A multiplicity result for a (p, q)-Schrödinger–Kirchhoff type equation

In this paper, we study a class of (p, q)-Schrödinger–Kirchhoff type equations involving a continuous positive potential satisfying del Pino–Felmer type conditions and a continuous nonlinearity with subcritical growth at infinity. By applying variational methods, penalization techniques and Lusternik–Schnirelman category theory, we relate the number of positive solutions with the topology of the set where the potential attains its minimum values.

(1)  [3,4,19,23,52]. Note that, in the case p = q = 2 , Eq. (3) reduces to the wellknown Schrödinger equation which has been widely studied in the last three decades; see for example [18,25,43,48,50]. When a = b ≠ 0 , p = q = 2 and N = 3 , problem (1) becomes a Kirchhoff equation of the form Equation (4) is related to the stationary analog of the Kirchhoff equation [32] where Ω ⊂ ℝ N is a smooth bounded domain, > 0 , ≥ 0 , and u satisfies some boundary conditions, which was proposed by Kirchhoff in 1883 as a nonlinear extension of D'Alembert's wave equation for free vibration of elastic strings Here, u = u(x, t) is the transverse string displacement at the space coordinate x and the time t, L is the length of the string, h is the area of the cross section, E is the Young's modulus of the material, is the mass density and p 0 is the initial axial tension. From a purely mathematical point of view, it is important to mention that the early studies dedicated to the Kirchhoff equation (5) were given by Bernstein [10] and Pohozaev [42]. However, the Kirchhoff equation (5) began to attract the attention of more researchers only after the paper by Lions [36], in which a functional analysis approach was proposed to attack it. For some interesting results on Kirchhoff problems, we refer to [13,24,30,41,49].
Due to the interest shared by the mathematical community toward quasilinear problems and Kirchhoff type equations, in [12,31], the authors studied Kirchhoff type equations involving the (p, q)-Laplacian operator with p ≠ q , in a bounded domain and in the whole of ℝ 3 , respectively.
Motivated by the above works, the purpose of this paper is to study the multiplicity and the concentration of solutions for (1).

Assumptions and main result
For simplicity, we assume that a = b = 1 in (1). Let us now introduce the hypotheses on the potential V and the nonlinearity f that we are going to consider throughout the paper.
Let V ∶ ℝ N → ℝ be a continuous function that satisfies the following assumptions due to del Pino-Felmer [18]: there exists an open bounded set Λ ⊂ ℝ N such that Assume that f ∶ ℝ → ℝ is a continuous function such that f (t) = 0 for t ≤ 0 and fulfills the following hypotheses: (f 4 ) the map t ↦ f (t) t 2q−1 is increasing for t > 0.
In order to give the precise statement of our main theorem, let us recall that, for any closed subset Y of a topological space X, the Lusternik-Schnirelman category of Y in X, cat X (Y) , stands for the least number of closed and contractible sets in X which cover Y; see [50]. Then, we can state the following result.

Theorem 1 Let V satisfy (V 1 ) and (V 2 )
, and let f be a continuous function such that the hypotheses (f 0 )-(f 4 ) hold. Then, for any > 0 such that there exists > 0 such that, for any ∈ (0, ) , problem (1) has at least cat M (M) positive solutions. Moreover, if u denotes one of these solutions and x ∈ ℝ N is a global maximum point of u , then

