Multiplicity and concentration results for a (p, q)-Laplacian problem in RN\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {R}}^{N}$$\end{document}

In this paper, we study the multiplicity and concentration of positive solutions for the following (p, q)-Laplacian problem: -Δpu-Δqu+V(εx)|u|p-2u+|u|q-2u=f(u)inRN,u∈W1,p(RN)∩W1,q(RN),u>0inRN,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta _{p} u -\Delta _{q} u +V(\varepsilon x) \left( |u|^{p-2}u + |u|^{q-2}u\right) = f(u) &{} \text{ in } {\mathbb {R}}^{N}, \\ u\in W^{1, p}({\mathbb {R}}^{N})\cap W^{1, q}({\mathbb {R}}^{N}), \quad u>0 \text{ in } {\mathbb {R}}^{N}, \end{array} \right. \end{aligned}$$\end{document}where ε>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varepsilon >0$$\end{document} is a small parameter, 1<p<q<N\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$1< p<q<N$$\end{document}, Δru=div(|∇u|r-2∇u)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta _{r}u={{\,\mathrm{div}\,}}(|\nabla u|^{r-2}\nabla u)$$\end{document}, with r∈{p,q}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r\in \{p, q\}$$\end{document}, is the r-Laplacian operator, V:RN→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$V:{\mathbb {R}}^{N}\rightarrow {\mathbb {R}}$$\end{document} is a continuous function satisfying the global Rabinowitz condition, and f:R→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f:{\mathbb {R}}\rightarrow {\mathbb {R}}$$\end{document} is a continuous function with subcritical growth. Using suitable variational arguments and Ljusternik–Schnirelmann category theory, we investigate the relation between the number of positive solutions and the topology of the set where V attains its minimum for small ε\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varepsilon $$\end{document}.


Introduction
In this paper we deal with the existence and multiplicity of solutions for the following (p, q)-Laplacian problem: where ε > 0 is a small parameter, 1 < p < q < N , ∆ r u = div(|∇u| r−2 ∇u), with r ∈ {p, q}, is the r-Laplacian operator, V : R N → R is a continuous potential and f : R → R is a continuous function with subcritical growth.We recall that this class of problems arises from a general reaction-diffusion system where D(u) = |∇u| p−2 + |∇u| q−2 .As pointed out in [9], this equation appears in several applications such as biophysics, plasma physics and chemical reaction design.In these applications, u describes a concentration, div(D(u)∇u) corresponds to the diffusion with a diffusion coefficient D(u), and the reaction term f (x, u) relates to source and loss processes.Classical (p, q)-Laplacian problems in bounded or unbounded domains have been studied by several authors; see for instance [3,[11][12][13][14][15][16]20] and references therein.
In order to precisely state our result, we introduce the assumptions on the potential V and the nonlinearity f .Along the paper we assume that V : R N → R is a continuous function satisfying the following condition introduced by Rabinowitz [21]: and the nonlinearity f : R → R fulfills the following hypotheses: (f 1 ) f ∈ C 0 (R, R) and f (t) = 0 for all t < 0; (f 3 ) there exists r ∈ (q, q * ), with q * = N q N −q , such that lim |t|→∞ |f (t)| |t| r−1 = 0; (f 4 ) there exists ϑ ∈ (q, q * ) such that 0 < ϑF (t) = ϑ t 0 f (τ ) dτ ≤ tf (t) for all t > 0; (f 5 ) the map t → f (t) t q−1 is increasing on (0, ∞).Since we deal with the multiplicity of solutions of (P ε ), we recall that if Y is a given closed subset of a topological space X, we denote by cat X (Y ) the Lyusternik-Shnirel'man category of Y in X, that is the least number of closed and contractible sets in X which cover Y (see [25] for more details).
Let us denote by Our main result can be stated as follows: Theorem 1.1.Assume that conditions (V ) and (f 1 )-(f 5 ) hold.Then for any δ > 0 there exists ε δ > 0 such that, for any ε ∈ (0, ε δ ), problem (P ε ) has at least cat M δ (M ) positive solutions.Moreover, if u ε denotes one of these solutions and x ε ∈ R N is a global maximum point of u ε , then and there exist C 1 , C 2 > 0 such that The proof of Theorem 1.1 will be obtained by using suitable variational techniques and category theory.We note that Theorem 1.1 improves Theorem 1.1 in [3], in which the authors assumed f ∈ C 1 and that there exist C > 0 and ν ∈ (p, q * ) such that f ′ (t)t 2 − (q − 1)f (t)t ≥ Ct ν for all t ≥ 0.
