Multiplicity and concentration behavior of positive solutions for a Schrödinger-Kirchhoff type problem via penalization method

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1 Multiplicity and concentration behavior of positive solutions for a Schrödinger-Kirchhoff type problem via penalization method Giovany M. Figueiredo and João R. Santos Júnior Faculdade de Matemática Universidade Federal do Pará Belém - Pará - Brazil Abstract In this paper we are concerned with questions of multiplicity and concentration behavior of positive solutions of the elliptic problem L εu = f(u) in, (P ε) u > 0 in, u H 1 ( ), where ε is a small positive parameter, f : R R is a continuous function, L ε is a nonlocal operator defined by ( 1 L εu = M u ) ε V (x)u 2 2 u + V (x)u ], ε ε 3 M : IR + IR + and V : IR are continuous functions which verify some hypotheses. Keywords: Penalization method; Schrödinger-Kirchhoff type problem; Lusternik- Schnirelmann 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 L ε u = f(u) in, (P ε ) u > 0 in, u H 1 ( ), Partially supported by CNPq/PQ / and / Partially supported by CAPES - Brazil /

2 where ε is a small positive parameter, f : R R is a continuous function, L ε is a nonlocal operator defined by ( 1 L ε u = M u ) ε ε ε 3 V (x)u 2 2 u + V (x)u ], M : IR + IR + and V : IR are continuous functions that satisfy some conditions which will be stated later on. Problem (P ε ) 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 ( ) M u 2 dx u = f(u) in R 3, R 3 u > 0 in, u H 1 ( ), 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 L 2 norm of the gradient of the displacement u. More precisely, we have M(t) = P 0 h + E t, t > 0, 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 P 0 is the initial tension. ( ) Moreover, problem (P ε ) is catch nonlocal because of the presence of the term M u 2 R 3 which implies that the equation in (P ε ) 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 (P ε ) 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 (1.1) { ε 2 u + V (x)u = f(u) in, u > 0 in, u H 1 ( ), 2

3 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 t = ε Ψ + (V (x) + E)Ψ f(ψ), x IRN, where f(t) = t p 2 u and 2 < p < 2 = 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. A considerable effort has been devoted during the last years in studying problems of the type (P ε ), 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: (M 1 ) There is m 0 > 0 such that M(t) m 0, t 0. (M 2 ) The function t M(t) is increasing. (M 3 ) For each t 1 t 2 > 0, where m 0 is given in (M 1 ). M(t 1 ) M(t ( 2) 1 m 0 1 ), t 1 t 2 t 1 t 2 The potential V is a continuous function satisfying: (V 1 ) There is V 0 > 0 such that V 0 = inf x V (x). (V 2 ) For each δ > 0 there is a bounded and Lipschitz domain Ω such that and V 0 < min Ω V, Π = {x Ω : V (x) = V 0} Π δ = {x : dist(x, Π) δ} Ω. Moreover, we assume that the continuous function f vanishes in (, 0) and verifies (f 1 ) (f 2 ) There is q (4, 6) such that f(t) lim t 0 + t 3 = 0. lim t f(t) = 0. tq 1 3

4 (f 3 ) There is θ (4, 6) such that (f 4 ) The application is non-decreasing in (0, ). The main result of this paper is: 0 < θf (t) f(t)t, t > 0. t f(t) t 3 Theorem 1.1 Suppose that the function M satisfies (M 1 ) (M 3 ), the potential V satisfies (V 1 ) (V 2 ) and the function f satisfies (f 1 ) (f 4 ). Then, given δ > 0 there is ε = ε(δ) > 0 such that the problem (P ε ) has at least Cat Πδ (Π) positive solutions, for all ε (0, ε). Moreover, if u ε denotes one of these positive solutions and η ε R 3 its global maximum, then lim ε 0 V (η ε) = V 0. A typical example of function verifying the assumptions (M 1 ) (M 3 ) is given by M(t) = m 0 + bt, where m 0 > 0 and b > 0. More generally, any function of the form M(t) = k m 0 + bt + b i t γi with b i 0 and γ i (0, 1) for all i {1, 2,..., k} verifies the hypotheses i=1 (M 1 ) (M 3 ). A typical example of function verifying the assumptions (f 1 ) (f 4 ) is given by f(t) = k c i (t + ) qi 1 with c i 0 not all null and q i θ, 6) for all i {1, 2,..., k}. i=1 Recently some authors have considered problems of the type (P ε ). 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 (1.5) (ε 2 a + bε u 2 ) u + V (x)u = f(u) in u > 0 in, u H 1 ( ), 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) > V 0 = inf V (x) > 0. x 3 In 27], Wang, Tian, Xu and Zhang have considered the problem (ε 2 a + bε u 2 ) u + V (x)u = λf(u) + u 4 u in IR (1.6) 3 u > 0 in IR 3, u H 1 ( ). IR 4

5 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 Lusternik- Schnirelmann theory and concentration behavior of solutions for the problem (P ε ). Here we use the hypotheses (V 1 ) (V 2 ) 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 (P ε ) we obtain the modified problem ( P ε ) L ε u = f(u) in, u > 0 in, u H 1 ( ), where ( ) L ε u = M u 2 + V (εx)u 2 u + V (εx)u], which is clearly equivalent to (P ε ). 5

6 Since (f 1 ) and (f 4 ) imply that and f(t) lim = 0 t 0 + t t 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]. For this, let K > 2, where m 0 is given in (M 1 ) and a > 0 such that f(a) = V 0 a. We m 0 K define { f(t) if t a, f(t) = V 0 K t if t > a and g(x, t) = χ Ω (x)f(t) + (1 χ Ω (x)) f(t), where χ is characteristic function of set Ω. From hypotheses (f 1 ) (f 4 ) we get that g is a Carathéodory function and the following conditions are observed: (g 1 ) (g 2 ) (g 3 ) (i) and (ii) g(x, t) lim t 0 + t 3 = 0, uniformly in x. g(x, t) lim t t q 1 = 0, uniformly in x R 3, 0 θg(x, t) < g(x, t)t, x Ω and t > 0 0 2G(x, t) g(x, t)t 1 K V 0t 2, x \Ω and t > 0. (g 4 ) For each x Ω, the application t g(x,t) t is increasing in (0, ) and for each x 3 \Ω, the application t g(x,t) t is increasing in (0, a). 3 Moreover, from definition of g, we have g(x, t) f(t), for all t (0, + ) and for all x, g(x, t) = 0 for all t (, 0) and for all x. Now we study the auxiliary problem (P ε,a ) L ε u = g(εx, u), u > 0, u H 1 ( ). 6