Main difficulties and ideas
Due to the lack of information about the behavior of the potential V at infinity and the fact that our problem is set in an unbounded domain, we adapt the local mountain pass argument introduced by del Pino and Felmer [18]. It consists in making a suitable modification on f, solving a modified problem, whose corresponding energy functional has a nice geometric structure, and then checking that, for > 0 small enough, the solutions of the new problem are indeed solutions of the original one. We note that, because of the presence of the (p, q)-Laplacian operators and Kirchhoff terms, even for the corresponding modified energy functional, it is hard to obtain compactness, and an accurate analysis will be done to prove a first existence result for the modified problem; see Lemmas 5, 6 and 7. Secondly, we make use of a technique given by Benci and Cerami [9] to establish a relationship between the category of the set M and the number of solutions for the modified problem. We underline that, since f is merely continuous, standard C 1 -Nehari manifold arguments as in [2-5, 30, 50] do not work in our setting, and so we take advantage of some abstract results due to Szulkin and Weth [46]. Note that, this type of approach has been also used in [24] where a Schrödinger-Kirchhoff elliptic equation was considered. Clearly, with respect to [24], a more careful analysis will be needed and some refined estimates will be used to overcome some technical difficulties. Finally, to obtain a uniform L ∞ -estimate for an appropriate translated sequence of solutions to the modified problem, we do not use the classical Moser iteration argument [38] as in [3,19,23,24,30], because such technique does not seem to work well in our situation, but we follow some arguments found in [2,21,26,33] which are inspired by the well-known method pioneered by De Giorgi [17]; see Lemma 15. As far as we know, all results presented in this work are new in the literature. Moreover, we believe that the ideas developed here can be applied in other situation to study (p, q)-Schrödinger-Kirchhoff type problems involving potentials satisfying local conditions and continuous nonlinearities.
The outline of the paper is the following. In Sect. 2, we introduce the modified problem. Section 3 is devoted to the study of the autonomous problem associated with (1). In Sect. 4, we prove a multiplicity result for the modified problem. The proof of Theorem 1 is given in Sect. 5.
The modified problem

Notations and preliminary results
In order to simplify the presentation, we denote by C a generic positive constant, which may change from line to line, but does not depend on crucial quantities. Let A be a measurable subset of ℝ N . By A c , we denote the complement of A. Let 1 ≤ r ≤ ∞ . We will use the notation | ⋅ | L r (A) for the norm in L r (A) , and when A = ℝ N , we simply write | ⋅ | r . By B r (x 0 ), we indicate the open ball in ℝ N centered at x 0 ∈ ℝ N and radius r > 0 . In the case x 0 = 0, we simply write B r . Let 1 < r < ∞ and N > r . By D 1,r (ℝ N ), we mean the closure of C ∞ c (ℝ N ) functions with respect to the norm By W 1,r (ℝ N ), we denote the Sobolev space equipped with the norm The following embeddings are well known.
For the reader's convenience, we also recall the following vanishing lemma. Lemma 1 [37] Let p ∈ (1, ∞) , N > p and r ∈ [p, p * ) . If {u n } n∈ℕ is a bounded sequence in Let p, q ∈ (1, ∞) and set endowed with the norm For any > 0 , we introduce the space endowed with the norm where Finally, we recall the following well-known elementary inequalities [45] which will be used in the sequel: for any , ∈ ℝ N , we have for some c 1 , c 2 > 0 constants. In particular, Note that, when 1 < r < 2 using (8) and the following elementary inequality we deduce that there exists c 3 > 0 such that for any , ∈ ℝ N , the following relation satisfies

The penalization approach
To deal with (1), we use a del Pino-Felmer penalization type approach [18]. Firstly, we note that the map t ↦ f (t) t p−1 +t q−1 is increasing in (0, ∞) . Indeed, once we write Take and let a > 0 be such that Consider the function f ∶ ℝ → ℝ given by Denote by A the characteristic function of A ⊂ ℝ N , and define the function Using the hypotheses on f, we infer that g is a Carathéodory function such that for all x ∈ Λ c and t > 0, is increasing in (0, ∞) , and for each is increasing in (0, a).
We point that, from (g 1 ) , (g 2 ) , (f 1 ) and (f 2 ) , for any > 0, there exists C > 0 such that Let us introduce the following auxiliary problem: Define the set Λ = {x ∈ ℝ N ∶ x ∈ Λ} . We underline that if u is a solution to (12) satisfying u (x) ≤ a for all x ∈ Λ c , then u is also a solution to (1). Let us introduce the functional L ∶ → ℝ associated with (12), that is We note that L ∈ C 1 ( , ℝ) and for any u, ∈ . The Nehari manifold associated with L is given by and let be the unit sphere in , and set + = ∩ + , where + stands for the open set Note that, + is an incomplete C 1,1 -manifold of codimension one. Hence, for all u ∈ + , First, we show that L has a mountain pass geometry [6].

