicity and concentration behavior of positive solutions for a Schrödinger-Kirchhoff type problem via penalization method Giovany M. Figueiredo∗ and João R. Santos J´ unior† Faculdade de Matemática Universidade Federal do Pará 66.075-110 - Belém - Pará - Brazil Abstract In this paper we are concerned with questions of multiplicity and concentration behavior of positive solutions of the elliptic problem  3  Lε u = f (u) in IR , u > 0 in IR3 , (Pε )  u ∈ H 1 (IR3 ), where ε is a small positive parameter, f : R → R is a continuous function, Lε is a nonlocal operator defined by Z Z 1 1 |∇u|2 + 3 V (x)u2 −ε2 ∆u + V (x)u , Lε u = M ε IR3 ε IR3 M : IR+ → IR+ and V : IR3 → IR are continuous functions which verify some hypotheses. Keywords: Penalization method; Schr¨ odinger-Kirchhoff type problem; LusternikSchnirelmann Theory; Moser iteration 1991 Mathematics Subject Classification. Primary 35J65, 34B15.

1

Introduction

In this paper we shall focus our attention on questions of multiplicity, concentration behavior and positivity of solutions for the following problem  3  Lε u = f (u) in IR , (Pε ) u > 0 in IR3 ,  u ∈ H 1 (IR3 ), ∗ Partially † Partially

supported by CNPq/PQ 301242/2011-9 and 200237/2012-8 supported by CAPES - Brazil - 7155123/2012-9

1

where ε is a small positive parameter, f : R → R is a continuous function, Lε is a nonlocal operator defined by Z Z 1 1 V (x)u2 −ε2 ∆u + V (x)u , Lε u = M |∇u|2 + 3 ε IR3 ε IR3 M : IR+ → IR+ and V : IR3 → IR are continuous functions that satisfy some conditions which will be stated later on. Problem (Pε ) is a natural extension of two classes of very important problems in applications, namely, Kirchhoff problems and Schrödinger problems. a) When ε = 1 and V = 0 we are dealing with problem  Z  |∇u|2 dx ∆u = f (u) in R3 , −M R3  u > 0 in IR3 , u ∈ H 1 (IR3 ), which represents the stationary case of Kirchhoff model [17] for small transverse vibrations of an elastic string by considering the effect of the changes in the length during the vibrations. In fact, since the length of string is variable during the vibrations, the tension in the string changes with time and depends of the L2 norm of the gradient of the displacement u. More precisely, we have P0 E M (t) = + t, t > 0, h 2L where L is the length of the string, h is the area of cross-section, E is the Young modulus of the material and P0 is the initial tension. Z 2 Moreover, problem (Pε ) is catch nonlocal because of the presence of the term M |∇u| R3

which implies that the equation in (Pε ) is no longer a pointwise identity. This phenomenon causes some mathematical difficulties which makes the study of such a class of problem particularly interesting. The version of problem (Pε ) in bounded domain began to call attention of several researchers especially after the work of Lions [20], where a functional analysis approach was proposed to attack it. We have to point out that nonlocal problems also appear in other fields as, for example, biological systems where u describes a process which depends on the average of itself (for example, population density). See, for example, [3] and its references. The reader may consult [1], [2], [3], [9], [10], [14], [21] and the references therein, for more informations on nonlocal problems. b) On the other hand, when M = 1 we have the problem −ε2 ∆u + V (x)u = f (u) in IR3 , (1.1) u > 0 in IR3 , u ∈ H 1 (IR3 ),

2

which appear in different models, for example, it is related to the existence of standing waves of the nonlinear Schrodinger equation (1.2)

iε

∂Ψ = −ε∆Ψ + (V (x) + E)Ψ − f (Ψ), ∀x ∈ IRN , ∂t

2N . A standing wave of (1.2) is a solution of N −2 the form Ψ(x, t) = exp(−iEt/ε)u(x). In this case, u is a solution of (1.1). Existence and concentration of positive solutions for the problem (1.1) have been extensively studied in recent years, see for example the papers [7], [8], [11], [12], [15], [24] and their references. where f (t) = |t|p−2 u and 2 < p < 2∗ =

A considerable effort has been devoted during the last years in studying problems of the type (Pε ), as can be seen in [4], [16], [18], [23], [27], [29] and references therein. This is due to their significance in applications as well as to their mathematical relevance. Before stating our main result, we need the following hypotheses on the function M : (M1 ) There is m0 > 0 such that M (t) ≥ m0 , ∀t ≥ 0. (M2 ) The function t 7→ M (t) is increasing. (M3 ) For each t1 ≥ t2 > 0, M (t1 ) M (t2 ) − ≤ m0 t1 t2

1 1 − t1 t2

,

where m0 is given in (M1 ). The potential V is a continuous function satisfying: (V1 ) There is V0 > 0 such that V0 = inf 3 V (x). x∈IR

(V2 ) For each δ > 0 there is a bounded and Lipschitz domain Ω ⊂ IR3 such that V0 < min V, ∂Ω

Π = {x ∈ Ω : V (x) = V0 } = 6 ∅

and Πδ = {x ∈ IR3 : dist(x, Π) ≤ δ} ⊂ Ω. Moreover, we assume that the continuous function f vanishes in (−∞, 0) and verifies (f1 ) lim

t→0+

f (t) = 0. t3

(f2 ) There is q ∈ (4, 6) such that lim

f (t)

t→∞ tq−1

3

= 0.

(f3 ) There is θ ∈ (4, 6) such that 0 < θF (t) ≤ f (t)t, ∀t > 0. (f4 ) The application t 7→

f (t) t3

is non-decreasing in (0, ∞). The main result of this paper is: Theorem 1.1 Suppose that the function M satisfies (M1 ) − (M3 ), the potential V satisfies (V1 ) − (V2 ) and the function f satisfies (f1 ) − (f4 ). Then, given δ > 0 there is ε = ε(δ) > 0 such that the problem (Pε ) has at least CatΠδ (Π) positive solutions, for all ε ∈ (0, ε). Moreover, if uε denotes one of these positive solutions and ηε ∈ R3 its global maximum, then lim V (ηε ) = V0 . ε→0

A typical example of function verifying the assumptions (M1 )−(M3 ) is given by M (t) = m0 + bt, where m0 > 0 and b > 0. More generally, any function of the form M (t) = k X bi tγi with bi ≥ 0 and γi ∈ (0, 1) for all i ∈ {1, 2, . . . , k} verifies the hypotheses m0 + bt + i=1

(M1 ) − (M3 ). A typical example of function verifying the assumptions (f1 ) − (f4 ) is given by f (t) = k X ci (t+ )qi −1 with ci ≥ 0 not all null and qi ∈ [θ, 6) for all i ∈ {1, 2, . . . , k}. i=1