7 Observe that positive solutions of (P ε,a ) with u(x) a for each x \Ω are also positive solutions of ( P ε ). We obtain solutions of (P ε,a ) as critical points of the energy functional where M(t) = t space H ε, given by J ε (u) = 1 ( ) 2 M u 2 + V (εx)u 2 0 M(s)ds and G(x, t) = t 0 G(εx, u), g(εx, s)ds, which is well defined on the Hilbert H ε = {u H 1 ( ) : V (εx)u 2 < }, provided of the inner product (u, v) ε = u v + V (εx)uv. The norm induced by inner product is denoted by u 2 ε = u 2 + V (εx)u 2. Since M and f are continuous we have that J ε C 1 (H ε, IR) and J ε(u)v = M( u 2 ε)(u, v) ε g(εx, u)v, u, v H ε. Now, we will fix some notations. We denote the Nehari manifold associated to J ε by and by Ω ε the set by H + ε the subset of H ε given by N ε = {u H ε \{0} : J ε(u)u = 0} Ω ε = {x : εx Ω}, H + ε = {u H ε : supp(u + ) Ω ε > 0} and by S + ε the intersection S ε H + ε, where S ε is the unit sphere of H ε. Lemma 2.1 The set H + ε is open in H ε. Proof. Suppose by contradiction there are a sequence {u n } H ε \H + ε and u H + ε such that u n u in H ε. Hence supp(u + n ) Ω ε = 0 for all n IN and u + n (x) u + (x) a.e. in x Ω ε. So, u + (x) = lim u+ n (x) = 0, a.e in x Ω ε. But, this contradicts the fact that u H ε +. Therefore H ε + is open. 7

8 From definition of S ε + and from Lemma 2.1 it follows that S ε + is a incomplete C 1,1 - manifold of codimension 1, modeled on H ε and contained in the open H ε +. Thus, H ε = T u S ε + IR u for each u S ε +, where T u S ε + = {v H ε : (u, v) ε = 0}. Finally, we mean by weak solution of (P ε,a ) a function u H ε such that M( u 2 ε)(u, v) ε = g(εx, u)v, v H ε. Therefore, critical points of J ε are weak solutions of (P ε,a ). Lemma 2.2 The functional J ε satisfies the following conditions: a) There are α, ρ > 0 such that b) There is e H ε \B ρ (0) with J ε (e) < 0. J ε (u) α, with u ε = ρ. Proof. The item a) follows directly from the hypotheses (M 1 ), (g 1 ) and (g 2 ). On the other hand, it follows from (M 3 ) that there is γ 1 > 0 such that M(t) γ 1 (1 + t) for all t 0. So, for each u H ε + and t > 0 we have J ε (tu) = 1 2 M( tu 2 ε) G(εx, tu) γ 1 2 t2 u 2 ε + γ 1 4 t4 u 4 ε G(εx, tu). Ω ε From (g 3 )(i), we obtain C 1, C 2 > 0 such that J ε (tu) γ 1 2 t2 u 2 ε + γ 1 4 t4 u 4 ε C 1 t θ Ω ε (u + ) θ + C 2 supp(u + ) Ω ε. 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 N ε (see Lemma 2.3 (A 3 ) and Proposition 2.1) and the incompleteness of S + ε (see Lemma 2.3 (A 4 )). Lemma 2.3 Suppose that the function M satisfies (M 1 ) (M 3 ), the potential V satisfies (V 1 ) (V 2 ) and the function f satisfies (f 1 ) (f 4 ). So: (A 1 ) For each u H + ε, let h : IR + IR be defined by h u (t) = J ε (tu). Then, there is a unique t u > 0 such that h u(t) > 0 in (0, t u ) and h u(t) < 0 in (t u, ). (A 2 ) there is τ > 0 independent on u such that t u τ for all u S + ε. Moreover, for each compact set W S + ε there is C W > 0 such that t u C W, for all u W. (A 3 ) The map m ε : H + ε N ε given by m ε (u) = t u u is continuous and m ε := m ε S + ε is a homeomorphism between S + ε and N ε. Moreover, m 1 ε (u) = u u ε. 8

9 (A 4 ) If there is a sequence (u n ) S + ε such that dist(u n, S + ε ) 0, then m ε (u n ) ε and J ε (m ε (u n )). Proof. To prove (A 1 ), it is sufficient to note that, from the Lemma 2.2, h u (0) = 0, h u (t) > 0 when t > 0 is small and h u (t) < 0 when t > 0 is large. Since h u C 1 (IR +, IR), there is t u > 0 global maximum point of h u and h u(t u ) = 0. Thus, J ε(t u u)(t u u) = 0 and t u u N ε. We see that t u > 0 is the unique positive number such that h u(t u ) = 0. Indeed, suppose by contradiction that there are t 1 > t 2 > 0 with h u(t 1 ) = h u(t 2 ) = 0. Then, for i = 1, 2 t i M( t i u 2 ε) u 2 ε = g(εx, t i u)u. So, Therefore, M( t 1 u 2 ε) t 1 u 2 ε M( t i u 2 ε) t i u 2 ε M( t 2u 2 ε) t 2 u 2 ε It follows from (M 3 ) and (g 4 ) that ( m 0 1 u 2 ε t 2 1 ) 1 1 t 2 2 u 4 ε 1 + u 4 ε = 1 u 4 ε g(εx, ti u) (t i u) 3 ] u 4. = 1 g(εx, t1 u) u 4 ε (t 1 u) 3 g(εx, t ] 2u) (t 2 u) 3 u 4. ( \Ω ε) {t 2u a<t 1u} ( \Ω ε) {a<t 2u} By using the definition of g we obtain ( m 0 1 u 2 ε t 2 1 ) 1 1 t 2 2 u 4 ε ( u 4 ε K t 2 1 ) 1 t 2 2 Multiplying both sides by m 0 u 2 ε + 1 K u ( 4 ε 1 t t 2 2 t 2 1t 2 2 t 2 2 t2 1 Therefore, ( ) t m 0 u 2 2 ε 2 1 t 2 1 t2 2 K ( ) t t 2 1 t2 2 ( \Ω ε) {t 2u a<t 1u} g(εx, t1 u) (t 1 u) 3 g(εx, t ] 2u) (t 2 u) 3 u 4 g(εx, t1 u) (t 1 u) 3 g(εx, t ] 2u) (t 2 u) 3 u 4. V0 1 K (t 1 u) 2 f(t ] 2u) (t 2 u) 3 u 4 V 0 u 2. ( \Ω ε) {a<t 2u} ) and using the hypothesis t 1 > t 2, it follows that ( \Ω ε) {t 2u a<t 1u} ( \Ω ε) {a<t 2u} V 0 u 2. ( \Ω ε) {t 2u a<t 1u} ( \Ω ε) {t 2u a<t 1u} 9 V0 1 K (t 1 u) 2 f(t ] 2u) (t 2 u) 3 u 4 V 0 u 2 f(t 2 u) t 2 u u2 + 1 K ( \Ω ε) {a<t 2u} V 0 u 2.