Lemma 2
The functional L has the following properties: (i) There exist , > 0 such that L (u) ≥ for ‖u‖ = .
(ii) By (f 3 ), we deduce that Then, for all u ∈ + and t > 0 , we have and observing that > 2q > 2p we deduce that L (tu) → −∞ as t → ∞ . ◻ In order to overcome the non-differentiability of N and the incompleteness of + , we prove the following results.

Lemma 3
Under the assumptions (V 1 )-(V 2 ) and (f 1 )-(f 4 ) , the following properties hold: (i) For each u ∈ + , there exists a unique t u > 0 such that if u (t) = L (tu) , then � u (t) > 0 for 0 < t < t u and � u (t) < 0 for t > t u . (ii) There exists > 0 independent of u such that t u ≥ for any u ∈ + and for each compact set W ⊂ + , there is a positive constant C W such that t u ≤ C W for any u ∈ W. (iii) The map m ∶ + → N given by m (u) = t u u is continuous and m =m | + is a homeomorphism between + and N . Moreover, Proof (i) Similar to the proof of Lemma 2, we can see that u (0) = 0 , u (t) > 0 for t > 0 small enough and u (t) < 0 for t > 0 sufficiently large. Then, there exists a global maximum point t u > 0 for u in [0, ∞) such that � u (t u ) = 0 and t u u ∈ N . We claim that t u > 0 is unique. We argue by contradiction and suppose that there exist t 1 > t 2 > 0 such that � u (t 1 ) = � u (t 2 ) = 0 . Therefore, and From the definition of g, (g 4 ) and (f 4 ) , we get Multiplying both sides by where we used the fact that (f 4 ) and our choice of the constant a give Since u ≠ 0 and K > 1 , we get a contradiction.
(ii) Let u ∈ + . Using (i), we can find t u > 0 such that � u (t u ) = 0 , that is Fix > 0 . By (11) and Theorem 2, we have Thanks to > q , we can find > 0 , independent of u, such that t u ≥ .
When t u > 1 , then t q−1 u > t p−1 u , and noting that 1 = ‖u‖ ≥ ‖u‖ V ,p and q > p imply ‖u‖ Now, let W ⊂ + be a compact set and assume by contradiction that there exists a sequence {u n } n∈ℕ ⊂ W such that t n = t u n → ∞ . Then, there exists u ∈ W such that u n → u in . From (ii) of Lemma 2, we have that Taking v n = t u n u n ∈ N in the above inequality, we find Since ‖v n ‖ = t n → ∞ and ‖v n ‖ = ‖v n ‖ ,p + ‖v n ‖ ,q , we can use (13) to get a contradiction. .
(iii) Let us observe that m , m and m −1 are well defined. Indeed, by (i), for each u ∈ + , there is a unique m (u) ∈ N . On the other hand, if u ∈ N , then u ∈ + . Otherwise, we have and by (g 3 )-(ii), we deduce that which is impossible due to K > 1 and u ≠ 0 . Therefore, m −1 (u) = u ‖u‖ ∈ + is well defined and continuous. From we infer that m is a bijection. To prove that m ∶ + → N is continuous, let {u n } n∈ℕ ⊂ + and u ∈ + be such that u n → u in . Since m(tu) =m(u) for all t > 0 , we may assume that ‖u n ‖ = ‖u‖ = 1 for all n ∈ ℕ . By (ii), there exists t 0 > 0 such that t n = t u n → t 0 . Since t n u n ∈ N , and passing to the limit as n → ∞, we obtain which yields t 0 u ∈ N . From (i), t u = t 0 , and this means that m (u n ) →m (u) in + . Thus, m and m are continuous functions.
(iv) Let {u n } n∈ℕ ⊂ + be a sequence such that dist(u n , + ) → 0 . Then, for each v ∈ + and n ∈ ℕ , we have u + n ≤ |u n − v| a.e. in Λ . Therefore, by (V 1 ) , (V 2 ) and Theorem 2, we can see that for each r ∈ [p, q * s ], there exists C r > 0 such that By virtue of (g 1 ) , (g 2 ) , (g 3 )-(ii) and q > p , we get, for all t > 0, Therefore, Then, for all t > 1 , we obtain that Bearing in mind the definition of m (u n ) and using (14), (15), we find By sending t → ∞ , we get L (m (u n )) → ∞ as n → ∞ . On the other hand, by the definition of L , we see that for all n ∈ ℕ (11) and Theorem 2 that Define the maps by ̂(u) = L (m (u)) and =̂| + . From Lemma 3 and arguing as in the proofs of Proposition 9 and Corollary 10 in [46], we may obtain the following result.
p . Now, we assume by contradiction that ‖u n ‖ → ∞ and consider the following cases: that is an absurd. In case (2), we have and consequently Since p > 1 and passing to the limit as n → ∞ , we obtain 0 <C ≤ 0 which is impossible. The last case is similar to the case (2), so we omit the details. Consequently, {u n } n∈ℕ is bounded in . ◻