Recently some authors have considered problems of the type (Pε ). For example, He and Zou [16], by using Lusternik-Schnirelmann theory and minimax methods, proved a result of multiplicity and concentration behavior for the following equation  Z  2  |∇u|2 )∆u + V (x)u = f (u) in IR3  −(ε a + bε IR3 (1.5) u > 0 in IR3 ,    u ∈ H 1 (IR3 ), assuming, between others hypotheses, that f ∈ C 1 (IR) has a subcritical growth 3-superlinear and the potential V verifies a assumption introduced by Rabinowitz [24], namely, (R)

V∞ = lim inf V (x) > V0 = inf3 V (x) > 0. |x|→∞

IR

In [27], Wang, Tian, Xu and Zhang have considered the problem  Z  2  −(ε a + bε |∇u|2 )∆u + V (x)u = λf (u) + |u|4 u in IR3  3 IR (1.6) u > 0 in IR3 ,    u ∈ H 1 (IR3 ). 4

Assuming that f is only continuous, has subcritical growth 3-superlinear and the potential verifies (R), the authors showed that (1.6) has multiple positive solutions when λ is large enough, by using Lusternik-Schnirelmann theory, minimax methods and a approach as in [26] (see also [25]). Other results for the problem Schödinger-Kirchhoff type can be seen in [4], [18], [23], [29] and references therein. Motivated by results found in [4], [12], [16] and [27], we study multiplicity via LusternikSchnirelmann theory and concentration behavior of solutions for the problem (Pε ). Here we use the hypotheses (V1 ) − (V2 ) that were first introduced by Del Pino and Felmer [12] for laplacian case. For p-laplacian case, see [5]. We emphasize that, at least in our knowledge, does not exist in the literature actually available results involving problems Schrödinger-Kirchhoff type, where the potential is like that introduced by Del Pino and Felmer [12]. This is a difficulty that occurs, possibly by competition between the growth of nonlocal term and the growth of nonlinearity. Here, we use the same type of truncation explored in [12], however, we make a new approach and some estimates are totally different, for example, we show that solution of truncated problem is solution of the original problem with distinct arguments. Moreover, we completed the results found in [4], [16] and [27] in the following sense: 1 - Since M is a function more general than those in [16] and [27], we have a additional difficulty. In general, the weak limit of the Palais-Smale sequences is not weak solution of the autonomous problem. We overcome this difficulty with assumptions different from those found in [4]. 2 - Since the function f is only continuous, we cannot use standard arguments on the Nehari manifold. Hence, our result is similar then those found in [27]. However, since the hypotheses on function V are different, our arguments are completely different. Moreover, our result is for all positive lambda. The paper is organized as follows. In the Section 2 we show that the auxiliary problem has a positive solution and we introduce some tools needed for the multiplicity result, namely, Lemma 2.3 and Proposition 2.1. In the Section 3 we study the autonomous problem associated. This study allows us to show that the auxiliary problem has multiple solutions. In the section 4 we prove the main result using Moser iteration method [22].

2

The auxiliary problem

Considering the change of variable x = εz in (Pε ) we obtain the modified problem   Leε u = f (u) in IR3 , u > 0 in IR3 ,  u ∈ H 1 (IR3 ),

(Peε ) where

Z Leε u = M

2

Z

|∇u| + IR3

2

V (εx)u IR3

which is clearly equivalent to (Pε ). 5

[−∆u + V (εx)u] ,

Since (f1 ) and (f4 ) imply that lim+

t→0

f (t) =0 t

and

f (t) t is a application increasing in (0, ∞) and unbounded, we can adapt to our case the penalization method introduced by Del Pino and Felmer [12]. 2 V0 For this, let K > , where m0 is given in (M1 ) and a > 0 such that f (a) = a. We m0 K define f (t) if t ≤ a, e f (t) = V0 if t > a Kt t 7→

and g(x, t) = χΩ (x)f (t) + (1 − χΩ (x))fe(t), where χ is characteristic function of set Ω. From hypotheses (f1 ) − (f4 ) we get that g is a Carathéodory function and the following conditions are observed: (g1 ) lim+

g(x, t) = 0, uniformly in x ∈ IR3 . t3

lim

g(x, t) = 0, uniformly in x ∈ R3 , tq−1

t→0

(g2 ) t→∞

(g3 ) (i) 0 ≤ θG(x, t) < g(x, t)t, ∀x ∈ Ω and ∀t > 0 and (ii) 0 ≤ 2G(x, t) ≤ g(x, t)t ≤

1 V0 t2 , ∀x ∈ IR3 \Ω and ∀t > 0. K

(g4 ) For each x ∈ Ω, the application t 7→ g(x,t) is increasing in (0, ∞) and for each x ∈ t3 g(x,t) 3 IR \Ω, the application t 7→ t3 is increasing in (0, a). Moreover, from definition of g, we have g(x, t) ≤ f (t), for all t ∈ (0, +∞) and for all x ∈ IR3 , g(x, t) = 0 for all t ∈ (−∞, 0) and for all x ∈ IR3 . Now we study the auxiliary problem

(Pε,A )

  Leε u = g(εx, u), IR3 u > 0, IR3  u ∈ H 1 (IR3 ). 6

Observe that positive solutions of (Pε,A ) with u(x) ≤ a for each x ∈ IR3 \Ω are also positive solutions of (Peε ). We obtain solutions of (Pε,A ) as critical points of the energy functional Z Z Z 1c 2 2 Jε (u) = M |∇u| + V (εx)u − G(εx, u), 2 IR3 IR3 IR3 Z

t

c(t) = where M

Z M (s)ds and G(x, t) =

t

g(εx, s)ds, which is well defined on the Hilbert

0

0

space Hε , given by

Hε = {u ∈ H 1 (IR3 ) :

Z

V (εx)u2 < ∞},

IR3

provided of the inner product Z

Z ∇u∇v +

(u, v)ε =

V (εx)uv.

IR3

IR3

The norm induced by inner product is denoted by Z Z 2 2 kukε = |∇u| + IR3

V (εx)u2 .

IR3

Since M and f are continuous we have that Jε ∈ C 1 (Hε , IR) and Z Jε0 (u)v = M (kuk2ε )(u, v)ε − g(εx, u)v, ∀u, v ∈ Hε . IR3

Now, we will fix some notations. We denote the Nehari manifold associated to Jε by Nε = {u ∈ Hε \{0} : Jε0 (u)u = 0} and by Ωε the set

Ωε = {x ∈ IR3 : εx ∈ Ω},

by Hε+ the subset of Hε given by Hε+ = {u ∈ Hε : |supp(u+ ) ∩ Ωε | > 0} and by Sε+ the intersection Sε ∩ Hε+ , where Sε is the unit sphere of Hε . Lemma 2.1 The set Hε+ is open in Hε . Proof. Suppose by contradiction there are a sequence {un } ⊂ Hε \Hε+ and u ∈ Hε+ such + + that un → u in Hε . Hence |supp(u+ n ) ∩ Ωε | = 0 for all n ∈ IN and un (x) → u (x) a.e. in x ∈ Ωε . So, u+ (x) = lim u+ n (x) = 0, a.e in x ∈ Ωε . n→∞