10 So, m 0 u 2 ε 1 K \Ω ε V 0 u 2 1 K u 2 ε. Since u 0, we have that m 0 1 K < m 0, but this is a contradiction. Thus, (A 1 ) is proved. (A 2 ) Now, let u S ε +. From (M 1 ), (g 1 ), (g 2 ) and from the Sobolev embeddings m 0 t u M(t 2 u)t u = g(εx, t u u)u ξ 4 C 1t 4 u + C ξ q C 2t q u. From previous inequality we obtain τ > 0, independent on u, such that t u τ. Finally, if W S ε + is compact, suppose by contradiction that there is {u n } W such that t n = t un. Since W is compact, there is u W with u n u in H ε. It follows from the arguments used in the proof of item b) of the Lemma 2.2 that On the other hand, note that if v N ε, then by (g 3 )(i) J ε (v) = J ε (v) 1 θ J ε(v)v 1 2 M( v 2 ε) 1 θ M( v 2 ε) v 2 ε + 1 θ J ε (t n u n ). (2.1) \Ω ε g(εx, v)v θg(εx, v)]. From (g 3 )(ii) we have J ε (v) 1 2 M( v 2 ε) 1 ( ) θ 2 1 θ M( v 2 ε) v 2 ε V (εx)v 2, 2θ K \Ω ε and so J ε (v) 1 2 M( v 2 ε) 1 ( ) θ 2 1 θ M( v 2 ε) v 2 ε 2θ K v 2 ε. By using the hypothesis (M 3 ), we derive M(t) M(t) + m 0] t, for all t 0. Then, 2 ( ) θ 4 J ε (v) M( v 2 4θ ε) v 2 ε + m ( ) 0 θ v 2 ε 2θ K v 2 ε. From (M 1 ), we conclude ( ) ( θ 2 J ε (v) m 0 1 ) v 2 2θ K ε. Once {t n u n } N ε the previous inequality contradicts (2.1). Therefore (A 2 ) is true. (A 3 ) First of all we observe that m ε, m ε and m 1 ε are well defined. In fact, by (A 1 ), for each u H ε +, there is a unique m ε (u) N ε. On the other hand, if u N ε then u H ε +. Otherwise, we have supp(u + ) Ω ε = 0 and by (g 3 )(ii) 0 < M( u 2 ε) u 2 ε = g(εx, u)u = g(εx, u + )u + 1 V (εx)u 2. \Ω ε K \Ω ε 10

11 Hence, from (M 1 ) 0 < ( m 0 1 ) u 2 ε 0, K a contradiction. Consequently m 1 ε (u) = u u ε continuous function. Since, S + ε, m 1 ε m 1 ε (m ε (u)) = m 1 ε (t u u) = t uu = u, u S ε +, t u u ε is well defined and it is a we conclude that m ε is a bijection. To show that m ε : H ε + N ε is continuous, let {u n } H ε + and u H ε + be such that u n u in H ε. Thus u n / u n ε u/ u ε in H ε and from (A 2 ), there is t 0 > 0 such that t ( un un ε ) t 0. Since, t ( un un ε ) (u n / u n ε ) N ε, we obtain ( M ) t 2 ( un un ε ) t ( un un ε ) = 1 u n ε g Passing to the limit n, it follows that M(t 2 0)t 0 = 1 u ε u g ( εx, t ( un un ε ) u n u n ε ( ) u εx, t 0 u. u ε ) u n, n IN. Hence t 0 u ε N ε and, by (A 1 ), t ( u u ε ) = t 0, showing that m ε (u n ) = m ε ( un u n ε ) m ε ( u u ε ) = m ε (u) in H ε. So, m ε and m ε are continuous functions and (A 3 ) is proved. (A 4 ) Finally, let {u n } S ε + be a sequence such that dist(u n, S ε + ) 0. Since, for each v S ε + and n IN, we have it follows that Ω ε (u + n ) s u + n (x) u n (x) v(x) a.e in x Ω ε, inf v S ε + u n v s, n IN and s 2, 6]. Ω ε (2.2) Hence, from (V 1 ), (V 2 ) and Sobolev s embedding, there is C(s) > 0 such that { (u + n ) s C(s) inf (un v) 2 + V (εx)(u n v) 2] Ω ε v S ε + Ω ε C(s)dist(u n, S ε + ) s, n IN. From (g 1 ), (g 2 ) and (g 3 )(ii), there are positive constants C 1 and C 2, such that, for each t > 0 G(εx, tu n ) F (tu n ) + t2 V (εx)u 2 n Ω ε K \Ω ε C 1 t 4 (u + n ) 4 + C 2 t q (u + n ) q + 1 Ω ε Ω ε K t2 u n 2 ε C 1 C(4)t 4 dist(u n, S ε + ) 4 + C 2 C(q)t q dist(u n, S + ε ) q + 1 K t2. } s/2 11

12 Therefore, From definition of m ε, we have lim sup G(εx, tu n ) 1 K t2, t > 0. lim inf J ε(m ε (u n )) lim inf J ε(tu n ) 1 2 M(t 2 ) 1 K t2, t > 0. It follows from (M 1 ) and from the particular choice of K, that lim J ε(m ε (u n )) =. Since 1 2 M(t 2 u n ) J ε (m ε (u n )), for each n IN, we conclude from (M 3 ) that m ε (u n ) ε as n. The Lemma is proved. We set the applications by Ψ ε (u) = J ε ( m ε (u)) and Ψ ε := ( Ψ ε ) S + ε. Ψ ε : H + ε IR and Ψ ε : S + ε IR, The next proposition is a direct consequence of the Lemma 2.3. The details can be seen in the relevant material from 26]. For the convenience of the reader, here we do a sketch of the proof. Proposition 2.1 Suppose that the function M satisfies (M 1 ) (M 3 ), the potential V satisfies (V 1 ) (V 2 ) and the function f satisfies (f 1 ) (f 4 ). Then: (a) Ψ ε C 1 (H + ε, IR) and (b) Ψ ε C 1 (S + ε, IR) and Ψ ε(u)v = m ε(u) ε u ε J ε( m ε (u))v, u H + ε and v H ε. Ψ ε(u)v = m ε (u) ε J ε(m ε (u))v, v T u S + ε. (c) If {u n } is a (P S) d sequence for Ψ ε then {m ε (u n )} is a (P S) d sequence for J ε. If {u n } N ε is a bounded (P S) d sequence for J ε then {m 1 ε (u n )} is a (P S) d sequence for Ψ ε. (d) u is a critical point of Ψ ε if, and only if, m ε (u) is a nontrivial critical point of J ε. Moreover, corresponding critical values coincide and inf Ψ ε = inf J ε. S ε + N ε 12

