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Multiplicity of concentrating solutions for a class of magnetic Schrödinger-Poisson type equation

  • Yueli Liu , Xu Li and Chao Ji EMAIL logo
Published/Copyright: May 17, 2020

Abstract

In this paper, we study the following nonlinear magnetic Schrödinger-Poisson type equation

(εiA(x))2u+V(x)u+ϵ2(|x|1|u|2)u=f(|u|2)uin R3,uH1(R3,C),

where ϵ > 0, V : ℝ3 → ℝ and A : ℝ3 → ℝ3 are continuous potentials. Under a local assumption on the potential V, by variational methods, penalization technique, and Ljusternick-Schnirelmann theory, we prove multiplicity and concentration properties of nontrivial solutions for ε > 0 small. In this problem, the function f is only continuous, which allow to consider larger classes of nonlinearities in the reaction.

MSC 2010: 35J60; 35J25

1 Introduction and main results

In this paper, we are concerned with multiplicity and concentration results for the following Schrödinger-Poisson type equation

(εiA(x))2u+V(x)u+ϵ2(|x|1|u|2)u=f(|u|2)uin R3, (1.1)

where uH1(ℝ3, ℂ), ε > 0 is a parameter, V : ℝ3 → ℝ is a continuous function, fC(ℝ, ℝ), the magnetic potential A : ℝ3 → ℝ3 is Hölder continuous with exponent α ∈ (0, 1], and the convolution potential is defined by ∣x−1 ∗ ∣u2 = 3xy−1u(y)∣2dy.

Problem (1.1) arises in quantum mechanics, abelian gauge theories, plasma physics, and so on which can be used to simulate the mutual interactions of many particles. In fact, the linear Schrödinger equation describes the behavior of a single particle. However, the interaction among particles can be simulated by adding a nonlinear term f. Moreover, the convolution potential is a solution of Poisson equation which implies that the particles move in their own gravitational field generated by the probability density of particles via classical Newton field equation. Therefore, problem (1.1) can be regarded as the coupling of the Schrödinger equation and Poisson equation.

There is a vast literature concerning the existence and multiplicity of solutions for nonlinear equation without magnetic field. We notice that Fiscella, Pucci and Zhang [16] studied the existence of solutions for p-fractional Hardy-Schrödinger-Kirchhoff systems with critical nonlinearities, Ji and Radulescu [17] considered the multiplicity of multi-bump solutions for quasilinear elliptic equations with variable exponents and critical growth in ℝN, for more results, we refer to the Monograph [25]. Recently, by using the method of Nehari manifold and Ljusternik-Schnirelmann theory, He [21] proved the multiplicity and concentration of solutions of problem (1.1) for fC1(ℝ, ℝ) and the potential satisfying a global condition introduced by Rabinowitz [26]. In [22], on the similar assumptions, He and Zou studied the existence and concentration behavior of ground state solutions for a class of Schrödinger-Poisson system with critical the nonlinearity fC1(ℝ, ℝ). Then, under a local assumption introduced by del Pino and Felmer [14], He and Zou [23] studied the multiplicity of concentrating positive solutions for Schrödinger-Poisson equations with critical nonlinear fC1(ℝ, ℝ). For further results about existence and nonexistence of solutions, multiplicity of solutions, ground states, semiclassical limit and concentrations of solutions for Schrödinger-Poisson system(see [1, 2, 3, 4, 11, 12, 27, 28, 31, 35] and the references therein).

On the other hand, the magnetic nonlinear Schrödinger equation (1.1) has been extensively investigated by many authors applying suitable variational and topological methods (see [5, 6, 7, 9, 10, 13, 15, 18, 19, 20, 32, 33, 34] and references therein). It is well known that the first result involving the magnetic field was obtained by Esteban and Lions [15]. They used the concentration-compactness principle and minimization arguments to obtain solutions for ε > 0 fixed. In [34], Xiang, Rădulescu and Zhang studied multiplicity and concentration of solutions for magnetic relativistic Schrödinger equations, Xia [32] studied a critical fractional Choquard-Kirchhoff problem with magnetic field. In particular, due to our scope, we want to mention [36] where the authors studied a Schrödinger-Poisson type equation with magnetic field by using the method of the Nehari manifold, the penalization method and Ljusternik-Schnirelmann category theory for subcritical nonlinearity fC1. If f is only continuous, then the arguments in [36] failed.

In this paper, motivated by [23, 29, 36], for the case f is only continuous, we intend to prove multiplicity and concentration of nontrivial solutions for problem (1.1). We note that, due to the appearance of magnetic field A(x), problem (1.1) will be more difficult in employing the methods and some estimates. On the other hand, due to the nonlocal term ∣x−1 ∗ ∣u2, some estimates are also more complicated.

Throughout the paper, we make the following assumptions on the potential V:

  1. There exists V0 > 0 such that V(x) ≥ V0 for all x ∈ ℝ3;

  2. There exists a bounded open set Λ ⊂ ℝ3 such that

    V0=minxΛV(x)<minxΛV(x).

Observe that

M:={xΛ:V(x)=V0}.

Moreover, let the nonlinearity fC(ℝ, ℝ) be a function satisfying:

  1. f(t) = 0 if t ≤ 0, and limt0+f(t)t=0;

  2. there exists q ∈ (4, 6) such that

    limt+f(t)tq22=0;
  3. there is a positive constant θ > 4 such that

    0<θ2F(t)tf(t),t>0,where F(t)=0tf(s)ds;
  4. f(t)t is strictly increasing in (0, ∞).

The main result of this paper is the following:

Theorem 1.1

Assume that V satisfies (V1), (V2) and f satisfies (f1)–(f4). Then, for any δ > 0 such that

Mδ:={xR3:dist(x,M)<δ}Λ,

there exists εδ > 0 such that, for any 0 < ε < εδ, problem (1.1) has at least catMδ(M) nontrivial solutions. Moreover, for every sequence {εn} such that εn → 0+ as n → +∞, if we denote by uεn one of these solutions of problem (1.1) for ε = εn and ηεn ∈ ℝ3 the global maximum point ofuεn∣, then

limεn0+V(ηεn)=V0.

The paper is organized as follows. In Section 2 we introduce the functional setting and give some preliminaries. In Section 3, we study the modified problem. We prove the Palais-Smale condition for the modified functional and provide some tools which are useful to establish a multiplicity result. In Section 4, we study the autonomous problem associated. It allows us to show the modified problem has the multiple soutions. Finally, in Section 5, we give the proof of Thereom 1.1.