Since we require that f is only continuous, the classical Nehari manifold arguments used in [3] do not work in our context, and in order to overcome the non-differentiability of the Nehari manifold, we take advantage of some variants of critical point theorems from [23].Clearly, with respect to [3], a more accurate and delicate analysis will be needed to implement our variational machinery.To obtain multiple solutions, we use a technique introduced by Benci and Cerami in [7], which consists of making precise comparisons between the category of some sublevel sets of the energy functional I ε associated with (P ε ) and the category of the set M .Since we aim to apply Lyusternik-Shnirel'man theory, we need to prove certain compactness property for the functional I ε .In particular, we will see that the levels of compactness are strongly related to the behavior of the potential V at infinity.This kind of argument has been recently employed by the first author for nonlocal fractional problems; see for example [5,6].Finally, we prove the exponential decay of solutions by following some ideas from [13].We would like to point out that our arguments are rather flexible and we believe that the ideas contained here can be applied in other situations to study problems driven by (p, q)-Laplacian operators, φ-Laplacian operator, or also fractional (p, q)-Laplacian problems, on the entire space.
The paper is organized as follows: in Section 2 we collect some facts about the involved Sobolev spaces and some useful lemmas.In Section 3 we provide some technical results which will be crucial to prove our main theorem.In Section 4 we deal with the autonomous problems associated to (P ε ).In Section 5 we obtain an existence result for (P ε ) for sufficiently small ε.Section 6 is devoted to the multiplicity result for (P ε ), and Section 7 to the concentration phenomenon.

Preliminaries
In this section we recall some facts about the Sobolev spaces and we prove some technical lemmas which we will use later.
Let p ∈ [1, ∞] and A ⊂ R N .We denote by |u| L p (A) the L p (A)-norm of a function u : R N → R belonging to L p (A).When A = R N , we simply write |u| p instead of |u| L p (R N ) .For p ∈ (1, ∞) and N > p, we define D 1,p (R N ) as the closure of C ∞ c (R N ) with respect to Let us denote by W 1,p (R N ) the set of functions u ∈ L p (R N ) such that |∇u| p < ∞, endowed with the natural norm u p 1,p = |∇u| p p + |u| p p .We begin by recalling the following embedding theorem for Sobolev spaces.
Theorem 2.1.(see [1]) Let N > p. Then there exists a constant S * > 0 such that, for any We recall the following Lions compactness lemma.
We also have the following useful lemma.
Lemma 2.2.(see [2,18]) Let ) fixed, and t ′ = t t−1 is the conjugate exponent of t.For ε > 0, we define the space Then the following embedding lemma hold.
Lemma 2.3.(see [3]) The space X ε is continuously embedded into W 1,p (R N )∩W 1,q (R N ).Therefore X ε is continuously embedded in L t (R N ) for any t ∈ [p, q * ] and compactly embedded in L t (B R ), for all R > 0 and any t ∈ [1, q * ).
Finally we have the following splitting lemma which will be very useful in this work.
Proof.It is clear that (i) and (ii) are consequences of the well-known Brezis-Lieb lemma [8].The proofs of (iii) and (iv) are given in [3] for f ∈ C 1 .Since here we are assuming f ∈ C 0 , we need to use different arguments.We start by proving (iii).Let us note that u n = v n + u and In view of (f 2 ) and (f 3 ), for any δ > 0 there exists c δ > 0 such that Fix η > 0. Applying the Young inequality ab ≤ ηa r + C(η)b r ′ for all a, b > 0, with r, r ′ ∈ (1, ∞) such that 1 r + 1 r ′ = 1, to the first and the third term on the right hand side of (2.4), we deduce that which together with (2.3) with δ = η implies that As a consequence of the dominated convergence theorem we get On the other hand, by the definition of G η,n , it follows that which together with the boundedness of By the arbitrariness of η > 0 we can deduce that (iii) holds.Finally, we prove (iv).For any fixed η > 0, by (f 2 ) we can choose r 0 = r 0 (η) ∈ (0, 1) such that On the other hand, by (f 3 ) we can pick By the continuity of f , there exists δ = δ(η) ∈ (0, r 0 ) satisfying Moreover, by (f 3 ) there exists a positive constant c = c(η) such that In what follows, we shall estimate the following term:

5) and applying the Hölder inequality we get
) and the Hölder inequality yield Putting together (2.9), (2.10) and (2.11), we obtain that (2.12) Next, we note that (2.8) implies so we can see that where we have used the generalized Hölder inequality.Therefore which combined with (2.12) yields Now, recalling that u n ⇀ u in W 1,p (R N ), we may assume that, up to a subsequence, u n → u strongly converges in L p (B R (0)) and there exists h Observing that |D n | → 0 as n → ∞, we can deduce that Since u ∈ W 1,p (R N ), we know that |{|u| ≥ L}| → 0 as L → ∞, so there exists L = L(η) > 0 such that for all n On the other hand, by the dominated convergence theorem we can infer Consequently, for n large enough.Putting together (2.15), (2.16) and (2.17), we have This and (2.14) yield Taking into account (2.13) and (2.18), we can conclude that for n large enough This completes the proof of lemma.