13 Proof. (a) Consider u H ε + value Theorem, and v H ε. From definition of Ψ ε, definition of t u and mean Ψ ε (u + sv) Ψ ε (u) = J ε (t u+sv (u + sv)) J ε (t u u) J ε (t u+sv (u + sv)) J ε (t u+sv u) = J ε(t u+sv (u + τsv))t u+sv sv, where s is small sufficient and τ (0, 1). On the other hand, Ψ ε (u + sv) Ψ ε (u) J ε (t u (u + sv)) J ε (t u u) = J ε(t u (u + ςsv))t u sv, where ς (0, 1). Since u t u is a continuous application, it follows from previous inequalities that Ψ ε (u + sv) lim Ψ ε (u) = t u J s 0 s ε(t u u)v = m ε(u) ε J u ε( m ε (u))v. ε Since J ε C 1, it follows that the Gateaux derivative of Ψ ε is linear, bounded on v and it is continuous on u. From 28], Proposition 1.3], Ψ ε C 1 (H + ε, IR) and Ψ ε(u)v = m ε(u) ε u ε J ε( m ε (u))v, u H + ε and v H ε. The item (a) is proved. (b) The item (b) is a direct consequence of the item (a). (c) Once H ε = T u S ε + IR u for each u S ε +, the linear projection P : H ε T u S ε + defined by P (v + tu) = v is continuous, namely, there is C > 0 such that From item (a), we obtain v ε C v + tu ε, u S + ε, v T u S + ε and t IR. (2.3) Ψ ε(u) = sup v TuS + ε v ε=1 Ψ ε(u)v = w ε where w = m ε (u). Since w N ε, we conclude that By (2.4), we have Hence, from (2.3) and (2.5) sup v TuS + ε v ε=1 J ε(w)v, (2.4) J ε(w)u = J ε(w) w = 0. (2.5) w ε Ψ ε(u) w ε J ε(w) = w ε Ψ ε(u) w ε J ε(w) C w ε sup v TuS + ε,t IR v+tu 0 sup v T us + ε \{0} J ε(w)(v + tu) v + tu ε. J ε(w)(v) v ε = C Ψ ε(u), 13

14 showing that, Ψ ε(u) w ε J ε(w) C Ψ ε(u), u S + ε. (2.6) Since w N ε, we have w τ > 0. Therefore, the inequality in (2.6) together with J ε (w) = Ψ ε (u) imply the item (c). (d) It follows from (2.6) that Ψ ε(u) = 0 if, and only if, J ε(w) = 0. The remainder follows from definition of Ψ ε. By using (M 1 ) (M 3 ) we have, as in 26], the following variational characterization of the infimum of J ε over N ε : c ε = inf J ε (u) = inf max J ε(tu) = inf max J ε(tu). (2.7) u N ε u H ε + t>0 u S ε + t>0 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 {u n } be a (P S) d sequence for J ε. Then {u n } is bounded. Proof. Since {u n } a (P S) d sequence for J ε, then there is C > 0 such that C + u n ε J ε (u n ) 1 θ J ε(u n )u n, n IN. From (M 3 ) and (g 3 ), we obtain ( ) ( θ 2 C + u n ε m 0 1 ) u n 2 2θ K ε, n IN. Therefore {u n } is bounded in H ε. Lemma 2.5 Let {u n } be a (P S) d sequence for J ε. Then for each ξ > 0, there is R = R(ξ) > 0 such that lim sup un 2 + V (εx)u 2 ] n < ξ. \B R Proof. Let η R C ( ) such that 0 se x B R (0) η R (x) = 1 se x B 2R (0), where 0 η R (x) 1, η R C R and C is a constant independent on R. Note that {η Ru n } is bounded in H ε. From definition of J ε η R M( u n 2 ε) u n 2 + V (εx)u 2 ] n = J ε(u n )(u n η R ) + g(εx, u n )u n η R IR 3 M( u n 2 ε)u n u n η R. 14

15 Choosing R > 0 such that Ω ε B R (0) and by using (M 1 ) and (g 3 )(ii), we have m 0 η R un IR 2 + V (εx)u 2 ] n 3 J ε(u n )(u n η R ) 1 + K V (εx)u2 nη R M( u n 2 ε)u n u n η R. Therefore, ( m 0 1 ) η R un 2 + V (εx)u 2 ] n J K ε (u n )(u n η R ) + M( u n 2 ε)u n u n ( η R ). By using Cauchy-Schwarz inequality in, definition of η R, Holder s inequality and the boundedness of {u n } in H ε, we conclude that un 2 + V (εx)u 2 ] n C J ε (u n )(u n η R ) + C \B R R. Since {u n η R } is bounded in H ε and {u n } is a (P S) d sequence for J ε, passing to the upper limit of n, we obtain lim sup un 2 + V (εx)u 2 ] C n \B R R < ξ, 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 {u n } be a (P S) d sequence for J ε such that u n u, then lim un 2 + V (εx)u 2 ] n = u 2 + V (εx)u 2], B R B R for all R > 0. Proof. We can assume that u n ε t 0, thus, we have u ε t 0. Let η ρ C ( ) such that 1 se x B ρ (0) η ρ (x) = 0 se x B 2ρ (0). with 0 η ρ (x) 1. Let, P n (x) = M( u n 2 ε) u n u 2 + V (εx)(u n u) 2]. For each R > 0 fixed, choosing ρ > R we obtain P n = P n η ρ M( u n 2 ε) un u 2 + V (εx)(u n u) 2] η ρ. B R B R 15

16 By expanding the inner product in, P n M( u n 2 ε) un 2 + V (εx)(u n ) 2] η ρ B R IR 3 2M( u n 2 ε) u n u + V (εx)u n u] η ρ IR 3 + M( u n 2 ε) u 2 + V (εx)u 2] η ρ. Setting and In,ρ 1 = M( u n 2 ε) un 2 + V (εx)(u n ) 2] η ρ In,ρ 2 = M( u n 2 ε) In,ρ 3 = M( u n 2 ε) We have that, g(εx, u n )u n η ρ, u n u + V (εx)u n u] η ρ g(εx, u n )uη ρ, IR 3 u n u + V (εx)u n u] η ρ + M( u n 2 ε) u 2 + V (εx)u 2] η ρ In,ρ 4 = g(εx, u n )u n η ρ g(εx, u n )uη ρ. 0 P n In,ρ 1 + In,ρ 2 + In,ρ 3 + In,ρ. 4 B R (2.8) Observe that In,ρ 1 = J ε(u n )(u n η ρ ) M( u n 2 ε) u n u n η ρ. Since {u n η ρ } is bounded in H ε, we have J ε(u n )(u n η ρ ) = o n (1). Moreover, from a straightforward computation ] lim lim sup ρ M( u n 2 ε) u n u n η ρ = 0. Then, We see also that ] lim lim sup In,ρ 1 ρ In,ρ 2 = J ε(u n )(uη ρ ) M( u n 2 ε) u u n η ρ. By arguing in the same way as in the previous case, J ε(u n )(uη ρ ) = o n (1) = 0. (2.9) and ] lim lim sup ρ M( u n 2 ε) u u n η ρ = 0. 16