Notation

  • C, C1, C2, … denote positive constants whose exact values are inessential and can change from line to line;

  • BR(y) denotes the open disk centered at y ∈ ℝ3 with radius R > 0 and BRc (y) denotes the complement of BR(y) in ℝ3;

  • ∥ ⋅ ∥, ∥ ⋅ ∥q, and ∥⋅ ∥L(Ω) denote the usual norms of the spaces H1(ℝ3, ℝ), Lq(ℝ3, ℝ), and L(Ω, ℝ), respectively, where Ω ⊂ ℝ3.

2 Abstract setting and preliminary results

In this section, we present the functional spaces and some useful preliminary remarks which will be useful for our arguments.

For u : ℝ3 → ℂ, let us denote by

Au:=(iA)u,

and

DA1(R3,C):={uL6(R3,C):|Au|L2(R3,R)}.

and

HA1(R3,C):={uDA1(R3,C):uL2(R3,C))}.

The space HA1 (ℝ3, ℂ) is an Hilbert space endowed with the scalar product

u,v:=ReR3(AuAv¯+uv¯)dx,for any u,vHA1(R3,C),

where Re and the bar denote the real part of a complex number and the complex conjugation, respectively. Moreover we denote by ∥uA the norm induced by this inner product.

On HA1 (ℝ3, ℂ) we will frequently use the following diamagnetic inequality (see e.g. [24, Theorem 7.21])

|Au(x)|||u(x)||. (2.1)

Moreover, making a simple change of variables, we can see that (1.1) is equivalent to

(1iAε(x))2u+Vε(x)u+(|x|1|u|2)u=f(|u|2)uin R3, (2.2)

where Aε(x) = A(εx) and Vε(x) = V(εx).

Let Hε be the Hilbert space obtained as the closure of Cc (ℝ3, ℂ) with respect to the scalar product

u,vϵ:=ReR3(AεuAεv¯+Vε(x)uv¯)dx

and let us denote by ∥ ⋅ ∥ε the norm induced by this inner product.

The diamagnetic inequality (2.1) implies that, if u HAε1 (ℝ3, ℂ), then ∣u∣ ∈ H1(ℝ3, ℝ) and ∥u∥ ≤ Cuε. Therefore, the embedding HεLr(ℝ3, ℂ) is continuous for 2 ≤ r ≤ 6 and the embedding Hε Llocr (ℝ3, ℂ) is compact for 1 ≤ r < 6.

By using the continuous embedding H1(ℝ3, ℝ) ↪ Lr(ℝ3, ℝ) for 2 ≤ r ≤ 6, we can see that

H1(R3,R)L125(R3,R). (2.3)

For any uHε, we get ∣u∣ ∈ H1(ℝ3, ℝ), and the linear functional 𝓛u : D1,2(ℝ3, ℝ) → ℝ given by

L|u|(v)=R3|u|2vdx

is well defined and continuous in view of the Hölder inequality and (2.4). Indeed, we can see that

|L|u|(v)|(R3|u|125dx)56(R3|v|6dx)16CuD1,22vD1,2, (2.4)

where

vD1,22=R3(|x|1|v|2)|v|2dx=R3R3|xy|1|u(x)u(y)|2dxdy.

Then, by the Lax-Milgram Theorem, there exists a unique ϕuD1,2(ℝ3, ℝ) such that

Δϕ|u|=|u|2,inR3.

Therefore we obtain the following t-Riesz formula

ϕ|u|(x)=cR3|xy|1|u(y)|2dy.

In the sequel, we will omit the constant for simplicity. The function ϕu possesses the following properties.

Lemma 2.1

For any uHε, we have

  1. ϕu: H1(ℝ3, ℝ) → D1,2(ℝ3, ℝ) is continuous and maps bounded sets into bounded sets;

  2. if unu in Hε, then ϕunϕu in D1,2(ℝ3, ℝ), and

    lim infnR3ϕ|un|2|un|2dxR3ϕ|u|2|u|2dx;
  3. ϕru = r2ϕu for all r ∈ ℝ and ϕu(⋅+y)∣ = ϕu(x + y);

  4. ϕu ≥ 0 for all uHε and we have

    ϕ|u|D1,2CuL125(R3)2Cuε2,andR3ϕ|u||u|2dxCuL125(R3)4Cuε4.

The proof of Lemma 2.1 is similar to one in [27, 35], so we omit it.

3 The modified problem

To study problem (1.1), or equivalently (2.2) by variational methods, we shall modify suitably the nonlinearity f so that, for ε > 0 small enough, the solutions of such modified problem are also solutions of the original one. More precisely, we choose K > 2. By (f4) there exists a unique number a > 0 verifying K f(a) = V0, where V0 is given in (V1). Hence we consider the function

f~(t):=f(t),ta,V0/K,t>a.

Now we introduce the penalized nonlinearity g : ℝ3 × ℝ → ℝ

g(x,t):=χΛ(x)f(t)+(1χΛ(x))f~(t), (3.1)

where χΛ is the characteristic function on Λ and G(x,t):=0tg(x,s)ds.

In view of (f1)–(f4), we have that g is a Carathéodory function satisfying the following properties:

  1. g(x, t) = 0 for each t ≤ 0;

  2. limt0+g(x,t)t=0 uniformly in x ∈ ℝ3, and there exists q ∈ (4, 6) such that

    limt+g(x,t)tq22=0uniformly inxR3;
  3. g(x, t) ≤ f(t) for all t ≥ 0 and uniformly in x ∈ ℝ3;

  4. 0 < θ G(x, t) ≤ 2 g(x, t)t, for each xΛ, t > 0;

  5. 0 < G(x, t) ≤ g(x, t)tV0t/K, for each xΛc, t > 0;

  6. for each xΛ, the function tg(x,t)t is strictly increasing in t ∈ (0, +∞) and for each xΛc, the function tg(x,t)t is strictly increasing in (0, a).

Then we consider the modified problem

(1iAε(x))2u+Vε(x)u+(|x|1|u|2)u=g(εx,|u|2)uin R3. (3.2)

Note that, if u is a solution of problem (3.2) with

|u(x)|2afor all xΛεc,Λε:={xR3:εxΛ},

then u is a solution of problem (2.2).

The functional associated to problem (3.2) is

Jε(u):=12R3(|Aεu|2+Vε(x)|u|2)dx+14R3(|x|1|u|2)|u|2dx12R3G(εx,|u|2)dx

defined in Hε. It is standard to prove that JεC1(Hε, ℝ) and its critical points are the weak solutions of the modified problem (3.2).

We denote by 𝓝ε the Nehari manifold of Jε, that is

Nε:={uHε{0}:Jε(u)[u]=0},

and define the number cε by

cε=infuNεJε(u).