Functional setting
In this section we consider the following problem In order to study (P ε ), we look for critical points of the functional I ε : X ε → R defined as It is easy to see that I ε ∈ C 1 (X ε , R) and its differential is given by for any u, ϕ ∈ X ε .Now, let us introduce the Nehari manifold associated to I ε , that is Let us note that I ε possesses a mountain pass geometry [4].
Proof.(i) Using (f 2 ) and (f 3 ), for any given ξ > 0 there exists C ξ > 0 such that Hence, taking ξ ∈ (0, V 0 ), we have Choosing u ε = ρ ∈ (0, 1) and using 1 < p < q, we have u V,p < 1 and therefore u p V,p ≥ u q V,p which combined with a t + b t ≥ C t (a + b) t for any a, b ≥ 0 and t > 1, yields Now, in view of Lemma 3.1, we can use a version of mountain pass theorem without the Palais-Smale condition [25] to deduce the existence of a (P S)-sequence and Lemma 3.2.The following holds Proof.For each u ∈ X ε \ {0} and t > 0, let us introduce the function h(t) = I ε (tu).Following the same arguments as in the proof of Lemma 3.1 we deduce that h(0) = 0, h(t) < 0 for t sufficiently large and h(t) > 0 for t sufficiently small.Hence, max t≥0 h(t) is achieved at Now, if u + ≡ 0, then u p V,p + u q V,q = 0, that is u ≡ 0, and this is a contradiction in view of u ∈ N ε .Next, we prove that t u is the unique critical point of h.Assume by contradiction that there exist t 1 and t 2 such that t 1 u, t 2 u ∈ N ε , that is Subtracting term by term in the above equalities we get ) and recalling that p < q, we can infer which gives a contradiction.Now we can argue as in [25] to complete the proof.
Next, we prove the following useful result.
Now, assume by contradiction that u n ε → ∞.We shall distinguish among the following cases: Case 1. u n V,p → ∞ and u n V,q → ∞.Since p < q, we have, for n sufficiently large, that u n q−p V,q ≥ 1, that is u n q V,q ≥ u n p V,q , and thus and letting n → ∞, we get 0 ≥ 1 q − 1 ϑ > 0, which yields a contradiction.u n V,p is bounded and u n V,q → ∞.Case 3. We can proceed similarly as in the case (2).
Hence, {u n } is bounded in X ε and we may assume that u n ⇀ u in X ε and u n → u a.e. in R N . (ii , where u − n = min{u n , 0}, and f (t) = 0 for t ≤ 0, we have that for t ∈ {p, q}.On the other hand, by (3.2), the mean value theorem, and since .