17 Therefore, ] lim lim sup In,ρ 2 ρ On the other hand, from the weak convergence Finally, from we conclude that = 0. (2.10) lim I3 n,ρ = 0, ρ > R. (2.11) u n u, in L s loc( ), 1 s < 6, From (2.8), (2.9), (2.10), (2.11) and (2.12), we obtain 0 lim sup P n 0. B R Hence, lim P n = 0 and consequently B R lim lim I4 n,ρ = 0, ρ > R. (2.12) un 2 + V (εx)u 2 ] n = u 2 + V (εx)u 2]. B R B R Proposition 2.2 The functional J ε verifies the (P S) d condition in H ε. Proof. Let {u n } be a (P S) d sequence for J ε. From Lemma 2.4 we know that {u n } is bounded in H ε. Passing to a subsequence, we obtain u n u, in H ε. 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 lim sup un 2 + V (εx)u 2 ] n < ξ. \B R Therefore, from Lemma 2.6, u 2 ε lim inf u n 2 ε lim sup u n 2 ε { = lim sup un 2 + V (εx)u 2 ] n + un 2 + V (εx)u 2 ] } n B R \B R = u 2 + V (εx)u 2] + lim sup un 2 + V (εx)u 2 ] n B R \B R < u 2 + V (εx)u 2] + ξ, B R 17

18 where R = R(ξ) > C/ξ. Passing to the limit of ξ 0 we have R, which implies u 2 ε lim inf u n 2 ε lim sup u n 2 ε u 2 ε, and so u n ε u ε and consequently u n u in H ε. 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 Ψ ε verifies the (P S) d condition on S + ε. Proof. Let {u n } S + ε be a (P S) d sequence for Ψ ε. Thus, Ψ ε (u n ) d and Ψ ε(u n ) 0, where. is the norm in the dual space (T un S ε + ). It follows from Proposition 2.1(c) that {m ε (u n )} is a (P S) d sequence for J ε in H ε. From Proposition 2.2 we conclude there is u S ε + such that, passing to a subsequence, From Lemma 2.3(A 3 ), it follows that m ε (u n ) m ε (u) in H ε. u n u in S + ε. Theorem 2.1 Suppose that the function M satisfies (M 1 ) (M 3 ), the potential V satisfies (V 1 ) (V 2 ) and the function f satisfies (f 1 ) (f 4 ). Then, the auxiliary problem (P ε,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 Multiplicity of solutions of auxiliary problem 3.1 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 ( P ε ), namely, the problem (P 0 ) L 0 u = f(u), u > 0, u H 1 ( ) 18

19 where ( ) L 0 u = M u 2 + V 0 u 2 u + V 0 u], which has the following associated functional I 0 (u) = 1 ( ) 2 M u 2 + V 0 u 2 F (u). This functional is well defined on the Hilbert space H 0 = H 1 ( ) with the inner product (u, v) 0 = u v + V 0 uv and norm u 2 0 = u 2 + V 0 u 2 fixed. We denote the Nehari manifold associated to I 0 by N 0 = {u H 0 \{0} : I 0(u)u = 0}. We denote by H + 0 the open subset of H 0 given by H + 0 = {u H 0 : supp(u + ) > 0}, and S 0 + = S 0 H 0 +, where S 0 is the unit sphere of H 0. As in the section 2, S 0 + is a incomplete C1,1 -manifold of codimension 1, modeled on H 0 and contained in the open H 0 +. Thus, H 0 = T u S 0 + IR u for each u S+ 0, where T u S 0 + = {v H 0 : (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 (M 1 ) (M 3 ) and the function f satisfies (f 1 ) (f 4 ). So: (A 1 ) For each u H + 0, let h : IR + IR be defined by h u (t) = I 0 (tu). Then, there is a unique t u > 0 such that h u(t) > 0 in (0, t u ) and h u(t) < 0 in (t u, ). (A 2 ) there is τ > 0 independent on u such that t u τ for all u S 0 +. Moreover, for each compact set W S 0 + there is C W > 0 such that t u C W, for all u W. (A 3 ) The map m : H 0 + N 0 given by m(u) = t u u is continuous and m := m S + 0 homeomorphism between S 0 + and N 0. Moreover, m 1 (u) = u u 0. is a (A 4 ) If there is a sequence (u n ) S + 0 such that dist(u n, S + 0 ) 0, then m(u n) 0 and I 0 (m(u n )). 19

20 We set the applications by Ψ 0 (u) = I 0 ( m(u)) and Ψ 0 := ( Ψ 0 ) S. + 0 Ψ 0 : H + 0 IR and Ψ 0 : S + 0 IR, Proposition 3.1 Suppose that the function M satisfies (M 1 ) (M 3 ) and the function f satisfies (f 1 ) (f 4 ). So: (a) Ψ 0 C 1 (H 0 +, IR) and (b) Ψ 0 C 1 (S 0 +, IR) and Ψ 0(u)v = m(u) 0 u 0 I 0( m(u))v, u H + 0 and v H 0. Ψ 0(u)v = m(u) 0 I 0(m(u))v, v T u S + 0. (c) If {u n } is a (P S) d sequence for Ψ 0 then {m(u n )} is a (P S) d sequence for I 0. If {u n } N 0 is a bounded (P S) d sequence for I 0 then {m 1 (u n )} is a (P S) d sequence for Ψ 0. (d) u is a critical point of Ψ 0 if, and only if, m(u) is a nontrivial critical point of I 0. Moreover, corresponding critical values coincide and Remark 3.1 As in the section 2, there holds inf Ψ 0 = inf I 0. S + N 0 0 c 0 = inf I 0 (u) = inf max I 0(tu) = inf max I 0(tu). (3.1) u N 0 u H + t>0 0 u S + t>0 0 The next Lemma allows us to assume that the weak limit of a (P S) d sequence is nontrivial. Lemma 3.2 Let {u n } H 0 be a (P S) d sequence for I 0 with u n 0. Then, only one of the alternatives below holds: a) u n 0 in H 0 b) there is a sequence (y n ) and constants R, β > 0 such that lim inf u 2 n β > 0. B R (y n) Proof. Suppose that b) doesn t hold. It follows that for all R > 0 we have lim sup u 2 n = 0. y B R (y) 20