Let Hε+ be open subset Hε given by

Hε+={uHε:|supp(u)Λε|>0},

and Sε+=SεHε+, where Sε is the unit sphere of Hε. Note that Sε+ is a non-complete C1,1-manifold of codimension 1, modeled on Hε and contained in Hε+ . Therefore, Hε = Tu Sε+ ⨁ ℝu for each uTu Sε+ , where Tu Sε+ = {vHε : 〈u, vϵ = 0}.

Now we show that the functional Jε satisfies the Mountain Pass Geometry.

Lemma 3.1

For any fixed ε > 0, the functional Jε satisfies the following properties:

  1. there exist β, r > 0 such that Jε(u) ≥ β ifuε = r;

  2. there exists eHε witheε > r such that Jε(e) < 0.

Proof

  1. By (g3), (f1) and (f2), for any ζ > 0 small, there exists Cζ > 0 such that

    G(εx,|u|2)ζ|u|4+Cζ|u|qfor all xR3.

    By the Sobolev embedding it follows

    Jε(u)12R3(|Aεu|2+Vε(x)|u|2)dxζ2R3|u|4dxCζ2R3|u|qdx12unε2C1ζunε4C2Cζunεq.

    Hence we can choose some β, r > 0 such that Jε(u) ≥ β if ∥uε = r since q > 4.

  2. For each u Hε+ and t > 0, by the definition of g and (f3), one has

    Jε(tu)t22R3(|Aεu|2+Vε(x)|u|2)dx+t44R3(|x|1|u|2)|u|2dx12ΛεG(εx,t2|u|2)dx,t22uε2+t44R3(|x|1|u|2)|u|2dxC1tθΛε|u|θdx+C2|supp(u)Λε|.

    Since θ > 4, we can get the conclusion.□

    Since f is only continuous, the next results are very important because they allow us to overcome the non-differentiability of 𝓝ε and the incompleteness of Sε+ .

Lemma 3.2

Assume that (V1)–(V2) and (f1)–(f4) are satisfied, then the following properties hold:

  1. For any u Hε+ , let gu : ℝ+ → ℝ be given by gu(t) = Jε(tu). Then there exists a unique tu > 0 such that gu (t) > 0 in (0, tu) and gu (t) < 0 in (tu, ∞);

  2. There is a τ > 0 independent on u such that tuτ for all u Sε+ . Moreover, for each compact 𝓦 ⊂ Sε+ there is such that tuC𝓦, for all u ∈ 𝓦;

  3. The map ε : Hε+ → 𝓝ε given by ε(u) = tuu is continuous and mε = m^ε|Sε+ is a homeomorphism between Sε+ and 𝓝ε. Moreover, mε1(u)=uuε;

  4. If there is a sequence {un} ⊂ Sε+ such that dist(un, Sε+ ) → 0, thenmε(un)∥ε → ∞ and Jε(mε(un)) → ∞.

Proof

(A1) As in the proof of Lemma 3.1, we have gu(0) = 0, gu(t) > 0 for t > 0 small and gu(t) < 0 for t > 0 large. Therefore, maxt≥0 gu(t) is achieved at a global maximum point t = tu verifying gu (tu) = 0 and tuu ∈ 𝓝ε. From (f4), the definition of g and ∣supp(u) ∩ Λε∣ > 0, we may obtain the uniqueness of tu. Therefore, maxt≥0 gu(t) is achieved at a unique t = tu so that gu (t) = 0 and tuu ∈ 𝓝ε.

(A2) For ∀ u Sε+ , we have

tu+tu3R3(|x|1|u|2)|u|2dx=R3g(εx,tu2|u|2)tu|u|2dx.

From (g2), the Sobolev embeddings and q > 4, we get

tuζtu3R3|u|4dx+Cζtuq1R3|u|qdxC1ζtu3+C2Cζtuq1,

which implies that tuτ for some τ > 0. If 𝓦 ⊂ Sε+ is compact, and suppose by contradiction that there is {un} ⊂ 𝓦 with tn := tun → ∞. Since 𝓦 is compact, there exists a u ∈ 𝓦 such that unu in Hε. Moreover, using the proof of Lemma 3.1(ii), we have that Jε(tn un) → −∞.

On the other hand, let vn : = tnun ∈ 𝓝ε, from (g4), (g5), (g6) and θ > 4, it yields that

Jε(vn)=Jε(vn)1θJε(vn)[vn](121θ)vnε2+(141θ)R3(|x|1|vn|2)|vn|2dx+Λεc(1θg(εx,|vn|2)|un|212G(εx,|vn|2))dx(121θ)(vnε21KR3V(εx)|vn|2dx)(121θ)(11K)vnε2.

Thus, substituting vn : = tnun and ∥ vnε = tn, we obtain

0<(121θ)(11K)Jε(vn)tn20

as n → ∞, which yields a contradiction. This proves (A2).

(A3) First of all, we note that ε, mε and mε1 are well defined. Indeed, by (A2), for each u Hε+ , there is a unique ε(u) ∈ 𝓝ε. On the other hand, if u ∈ 𝓝ε, then u Hε+ . Otherwise, we have ∣supp(u) ∩ Λε∣ = 0 and by (g5) we have

uε2uε2+R3(|x|1|u|2)|u|2dx=R3g(εx,|u|2)|u|2dx=Λεcg(εx,|u|2)|u|2dx1KR3V(εx)|u|2dx1Kuε2

which is impossible since K > 1 and u ≠ 0. Therefore, mε1(u)=uuεSε+ is well defined and continuous. From

mε1(mε(u))=mε1(tuu)=tuutuuε=u,uSε+,

we conclude that mε is a bijection. Now we prove ε : Hε+ → 𝓝ε is continuous, let {un} ⊂ Hε+ and u Hε+ such that unu in Hε. By (A2), there is a t0 > 0 such that tn := tunt0. Using tnun ∈ 𝓝ε, i.e.,

tn2unε2+tn2R3(|x|1|un|2)|un|2dx=R3g(εx,tn2|un|2)tn2|un|2dx,nN,

and passing to the limit as n → ∞ in the last inequality, we obtain

t02uε2+t02R3(|x|1|u|2)|u|2dx=R3g(εx,t02|u|2)t02|u|2dx,

which implies that t0u ∈ 𝓝ε and tu = t0. This proves ε(un) → ε(u) in Hε+ . Thus, ε and mε are continuous functions and (A3) is proved.