Now, recalling that for all ξ > 0 there exists C ξ > 0 such that we see that for t ∈ {p, q} the following holds and by the arbitrariness of ξ > 0 we get A similar argument shows that . Since f is only continuous, the next results are very important because they allow us to overcome the non-differentiability of N ε .We begin by proving some properties of the functional I ε .Lemma 3.4.Under assumptions (V ) and (f 1 )-(f 5 ), for any ε > 0 we have: Proof.(i) Let {u n } be a bounded sequence in X ε and v ∈ X ε .Then from assumptions (f 2 ) and (f 3 ) we can deduce that , it follows from (3.1) and the dominated convergence theorem that Note that (3.1) and Lemma 2.3 yield for all n ≥ n 0 and this shows that I ′ ε is weakly sequentially continuous in X ε .(iii) Without loss of generality, we may assume that u ε ≤ 1 for each u ∈ K.For u n ∈ K, after passing to a subsequence, we obtain that u n → u ∈ S ε .Then, using (f 4 ) and Fatou's lemma, we can see that Lemma 3.5.Under the assumptions of Lemma 3.4, for ε > 0 we have: (i) for all u ∈ S ε , there exists a unique (iii) There exists α > 0 such that t u ≥ α for each u ∈ S ε and, for each compact subset W ⊂ S ε , there exists Proof.(i) The proof follows the same lines as the proof of Lemma 3.2.
(ii) Using (3.1) and Lemma 2.3, for any u ∈ N ε we have Taking ξ > 0 sufficiently small we can deduce that (iii) For each u ∈ S ε there exists t u > 0 such that t u u ∈ N ε .Then, using u ε ≥ κ, we also have t u = t u u ε ≥ κ.It remains we prove that t u ≤ C W for all u ∈ W ⊂ S ε .We argue by contradiction: we suppose that there exists a sequence {u n } ⊂ W ⊂ S ε such that t un → ∞.Since W is compact, we can find u ∈ W such that u n → u in X ε and u n → u a.e. in R N .Now, using (f 4 ) we have In view of (i)-(iii) and Proposition 3.1 in [23] we can deduce that m ε is a homeomorphism between S ε and N ε and the inverse of m ε is given by m −1 ε (u) = u u ε .Therefore N ε is a regular manifold diffeomorphic to S ε .
(v) For ε > 0, t > 0 and u ∈ X ε \ {0}, we can see that (3.2) yields t q q u q V,q − C ξ t r u r ε so we can find ρ > 0 such that I ε (tu) ≥ ρ > 0 for t > 0 small enough.On the other hand, by using (i)-(iii), we get (see [23]) that Now we introduce the following functionals Ψε : where mε (u) = t u u is given in (3.4).As in [23], we have the following result: Lemma 3.6.Under the assumptions of Lemma 3.4, we have that for ε > 0: corresponding critical values coincide and

The autonomous problem
In this section we deal with the autonomous problem associated with (P ε ), that is (AP µ ) The functional associated with (AP µ ) is given by which is well-defined on the space It is easy to check that J µ ∈ C 1 (Y µ , R) and its differential is given by for any u, ϕ ∈ Y µ .Let us define the Nehari manifold associated with J µ Arguing as in the previous section and using (4.1), it is easy to prove the following lemma.
Lemma 4.1.Under the assumptions of Lemma 3.4, for µ > 0 we have: (i) for all u ∈ S µ , there exists a unique t u > 0 such that t u u ∈ M µ .Moreover, m µ (u) = t u u is the unique maximum of J µ on Y µ , where (iii) There exists α > 0 such that t u ≥ α for each u ∈ S µ and, for each compact subset W ⊂ S µ , there exists C W > 0 such that t u ≤ C W for all u ∈ W . (iv) M µ is a regular manifold diffeomorphic to the sphere in Y µ .(v) d µ = inf Mµ J µ > 0 and J µ is bounded below on M µ by some positive constant.(vi) J µ is coercive on M µ .

Now we define the following functionals
Then we obtain the following result: Lemma 4.2.Under the assumptions of Lemma 3.4, we have that for µ > 0: (i) Ψ µ ∈ C 1 (S µ , R), and Proof.Arguing as in the proof of Lemma 3.3, we can see that {u n } is bounded in Y µ .Now, in order to prove the other assertion of this lemma, we argue by contradiction.Assume that for any R > 0 it holds |u n | q dx = 0.
Since {u n } is bounded in Y µ , it follows by Lemma 2.1 that Fix ξ ∈ (0, µ).Then, taking into account that {u n } ⊂ M µ and (3.1), we have , and in view of (4.3), we have that u n µ → 0.