21 Since {u n } is bounded in H 0, we conclude from 28], Lemma 1.21] that u n 0 in L s ( ), 2 < s < 6. From (M 1 ), (f 1 ) and (f 2 ), 0 m 0 u n 0 f(u n )u n + o n (1) = o n (1). 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 {u n } for the functional I 0, then we can assume u 0, otherwise we would have u n 0 and, once it doesn t occur u n 0, we conclude from the Lemma 3.2 that there are {y n } and R, β > 0 such that lim inf u 2 n β > 0. B R (y n) Set v n (x) = u n (x + y n ), making a change of variable, we can prove that {v n } is a (P S) c0 sequence for the functional I 0, it is bounded in H 0 and there is v H 0 with v n v in H 0 with v 0. In the next Proposition we obtain a positive ground-state solution for the autonomous problem (P 0 ). Theorem 3.1 Let {u n } H 0 be a (P S) c0 sequence for I 0. Then there is u H 0 \{0} with u 0 such that, passing a subsequence, we have u n u in H 0. Moreover, u is a positive ground-state solution for the problem (P 0 ). Proof. Arguing as Lemma 2.4, we have that {u n } is bounded in H 0. Thus, passing a subsequence if necessary, we obtain and So, from (3.2) we conclude that u n u em H 0, (3.2) u n u em L s loc( ), 1 s < 6 (3.3) u n 0 t 0. (3.4) (u n, v) 0 (u, v) 0, v H 0. (3.5) On the other hand, due to density of C0 ( ) in H 0 and from convergence in (3.3), it results that f(u n )v f(u)v, v H 0. (3.6) Now, from convergence in (3.2) and (3.4), occurs u 2 0 lim inf u n 2 0 = t 2 0, 21

22 and from (M 2 ) it follows that M( u 2 0) M(t 2 0). Since (M 3 ) implies that the function t 1 M(t) 2 1 4M(t)t is non-decreasing, we can argue as in 4] and to prove that M(t 2 0) = M( u 2 0) and the theorem now follows from fact that functional I 0 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 {u n } be a sequence in H 1 ( ) such that I 0 (u n ) c 0 and {u n } N 0. Then, {u n } has a convergent subsequence in H 1 ( ). Proof. Since {u n } N 0, it follows from Lemma 3.1(A 3 ), Proposition 3.1(d) and Remark 3.1 that v n = m 1 (u n ) = u n S 0 +, n IN (3.7) u n 0 and Ψ 0 (v n ) = I 0 (u n ) c 0 = inf Ψ 0. S + 0 Although S 0 + is incomplete, due to Lemma 3.1(A 4), we can still apply the Ekeland s variational principle 13], Theorem 1.1] to the functional ξ 0 : V IR { } defined by ξ 0 (u) = Ψ 0 (u) if u S 0 + and ξ 0(u) = if u S 0 +, where V = S+ 0 is a complete metric space equipped with the metric d(u, v) = u v 0. In fact, from Lemma 3.1(A 4 ), ξ 0 C(V, IR { }) and, from Proposition 3.1(d), ξ 0 is bounded from below. Thus, we can conclude there is a sequence { v n } S 0 + such that { v n} is a (P S) c0 sequence for Ψ 0 on S 0 + and v n v n 0 = o n (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 (P ε,a ) to topology of Π, for this we need some preliminary results. 3.2 Technical results Let δ > 0 fixed and Π δ 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 (P 0 ) (see Theorem 3.1). For each y Π = {x Ω : V (x) = V 0 }, we define the function ( ) εx y Υ ε,y (x) = η( εx y )w. ε Let t ε > 0 be the unique positive number such that max t 0 J ε(t Υ ε,y ) = J ε (t ε Υε,y ). 22

23 By noticing that t ε Υε,y N ε, we can now define the continuous function Lemma 3.4 Let Π Ω. Then, Υ ε : Π N ε y Υ ε (y) = t ε Υε,y. lim J ε(υ ε (y)) = c 0 uniformly in y Π. ε 0 Proof. Arguing by contradiction, we suppose that there exist δ 0 > 0 and a sequence {y n } Π verifying J εn (Υ εn (y n )) c 0 δ 0 where ε n 0 when n. (3.9) From definition of Υ εn (y n ), we have and J εn (Υ εn (y n )) = 1 2 M (t 2 εn Υ εn,y n 2 εn ) G (ε n x, t εn Υεn,y n ) (3.10) J ε n (Υ εn (y n ))Υ εn (y n ) = 0. (3.11) Using definition of Υ εn (y n ) again and making the change of variable z = εnx yn ε n, we have J εn (Υ εn (y n )) = ( )) 1 (t 2 M 2εn (η( ε n z )w(z)) 2 + V (ε n z + y n ) (η( ε n z )w(z)) 2 IR 3 G (ε n z + y n, t εn η( ε n z )w(z)). Moreover, putting Λ 2 n = (η( ε n z )w(z)) 2 + V (ε n z + y n ) (η( ε n z )w(z)) 2, the equality in (3.11) yields M(t 2 ε n Λ 2 n) t 2 ε n Λ 2 n = 1 Λ 4 n ] g(ε n z + y n, t εn η( ε n z )w(z)) (t εn η( ε n z )w(z)) 3 (η( ε n z )w(z)) 4. For each n IN and for all z B δ (0), we have ε n z B δ (0). So, εn ε n z + y n B δ (y n ) Π δ Ω. Since G = F in Ω, it follows from (3.10) that J εn (Υ εn (y n )) = 1 2 M(t 2 ε n Λ 2 n) F (t εn η( ε n z )w(z)) (3.12) 23