Lemma 5 Let {u n } n∈ℕ ⊂
be a (PS) c sequence for L . Then, for any > 0, there exists R = R( ) > 0 such that . From the definition of R and (g 3 )-(ii), we see that Now, using the Hölder inequality and the boundedness of {u n } n∈ℕ in , we have, for t ∈ {p, q}, which implies that Thanks to (17) and (18) with 0 ≤ (x) ≤ 1 and |∇ | ≤ 2 . Since {u n } n∈ℕ is bounded in (by Lemma 4), we may assume that Fix R > 0 and take > R . For t ∈ {p, q} and n ∈ ℕ , define By (9), we note that A t n ≥ 0 . Moreover, we see that |∇u n | p p → T p and |∇u n | q q → T q as n → ∞. On the other hand, the weak convergence and (20) imply (21) 0 ≤ A p n + A q n ≤ |I 1 n, | + |I 2 n, | + |I 3 n, | + |I 4 n, |.  In particular, if t ≥ 2 , from (7), we have When 1 < t < 2 , by (10) and Hölder inequality, we obtain Arguing as before, we deduce that for t ∈ {p, q} Accordingly, for t ∈ {p, q} , we get which gives (19). ◻ Now, we show that L verifies the Palais-Smale compactness condition.

Lemma 7 L satisfies the Palais-Smale condition at any level c ∈ ℝ.
Proof Let {u n } n∈ℕ ⊂ be a (PS) c sequence for L . From Lemma 4, we know that {u n } n∈ℕ is bounded in . Up to a subsequence, we may assume that u n ⇀ u in and u n → u in L r loc (ℝ N ) for all r ∈ [1, q * ) . By Lemma 5, for each > 0, there exists R = R( ) > C , with C > 0 independent of , such that (19) is satisfied. This together with Lemma 6 implies that Since R → ∞ when → 0 , it follows that

and thus
From the Brezis-Lieb lemma [11], we have and Therefore, which yields u n → u in as n → ∞ . ◻

Corollary 1 The functional satisfies the Palais-Smale condition on + at any level c ∈ ℝ.
Proof Let {u n } n∈ℕ ⊂ + be a Palais-Smale sequence for at the level c. Then, By Proposition 1-(c), we see that {m (u n )} n∈ℕ ⊂ is a Palais-Smale sequence for L at the level c. From Lemma 7, we deduce that L satisfies the (PS) c condition in . Then, up to a subsequence, we can find u ∈ + such that In view of Lemma 3-(iii), we conclude that u n → u in + . ◻

The autonomous problem
In this section, we consider the following autonomous problem related to (1):

the energy functional associated with (25), then
It is easy to check that L V 0 ∈ C 1 ( V 0 , ℝ) and that for any u, ∈ V 0 . The Nehari manifold M V 0 associated with L V 0 is and we set Denote by V 0 the unit sphere of V 0 and set + V 0 is an incomplete C 1,1 -manifold of codimension one contained in + V 0 . Thus, .

