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Concentration results for a magnetic Schrödinger-Poisson system with critical growth

  • Jingjing Liu and Chao Ji EMAIL logo
Published/Copyright: November 27, 2020

Abstract

This paper is concerned with the following nonlinear magnetic Schrödinger-Poisson type equation

(ϵiA(x))2u+V(x)u+ϵ2(|x|1|u|2)u=f(|u|2)u+|u|4uin R3,uH1(R3,C),

where ϵ > 0, V : ℝ3 → ℝ and A : ℝ3 → ℝ3 are continuous potentials, f : ℝ → ℝ is a subcritical nonlinear term and is only continuous. Under a local assumption on the potential V, we use variational methods, penalization technique and Ljusternick-Schnirelmann theory to prove multiplicity and concentration of nontrivial solutions for ϵ > 0 small.

MSC 2010: 35J60; 35J25

1 Introduction and main results

In this paper, we study multiplicity and concentration of the nontrivial solutions of the following Schrödinger-Poisson type equations with critical growth

(ϵiA(x))2u+V(x)u+ϵ2(|x|1|u|2)u=f(|u|2)u+|u|4uin R3, (1.1)

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

In recent years a considerable amount of work has been devoted to investigating the existence and multiplicity of solutions for nonlinear Schrödinger-Poisson system without magnetic field. We notice that, by using minimax theorema and the Ljusternik-Schnirelmann theory, He [25] gave multiplicity and concentration of positive solutions of the following problem

ϵ2Δu+V(x)u+ϕ(x)u=f(u),in R3,ϵ2Δϕ=u2,in R3,uH1(R3),u(x)>0,in R3.

where fC1(ℝ) has the subcritical growth and the potential V satisfies a global condition introduced by Rabinowitz [31]. In [26], He and Zou studied the existence and concentration of ground state solutions for the following Schrödinger-Poisson system with the critical growth

ϵ2Δu+V(x)u+ϕ(x)u=f(u)+|u|4u,in R3,ϵ2Δϕ=u2,u(x)>0,in R3, (1.2)

where fC1(ℝ) and the potential V satisfies a global condition. Then, He [27] studied the multiplicity of concentrating positive solutions for Schrödinger-Poisson system (1.2) with nonlinear term fC(ℝ) under a local assumption introduced by del Pino and Felmer [17]. For further results on Schrödinger-Poisson system without magnetic field, we refer to[1, 4, 5, 14, 15, 22, 32, 33, 36, 40] and the references therein(see also [21] for the fractional case).

Concerning the magnetic nonlinear Schrödinger equation (1.1), we refer to [6, 7, 8, 10, 11, 12, 13, 16, 19, 23, 24, 29, 38, 39] and references therein. It is well known that the first result involving the magnetic field was obtained by Esteban and Lions [19]. They used the concentration-compactness principle and minimization arguments to obtain solutions for ε > 0 fixed. In [39], the authors studied multiplicity and concentration of solutions for magnetic relativistic Schrödinger equations, Xia [38] studied a critical fractional Choquard-Kirchhoff problem with magnetic field. In particular, due to our scope, we want to mention [41] 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 [41] failed. Recently, by variational methods, penalization technique, and Ljusternick-Schniremann theory, for the magnetic Schrödinger-Poisson system with subcritical growth nonlinearity f which is only continuous, in [29] we proved multiplicity and concentration properties of nontrivial solutions for ϵ > 0 small. For the fractional Schrödinger-Poisson type equations with magnetic field, we refer to [2, 3].

Inspried by [27, 29], we intend to prove multiplicity and concentration of nontrivial solutions for problem (1.1) with critical growth. Since the probem we deal with has the critical growth, we need more refined estimates to overcome the lack of compactness. On the other hand, due to the appearance of magnetic field A(x) and the nonlocal term |x|–1 * |u|2, problem (1.1) will be more difficult, and some estimates are also more complicated.

In this 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}.

On the nonlinearity fC(ℝ, ℝ), we require that:

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

  2. There exist σ, q ∈ (4, 6) and μ > 0 such that

    f(t)μtσ22t>0,andlimt+f(t)tq22=0;
  3. there is a positive constant θ ∈ (4, 6) 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 listed as follows:

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 (1.1) for ϵ = ϵn and ηϵn ∈ ℝ3 the global maximum point of |uϵn|, then

limnV(ηϵn)=V0.

The paper is organized as follows. In Section 2 we indicate the functional setting and give some preliminary results. In Section 3, we study the modified problem, and 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 limit problem associated. It allows us to show the modified problem has the multiple solutions. Finally, the proof of Thereom 1.1 is derived in Section 5.

Notation

  • C, C1, C2, … denote any positive constants, whose exact values are not relevant;

  • 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. 〈⋅, ⋅〉0 denotes the inner product of the space H1(ℝ3, ℝ).

2 Abstract setting and preliminary results

For u: ℝ3 → ℂ, let us denote by

Au:=(iA)u,

and

DA1(R3,C):={uL6(R3,C):|Au|L2(R3,R)},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), an important tool is the following diamagnetic inequality (see e.g. [28, Theorem 7.21])

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

Now, by 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.