24 and M(t 2 ε n Λ 2 n) t 2 ε n Λ 2 n = 1 f(t εn η( ε n z )w(z)) Λ 4 n (t εn η( ε n z )w(z)) 3 From the Lebesgue s theorem, when n and ] (η( ε n z )w(z)) 4. (3.13) Υ εn,y n 2 ε n = Λ 2 n w 2 0, (3.14) f(η( ε n z )w(z))η( ε n z )w(z) f(w)w F (η( ε n z )w(z)) F (w). (3.15) We see that there is a subsequence, still denoted by {t εn }, with t εn 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(t 2 ε n Λ 2 n) t 2 ε n Λ 2 n 1 Λ 4 n B δ2 (0) f(t εn w(z)) (t εn w(z)) 3 ] w(z) 4. From continuity of w (which follows from standard regularity theory), there is ẑ such that w(ẑ) = min w(z). So, from (f 4 ) B δ2 (0) 1 Λ 4 n f(t εn w(ẑ)) (t εn w(ẑ)) 3 B δ2 (0) w(z) 4 M(t2 ε n Λ 2 n) t 2 ε n Λ 2. (3.16) n Suppose by contradiction that there is a subsequence {t εn } with t εn. Thus, passing to the limit as n in (3.16), we conclude, from (M 3 ) and (f 3 ), that the left side converges to infinity and the right side is bounded, which is a contradiction. Therefore, {t εn } is bounded and passing to a subsequence we have t εn t 0 with t 0 0. From (3.13), (3.14), (M 1 ) and (f 4 ) we have that t 0 > 0. Thus, passing to the limit as n in (3.13), we have M(t 2 0 w 2 0) w 2 0t 0 = f(t 0 w)w. (3.17) Since w N 0, we obtain t 0 = 1. So, passing to the limit of n in (3.12) and using (3.14) and (3.15) we obtain lim J ε n (Υ εn (y n )) = I 0 (w) = c 0, which is a contradiction with (3.9). Let s consider the specific subset of the Nehari manifold Ñ ε = {u N ε : J ε (u) c 0 + h 1 (ε)}, 24

25 where h 1 : IR + IR + is a function such that Υ ε (Π) Ñε and lim ε 0 h 1 (ε) = 0. Observe that h 1 exists due to the Lemma 3.4. In particular, Ñ ε for all small ε > 0. Now we consider ρ > 0 such that Π δ B ρ (0) and χ : defined by χ(x) = x se x ρ ρx x se x ρ. We also consider the barycenter map β ε : N ε given by χ(εx)u(x) 2 IR β ε (u) = 3. u(x) 2 Since Π B ρ (0), the definition of χ and Lebesgue s theorem imply that lim β ε(υ ε (y)) = y uniformly in y Π. (3.18) ε 0 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 ( \Ω ε). Proposition 3.2 Let {u n } be a sequence in H 1 ( ) such that and J εn (u n ) c 0 J ε n (u n )(u n ) = 0, n IN with ε n 0 when n. Then, there is a subsequence {ỹ n } such that the sequence v n (x) = u n (x + ỹ n ) has a convergent subsequence in H 1 ( ). Moreover, passing to a subsequence, where y n = ε n ỹ n. y n ỹ with y Π, Proof. We can always consider u n 0 and u n 0. As in Lemma 2.4 and arguing as Remark 3.2 we have that {u n } is bounded in H 1 ( ) and there are (ỹ n ) and positive constants R and α such that lim inf u 2 n α > 0. (3.19) B R (ỹ n) 25

26 Considering v n (x) = u n (x + ỹ n ) we conclude that {v n } is bounded in H 1 ( ) and therefore, passing to a subsequence, we get v n v, in H 1 ( ) with v 0. For each n IN, let t n > 0 such that ṽ n = t n v n N 0 (see Lemma 3.1(A 1 )). We have that c 0 I 0 (ṽ n ) = 1 2 M(t 2 n u n 2 0) F (t n u n ) 1 2 M(t 2 n u n 2 ε n ) G(ε n x, t n u n ). Hence, which implies, c 0 I 0 (ṽ n ) J ε (t n u n ) J ε (u n ) = c 0 + o n (1), (3.20) I 0 (ṽ n ) c 0 and {ṽ n } N 0. (3.21) Thus, {ṽ n } is bounded in H 1 ( ) and ṽ n ṽ. From well-known arguments we can assume that t n t 0 with t 0 > 0. So, from uniqueness of the weak limit we have ṽ = t 0 v, v 0. From Lemma 3.3 we obtain, and This convergence implies ṽ n ṽ in H 1 ( ). (3.22) v n ṽ t 0 = v in H 1 ( ) I 0 (ṽ) = c 0 and I 0(ṽ)ṽ = 0. (3.23) Now, we will show that {y n } is bounded, where y n = ε n ỹ n. In fact, otherwise, there exists a subsequence {y n } with y n. Observe that m 0 v n 2 0 g(ε n z + y n, v n )v n. Let R > 0 such that Ω B R (0). Since we may suppose that y n 2R, for each z B R (0) εn we have ε n z + y n y n ε n z 2R R = R. Thus, m 0 v n 2 0 B Rεn (0) f(v n )v n + f(v n )v n. \B Rεn (0) Since v n v in H 1 ( ), it follows from Lebesgue s theorem that f(v n )v n = o n (1). \B Rεn (0) 26

27 On the other hand, since f(v n ) V0 K v n, we obtain m 0 v n V 0 vn 2 + o n (1), K B Rεn (0) and therefore, ( m 0 1 ) v n 0 o n (1), K which is a contradiction. Hence, {y n } is bounded and we can assume y n y in. We see that y Ω because if y / Ω, we can proceed as above and conclude that v n 0 o n (1). In order to prove that V (y) = V 0, we suppose by contradiction that V 0 < V (y). Consequently, from (3.22), Fatou s Lemma and the invariance of R 3 by translations, we obtain c 0 < lim inf 1 2 M ( ) ] ṽ n 2 + V (ε n z + y n )ṽn 2 F (ṽ n ) lim inf ε n (t n u n ) lim inf ε n (u n ) = c 0, which is a contradiction and the proof is finished. Corollary 3.1 Assume the same hypotheses of Proposition 3.2. Then, for any given γ 2 > 0, there exists R > 0 and n 0 N such that B R (ỹ n) c ( un 2 + u n 2) < γ 2, for all n n 0. Proof. By using the same notation of the proof of Proposition 3.2, we have for any R > 0 ( un 2 + u n 2) = ( v n 2 + v n 2 ). B R (ỹ n) c B R (0) c Since (v n ) strongly converges in H 1 (R N ) the result follows. Lemma 3.5 Let δ > 0 and Π δ = {x : dist(x, M) δ}. Then, lim sup ε 0 u Ñε inf β ε (u) y = lim y Π δ ε 0 sup u Ñε dist(β ε (u), Π δ ) = 0. 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 (P ε,a ) Next we prove our multiplicity result for the problem (P ε,a ), by using arguments slightly different to those in 27], in fact, since S ε + 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