(A4) Let {un} ⊂ Sε+ be a subsequence such that dist(un, Sε+ ) → 0, then for each v Sε+ and nN, we have ∣ un∣ = ∣ unv∣ a.e. in Λε. Therefore, by (V1), (V2) and the Sobolev embedding, there exists a constant Ct > 0 such that

unLt(Λε)infvSε+unvLt(Λε)Ct(infvSε+Λε(|Aεunv|2+Vε(x)|unv|2)dx)12Ctdist(un,Sε+)

for all nN, t ∈ [2, 6]. By (g2), (g3) and (g5), for each t > 0, we have

R3G(εx,t2|un|2)dxΛεF(t2|un|2)dx+t2KΛεcV(εx)|un|2dxC1t4Λε|un|4dx+C2tqΛε|un|qdx+t2Kunε2C3t4dist(un,Sε+)4+C4tqdist(un,Sε+)q+t2K.

Therefore,

lim supnR3G(εx,t2|un|2)dxt2K,t>0.

On the other hand, from the definition of mε and the last inequality, for all t > 0, one has

lim infnJε(mε(un))lim infnJε(tun)liminfnt22unε2t2K=K22Kt2,

this implies that

lim infn12mε(un)ε2K22Kt2,t>0.

From the arbitrary of t > 0, it is easy to see that ∥ mε(un)∥ε → ∞ and Jε(mε(un)) → ∞ as n → ∞. This completes the proof of Lemma 3.2.□

Now we define the function

Ψ^ε:Hε+R,

by Ψ͡ε(u) = Jε(ε(u)) and denote by Ψε:=(Ψ^ε)|Sε+.

We may obtain the following result from Lemma 3.2 directly, and its proof is similar to that of Corollary 10 in [30], so we omit it.

Lemma 3.3

Assume that (V1)–(V2) and (f1)–(f4) are satisfied, then

  1. Ψ͡εC1( Hε+ , ℝ) and

    Ψ^ε(u)v=m^ε(u)ϵuϵJε(m^ε(u))[v],uHε+andvHε;
  2. ΨεC1( Sε+ , ℝ) and

    Ψε(u)v=mε(u)ϵJε(m^ε(u))[v],vTuSε+;
  3. If {un} is a (PS)c sequence of Ψε, then {mε(un)} is a (PS)c sequence of Jε. If {un} ⊂ 𝓝ε is a bounded (PS)c sequence of Jε, then {mε1(un)} is a (PS)c sequence of Ψε;

  4. u is a critical point of Ψε if and only if mε(u) is a critical point of Jε. Moreover, the corresponding critical values coincide and

    infSε+Ψε.=infNεJε.

As in [30], we have the following variational characterization of the infimum of Jε over 𝓝ε:

cε=infuNεJε(u)=infuHε+supt>0Jε(tu)=infuSε+supt>0Jε(tu)

Lemma 3.4

Let c > 0 and {un} is a (PS)c sequence for Jε, then {un} is bounded in Hε.

Proof

Assume that {un} ⊂ Hε is a (PS)c sequence for Jε, that is, Jε(un) → c and Jε (un) → 0. By using (g4), (g5) and θ > 4, we have

d+on(1)+on(1)unεJε(un)1θJε(un)[un]=(121θ)unε2+(141θ)R3(|x|1|un|2)|un|2dx+R3(1θg(εx,|un|2)|un|212G(εx,|un|2))dx(121θ)unε2+Λεc(1θg(εx,|un|2)|un|212G(εx,|un|2))dx(121θ)unε212ΛεcG(εx,|un|2)dx(121θ)unε212KR3V(εx)|un|2dx(121θ12K)unε2.

Since K > θ/(θ − 2), from the above inequalities we obtain that {un} is bounded in Hε.□

The following result is important to prove the (PS)cε condition for the functional Jε.

Lemma 3.5

The functional Jε satisfies the (PS)c condition at any level c > 0.

Proof

Let (un) ⊂ Hε be a (PS)c for Jε. By Lemma 3.4, (un) is bounded in Hε. Thus, up to a subsequence, unu in Hε and unu in Llocr (ℝ3, ℂ) for all 1 ≤ r < 6 as n → +∞. Moreover, Lemma 2.1(ii) and the subcritical growth of g imply that Jε (u) = 0, and

uε2+R3(|x|1|u|2)|u|2dx=R3g(εx,|u|2)|u|2dx.

Let R > 0 be such that ΛεBR/2(0). We show that for any given ζ > 0, for R large enough,

lim supnBRc(0)(|Aεun|2+Vε(x)|un|2)dxζ. (3.3)

Let ϕRC(ℝ3, ℝ) be a cut-off function such that

ϕR=0xBR/2(0),ϕR=1xBRc(0),0ϕR1,and|ϕR|C/R

where C > 0 is a constant independent of R. Since the sequence (ϕRun) is bounded in Hε, we have

Jε(un)[ϕRun]=on(1),

that is

ReR3AεunAε(ϕRun)¯dx+R3Vε(x)|un|2ϕRdx+R3(|x|1|un|2)|un|2ϕRdx=R3g(εx,|un|2)|un|2ϕRdx+on(1).

Since Aε(unϕR) = iunϕR + ϕRAεun, using (g5), we have

R3(|Aεun|2+Vε(x)|un|2)ϕRdxR3g(εx,|un|2)|un|2ϕRdxReR3iun¯AεunϕRdx+on(1)1KR3Vε(x)|un|2ϕRdxReR3iun¯AεunϕRdx+on(1).

By the definition of ϕR, the Hölder inequality and the boundedness of (un) in Hε, we obtain

(11K)R3(|Aεun|2+Vε(x)|un|2)ϕRdxCRun2Aεun2+on(1)C1R+on(1)

and so (3.3) holds.

Using unu in Llocr (ℝ3, ℂ), for all 1 ≤ r < 6 again, up to a subsequence, we have that

|un||u| a.e. in R3 as n+,

then

g(εx,|un|2)|un|2g(εx,|u|2)|u|2 a.e. in R3 as n+.

Moreover, from the subcritical growth of g and and the Lebesgue Dominated Convergence Theorem, we can infer

limnBR(0)|g(εx,|un|2)|un|2g(εx,|u|2)|u|2|dx=0.

Now, by (g5) and (3.3) we have

BRc(0)|g(εx,|un|2)|un|2g(εx,|u|2)|u|2|dx2KBRc(0)(|Aεun|2+V(εx)|un|2)dx<2ζK

for every ζ > 0.

Hence

R3g(εx,|un|2)|un|2dxR3g(εx,|u|2)|u|2dx as n+.

Finally, since Jε (u) = 0, we have

on(1)=Jε(un)[un]=unε2+R3(|x|1|un|2)|un|2dxR3g(εx,|un|2)|un|2dx=unε2+R3(|x|1|un|2)|un|2dxuε2R3(|x|1|u|2)|u|2dx+on(1).