Next, we prove the following useful compactness result for the autonomous problem.For completeness, we recall that a critical point u = 0 of J µ satisfying J µ (u) = inf Mµ J µ = d µ is called a ground state solution to (AP µ ); see chapter 4 in [25] for more details.Proof.By virtue of (v) of Lemma 4.1, we know that In view of Lemma 4.2, we can see that u is a critical point of J µ .Now we show that there exists a minimizer of J µ | Mµ .By Ekeland's variational principle [25] there exists a sequence {ν n } ⊂ S µ such that Then, thanks to Lemma 4.2, J µ (u n ) → d µ and J ′ µ (u n ) → 0 as n → ∞.Therefore, arguing as in the proof of Lemma 3.3, {u n } is bounded in Y µ which is a reflexive space, so we may assume that u n ⇀ u in Y µ for some u ∈ Y µ .
It is clear that and using the fact that we obtain that u is a critical point of J µ .Now, if u = 0, then u is a nontrivial solution to (AP µ ).Assume that u = 0. Then u n µ → 0 in Y µ .Hence, arguing as in the proof of Lemma 4.3 we can find a sequence {y n } ⊂ R N and constants R, β > 0 such that Now, let us define Due to the invariance by translations of R N , it is clear that ṽn µ,t = u n µ,t , with t ∈ {p, q}, so {ṽ n } is bounded in Y µ and there exists ṽ such that ṽn ⇀ ṽ in Y µ , ṽn → ṽ in L m loc (R N ) for any m ∈ [1, q * ) and ṽ = 0 in view of (4.4).Moreover, J µ (ṽ n ) = J µ (u n ) and J ′ µ (ṽ n ) = o n (1), and arguing as before it is easy to check that J ′ µ (ṽ) = 0. Now, say u be the solution obtained before, and we prove that u is a ground state solution.It is clear that d µ ≤ J µ (u).On the other hand, by Fatou's lemma we can see that which implies that d µ = J µ (u).
Finally, we prove that the ground state obtained earlier is positive.Indeed, taking u − = min{u, 0} as test function in (AP µ ), and applying (f 1 ) and invoking the following inequality we can see that By the regularity results in [13], we have that ) and u(x) → 0 as |x| → ∞ (in the exponential way).Applying the Harnack inequality in [24], we can see that u > 0 in R N .This completes the proof of the lemma.

A first existence result for (P ε )
In this section we focus on the existence of a solution to (P ε ) provided that ε is sufficiently small.Let us start with the following useful lemma.Lemma 5.1.Let {u n } ⊂ N ε be a sequence such that I ε (u n ) → c and u n ⇀ 0 in X ε .Then one of the following alternatives occurs: Proof.Assume that (b) does not hold.Then, for any R > 0, the following holds Since {u n } is bounded in X ε , it follows by Lemma 2.1 that u n → 0 in L t (R N ) for any t ∈ (q, q * ). (5.1) Now, we can argue as in the proof of Lemma 4.3 and deduce that u n ε → 0 as n → ∞.
In order to get a compactness result for I ε , we need to prove the following auxiliary lemma.
Lemma 5.2.Assume that V ∞ < ∞ and let {v n } ⊂ N ε be a sequence such that Case 1: Assume that lim sup n→∞ t n = 1.Thus there exists {t n } such that t n → 1. Taking into account that (5.10) Now, let us point out that (5.11) Using condition (V ), v n → 0 in L p (B R (0)), t n → 1, (5.6), and the fact that (5.12) In a similar fashion we can prove that Since {v n } is bounded in X ε , we can conclude that and Thus, putting together (5.11), (5.12), (5.13) and (5.14), we obtain At this point, we aim to show that Applying the mean value theorem and (3.1), we can deduce that Exploiting the boundedness of {v n }, we get the assertion.Gathering (5.10), (5.15) and (5.16), we can infer that and taking the limit as ζ → 0 we get c ≥ d V∞ .
Case 2: Assume that lim sup n→∞ t n = t 0 < 1.Then there is a subsequence, still denoted by {t n }, such that t n → t 0 (< 1) and t n < 1 for any n ∈ N. Let us observe that (5.17) Recalling that t n v n ∈ M V∞ , and using (f 5 ) and (5.17), we obtain Taking the limit as n → ∞, we get c ≥ d V∞ .
At this point we are able to prove the following compactness result.