28 Theorem 3.2 Suppose that the function M satisfies (M 1 ) (M 3 ), the potential V satisfies (V 1 ) (V 2 ) and the function f satisfies (f 1 ) (f 4 ). Then, given δ > 0 there is ε = ε(δ) > 0 such that the auxiliary problem (P ε,a ) has at least Cat Πδ (Π) positive solutions, for all ε (0, ε). Proof. For each ε > 0, we define the function ζ ε : Π S + ε by ζ ε (y) = m 1 ε (Υ ε (y)), y Π. From the Lemma 3.4, we have lim Ψ ε(ζ ε (y)) = lim J ε (Υ ε (y)) = c 0, uniformly in y Π. ε 0 ε 0 Thus, the set S + ε = {u S + ε : Ψ ε (u) c 0 + h 1 (ε)}, is nonempty, for all ε (0, ε), because ζ ε (Π) S + ε, where the function h 1 was already introduced in the definition of the set Ñε. From up above considerations, together with Lemma 3.4, Lemma 2.3(A 3 ), 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) Π Υε Υ ε (Π) m 1 ε ζ ε (Π) mε Υ ε (Π) βε Π δ. We conclude from (3.18) that there is a function λ(ε, y) with λ(ε, y) < δ 2 uniformly in y Π, for all ε (0, ε), such that β ε (Υ ε (y)) = y + λ(ε, y) for all y Π. Hence, the application H : 0, 1] Π Π δ defined by H(t, y) = y + (1 t)λ(ε, y) is a homotopy between α ε ζ ε = β ε Υ ε and the inclusion i : Π Π δ, where α ε = β ε m ε. Therefore, cat ζε(π)ζ ε (Π) cat Πδ (Π). (3.24) It follows from Corollary 2.1 and from category abstract theorem in 26], with c = c ε c 0 + h 1 (ε) = d and K = ζ ε (Π), that Ψ ε has at least cat ζε(π)ζ ε (Π) critical points on S + ε. So, from item (d) of the Proposition 2.1 and from (3.24), we conclude that J ε has at least cat Πδ (Π) critical points in Ñε. 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 u ε (x) a x Ω c ε for ε small enough. This fact implies that these solutions are in fact solutions of the original problem ( P ε ). 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

29 Lemma 4.1 Let ε n 0 + and u n Ñε n be a solution of (P εn,a). Then J εn (u n ) c 0 and u n L (R 3 ). Moreover, for any given γ > 0, there exists R > 0 and n 0 N such that where ỹ n is given by Proposition 3.2. u n L (B R (ỹ n) c ) < γ, for all n n 0, (4.1) Proof. Since J εn (u n ) c 0 + h 1 (ε n ) with lim h 1(ε n ) = 0, we can argue as in the proof of the inequality (3.20) to conclude that J εn (u n ) c 0. Thus, we may invoke Proposition 3.2 to obtain a sequence (ỹ n ) R 3 satisfying the conclusions of that proposition. Fix R > 1 and consider η R C (R 3 ) such that 0 η R 1, η R 0 in B R/2 (0), η R 1 in B R (0) c and η R C/R. For each n N and L > 0, we define η n (x) := η R (x ỹ n ), u L,n H 1 (R 3 ) and z L,n H ε by u L,n (x) := min{u n (x), L}, z L,n := η 2 nu 2(β 1) L,n u n, with β > 1 to be determined later. From definition of z L,n and J ε n (u n )z L,n = 0, we have ] m 0 ηnu 2 2(β 1) L,n u n η n u n u 2(β 1) L,n η n u n R 3 R 3 (g(ε n x, u n ) m 0 V (ε n x)u n ) ηnu 2 n u 2(β 1) L,n. R 3 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 Π δ Ω. We first claim that there exists ε δ > 0 such that, for any 0 < ε < ε δ and any solution u Ñε of the problem (P ε,a ), there holds u L (R 3 \Ω ε) < a. (4.2) In order to prove the claim we argue by contradiction. So, suppose that for some sequence ε n 0 + we can obtain u n Ñε n such that J ε n (u n ) = 0 and u n L (R 3 \Ω εn ) a. (4.3) As in Lemma 4.1, we have that J εn (u n ) c 0 and therefore we can use Proposition 3.2 to obtain a sequence (ỹ n ) R 3 such that ε n ỹ n y 0 Π. If we take r > 0 such that B r (y 0 ) B 2r (y 0 ) Ω we have that B r/εn (y 0 /ε n ) = (1/ε n )B r (y 0 ) Ω εn. Moreover, for any z B r/εn (ỹ n ), there holds z y 0 z ỹ n + ỹn y 0 < 1 (r + o n (1)) < 2r, ε n ε n ε n 29 ε n

30 for n large. For these values of n we have that B r/εn (ỹ n ) Ω εn or, equivalently, R 3 \Ω εn R 3 \ B r/εn (ỹ n ). On the other hand, it follows from Lemma 4.1 with γ = a that, for any n n 0 such that r/ε n > R, there holds u n L (R 3 \Ω εn ) u n L (R 3 \B r/εn (ỹ n)) u n L (R 3 \B R (ỹ n)) < a, which contradicts (4.3) and proves the claim. Let ε δ > 0 given by Theorem 3.2 and set ε δ := min{ ε δ, ε δ }. We shall prove the theorem for this choice of ε δ. Let 0 < ε < ε δ be fixed. By applying Theorem 3.2 we obtain cat Πδ (Π) nontrivial solutions of the problem (P ε,a ). If u H ε is one of these solutions we have that u Ñε, and therefore we can use (4.2) and the definition of g to conclude that g ε (, u) f(u). Hence, u is also a solution of the problem ( P ε ). An easy calculation shows that û(x) := u(x/ε) is a solution of the original problem (P ε ). Then, (P ε ) has at least cat Πδ (Π) nontrivial solutions. We now consider ε n 0 + and take a sequence u n H εn of solutions of the problem ( P εn ) as above. In order to study the behavior of the maximum points of u n, we first notice that, by (g 1 ), there exists γ > 0 such that g(εx, s)s V 0 K s2, for all x R 3, s γ. (4.4) By applying Lemma 4.1 we obtain R > 0 and (ỹ n ) R 3 such that Up to a subsequence, we may also assume that u n L (B R (ỹ n)) c < γ, (4.5) u n L (B R (ỹ n)) γ. (4.6) Indeed, if this is not the case, we have u n L (R 3 ) < γ, and therefore it follows from J ε n (u n ) = 0 and (4.4) that m 0 u n 2 ε n g(ε n x, u n )u n V 0 u 2 K n. R 3 R 3 The above expression implies that u n εn = 0, which does not make sense. Thus, (4.6) holds. By using (4.5) and (4.6) we conclude that the maximum point p n R 3 of u n belongs to B R (ỹ n ). Hence p n = ỹ n + q n, for some q n B R (0). Recalling that the associated solution of (P εn ) is of the form û n (x) = u n (x/ε n ), we conclude that the maximum point η n of û n is η n := ε n ỹ n + ε n q n. Since (q n ) B R (0) is bounded and ε n ỹ n y 0 Π (according to Proposition 3.2), we obtain which concludes the proof of the theorem. lim V (η ε n ) = V (y 0 ) = V 0, 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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