Thus, from Lemma 2.1, the sequence (un) strong converges to u in Hε and 3(∣x−1 ∗ ∣ un2)∣ un2dx3(∣x−1 ∗ ∣u2)∣u2dx as n → ∞.□

Since f is only assumed to be continuous, the following result is required for the multiplicity result in the next section.

Corollary 3.1

The functional Ψε satisfies the (PS)c condition on Sε+ at any level c > 0.

Proof

Let {un} ⊂ Sε+ be a (PS)c sequence for Ψε. Then Ψε(un) → c and ∥ Ψε (un)∥* → 0, where ∥ ⋅ ∥* is the norm in the dual space (TunSε+). By Lemma 3.3(B3), we know that {mε(un)} is a (PS)c sequence for Jε in Hε. From Lemma 3.5, we know that there exists a u Sε+ such that, up to a subsequence, mε(un) → mε(u) in Hε. By Lemma 3.2(A3), we obtain

unuinSε+,

and the proof is complete.□

Proposition 3.1

Assume that (V1)–(V2) and (f1)–(f4) hold, then problem (3.2) has a ground state solution for any ϵ > 0.

Proof

Since

cε=infuNεJε(u)=infuHε+supt>0Jε(tu)=infuSε+supt>0Jε(tu),

by the Ekeland variational principle [37], we obtain a minimizing (PS)cε sequence on Sε+ for the functional Ψε. Moreover, by Corollary 3.1, we deduce the existence of a ground state uHε for problem (3.2).□

4 Multiple solutions for the modified problem

4.1 The autonomous problem

For our scope, we need also to study the following limit problem

Δu+V0u+(|x|1|u|2)u=f(u2)u,u:R3R, (4.1)

whose associated C1-functional, defined in H1(ℝ3, ℝ), is

I0(u):=12R3(|u|2+V0u2)dx+14R3(|x|1|u|2)|u|2dx12R3F(u2)dx.

Let

N0:={uH1(R3,R){0}:I0(u)[u]=0}

and

cV0:=infuN0I0(u).

Let S0 be the unit sphere of H0 := H1(ℝ3, ℝ) and is complete and smooth manifold of codimension 1. Therefore, H0 = TuS0 ⨁ ℝu for each uTuS0, where TuS0 = {vH0 : 〈u, v0 = 0}.

Lemma 4.1

Let V0 be given in (V1) and suppose that (f1)–(f4) are satisfied, then the following properties hold:

  1. For any uH0 ∖ {0}, let gu : ℝ+ → ℝ be given by gu(t) = I0(tu). Then there exists a unique tu > 0 such that gu (t) > 0 in (0, tu) and gu (t) < 0 in (tu, ∞);

  2. There is a τ > 0 independent on u such that tu > τ for all uS0. Moreover, for each compact 𝓦 ⊂ S0 there is such that tuC𝓦, for all u ∈ 𝓦;

  3. The map : H0 ∖ {0} → 𝓝_{0} given by (u) = tuu is continuous and m0 = 0S0 is a homeomorphism between S0 and 𝓝0. Moreover, m1(u)=uu0.

The proof of Lemma 4.1 is similar to that of Lemma 3.2, we omit it.

Lemma 4.2

Let V0 be given in (V1) and suppose that (f1)–(f4) are satisfied, then

  1. Ψ͡0C1(H0 ∖ {0}, ℝ) and

    Ψ^0(u)v=m^(u)0u0I0(m^(u))[v],uH0{0}andvH0;
  2. Ψ0C1(S0, ℝ) and

    Ψ0(u)v=m(u)0I0(m^(u))[v],vTuS0;
  3. If {un} is a (PS)c sequence of Ψ0, then {m(un)} is a (PS)c sequence of I0. If {un} ⊂ 𝓝0 is a bounded (PS)c sequence of I0, then {m−1(un)} is a (PS)c sequence of Ψ0;

  4. u is a critical point of Ψ0 if and only if m(u) is a critical point of I0. Moreover, the corresponding critical values coincide and

    infS0Ψ0=infN0I0.

The proof of Lemma 4.2 can be found in the proofs of Proposition 9 and Corollary 10 of Szulkin and Weth [30], so we omit it.

Similar to the previous argument, we have the following variational characterization of the infimum of I0 over 𝓝0:

cV0=infuN0I0(u)=infuH0{0}supt>0I0(tu)=infuS0supt>0I0(tu)

The next result is useful in later arguments.

Lemma 4.3

Let {un} ⊂ H0 be a (PS)c sequence for I0 such that un ⇀ 0. Then, one of the following alternatives occurs:

  1. un → 0 in H0 as n → +∞;

  2. there are a sequence {yn} ⊂ ℝ3 and constants R, β > 0 such that

    lim infnBR(yn)|un|2dxβ.

Proof

Assume that (ii) does not hold. Then, for every R > 0, we have

limnsupyR3BR(y)|un|2dx=0.

Being {un} bounded in H0, by the Lion’s lemma [37], it follows that

un0inLr(R3,R),2<r<6.

From the subcritical growth of f, we have

R3F(un2)dx=on(1)=R3f(un2)un2dx.

Moreover, from I0 (un)[un] → 0, it follows that

RN(|un|2+V0un2)dx+R3(|x|1|un|2)|un|2dx=R3f(un2)un2dx+on(1)=on(1).

Thus (i) holds.□

Remark 4.1

From Lemma 4.3 we see that if u is the weak limit of (PS)cV0 sequence {un} of the functional I0, then we have u ≠ 0. Otherwise we have that un ⇀ 0 and if un ↛ 0, from Lemma 4.3 it follows that there are a sequence {yn} ⊂ ℝ3 and constants R, β > 0 such that

lim infnBR(yn)|un|2dxβ>0.

Then set vn(x) = un(x + zn), it is easy to see that {vn} is also a (PS)cV0 sequence for the functional I0, it is bounded, and there exists vH0 such that vnv in H0 with v ≠ 0.

Lemma 4.4

Assume that V satisfies (V1), (V2) and f satisfies (f1)–(f4), then problem (4.1) has a positive ground state solution.