But, this contradicts the fact that u ∈ Hε+ . Therefore Hε+ is open. 7

From definition of Sε+ and from Lemma 2.1 it follows that Sε+ is a incomplete C 1,1 manifold of codimension 1, modeled on Hε and contained in the open Hε+ . Thus, Hε = Tu Sε+ ⊕ IR u for each u ∈ Sε+ , where Tu Sε+ = {v ∈ Hε : (u, v)ε = 0}. Finally, we mean by weak solution of (Pε,A ) a function u ∈ Hε such that Z M (kuk2ε )(u, v)ε = g(εx, u)v, ∀v ∈ Hε . IR3

Therefore, critical points of Jε are weak solutions of (Pε,A ). Lemma 2.2 The functional Jε satisfies the following conditions: a) There are α, ρ > 0 such that Jε (u) ≥ α, with kukε = ρ. b) There is e ∈ Hε \Bρ (0) with Jε (e) < 0. Proof. The item a) follows directly from the hypotheses (M1 ), (g1 ) and (g2 ). On the other hand, it follows from (M3 ) that there is γ1 > 0 such that M (t) ≤ γ1 (1 + t) for all t ≥ 0. So, for each u ∈ Hε+ and t > 0 we have Z 1c 2 M (ktukε ) − G(εx, tu) Jε (tu) = 2 IR3 Z γ1 2 γ1 G(εx, tu). ≤ t kuk2ε + t4 kuk4ε − 2 4 Ωε From (g3 )(i), we obtain C1 , C2 > 0 such that Jε (tu) ≤

γ1 2 γ1 t kuk2ε + t4 kuk4ε − C1 tθ 2 4

Z

(u+ )θ + C2 |supp(u+ ) ∩ Ωε |.

Ωε

Since θ ∈ (4, 6) we conclude b). Once f and M are only continuous the next two results are very important, because allow us to overcome the non-differentiability of Nε (see Lemma 2.3 (A3 ) and Proposition 2.1) and the incompleteness of Sε+ (see Lemma 2.3 (A4 )). Lemma 2.3 Suppose that the function M satisfies (M1 ) − (M3 ), the potential V satisfies (V1 ) − (V2 ) and the function f satisfies (f1 ) − (f4 ). So: (A1 ) For each u ∈ Hε+ , let h : IR+ → IR be defined by hu (t) = Jε (tu). Then, there is a unique tu > 0 such that h0u (t) > 0 in (0, tu ) and h0u (t) < 0 in (tu , ∞). (A2 ) there is τ > 0 independent on u such that tu ≥ τ for all u ∈ Sε+ . Moreover, for each compact set W ⊂ Sε+ there is CW > 0 such that tu ≤ CW , for all u ∈ W. (A3 ) The map m b ε : Hε+ → Nε given by m b ε (u) = tu u is continuous and mε := m b ε is a Sε+ homeomorphism between Sε+ and Nε . Moreover, m−1 ε (u) =

8

u kukε .

(A4 ) If there is a sequence (un ) ⊂ Sε+ such that dist(un , ∂Sε+ ) → 0, then kmε (un )kε → ∞ and Jε (mε (un )) → ∞. Proof. To prove (A1 ), it is sufficient to note that, from the Lemma 2.2, hu (0) = 0, hu (t) > 0 when t > 0 is small and hu (t) < 0 when t > 0 is large. Since hu ∈ C 1 (IR+ , IR), there is tu > 0 global maximum point of hu and h0u (tu ) = 0. Thus, Jε0 (tu u)(tu u) = 0 and tu u ∈ Nε . We see that tu > 0 is the unique positive number such that h0u (tu ) = 0. Indeed, suppose by contradiction that there are t1 > t2 > 0 with h0u (t1 ) = h0u (t2 ) = 0. Then, for i = 1, 2 Z ti M (kti uk2ε )kuk2ε =

g(εx, ti u)u. IR3

So, 1 M (kti uk2ε ) = 2 kti ukε kuk4ε

Z

M (kt1 uk2ε ) M (kt2 uk2ε ) 1 − = kt1 uk2ε kt2 uk2ε kuk4ε

Z

IR3

g(εx, ti u) 4 u . (ti u)3

Therefore,

IR3

g(εx, t1 u) g(εx, t2 u) 4 − u . (t1 u)3 (t2 u)3

It follows from (M3 ) and (g4 ) that Z 1 1 g(εx, t1 u) g(εx, t2 u) 4 1 m0 − ≥ − u kuk2ε t21 t22 kuk4ε (IR3 \Ωε )∩{t2 u≤a0

u∈Sε t>0

(2.7)

The main feature of the modified functional is that it satisfies the Palais-Smale condition, as we can see from the next results. Lemma 2.4 Let {un } be a (P S)d sequence for Jε . Then {un } is bounded. Proof. Since {un } a (P S)d sequence for Jε , then there is C > 0 such that 1 C + kun kε ≥ Jε (un ) − Jε0 (un )un , ∀n ∈ IN. θ From (M3 ) and (g3 ), we obtain C + kun kε ≥

θ−2 2θ

1 m0 − kun k2ε , ∀n ∈ IN. K

Therefore {un } is bounded in Hε . Lemma 2.5 Let {un } be a (P S)d sequence for Jε . Then for each ξ > 0, there is R = R(ξ) > 0 such that Z |∇un |2 + V (εx)u2n < ξ. lim sup n→∞

IR3 \BR

Proof. Let ηR ∈ C ∞ (IR3 ) such that  0    ηR (x) = 1   

se

x ∈ BR (0)

se

x 6∈ B2R (0),

C where 0 ≤ ηR (x) ≤ 1, |∇ηR | ≤ and C is a constant independent on R. Note that {ηR un } R is bounded in Hε . From definition of Jε Z Z ηR M (kun k2ε ) |∇un |2 + V (εx)u2n = Jε0 (un )(un ηR ) + g(εx, un )un ηR IR3 IR3 Z − M (kun k2ε )un ∇un ∇ηR . IR3

14

Choosing R > 0 such that Ωε ⊂ BR (0) and by using (M1 ) and (g3 )(ii), we have Z m0 ηR |∇un |2 + V (εx)u2n ≤ Jε0 (un )(un ηR ) 3 IR Z Z 1 V (εx)u2n ηR − + M (kun k2ε )un ∇un ∇ηR . 3 3 K IR IR Therefore, Z Z 1 ηR |∇un |2 + V (εx)u2n ≤ |Jε0 (un )(un ηR )| + M (kun k2ε )un |∇un (∇ηR )|. m0 − K IR3 IR3 By using Cauchy-Schwarz inequality in IR3 , definition of ηR , Holder’s inequality and the boundedness of {un } in Hε , we conclude that Z C |∇un |2 + V (εx)u2n ≤ C|Jε0 (un )(un ηR )| + . 3 R IR \BR Since {un ηR } is bounded in Hε and {un } is a (P S)d sequence for Jε , passing to the upper limit of n → ∞, we obtain Z C lim sup |∇un |2 + V (εx)u2n ≤ < ξ, R n→∞ IR3 \BR whenever R = R(ξ) > C/ξ. The next result does not appear in [12], however, since we are working with the Kirchhoff problem type, it is required here. Lemma 2.6 Let {un } be a (P S)d sequence for Jε such that un * u, then Z Z lim |∇un |2 + V (εx)u2n = |∇u|2 + V (εx)u2 , n→∞