Proof.It is easy to see that {u n } is bounded in X ε .Then, up to a subsequence, we may assume that (5.18)By using assumptions (f 2 )-(f 3 ), (5.18) and the fact that Now, we prove that . For t ∈ {p, q}, by using Lemma 2.2 with η n = v n and w = u, we get and arguing as in the proof of Lemma 3.3 in [18], we can see that Hence, by using the Hölder inequality, for any ϕ ∈ X ε such that ϕ ε ≤ 1, we get and in view of (iv) of Lemma 2.5, (5.20), (5.21),I ′ ε (u n ) = 0 and I ′ ε (u) = 0 we obtain the assertion.Now, we note that by using (f 4 ) we can see that Let us consider the case V ∞ = ∞.Then, we can use Lemma 2.4 to deduce that v n → 0 in L m (R N ) for all m ∈ [p, q * ).This, combined with assumptions (f 2 ) and (f 3 ), implies that Since , and applying (5.23) we can infer that We conclude this section by giving the proof of the existence of a ground state solution to (P ε ) (that is a nontrivial critical point u of I ε such that I ε (u) = inf Nε I ε = c ε ) whenever ε > 0 is small enough.Theorem 5.1.Assume that (V ) and (f 1 )-(f 5 ) hold.Then there exists ε 0 > 0 such that, for any ε ∈ (0, ε 0 ), problem (P ε ) admits a ground state solution.
Proof.By (v) of Lemma 3.5, we know that ) is a minimizer of Ψ ε and it is a critical point of Ψ ε .By virtue of Lemma 3.6, we can see that u ε is a critical point of I ε .It remains to show that there exists a minimizer of I ε | Nε .By Ekeland's variational principle [25], there exists a sequence Therefore, {u n } is a Palais-Smale sequence for I ε at level c ε .It is easy to check that {u n } is bounded in X ε and we denote by u its weak limit.It is also easy to verify that I ′ ε (u) = 0.When V ∞ = ∞, by using Lemma 2.4, we have I ε (u) = c ε and I ′ ε (u) = 0. Now, we deal with the case V ∞ < ∞.In view of Proposition 5.1 it is enough to show that c ε < d V∞ for small ε.Without loss of generality, we may suppose that Let us prove that there exists a function w ∈ Y µ with compact support such that J µ (w) = max t≥0 J µ (tw) and J µ (w) < d V∞ . (5.24) , where w µ is a ground state solution to (AP µ ).By the dominated convergence theorem we can see that and w µ is a ground state, we can deduce that t R → 1 and which gives a contradiction.Then, taking w = ψ rw µ , we can conclude that (5.24) holds.Now, by (V ), we obtain that for some ε > 0 Then, in the light of (5.24) and (5.26), we have for all ε ∈ (0, ε) It follows from (3.5) that c ε < d V∞ for all ε ∈ (0, ε).

Multiple solutions for (P ε )
This section is devoted to the study of the multiplicity of solutions to (P ε ).We begin by proving the following result which will be needed to implement the barycenter machinery.Proposition 6.1.Let ε n → 0 and {u n } ⊂ N εn be such that I εn (u n ) → d V 0 .Then there exists {ỹ n } ⊂ R N such that the translated sequence has a subsequence which converges in Y V 0 .Moreover, up to a subsequence, {y n } = {ε n ỹn } is such that y n → y ∈ M .
Proof.Since I ′ εn (u n ), u n = 0 and I εn (u n ) → d V 0 , we know that {u n } is bounded in X ε .Since d V 0 > 0, we can infer that u n εn → 0. Therefore, as in the proof of Lemma 5.1, we can find a sequence {ỹ n } ⊂ R N and constants R, β > 0 such that Let us define v n (x) = u n (x + ỹn ).