Proof

First of all, it is easy to show that cV0 > 0. Moreover, if u0 ∈ 𝓝0 satisfies I0(u0) = cV0, then m−1(u0) ∈ S0 is a minimizer of Ψ0, so that u0 is a critical point of I0 by Lemma 4.2. Now, we show that there exists a minimizer u ∈ 𝓝0 of I0𝓝0. Since infS0 Ψ0 = inf𝓝0 I0 = cV0 and S0 is a C1 manifold, by Ekelanďs variational principle, there exists a sequence ωnS0 with Ψ0(ωn) → cV0 and Ψ0 (ωn) → 0 as n → ∞. Put un = m(ωn) ∈ 𝓝0 for nN. Then I0(un) → cV0 and I0 (un) → 0 as n → ∞ by Lemma 4.2(b3). Similar to the proof of Lemma 3.4, it is easy to know that {un} is bounded in H0. Thus, we have unu in H0, unu in Llocr (ℝ3, ℝ), 1 ≤ r < 6 and unu a.e. in ℝ3, thus I0 (u) = 0. From Remark 4.1, we know that u ≠ 0. Moreover, by Lemma 2.1,

cV0I0(u)=I0(u)1θI0(u)[u]=(121θ)u02+(141θ)R3(|x|1|u|2)|u|2dx+R3(1θf(u2)u212F(u2))dxlim infn{(121θ)un02+(141θ)R3(|x|1|un|2)|un|2dx+R3(1θf(un)un212F(un2))dx}=lim infn{I0(un)1θI0(un)[un]}=cV0,

thus, u is a ground state solution. From the assumption of f, u ≥ 0, moreover, by [8, Proposition 6 and Proposition 7], we know that u(x) > 0 for x ∈ ℝN. The proof is complete.□

Lemma 4.5

Let (un) ⊂ 𝓝0 be such that I0(un) → cV0. Then (un) has a convergent subsequence in H0.

Proof

Since (un) ⊂ 𝓝0, from Lemma 4.1(a3), Lemma 4.2(b4) and the definition of cV0, we have

vn=m1(un)=unun0S0,nN,

and

Ψ0(vn)=I0(un)cV0=infuS0Ψ0(u).

Since S0 is a complete C1 manifold, by Ekelanďs variational principle, there exists a sequence {n} ⊂ S0 such that {n} is a (PS)cV0 sequence for Ψ0 on S0 and

v~nvn0=on(1).

Similar to the proof of Lemma 4.4, we may obtain the conclusion of this lemma.□

4.2 The technical results

In this subsection, we prove a multiplicity result for the modified problem (3.2) using the Ljusternik-Schnirelmann category theory. In order to get it, we first provide some useful preliminaries.

Let δ > 0 be such that MδΛ, ωH1(ℝ3, ℝ) be a positive ground state solution of the limit problem (4.1), and ηC(ℝ+, [0, 1]) be a nonincreasing cut-off function defined in [0, +∞) such that η(t) = 1 if 0 ≤ tδ/2 and η(t) = 0 if tδ.

For any yM, let us introduce the function

Ψε,y(x):=η(|εxy|)ω(εxyε)exp(iτy(εxyε)),

where

τy(x):=i3Ai(y)xi.

Let tε > 0 be the unique positive number such that

maxt0Jε(tΨε,y)=Jε(tεΨε,y).

Note that tεΨε,y ∈ 𝓝ε.

Let us define Φε : M → 𝓝ε as

Φε(y):=tεΨε,y.

By construction, Φε(y) has compact support for any yM.

Moreover, the energy of the above functions has the following behavior as ε → 0+.

Lemma 4.6

The limit

limε0+Jε(Φε(y))=cV0

holds uniformly in yM.

Proof

Assume by contradiction that the statement is false. Then there exist δ0 > 0, (yn) ⊂ M and εn → 0+ satisfying

|Jεn(Φεn(yn))cV0|δ0.

For simplicity, we write Φn, Ψn and tn for Φεn(yn), Ψεn,yn and tεn, respectively.

Similar to the proof of Lemma 3.4 in [36], by the Lebesgue Dominated Convergence Theorem, we have that

Ψnεn2R3(|ω|2+V0ω2)dx as n+. (4.2)
R3(|x|1|Ψn|2)|Ψn|2dxR3(|x|1|ω|2)|ω|2dx as n+. (4.3)

Since Jεn (tnΨn)(tnΨn) = 0, by the change of variables z = (εn xyn)/εn, observe that, if zBδ/εn(0), then εn z+ynBδ(yn) ⊂ MδΛ, we have

Ψnεn2+tn2R3(|x|1|Ψn|2)|Ψn|2dx=R3g(εnz+yn,tn2η2(|εnz|)ω2(z))η2(|εnz|)ω2(z)dz=R3f(tn2η2(|εnz|)ω2(z))η2(|εnz|)ω2(z)dzBδ/(2εn)(0)f(tn2ω2(z))ω2(z)dzBδ/2(0)f(tn2ω2(z))ω2(z)dzf(tn2y2)Bδ/2(0)ω4(z)dz

for all n large enough and where y = min{ω(z) : ∣z∣ ≤ δ/2}. Moreover, we have

tn2Ψnεn2+R3(|x|1|Ψn|2)|Ψn|2dxf(tn2y2)f(tn2y2)y2Bδ/2(0)ω4(z)dz.

If tn → +∞, by (f4) we derive a contradiction.

Therefore, up to a subsequence, we may assume that tnt0 ≥ 0.

If tn → 0, using the fact that f is increasing and the Lebesgue Dominated Convergence Theorem, we obtain that

Ψnεn2+tn2R3(|x|1|Ψn|2)|Ψn|2dx=R3f(tn2η2(|εnz|)ω2(z))η2(|εnz|)ω2(z)dz0, as n+,

which contradicts (4.2). Thus, from (4.2) and (4.3), we have t0 > 0 and

R3(|ω|2+V0ω2)dx+t02R3(|x|1|ω|2)|ω|2dx=R3f(t0ω2)ω2dx,

so that t0ω ∈ 𝓝V0. Since ω ∈ 𝓝V0, we obtain that t0 = 1 and so, using the Lebesgue Dominated Convergence Theorem, we get

limnR3F(|tnΨn|2)dx=R3F(ω2)dx.

Hence

limnJεn(Φεn(yn))=I0(ω)=cV0

which is a contradiction and the proof is complete.□

Now we define the barycenter map.

Let ρ > 0 be such that MδBρ and consider Y : ℝ3 → ℝ3 defined by setting

Y(x):=x,if|x|<ρ,ρx/|x|,if|x|ρ.

The barycenter map βε : 𝓝ε → ℝ3 is defined by

βε(u):=1u44R3Y(εx)|u(x)|4dx.

We have the following lemma.

Lemma 4.7

The limit

limε0+βε(Φε(y))=y

holds uniformly in yM.

Proof

Assume by contradiction that there exists κ > 0, (yn) ⊂ M and εn → 0 such that

|βεn(Φεn(yn))yn|κ. (4.4)

Using the change of variable z = (εn xyn)/εn, we can see that

βεn(Φεn(yn))=yn+R3(Y(εnz+yn)yn)η4(|εnz|)ω4(z)dzR3η4(|εnz|)ω4(z)dz.