BR

BR

for all R > 0. Proof. We can assume that kun kε → t0 , thus, we have kukε ≤ t0 . Let ηρ ∈ C ∞ (IR3 ) such that    1 se x ∈ Bρ (0)  ηρ (x) = 0 se x 6∈ B2ρ (0).    with 0 ≤ ηρ (x) ≤ 1. Let, Pn (x) = M (kun k2ε ) |∇un − ∇u|2 + V (εx)(un − u)2 . For each R > 0 fixed, choosing ρ > R we obtain Z Z Z Pn = Pn ηρ ≤ M (kun k2ε ) |∇un − ∇u|2 + V (εx)(un − u)2 ηρ . BR

BR

IR3

15

By expanding the inner product in IR3 , Z Z Pn ≤ M (kun k2ε ) |∇un |2 + V (εx)(un )2 ηρ BR IR3 Z − 2M (kun k2ε ) [∇un ∇u + V (εx)un u] ηρ IR3 Z + M (kun k2ε ) |∇u|2 + V (εx)u2 ηρ . IR3

Setting 1 In,ρ

=

2 In,ρ

M (kun k2ε )

=

Z IR3

M (kun k2ε )

3 In,ρ = −M (kun k2ε )

Z IR3

|∇un |2 + V (εx)(un )2 ηρ −

Z g(εx, un )un ηρ , IR3

Z

Z [∇un ∇u + V (εx)un u] ηρ −

IR3

g(εx, un )uηρ , IR3

[∇un ∇u + V (εx)un u] ηρ + M (kun k2ε )

and 4 In,ρ

Z

Z

|∇u|2 + V (εx)u2 ηρ

IR3

Z g(εx, un )un ηρ −

=

g(εx, un )uηρ .

IR3

IR3

We have that, Z

1 2 3 4 Pn ≤ |In,ρ | + |In,ρ | + |In,ρ | + |In,ρ |.

0≤

(2.8)

BR

Observe that 1 In,ρ = Jε0 (un )(un ηρ ) − M (kun k2ε )

Z un ∇un ∇ηρ . IR3

Since {un ηρ } is bounded in Hε , we have Jε0 (un )(un ηρ ) = on (1). Moreover, from a straightforward computation Z 2 lim lim sup M (kun kε ) un ∇un ∇ηρ = 0. ρ→∞

n→∞

IR3

Then, 1 lim lim sup |In,ρ | = 0.

ρ→∞

(2.9)

n→∞

We see also that 2 In,ρ = Jε0 (un )(uηρ ) − M (kun k2ε )

Z u∇un ∇ηρ . IR3

By arguing in the same way as in the previous case, Jε0 (un )(uηρ ) = on (1) and

Z 2 lim lim sup M (kun kε )

ρ→∞

n→∞

u∇un ∇ηρ = 0. 3

IR

16

Therefore,

2 lim lim sup |In,ρ | ρ→∞ n→∞

= 0.

(2.10)

3 lim |In,ρ | = 0, ∀ ρ > R.

(2.11)

On the other hand, from the weak convergence n→∞

Finally, from un → u, in Lsloc (IR3 ), 1 ≤ s < 6, we conclude that 4 lim |In,ρ | = 0, ∀ ρ > R.

(2.12)

n→∞

From (2.8), (2.9), (2.10), (2.11) and (2.12), we obtain Z Pn ≤ 0. 0 ≤ lim sup n→∞

BR

Z Pn = 0 and consequently

Hence, lim

n→∞

BR

Z lim

n→∞

|∇un |2 + V (εx)u2n =

Z

BR

|∇u|2 + V (εx)u2 .

BR

Proposition 2.2 The functional Jε verifies the (P S)d condition in Hε . Proof. Let {un } be a (P S)d sequence for Jε . From Lemma 2.4 we know that {un } is bounded in Hε . Passing to a subsequence, we obtain un * u, in Hε . From Lemma 2.5, it follows that for each ξ > 0 given there is R = R(ξ) > C/ξ with C > 0 independent on ξ such that Z |∇un |2 + V (εx)u2n < ξ. lim sup n→∞

IR3 \BR

Therefore, from Lemma 2.6, kuk2ε

≤ =

lim inf kun k2ε ≤ lim sup kun k2ε n→∞ n→∞ (Z Z 2 2 lim sup |∇un | + V (εx)un + n→∞

Z =

|∇u|2 + V (εx)u2 + lim sup

|∇u|2 + V (εx)u2 + ξ,

n→∞

BR

Z
C/ξ. Passing to the limit of ξ → 0 we have R → ∞, which implies kuk2ε ≤ lim inf kun k2ε ≤ lim sup kun k2ε ≤ kuk2ε , n→∞

n→∞

and so kun kε → kukε and consequently un → u in Hε . Since f is only continuous and V has geometry of the Del Pino and Felmer type [12], in the next result (which is required for the multiplicity result) we use arguments that don’t appear in [12] and [27]. Corollary 2.1 The functional Ψε verifies the (P S)d condition on Sε+ . Proof. Let {un } ⊂ Sε+ be a (P S)d sequence for Ψε . Thus, Ψε (un ) → d and kΨ0ε (un )k∗ → 0, 0

where k.k∗ is the norm in the dual space (Tun Sε+ ) . It follows from Proposition 2.1(c) that {mε (un )} is a (P S)d sequence for Jε in Hε . From Proposition 2.2 we conclude there is u ∈ Sε+ such that, passing to a subsequence, mε (un ) → mε (u) in Hε . From Lemma 2.3(A3 ), it follows that un → u in Sε+ . Theorem 2.1 Suppose that the function M satisfies (M1 ) − (M3 ), the potential V satisfies (V1 ) − (V2 ) and the function f satisfies (f1 ) − (f4 ). Then, the auxiliary problem (Pε,A ) has a positive ground-state solution for all ε > 0. Proof. This result follows from Lemma 2.2, Proposition 2.2 and maximum principle.