3
Arguing as in Sect. 2, we can see that the following results hold.

Lemma 8
Under the assumptions (f 1 )-(f 4 ) , the following properties hold: Let us consider the maps defined as ̂V

Proposition 2
Assume that (f 1 )-(f 4 ) are satisfied. Then, Moreover, the corresponding critical values coincide and Remark 4 As in [46], we have the following characterization of the infimum of L V 0 over

3
The next lemma allows us to assume that the weak limit of a (PS) d V 0 sequence of L V 0 is nontrivial.
Then, one and only one of the following alternatives occurs:

Remark 5
From the above result, we deduce that if u is the weak limit of a (PS) d V 0 sequence for L V 0 , then we can assume u ≠ 0 . In fact, if u n ⇀ 0 in V 0 and, if u n ↛ 0 in V 0 , by Lemma 9, we can find {y n } n∈ℕ ⊂ ℝ N and R, > 0 such that Set v n (x) = u n (x + y n ) . Then, using the invariance of ℝ N by translation, we see that In what follows, we prove the existence of a positive ground-state solution for (25).
there exists u ∈ V 0 ⧵ {0} with u ≥ 0 such that, up to a subsequence, u n → u in V 0 . Moreover, u is a positive ground-state solution to (25).
Proof As in the proof of Lemma 7, we can see that {u n } n∈ℕ is a bounded sequence in V 0 so, going if necessary to a subsequence, we may assume that .
From Remark 5, we may suppose that u ≠ 0 . Moreover, we may assume that Step 1 ∇u n → ∇u a.e. in ℝ N . (1) . Therefore, By using (26) and the boundedness of {u n } n∈ℕ in V 0 , it is easy to check that, for t ∈ {p, q}, |∇u n | p p → t 1 and |∇u n | q q → t 2 .

3
Then, for some subsequence of {u n } n∈ℕ , we have ∇u n → ∇u a.e. in ℝ N .
Step 2 |∇u n | t → |∇u| t for t ∈ {p, q}. By Step 1 and Fatou's lemma, we know that |∇u| p p ≤ t 1 and |∇u| q q ≤ t 2 . Now, we show that Assume by contradiction that |∇u| and (f 4 ) ), the Fatou's lemma gives and this is an absurd. Consequently, |∇u n | t → |∇u| t for t ∈ {p, q} and we have L � V 0 (u) = 0.
Then, {u n } n∈ℕ has a convergent subsequence in V 0 .
Proof By Lemma 8-(iii), Proposition 2-(d) and the definition of d V 0 we have that |∇u| p p = t 1 and |∇u| q q = t 2 . ( and Note that, ( , is a complete metric space. Consider the , and by Proposition 2-(d), we have that G is bounded below. Then, we can apply Ekeland's variational principle [20] to deduce that there exists a sequence {v n } n∈ℕ ⊂ + (1) . At this point, the proof follows from Proposition 2, Theorem 3, and arguing as in the proof of Corollary 1. ◻ For the minimax levels c and d V 0 , we have the following relation.
Finally, we observe that Since V( ⋅) is bounded on the support of , we use the dominated convergence theorem, (34) and the above inequality to see that lim sup →0 c ≤ d V 0 . On the other hand, it follows from

Multiplicity of solutions to (12)
In this section, we collect some technical results which will be used to implement the barycenter machinery below. Take > 0 such that and choose a non-increasing function For any y ∈ M , we define and take t > 0 satisfying where w ∈ V 0 is a positive ground-state solution to (25) whose existence is guaranteed by Theorem 3.
Let Φ ∶ M → N be given by By construction, Φ (y) has compact support for any y ∈ M.

Lemma 12
The functional Φ verifies the following limit: Proof Suppose that the thesis of the lemma is false. Then, we can find 0 > 0 , {y n } n∈ℕ ⊂ M and n → 0 such that Now, for each n ∈ ℕ and for all z ∈ B n , we have n z ∈ B , and so Using the definition of Φ n (y n ) , that G = F in Λ × ℝ and taking the change of variable z = n x−y n n , we have Our purpose is to show that t n → 1 as n → ∞ . First, we prove that t n → t 0 ∈ [0, ∞) . Note that, ⟨L � Then, using (f 4 ) , we obtain where w(ẑ) = min z∈B In view of (39), (41) and (42), we obtain an absurd. Therefore, {t n } n∈ℕ is bounded, and we may suppose that t n → t 0 for some t 0 ≥ 0 . Taking into account (38), (40), (f 1 )-(f 2 ) , we deduce that t 0 ∈ (0, ∞) . Now, letting n → ∞ in (38), and using (40) and the dominated convergence theorem, we obtain that Since w ∈ M V 0 , then Combining the above identities, we find and using 2q > q > p and (f 4 ) , we deduce that t 0 = 1 and the claim is proved.
By sending n → ∞ in (37), we have that which is in contrast with (36). This completes the proof of the lemma. ◻ Let = ( ) > 0 be such that M ⊂ B . Define ∶ ℝ N → ℝ N as (x) = x for |x| < and (x) = x |x| for |x| > . Finally, we introduce the barycenter map ∶ N → ℝ N given by Since M ⊂ B , by the definition of and applying the dominated convergence theorem, we conclude that The next compactness result is fundamental for showing that the solutions of (12) are solutions of (1).