Taking into account (yn) ⊂ MMδBρ and the Lebesgue Dominated Convergence Theorem, we can obtain that

|βεn(Φεn(yn))yn|=on(1),

which contradicts (4.4).□

Now, we prove the following useful compactness result.

Proposition 4.1

Let εn → 0+ and (un) ⊂ 𝓝εn be such that Jεn(un) → cV0. Then there exists (n) ⊂ ℝ3 such that the sequence (∣ vn∣) ⊂ H1(ℝ3, ℝ), where vn(x) := un(x + n), has a convergent subsequence in H1(ℝ3, ℝ). Moreover, up to a subsequence, yn := εnnyM as n → +∞.

Proof

Since Jεn (un) [un] = 0 and Jεn(un) → cV0, arguing as in the proof of Lemma 3.4, we can prove that there exists C > 0 such that ∥ unεnC for all n ∈ ℕ.

Arguing as in the proof of Lemma 3.2 and recalling that cV0 > 0, we have that there exist a sequence {n} ⊂ ℝ3 and constants R, β > 0 such that

lim infnBR(y~n)|un|2dxβ. (4.5)

Now, let us consider the sequence {∣vn∣} ⊂ H1(ℝ3, ℝ), where vn(x) := un(x + n). By the diamagnetic inequality (2.1), we get that {∣vn∣} is bounded in H1(ℝ3, ℝ), and using (4.5), we may assume that ∣ vn∣ ⇀ v in H1(ℝ3, ℝ) for some v ≠ 0.

Let now tn > 0 be such that n := tnvn∣ ∈ 𝓝V0, and set yn := εnn.

By the diamagnetic inequality (2.1), we have

cV0I0(v~n)maxt0Jεn(tun)=Jεn(un)=cV0+on(1),

which yields I0(n) → cV0 as n → +∞.

Since the sequences {∣vn∣} and {n} are bounded in H1(ℝ3, ℝ) and ∣ vn∣ ↛ 0 in H1(ℝ3, ℝ), then (tn) is also bounded and so, up to a subsequence, we may assume that tnt0 ≥ 0.

We claim that t0 > 0. Indeed, if t0 = 0, then, since (∣vn∣) is bounded, we have n → 0 in H1(ℝ3, ℝ), that is I0(n) → 0, which contradicts cV0 > 0.

Thus, up to a subsequence, we may assume that n := t0 v ≠ 0 in H1(ℝ3, ℝ), and, by Lemma 4.5, we can deduce that n in H1(ℝ3, ℝ), which gives ∣ vn∣ → v in H1(ℝ3, ℝ).

Now we show the final part, namely that {yn} has a subsequence such that ynyM. Assume by contradiction that {yn} is not bounded and so, up to a subsequence, ∣ yn∣ → +∞ as n → +∞. Choose R > 0 such that ΛBR(0). Then for n large enough, we have ∣ yn∣ > 2R, and, for any xBR/εn(0),

|εnx+yn||yn|εn|x|>R.

Since un ∈ 𝓝εn, using (V1) and the diamagnetic inequality (2.1), we get that

R3(||vn||2+V0|vn|2)dxR3g(εnx+yn,|vn|2)|vn|2dxBR/εn(0)f~(|vn|2)|vn|2dx+BR/εnc(0)f(|vn|2)|vn|2dx. (4.6)

Since ∣ vn∣ → v in H1(ℝ3, ℝ) and (t) ≤ V0/K, we can see that (4.6) yields

min{1,V0(11K)}R3(||vn||2+|vn|2)dx=on(1),

that is ∣ vn∣ → 0 in H1(ℝ3, ℝ), which contradicts to v ≢ 0.

Therefore, we may assume that yny0 ∈ ℝ3. Assume by contradiction that y0Λ. Then there exists r > 0 such that for every n large enough we have that ∣yny0∣ < r and B2r(y0) ⊂ Λc. Then, if xBr/εn(0), we have that ∣εn x + yny0∣ < 2r so that εn x + ynΛc and so, arguing as before, we reach a contradiction. Thus, y0Λ.

To prove that V(y0) = V0, we suppose by contradiction that V(y0) > V0. Using the Fatou’s lemma, the change of variable z = x + n and maxt≥0 Jεn(tun) = Jεn(un), we obtain

cV0=I0(v~)<12R3(|v~|2+V(y0)|v~|2)dx+14R3(|x|1|v~|2)|v~|2dx12R3F(|v~|2)dxlim infn(12R3(|v~n|2+V(εnx+yn)|v~n|2)dx+14R3(|x|1|v~n|2)|v~n|2dx12R3F(|v~n|2)dx)=lim infn(tn22R3(||un||2+V(εnz)|un|2)dz+tn44R3(|x|1|un|2)|un|2dx12R3F(|tnun|2)dz)lim infnJεn(tnun)lim infnJεn(un)=cV0

which is impossible and the proof is complete.□

Let now

N~ε:={uNε:Jε(u)cV0+h(ε)},

where h : ℝ+ → ℝ+, h(ε) → 0 as ε → 0+.

Fixed yM, since, by Lemma 4.6, ∣ Jε(Φε(y)) − cV0∣ → 0 as ε → 0+, we get that 𝓝̃ε ≠ ∅ for any ε > 0 small enough.

We have the following relation between 𝓝̃ε and the barycenter map.

Lemma 4.8

We have

limε0+supuN~εdist(βε(u),Mδ)=0.

Proof

Let εn → 0+ as n → +∞. For any n ∈ ℕ, there exists un ∈ 𝓝̃εn such that

supuN~εninfyMδ|βεn(u)y|=infyMδ|βεn(un)y|+on(1).

Therefore, it is enough to prove that there exists (yn) ⊂ Mδ such that

limn|βεn(un)yn|=0.

By the diamagnetic inequality (2.1), we can see that I0(t∣ un∣) ≤ Jεn(tun) for any t ≥ 0. Therefore, recalling that {un} ⊂ 𝓝̃εn ⊂ 𝓝εn, we can deduce that

cV0maxt0I0(t|un|)maxt0Jεn(tun)=Jεn(un)cV0+h(εn) (4.7)

which implies that Jεn(un) → cV0 as n → +∞.

Then, Proposition 4.1 implies that there exists {n} ⊂ ℝ3 such that yn = εnnMδ for n large enough.

Thus, making the change of variable z = xn, we get

βεn(un)=yn+R3(Y(εnz+yn)yn)|un(z+y~n)|4dzR3|un(z+y~n)|4dz.