3 3.1

Multiplicity of solutions of auxiliary problem The autonomous problem

Since we are interested in giving a multiplicity result for the auxiliary problem, we start by considering the limit problem associated to (Peε ), namely, the problem

(P0 )

 3  L0 u = f (u), IR u > 0, IR3  u ∈ H 1 (IR3 ) 18

where Z L0 u = M

|∇u|2 +

IR3

Z

V0 u2 [−∆u + V0 u] ,

IR3

which has the following associated functional Z Z Z 1c 2 2 |∇u| + I0 (u) = M V0 u − F (u). 2 IR3 IR3 IR3 This functional is well defined on the Hilbert space H0 = H 1 (IR3 ) with the inner product Z Z (u, v)0 = ∇u∇v + V0 uv IR3

and norm kuk20 =

Z

IR3

|∇u|2 +

IR3

Z

V0 u2

IR3

fixed. We denote the Nehari manifold associated to I0 by N0 = {u ∈ H0 \{0} : I00 (u)u = 0}. We denote by H0+ the open subset of H0 given by H0+ = {u ∈ H0 : |supp(u+ )| > 0}, and S0+ = S0 ∩ H0+ , where S0 is the unit sphere of H0 . As in the section 2, S0+ is a incomplete C 1,1 -manifold of codimension 1, modeled on H0 and contained in the open H0+ . Thus, H0 = Tu S0+ ⊕ IR u for each u ∈ S0+ , where Tu S0+ = {v ∈ H0 : (u, v)0 = 0}. Next we enunciate without proof one Lemma and one Proposition, which allow us to prove the Lemma 3.3. The proofs follow from a similar argument to that used in the proofs of Lemma 2.3 and Proposition 2.1. Lemma 3.1 Suppose that the function M satisfies (M1 )−(M3 ) and the function f satisfies (f1 ) − (f4 ). So: (A1 ) For each u ∈ H0+ , let h : IR+ → IR be defined by hu (t) = I0 (tu). Then, there is a unique tu > 0 such that h0u (t) > 0 in (0, tu ) and h0u (t) < 0 in (tu , ∞). (A2 ) there is τ > 0 independent on u such that tu ≥ τ for all u ∈ S0+ . Moreover, for each compact set W ⊂ S0+ there is CW > 0 such that tu ≤ CW , for all u ∈ W. (A3 ) The map m b : H0+ → N0 given by m(u) b = tu u is continuous and m := m b + is a S0 homeomorphism between S0+ and N0 . Moreover, m−1 (u) =

u kuk0 .

(A4 ) If there is a sequence (un ) ⊂ S0+ such that dist(un , ∂S0+ ) → 0, then km(un )k0 → ∞ and I0 (m(un )) → ∞.

19

We set the applications b 0 : H + → IR and Ψ0 : S + → IR, Ψ 0 0 b 0 (u) = I0 (m(u)) b 0 )| by Ψ b and Ψ0 := (Ψ

+ S0

.

Proposition 3.1 Suppose that the function M satisfies (M1 ) − (M3 ) and the function f satisfies (f1 ) − (f4 ). So: b 0 ∈ C 1 (H + , IR) and (a) Ψ 0 b 0 0 b 00 (u)v = km(u)k Ψ I (m(u))v, b ∀u ∈ H0+ and ∀v ∈ H0 . kuk0 0 (b) Ψ0 ∈ C 1 (S0+ , IR) and Ψ00 (u)v = km(u)k0 I00 (m(u))v, ∀v ∈ Tu S0+ . (c) If {un } is a (P S)d sequence for Ψ0 then {m(un )} is a (P S)d sequence for I0 . If {un } ⊂ N0 is a bounded (P S)d sequence for I0 then {m−1 (un )} is a (P S)d sequence for Ψ0 . (d) u is a critical point of Ψ0 if, and only if, m(u) is a nontrivial critical point of I0 . Moreover, corresponding critical values coincide and inf Ψ0 = inf I0 . + N0

S0

Remark 3.1 As in the section 2, there holds c0 = inf I0 (u) = inf max I0 (tu) = inf max I0 (tu). u∈N0

u∈S0+ t>0

u∈H0+ t>0

(3.1)

The next Lemma allows us to assume that the weak limit of a (P S)d sequence is nontrivial. Lemma 3.2 Let {un } ⊂ H0 be a (P S)d sequence for I0 with un * 0. Then, only one of the alternatives below holds: a) un → 0 in H0 b) there is a sequence (yn ) ⊂ IR3 and constants R, β > 0 such that Z lim inf u2n ≥ β > 0. n→∞

BR (yn )

Proof. Suppose that b) doesn’t hold. It follows that for all R > 0 we have Z lim sup u2n = 0. n→∞ y∈IR3

BR (y)

20

Since {un } is bounded in H0 , we conclude from [[28], Lemma 1.21] that un → 0 in Ls (IR3 ), 2 < s < 6. From (M1 ), (f1 ) and (f2 ), Z 0 ≤ m0 kun k0 ≤

f (un )un + on (1) = on (1). IR3

Therefore the item a) is true. Remark 3.2 As it has been mentioned, if u is the weak limit of a (P S)c0 sequence {un } for the functional I0 , then we can assume u 6= 0, otherwise we would have un * 0 and, once it doesn’t occur un → 0, we conclude from the Lemma 3.2 that there are {yn } ⊂ IR3 and R, β > 0 such that Z u2n ≥ β > 0.

lim inf n→∞

BR (yn )

Set vn (x) = un (x + yn ), making a change of variable, we can prove that {vn } is a (P S)c0 sequence for the functional I0 , it is bounded in H0 and there is v ∈ H0 with vn * v in H0 with v 6= 0. In the next Proposition we obtain a positive ground-state solution for the autonomous problem (P0 ). Theorem 3.1 Let {un } ⊂ H0 be a (P S)c0 sequence for I0 . Then there is u ∈ H0 \{0} with u ≥ 0 such that, passing a subsequence, we have un → u in H0 . Moreover, u is a positive ground-state solution for the problem (P0 ). Proof. Arguing as Lemma 2.4, we have that {un } is bounded in H0 . Thus, passing a subsequence if necessary, we obtain un * u em H0 ,

(3.2)

un → u em Lsloc (IR3 ), 1 ≤ s < 6

(3.3)

kun k0 → t0 .

(3.4)

(un , v)0 → (u, v)0 , ∀v ∈ H0 .

(3.5)

and So, from (3.2) we conclude that

On the other hand, due to density of C0∞ (IR3 ) in H0 and from convergence in (3.3), it results that Z Z f (un )v → f (u)v, ∀v ∈ H0 . (3.6) IR3

IR3

Now, from convergence in (3.2) and (3.4), occurs kuk20 ≤ lim inf kun k20 = t20 , n→∞

21

and from (M2 ) it follows that M (kuk20 ) ≤ M (t20 ). c(t) − 1 M (t)t is non-decreasing, we can Since (M3 ) implies that the function t 7→ 12 M 4 2 argue as in [4] and to prove that M (t0 ) = M (kuk20 ) and the theorem now follows from fact that functional I0 has the mountain pass geometry and from [[28], Theorem 1.15]. The next lemma is a compactness result on the autonomous problem which we will use later. Lemma 3.3 Let {un } be a sequence in H 1 (IR3 ) such that I0 (un ) → c0 and {un } ⊂ N0 . Then, {un } has a convergent subsequence in H 1 (IR3 ). Proof. Since {un } ⊂ N0 , it follows from Lemma 3.1(A3 ), Proposition 3.1(d) and Remark 3.1 that un ∈ S0+ , ∀n ∈ IN (3.7) vn = m−1 (un ) = kun k0 and Ψ0 (vn ) = I0 (un ) → c0 = inf Ψ0 . S0+

Although S0+ is incomplete, due to Lemma 3.1(A4 ), we can still apply the Ekeland’s variational principle [[13], Theorem 1.1] to the functional ξ0 : V → IR ∪ {∞} defined by ξ0 (u) = Ψ0 (u) if u ∈ S0+ and ξ0 (u) = ∞ if u ∈ ∂S0+ , where V = S0+ is a complete metric space equipped with the metric d(u, v) = ku − vk0 . In fact, from Lemma 3.1(A4 ), ξ0 ∈ C(V, IR∪ {∞}) and, from Proposition 3.1(d), ξ0 is bounded from below. Thus, we can conclude there is a sequence {b vn } ⊂ S0+ such that {b vn } is a (P S)c0 sequence for Ψ0 on S0+ and kb vn − vn k0 = on (1).