Lemma 13 Let
Proof As in the proof of Lemma 7, we can prove that {u n } n∈ℕ is bounded in V 0 . In view of d V 0 > 0 , we have that ‖u n ‖ n ↛ 0 . Arguing as in the proof of Lemma 9 and Remark 5, we can find {ỹ n } n∈ℕ ⊂ ℝ N and R, > 0 such that Moreover, {ṽ n } n∈ℕ is bounded in V 0 and we may suppose that ṽ n ⇀ṽ . We may assume that t n → t 0 ∈ (0, ∞) . From the uniqueness of the weak limit, we have that ṽ = t 0 v ≢ 0 . By (44) and Lemma 10, ṽ n →ṽ in V 0 and thus v n → v in V 0 . In particular, Next, we prove that {y n } n∈ℕ admits a bounded subsequence. Assume, by contradiction, that there exists a subsequence of {y n } n∈ℕ , still denoted by itself, such that |y n | → ∞ . Let R > 0 be such that Λ ⊂ B R . Then, for n large enough, we have |y n | > 2R , and for each x ∈ B R n we obtain Then, taking into account that v n → v in V 0 , the definition of g, and the dominated convergence theorem, we deduce that which implies that and this gives a contradiction because of v n → v ≢ 0 in V 0 . Therefore, {y n } n∈ℕ is bounded in ℝ N and, up to a subsequence, we may assume that y n → y 0 . If y 0 ∉ Λ , we can proceed as above to conclude v n → 0 in V 0 . Hence, y ∈ Λ . In order to prove that V(y 0 ) = V 0 , we suppose, by contradiction, that V(y 0 ) > V 0 . Then, using ṽ n →ṽ in V 0 , Fatou's lemma and the invariance of ℝ N by translations, we deduce that which yields a contradiction. Thus, V(y 0 ) = V 0 and y 0 ∈ M . From (V 2 ), we get that y 0 ∉ M and so y 0 ∈ M . ◻ We now define the following subset of the Nehari manifold where ( ) = sup y∈M |L (Φ (y)) − d V 0 | . By Lemma 12, we deduce that ( ) → 0 as → 0 . By the definition of ( ) , we have that, for all y ∈ M and > 0 , Φ (y) ∈Ñ and Ñ ≠ ∅.
In what follows, we provide an interesting relation between Ñ and the barycenter map.

Lemma 14
Then, we can apply Lemma 13 to find {ỹ n } ⊂ ℝ N such that y n = nỹn ∈ M for n large enough. Hence, Since u n (⋅ +ỹ n ) strongly converges in V 0 and n z + y n → y ∈ M for all z ∈ ℝ N , we deduce that n (u n ) = y n + o n (1) . Therefore, {y n } n∈ℕ satisfies the required property and the lemma is proved. ◻

Proof of the main result
In this section, we give the proof of the main result of this work. We start by proving a multiplicity result for (12). Note that, since + is not a complete metric space, we cannot use directly an abstract result as in [2][3][4][5]23]. However, we can apply the abstract category result in [46] to deduce the following result. Proof For each > 0 , we consider ∶ M → + given bỹ n (u n ) = y n + ∫ ℝ N [ ( n z + y n ) − y n ](|u n (z +ỹ n )| p + |u n (z +ỹ n )| q ) dz ∫ ℝ N (|u n (z +ỹ n )| p + |u n (z +ỹ n )| q ) dz The next result will be crucial to study the behavior of the maximum points of the solutions. The proof is based on some arguments found in [2,21,26,33].