In view of the boundedness of {u n } and (6.1), we may assume that v n ⇀ v in Y V 0 for some v = 0. Let {t n } ⊂ (0, ∞) be such that w n = t n v n ∈ M V 0 , and we set y n = ε n ỹn .Thus, by using the change of variables z → x + ỹn , V (x) ≥ V 0 and the invariance by translation, we can see that Hence we can infer J V 0 (w n ) → d V 0 .This fact and {w n } ⊂ M V 0 imply that there exists K > 0 such that w n V 0 ≤ K for all n ∈ N.Moreover, we can prove that the sequence {t n } is bounded in R. In fact, v n → 0 in Y V 0 , so there exists α > 0 such that v n V 0 ≥ α.Consequently, for all n ∈ N, we have which yields |t n | ≤ K α for all n ∈ N. Therefore, up to a subsequence, we may suppose that t n → t 0 ≥ 0. Let us show that t 0 > 0. Otherwise, if t 0 = 0, by the boundedness of {v n }, we get w n = t n v n → 0 in Y V 0 , that is J V 0 (w n ) → 0 which is in contrast with the fact d V 0 > 0. Thus t 0 > 0 and, up to a subsequence, we may assume that w n ⇀ w = t 0 v = 0 in Y V 0 .Therefore From Lemma 4.4, we can deduce that w n → w in Y V 0 , that is v n → v in Y V 0 .Now, we show that {y n } has a subsequence satisfying y n → y ∈ M .First, we prove that {y n } is bounded in R N .Assume by contradiction that {y n } is not bounded, that is there exists a subsequence, still denoted by {y n }, such that |y n | → ∞.First, we deal with the case V ∞ = ∞.By using {u n } ⊂ N εn and by changing the variable, we can see that By applying Fatou's lemma and v n → v in Y V 0 , we deduce that which gives a contradiction.Let us consider the case V ∞ < ∞.Taking into account that w n → w strongly converges in Y V 0 , condition (V ) and using the change of variable z = x + ỹn , we have which is a contradiction.Thus {y n } is bounded and, up to a subsequence, we may assume that y n → y.If y / ∈ M , then V 0 < V (y) and we can argue as in (6.2) to get a contradiction.Therefore, we can conclude that y ∈ M .
At this point, we introduce a subset N ε of N ε by taking a function h : R + → R + such that h(ε) → 0 as ε → 0, and setting . By Lemma 6.1, we know that h(ε) → 0 as ε → 0. By definition of h(ε), we can deduce that for all y ∈ M and ε > 0, Φ ε (y) ∈ N ε and N ε = ∅.Moreover, we have the following lemma.Lemma 6.3.For any δ > 0, the following holds Therefore, it suffices to prove that there exists {y n } ⊂ M δ such that Thus, recalling that {u n } ⊂ N εn ⊂ N εn , we can deduce that which implies that I εn (u n ) → d V 0 .By Proposition 6.1, there exists {ỹ n } ⊂ R N such that y n = ε n ỹn ∈ M δ for n sufficiently large.Thus , that is (6.15) holds.Now we show that (P ε ) admits at least cat M δ (M ) solutions.In order to achieve our aim, we recall the following result for critical points involving Lyusternik-Shnirel'man category.For more details one can see [10].
Theorem 6.1.Let U be a C 1,1 complete Riemannian manifold (modelled on a Hilbert space).Assume that h ∈ C 1 (U, R) is bounded from below and satisfies −∞ < inf U h < d < k < ∞.Moreover, suppose that h satisfies the Palais-Smale condition on the sublevel {u ∈ U : h(u) ≤ k} and that d is not a critical level for h.Then With a view to apply Theorem 6.1, the following abstract lemma provides a very useful tool since relates the topology of some sublevel of a functional to the topology of some subset of the space R N ; see [10].
For each τ > 0 we can use Young's inequality to obtain dx and taking τ > 0 sufficiently small, we get On the other hand, using the Sobolev inequality and the Hölder inequality, we can infer Combining (7.1) and (7.2), we find We claim that v n ∈ L (q * ) 2 q (|x| ≥ R) for R large enough and uniformly in n.Let β = q * q .From (7.3) we have |w L,n | q q * ≤ Cβ q R N |∇η| q v q n v q * −q L,n dx + R N v q * n η q v q * −q L,n dx or equivalently |w L,n | q q * ≤ Cβ q R N |∇η| q v q n v q * −q L,n dx + R N v q n η q v q * −q L,n v q * −q n dx .
Using the Hölder inequality with exponents q * q and q * q * −q , we obtain From the definition of w L,n , we have R N (v n ηv q * −q q L,n ) q * dx q q * ≤ Cβ q R N |∇η| q v q n v q * −q L,n dx + Cβ q R N (v n ηv q * −q q L,n ) q * dx q q * |x|≥ R 2 v q * n dx q * −q q * .Since v n → v in W 1,p (R N ) ∩ W 1,q (R N ), for R > 0 sufficiently large, we get |x|≥ R 2 v q * n dx ≤ ǫ uniformly in n ∈ N.
Hence, |x|≥R (v n ηv q * −q q L,n ) q * dx q q * ≤ Cβ q R N v q n v q * −q L,n dx ≤ Cβ q R N v q n dx ≤ K < ∞.

Lemma 4 . 4 .
The problem (AP µ ) has a positive ground state solution.