(3.8)

The remainder of the proof follows by using Proposition 3.1, Theorem 3.1 and arguing as in the proof of Corollary 2.1. In the next subsection we will relate the number of positive solutions of (Pε,A ) to topology of Π, for this we need some preliminary results.

3.2

Technical results

Let δ > 0 fixed and Πδ ⊂ Ω. Let η ∈ C0∞ ([0, ∞)) be such that 0 ≤ η(t) ≤ 1, η(t) = 1 if 0 ≤ t ≤ δ/2 and η(t) = 0 if t ≥ δ. We denote by w a positive ground-state solution of the problem (P0 ) (see Theorem 3.1). For each y ∈ Π = {x ∈ Ω : V (x) = V0 }, we define the function εx − y e Υε,y (x) = η(|εx − y|)w . ε Let tε > 0 be the unique positive number such that e ε,y ) = Jε (tε Υ e ε,y ). max Jε (tΥ t≥0

22

e ε,y ∈ Nε , we can now define the continuous function By noticing that tε Υ Υε :

Π

−→ Nε

y

e ε,y . 7−→ Υε (y) = tε Υ

Lemma 3.4 Let Π ⊂ Ω. Then, lim Jε (Υε (y)) = c0 uniformly in y ∈ Π.

ε→0

Proof. Arguing by contradiction, we suppose that there exist δ0 > 0 and a sequence {yn } ⊂ Π verifying | Jεn (Υεn (yn )) − c0 |≥ δ0 where εn → 0 when n → ∞.

(3.9)

From definition of Υεn (yn ), we have 1 c 2 e M tεn kΥεn ,yn k2εn − 2

Jεn (Υεn (yn )) =

Z IR3

e ε ,y G εn x, tεn Υ n n

(3.10)

and Jε0 n (Υεn (yn ))Υεn (yn ) = 0.

(3.11)

Using definition of Υεn (yn ) again and making the change of variable z =

εn x−yn , εn

we have

Jεn (Υεn (yn )) = Z Z 1c 2 2 2 M t εn |∇ (η(|εn z|)w(z))| + V (εn z + yn ) (η(|εn z|)w(z)) 2 IR3 IR3 Z − G (εn z + yn , tεn η(|εn z|)w(z)) . IR3

Moreover, putting Λ2n

Z

2

Z

|∇ (η(|εn z|)w(z))| +

=

2

V (εn z + yn ) (η(|εn z|)w(z)) ,

IR3

IR3

the equality in (3.11) yields M (t2εn Λ2n ) 1 = 4 t2εn Λ2n Λn

"

Z

g(εn z + yn , tεn η(|εn z|)w(z)) (tεn η(|εn z|)w(z))

IR3

For each n ∈ IN and for all z ∈ B

δ εn

3

# (η(|εn z|)w(z))4 .

(0), we have εn z ∈ Bδ (0). So,

εn z + yn ∈ Bδ (yn ) ⊂ Πδ ⊂ Ω. Since G = F in Ω, it follows from (3.10) that Jεn (Υεn (yn )) =

1c 2 2 M (tεn Λn ) − 2 23

Z IR3

F (tεn η(|εn z|)w(z))

(3.12)

and M (t2εn Λ2n ) 1 = 4 2 2 t εn Λ n Λn

Z

"

f (tεn η(|εn z|)w(z))

IR3

(tεn η(|εn z|)w(z))

# (η(|εn z|)w(z))4 .

3

(3.13)

From the Lebesgue’s theorem, when n → ∞ e ε ,y k2ε = Λ2n → kwk20 , kΥ n n n

(3.14)

Z

Z f (η(|εn z|)w(z))η(|εn z|)w(z) →

IR3

and

f (w)w IR3

Z

Z F (η(|εn z|)w(z)) →

IR3

F (w).

(3.15)

IR3

We see that there is a subsequence, still denoted by {tεn }, with tεn → 1. In fact, since η = 1 in B δ (0) and B δ (0) ⊂ B δ (0) for n large enough, it follows from (3.13) that 2

2

2εn

M (t2εn Λ2n ) 1 ≥ 4 t2εn Λ2n Λn

"

Z B δ (0) 2

f (tεn w(z)) (tεn w(z))

#

3

w(z)4 .

From continuity of w (which follows from standard regularity theory), there is zb ∈ IR3 such that w(b z ) = min w(z). So, from (f4 ) B δ (0) 2

1 f (tεn w(b z )) 3 Λ4n (tεn w(b z ))

Z

w(z)4 ≤

B δ (0) 2

M (t2εn Λ2n ) . t2εn Λ2n

(3.16)

Suppose by contradiction that there is a subsequence {tεn } with tεn → ∞. Thus, passing to the limit as n → ∞ in (3.16), we conclude, from (M3 ) and (f3 ), that the left side converges to infinity and the right side is bounded, which is a contradiction. Therefore, {tεn } is bounded and passing to a subsequence we have tεn → t0 with t0 ≥ 0. From (3.13), (3.14), (M1 ) and (f4 ) we have that t0 > 0. Thus, passing to the limit as n → ∞ in (3.13), we have Z 2 2 2 M (t0 kwk0 )kwk0 t0 = f (t0 w)w. (3.17) IR3

Since w ∈ N0 , we obtain t0 = 1. So, passing to the limit of n → ∞ in (3.12) and using (3.14) and (3.15) we obtain lim Jεn (Υεn (yn )) = I0 (w) = c0 ,

n→∞

which is a contradiction with (3.9). Let’s consider the specific subset of the Nehari manifold eε = {u ∈ Nε : Jε (u) ≤ c0 + h1 (ε)}, N

24

eε and lim h1 (ε) = 0. Observe that where h1 : IR+ → IR+ is a function such that Υε (Π) ⊂ N ε→0

eε 6= ∅ for all small ε > 0. h1 exists due to the Lemma 3.4. In particular, N Now we consider ρ > 0 such that Πδ ⊂ Bρ (0) and χ : IR3 −→ IR3 defined by  x se |x| ≤ ρ     ρx χ(x) = se |x| ≥ ρ.     |x| We also consider the barycenter map βε : Nε −→ IR3 given by Z χ(εx)u(x)2 IR3Z . βε (u) = u(x)2 IR3

Since Π ⊂ Bρ (0), the definition of χ and Lebesgue’s theorem imply that lim βε (Υε (y)) = y uniformly in y ∈ Π.