Lemma 15
Let n → 0 and u n ∈Ñ n be a solution to (12). Then, L n (u n ) → d V 0 , and there exists {ỹ n } n∈ℕ ⊂ ℝ N such that v n = u n (⋅ +ỹ n ) ∈ L ∞ (ℝ N ) and for some C > 0 it holds Moreover, Proof Observing that L n (u n ) ≤ d V 0 + ( n ) with ( n ) → 0 as n → ∞ , we can repeat the same arguments used at the beginning of the proof of Lemma 13 to show that L n (u n ) → d V 0 . Then, applying Lemma 13, there exists {ỹ n } n∈ℕ ⊂ ℝ N such that v n = u n (⋅ +ỹ n ) → v in V 0 for some v ∈ V 0 ⧵ {0} and nỹn → y 0 ∈ M.
Let x 0 ∈ ℝ N , R 0 > 1 , 0 < t < s < 1 < R 0 and ∈ C ∞ c (ℝ N ) such that    Setting C = max 0 2 , |v 1 | ∞ , … , |v n 0 −1 | ∞ , we find that |v n | ∞ ≤C for all n ∈ ℕ . Combining this estimate with the regularity results in [28], we can see that {v n } n∈ℕ ⊂ C 1, loc (ℝ N ). Finally, we show that v n (x) → 0 as |x| → ∞ uniformly in n ∈ ℕ . Arguing as before, we can see that for each > 0, we have that Therefore, applying lemma Lemma 4.7 in [33], there exist R * > 0 and n 0 ∈ ℕ such that which yields Now, increasing R * if necessary, it holds This completes the proof of the lemma. ◻ We are now ready to provide the main result of this section.

Proof of Theorem 1
Fix > 0 such that M ⊂ Λ . We first claim that there exists ̃> 0 such that for any ∈ (0,̃) and any solution u ∈Ñ of (12), it holds Suppose, by contradiction, that for some sequence n → 0 we can find u n = u n ∈Ñ n such that L � n (u n ) = 0 and As in Lemma 13, we have that L n (u n ) → d V 0 and therefore we can use Lemma 13 to find a sequence {ỹ n } n∈ℕ ⊂ ℝ N such that v n = u n (⋅ +ỹ n ) → v in V 0 and nỹn → y 0 ∈ M.
Take r > 0 such that B r (y 0 ) ⊂ B 2r (y 0 ) ⊂ Λ , and so B On the other hand, there exists ∈ ℕ such that for any n ≥ it holds Consequently, u n (x) < a for any x ∈ Λ c n and n ≥ , which contradicts (48). Let ̄> 0 be given by Theorem 4 and set = min{̃,̄} . Take ∈ (0, ) . By Theorem 4, we obtain at least cat M (M) positive solutions to (12). If u is one of these solutions, we have that u ∈Ñ , and we can use (47) and the definition of g to deduce that g( x, u ) = f (u ) . This means that u is also a solution of (1). Therefore, (1) has at least cat M (M) solutions. Now, we consider n → 0 and take a sequence {u n } n∈ℕ ⊂ n of solutions to (12) as above. In order to study the behavior of the maximum points of u n , we first note that, by the definition of g and (g 1 ) , there exists ∈ (0, a) sufficiently small such that As before, we can take R > 0 such that Up to a subsequence, we may also assume that Otherwise, if this is not the case, we see that |u n | ∞ < . Then, it follows from ⟨L � n (u n ), u n ⟩ = 0 and (49) that which implies that ‖u n ‖ p V n ,p , ‖u n ‖ q V n ,q → 0 , and thus L (u n ) → 0 . This last fact is impossible because L (u n ) → d V 0 > 0 . Hence, (51) holds.
By virtue of (50) and (51), we can see that if p n is a global maximum point of u n , then p n =ỹ n + q n for some q n ∈ B R . Recalling that nỹn → y 0 ∈ M and using the fact that {q n } n∈ℕ ⊂ B R , we obtain that n p n → y 0 . Since V is a continuous function, we deduce that This completes the proof of Theorem 1. ◻ as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http:// creat iveco mmons. org/ licen ses/ by/4. 0/.