Since, up to a subsequence, ∣ un∣(⋅+n) converges strongly in H1(ℝ3, ℝ) and εnz+ynyM for any z ∈ ℝ3, we conclude.□

4.3 Multiplicity of solutions for problem (3.2)

Finally, we present a relation between the topology of M and the number of solutions of the modified problem (3.2).

Theorem 4.1

For any δ > 0 such that MδΛ, there exists ε̃δ > 0 such that, for any ε ∈ (0, ε̃δ), problem (3.2) has at least catMδ(M) nontrivial solutions.

Proof

For any ϵ > 0, we define the function πϵ : M Sε+ by

πϵ(y)=mε1(Φϵ(y)),yM.

By Lemma 4.6 and Lemma 3.3(B4), we obtain

limϵ0Ψϵ(πϵ(y))=limϵ0Jϵ(Φϵ(y))=cV0,uniformly inyM.

Hence, there is a number ϵ̂ > 0 such that the set S~ε+ := {u Sε+ : Ψε(u) ≤ cV0 + h(ε)} is nonempty, for all ϵ ∈ (0, ϵ̂), since πϵ(M) ⊂ S~ε+ . Here h is given in the definition of 𝓝̃ε.

Given δ > 0, by Lemma 4.6, Lemma 3.2(A3), Lemma 4.7, and Lemma 4.8, we can find ε̃δ > 0 such that for any ε ∈ (0, ε̃δ), the following diagram

MΦεΦε(M)mε1πϵ(M)mεΦε(M)βεMδ

is well defined and continuous. From Lemma 4.7, we can choose a function Θ(ϵ, z) with ∣ Θ(ϵ, z)∣ < δ2 uniformly in zM, for all ϵ ∈ (0, ϵ̂) such that βε(Φε(z)) = z + Θ(ϵ, z) for all zM. Define H(t, z) = z+(1−t)Θ(ϵ, z). Then H : [0, 1] × MMδ is continuous. Clearly, H(0, z) = βε(Φε(z)), H(1, z) = z for all zM. That is, H(t, z) is a homotopy between βεΦε = (βεmε) ∘ πϵ and the embedding ι : MMδ. Thus, this fact implies that

catπϵ(M)(πϵ(M))catMδ(M). (4.8)

By Corollary 3.1 and the abstract category theorem [30], Ψε has at least catπϵ(M)(πϵ(M)) critical points on Sε+ . Therefore, from Lemma 3.3(B4) and (4.8), we have that Jε has at least catMδ(M) critical points in 𝓝̃ε which implies that problem (3.2) has at least catMδ(M) solutions.□

5 Proof of Theorem 1.1

In this section we prove our main result. The idea is to show that the solutions uε obtained in Theorem 4.1 satisfy

|uε(x)|2a for xΛεc

for ε small. The key ingredient is the following result.

Lemma 5.1

Let εn → 0+ and un ∈ 𝓝̃εn be a solution of problem (3.2) for ε = εn. Then Jεn(un) → cV0. Moreover, there exists {n} ⊂ ℝN such that, if vn(x) := un(x + n), we have that {∣vn∣} is bounded in L(ℝN, ℝ) and

lim|x|+|vn(x)|=0uniformlyinnN.

We use the Moser iteration method to prove the theorem. Although there is more one term for problem, by the calculation, it is easy to know this term does not affect the procedure. We may refer to [36] for the details, so we omit it for simplicity.

Now, we are ready to give the proof of Theorem 1.1.

Proof

Proof of Theorem 1.1. Let δ > 0 be such that MδΛ. We want to show that there exists ε̂δ > 0 such that for any ε ∈ (0, ε̂δ) and any uε ∈ 𝓝̃ε solution of problem (3.2), it holds

uεL(Λεc)2a. (5.1)

We argue by contradiction and assume that there is a sequence εn → 0 such that for every n there exists un ∈ 𝓝̃εn which satisfies Jεn (un) = 0 and

unL(Λεnc)2>a. (5.2)

As in Lemma 5.1, we have that Jεn(un) → cV0, and therefore we can use Proposition 4.1 to obtain a sequence (n) ⊂ ℝ3 such that yn := εnny0 for some y0M. Then, we can find r > 0, such that Br(yn) ⊂ Λ, and so Br/εn(n) ⊂ Λεn for all n large enough.

Using Lemma 5.1, there exists R > 0 such that ∣vn2a in BRc (0) and n large enough, where vn = un(⋅+n). Hence ∣ un2a in BRc (n) and n large enough. Moreover, if n is so large that r/εn > R, then ΛεncBr/εnc (n) ⊂ BRc (n), which gives ∣ un2a for any x Λεnc . This contradicts (5.2) and proves the claim.

Let now εδ := min{ε̂δ, ε̃δ}, where ε̃δ > 0 is given by Theorem 4.1. Then we have catMδ(M) nontrivial solutions to problem (3.2). If uε ∈ 𝓝̃ε is one of these solutions, then, by (5.1) and the definition of g, we conclude that uε is also a solution to problem (2.2).

Finally, we study the behavior of the maximum points of ∣ûε∣, where ûε(x) := uε(x/ε) is a solution to problem (1.1), as ε → 0+.

Take εn → 0+ and the sequence (un) where each un is a solution of (3.2) for ε = εn. From the definition of g, there exists y ∈ (0, a) such that

g(εx,t2)t2V0Kt2,for allxRN,|t|y.

Arguing as above we can take R > 0 such that, for n large enough,

unL(BRc(y~n))<y. (5.3)

Up to a subsequence, we may also assume that for n large enough

unL(BR(y~n))y. (5.4)

Indeed, if (5.4) does not hold, up to a subsequence, if necessary, we have ∥ un < y. Thus, since Jεn (uεn) = 0, using (g5) and the diamagnetic inequality (2.1) that

R3(||un||2+V0|un|2)dxR3g(εnx,|un|2)|un|2dxV0KR3|un|2dx

and, being K > 1, ∥ un ∥ = 0, which is a contradiction.

Taking into account (5.3) and (5.4), we can infer that the global maximum points pn of ∣ uεn∣ belongs to BR(n), that is pn = qn + n for some qnBR. Recalling that the associated solution of problem (1.1) is ûn(x) = un(x/εn), we can see that a maximum point ηεn of ∣ ûn∣ is ηεn = εnn+εnqn. Since qnBR, εnny0 and V(y0) = V0, the continuity of V allows to conclude that

limnV(ηεn)=V0.

Acknowledgements

C. Ji was supported by Shanghai Natural Science Foundation(18ZR1409100).

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Received: 2020-02-19
Accepted: 2020-04-12
Published Online: 2020-05-17

© 2021 Yueli Liu et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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