ε→0

(3.18)

The next result is fundamental to show that the solutions of the auxiliary problem are solutions of the original problem. Moreover, it allows us to show the behavior of such solutions in the norm |.|L∞ (IR3 \Ωε ) . Proposition 3.2 Let {un } be a sequence in H 1 (IR3 ) such that Jεn (un ) → c0 and Jε0 n (un )(un ) = 0, ∀n ∈ IN with εn → 0 when n → ∞. Then, there is a subsequence {e yn } ⊂ IR3 such that the sequence vn (x) = un (x + yñ ) has a convergent subsequence in H 1 (IR3 ). Moreover, passing to a subsequence, yn → ye with y ∈ Π, where yn = εn yen . Proof. We can always consider un ≥ 0 and un 6= 0. As in Lemma 2.4 and arguing as Remark 3.2 we have that {un } is bounded in H 1 (IR3 ) and there are (e yn ) ⊂ IR3 and positive constants R and α such that Z lim inf u2n ≥ α > 0. (3.19) n→∞

BR (e yn )

25

Considering vn (x) = un (x + yen ) we conclude that {vn } is bounded in H 1 (IR3 ) and therefore, passing to a subsequence, we get vn * v, in H 1 (IR3 ) with v 6= 0. For each n ∈ IN, let tn > 0 such that ven = tn vn ∈ N0 (see Lemma 3.1(A1 )). We have that Z 1c 2 c0 ≤ I0 (e vn ) = M (tn kun k20 ) − F (tn un ) 2 IR3 Z 1c 2 M (tn kun k2εn ) − G(εn x, tn un ). ≤ 2 IR3 Hence, c0 ≤ I0 (e vn ) ≤ Jε (tn un ) ≤ Jε (un ) = c0 + on (1),

(3.20)

which implies, I0 (e vn ) → c0 and {e vn } ⊂ N0 . 1

(3.21)

3

Thus, {e vn } is bounded in H (IR ) and ven * ve. From well-known arguments we can assume that tn → t0 with t0 > 0. So, from uniqueness of the weak limit we have ve = t0 v, v 6= 0. From Lemma 3.3 we obtain, ven → ve in H 1 (IR3 ).

(3.22)

This convergence implies vn →

ve = v in H 1 (IR3 ) t0

and I0 (e v ) = c0 and I00 (e v )e v = 0.

(3.23)

Now, we will show that {yn } is bounded, where yn = εn yen . In fact, otherwise, there exists a subsequence {yn } with |yn | → ∞. Observe that Z m0 kvn k20 ≤ g(εn z + yn , vn )vn . IR3

Let R > 0 such that Ω ⊂ BR (0). Since we may suppose that |yn | ≥ 2R, for each z ∈ B R (0) εn we have |εn z + yn | ≥ |yn | − |εn z| ≥ 2R − R = R. Thus, m0 kvn k20

Z

Z

≤

fe(vn )vn + B

R εn

(0)

f (vn )vn . IR3 \B

R εn

(0)

Since vn → v in H 1 (IR3 ), it follows from Lebesgue’s theorem that Z f (vn )vn = on (1). IR3 \B

R εn

(0)

26

On the other hand, since fe(vn ) ≤

V0 K vn ,

we obtain Z 1 V0 vn2 + on (1), m0 kvn k20 ≤ K B R (0) εn

and therefore, 1 kvn k0 ≤ on (1), m0 − K which is a contradiction. Hence, {yn } is bounded and we can assume yn → y in IR3 . We see that y ∈ Ω because if y ∈ / Ω, we can proceed as above and conclude that kvn k0 ≤ on (1). In order to prove that V (y) = V0 , we suppose by contradiction that V0 < V (y). Consequently, from (3.22), Fatou’s Lemma and the invariance of R3 by translations, we obtain Z Z Z 1c 2 2 M |∇e vn | + V (εn z + yn )e vn − F (e vn ) c0 < lim inf n→∞ 2 IR3 IR3 IR3 ≤ lim inf Jεn (tn un ) n→∞

≤ lim inf Jεn (un ) = c0 , n→∞

which is a contradiction and the proof is finished. Corollary 3.1 Assume the same hypotheses of Proposition 3.2. Then, for any given γ2 > 0, there exists R > 0 and n0 ∈ N such that Z |∇un |2 + |un |2 < γ2 , for all n ≥ n0 . BR (e yn ) c

Proof. By using the same notation of the proof of Proposition 3.2, we have for any R > 0 Z Z 2 2 |∇un | + |un | = (|∇vn |2 + |vn |2 ). BR (e yn )c

BR (0)c

Since (vn ) strongly converges in H 1 (RN ) the result follows. Lemma 3.5 Let δ > 0 and Πδ = {x ∈ IR3 : dist(x, M ) ≤ δ}. Then, lim sup inf |βε (u) − y| = lim sup dist(βε (u), Πδ ) = 0.

ε→0

eε u∈N

y∈Πδ

ε→0

eε u∈N

Proof. The proof of this Lemma follows from well-known arguments and can be found in [5, Lemma 3.7].

3.3

Multiplicity of solutions for (Pε,A )

Next we prove our multiplicity result for the problem (Pε,A ), by using arguments slightly different to those in [27], in fact, since Sε+ is a incomplete metric space, we cannot use (directly) an abstract result as in [[11], Theorem 2.1], instead, we invoke the category abstract result in [26]. 27

Theorem 3.2 Suppose that the function M satisfies (M1 ) − (M3 ), the potential V satisfies (V1 ) − (V2 ) and the function f satisfies (f1 ) − (f4 ). Then, given δ > 0 there is ε = ε(δ) > 0 such that the auxiliary problem (Pε,A ) has at least CatΠδ (Π) positive solutions, for all ε ∈ (0, ε). Proof. For each ε > 0, we define the function ζε : Π → Sε+ by ζε (y) = m−1 ε (Υε (y)), ∀y ∈ Π. From the Lemma 3.4, we have lim Ψε (ζε (y)) = lim Jε (Υε (y)) = c0 , uniformly in y ∈ Π.

ε→0

ε→0

Thus, the set Seε+ = {u ∈ Sε+ : Ψε (u) ≤ c0 + h1 (ε)}, is nonempty, for all ε ∈ (0, ε), because ζε (Π) ⊂ Seε+ , where the function h1 was already eε . introduced in the definition of the set N From up above considerations, together with Lemma 3.4, Lemma 2.3(A3 ), equality (3.18) and Lemma 3.5, there is ε = ε(δ) > 0, such that the diagram of continuous applications bellow is well defined for ε ∈ (0, ε) Υ

m−1

m

βε

ε ε ε Π −→ Υε (Π) −→ ζε (Π) −→ Υε (Π) −→ Πδ .

We conclude from (3.18) that there is a function λ(ε, y) with |λ(ε, y)| < 2δ uniformly in y ∈ Π, for all ε ∈ (0, ε), such that βε (Υε (y)) = y + λ(ε, y) for all y ∈ Π. Hence, the application H : [0, 1] × Π → Πδ defined by H(t, y) = y + (1 − t)λ(ε, y) is a homotopy between αε ◦ ζε = βε ◦ Υε and the inclusion i : Π → Πδ , where αε = βε ◦ mε . Therefore, catζε (Π) ζε (Π) ≥ catΠδ (Π).

(3.24)

It follows from Corollary 2.1 and from category abstract theorem in [26], with c = cε ≤ c0 + h1 (ε) = d and K = ζε (Π), that Ψε has at least catζε (Π) ζε (Π) critical points on Seε+ . So, from item (d) of the Proposition 2.1 and from (3.24), we conclude that Jε has at least eε . catΠδ (Π) critical points in N

4

Proof of Theorem 1.1

In this section we prove our main theorem. The idea is to show that the solutions obtained in Theorem 3.2 verify the following estimate uε (x) ≤ a ∀x ∈ Ωcε for ε small enough. This fact implies that these solutions are in fact solutions of the original problem (Peε ). The key ingredient is the following result, whose proof uses an adaptation of the arguments found in [19], which are related to the Moser iteration method [22] .

28

eε be a solution of (Pε ,A ). Then Jε (un ) → c0 and Lemma 4.1 Let εn → 0+ and un ∈ N n n n un ∈ L∞ (R3 ). Moreover, for any given γ > 0, there exists R > 0 and n0 ∈ N such that |un |L∞ (BR (˜yn )c ) < γ,

for all n ≥ n0 ,

(4.1)

where yñ is given by Proposition 3.2. Proof. Since Jεn (un ) ≤ c0 + h1 (εn ) with lim h1 (εn ) = 0, we can argue as in the proof of n→∞

the inequality (3.20) to conclude that Jεn (un ) → c0 . Thus, we may invoke Proposition 3.2 to obtain a sequence (e yn ) ⊂ R3 satisfying the conclusions of that proposition. Fix R > 1 and consider ηR ∈ C ∞ (R3 ) such that 0 ≤ ηR ≤ 1, ηR ≡ 0 in BR/2 (0), ηR ≡ 1 in BR (0)c and |∇ηR | ≤ C/R. For each n ∈ N and L > 0, we define ηn (x) := ηR (x − yñ ), uL,n ∈ H 1 (R3 ) and zL,n ∈ Hε by uL,n (x) := min{un (x), L},

2(β−1)

zL,n := ηn2 uL,n

un ,

with β > 1 to be determined later. From definition of zL,n and Jε0 n (un )zL,n = 0, we have Z Z 2(β−1) 2 2(β−1) 2 m0 ηn uL,n |∇un | + 2 ηn un uL,n ∇ηn · ∇un R3 Z R3 2(β−1) ≤ (g(εn x, un ) − m0 V (εn x)un ) ηn2 un uL,n . R3

Now, the result follows by arguing as in [6, Lemma 4.1]. We are now ready to prove the main result of the paper.

4.1

Proof of Theorem 1.1

Suppose that δ > 0 is such that Πδ ⊂ Ω. We first claim that there exists εeδ > 0 such that, eε of the problem (Pε,A ), there holds for any 0 < ε < εeδ and any solution u ∈ N |u|L∞ (R3 \Ωε ) < a.

(4.2)

In order to prove the claim we argue by contradiction. So, suppose that for some sequence eε such that Jε0 (un ) = 0 and εn → 0+ we can obtain un ∈ N n n |un |L∞ (R3 \Ωεn ) ≥ a.

(4.3)

As in Lemma 4.1, we have that Jεn (un ) → c0 and therefore we can use Proposition 3.2 to obtain a sequence (e yn ) ⊂ R3 such that εn yen → y0 ∈ Π. If we take r > 0 such that Br (y0 ) ⊂ B2r (y0 ) ⊂ Ω we have that Br/εn (y0 /εn ) = (1/εn )Br (y0 ) ⊂ Ωεn . Moreover, for any z ∈ Br/εn (e yn ), there holds z − y0 ≤ |z − yen | + yñ − y0 < 1 (r + on (1)) < 2r , εn εn εn εn 29

for n large. For these values of n we have that Br/εn (e yn ) ⊂ Ωεn or, equivalently, R3 \ Ωεn ⊂ 3 R \ Br/εn (e yn ). On the other hand, it follows from Lemma 4.1 with γ = a that, for any n ≥ n0 such that r/εn > R, there holds |un |L∞ (R3 \Ωεn ) ≤ |un |L∞ (R3 \Br/εn (eyn )) ≤ |un |L∞ (R3 \BR (eyn )) < a, which contradicts (4.3) and proves the claim. Let εbδ > 0 given by Theorem 3.2 and set εδ := min{b εδ , εeδ }. We shall prove the theorem for this choice of εδ . Let 0 < ε < εδ be fixed. By applying Theorem 3.2 we obtain catΠδ (Π) nontrivial solutions of the problem (Pε,A ). If u ∈ Hε is one of these solutions eε , and therefore we can use (4.2) and the definition of g to conclude we have that u ∈ N that gε (·, u) ≡ f (u). Hence, u is also a solution of the problem (Peε ). An easy calculation shows that u b(x) := u(x/ε) is a solution of the original problem (Pε ). Then, (Pε ) has at least catΠδ (Π) nontrivial solutions. We now consider εn → 0+ and take a sequence un ∈ Hεn of solutions of the problem (Peεn ) as above. In order to study the behavior of the maximum points of un , we first notice that, by (g1 ), there exists γ > 0 such that g(εx, s)s ≤

V0 2 s , for all x ∈ R3 , s ≤ γ. K

(4.4)

By applying Lemma 4.1 we obtain R > 0 and (e yn ) ⊂ R3 such that |un |L∞ (BR (eyn ))c < γ,

(4.5)

Up to a subsequence, we may also assume that |un |L∞ (BR (eyn )) ≥ γ.

(4.6)

Indeed, if this is not the case, we have |un |L∞ (R3 ) < γ, and therefore it follows from Jε0 n (un ) = 0 and (4.4) that Z Z V0 m0 kun k2εn ≤ g(εn x, un )un ≤ u2n . K 3 3 R R The above expression implies that kun kεn = 0, which does not make sense. Thus, (4.6) holds. By using (4.5) and (4.6) we conclude that the maximum point pn ∈ R3 of un belongs to BR (e yn ). Hence pn = yen + qn , for some qn ∈ BR (0). Recalling that the associated solution of (Pεn ) is of the form u bn (x) = un (x/εn ), we conclude that the maximum point ηn of u bn is ηn := εn yen + εn qn . Since (qn ) ⊂ BR (0) is bounded and εn yen → y0 ∈ Π (according to Proposition 3.2), we obtain lim V (ηεn ) = V (y0 ) = V0 ,

n→∞

which concludes the proof of the theorem. Acknowledgement. This work has done while the authors were visiting the ”Departamento de ecuaciones diferenciales y análisis numérico” of the ”Universidad de Sevilla”. They would like to express their gratitude to the Prof. Antonio Suarez for his warm hospitality. 30

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