By the diamagnetic inequality (2.1), we have, 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.

Arguing as in [29], by the Lax-Milgram Theorem there exists a unique ϕ|u|D1,2(ℝ3, ℝ) such that

Δϕ|u|=|u|2,inR3.

We obtain the following t-Riesz formula

ϕ|u|(x)=cR3|xy|1|u(y)|2dy.

Arguing as in [14, 32, 40], the function ϕ|u| possesses the following properties.

Lemma 2.1

For any uHε, we have

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

  2. if unu in Hε, then ϕ|un|ϕ|u| in D1,2(ℝ3, ℝ), and

    lim infnR3ϕ|un||un|2dxR3ϕ|u||u|2dx;
  3. ϕ|ru| = r2ϕ|u| for all r ∈ ℝ and ϕ|u(⋅+y)| = ϕ|u|(x + y);

  4. ϕ|u| ≥ 0 for all uHϵ and we have

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

For compact supported functions in H1(ℝ3, ℝ), the following result will be very useful for some estimates below.

Lemma 2.2

If uH1(ℝ3, ℝ) and u has compact support, then ω := eiA(0)⋅xuHε.

Proof

Assume that supp(u) ⊂ BR(0). Since V is continuous, it is clear that

R3Vϵ(x)|ω|2dx=BR(0)Vϵ(x)|ω|2dxCu22<+.

Moreover, since V and A are continuous, we have

R3|Aϵω|2dx=R3|ω|2dx+R3|Aϵ(x)|2|ω|2dx+2ReR3iAϵ(x)ω¯ωdx2R3|ω|2dx+2R3|Aϵ(x)|2|ω|2dxCR3|u|2dx+R3|u|2dx<+

and we conclude.□

3 The modified problem

To study (1.1), we 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 f(a) + a2 = V0/K, where V0 is given in (V1). Hence we consider the function

f~(t):=f(t)+(t+)2,ta,V0/K,t>a.

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

g(x,t):=χΛ(x)(f(t)+(t+)2)+(1χΛ(x))f~(t), (3.1)

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

From (f1)–(f4), 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;

  3. g(x, t) ≤ f(t) + t2 for all t ≥ 0 and uniformly in x ∈ ℝ3;

  4. 0 < θG(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 t g(x,t)t is strictly increasing in t ∈ (0, +∞) and for each xΛc, the function t g(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).

We observe that the weak solutions of the modified problem (3.2) can be found as the critical points of the C1 functional

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

defined in Hϵ. Moreover, we denote by 𝓝ϵ the Nehari manifold of Jϵ,

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+ = Sϵ Hϵ+ , 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}.

Arguing as in [29, Lemma 3.1], we can show that the functional Jϵ satisfies the Mountain Pass Geometry [37].

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.

Due to 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 τ > 0 independent on u such that tuτ for all u Sϵ+ . Moreover, for each compact 𝓦 ⊂ Sϵ+ there is C𝓦 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) Arguing as in [29, Lemma 3.1], it follows that gu(0) = 0, gu(t) > 0 for t > 0 small and gu(t) < 0 for t > 0 large. Thus, maxt≥0 gu(t) is achieved at a global maximum point t = tu satisfying gu (tu) = 0 and tuu ∈ 𝓝ϵ. Now, we show that tu is unique. Arguing by contradiction, suppose that there exist t1 > t2 > 0 such that gu (t1) = gu (t2) = 0. Then, for i = 1, 2,

tiuϵ2+ti3R3(|x|1|u|2)|u|2dx=R3g(ϵx,ti2|u|2)ti|u|2dx.

Hence,

uϵ2ti2+R3(|x|1|u|2)|u|2dx=R3g(ϵx,ti2|u|2)|u|2ti2dx,

which implies that

(1t121t22)uϵ2=R3(g(ϵx,t12|u|2)t12|u|2g(ϵx,t22|u|2)t22|u|2)|u|4dxΛϵc{t22|u|2a0t12|u|2}(g(ϵx,t12|u|2)t12|u|2g(ϵx,t22|u|2)t22|u|2)|u|4dx+Λϵc{a0t22|u|2}(g(ϵx,t12|u|2)t12|u|2g(ϵx,t22|u|2)t22|u|2)|u|4dx=Λϵc{t22|u|2a0t12|u|2}(V0K1t12|u|2f(t22|u|2)+t24|u|4t22|u|2)|u|4dx+1K(1t121t22)Λϵc{a0t22|u|2}V0|u|2dx.

Since t1 > t2 > 0, we have

uϵ2t12t22t22t12Λϵc{t22|u|2a0t12|u|2}(V0K1t12|u|2f(t22|u|2)+t24|u|4t22|u|2)|u|4dx+1KΛϵc{a0t22|u|2}V0|u|2dx1KΛϵcV0|u|2dx1Kuϵ2,

which is a contradiction. Therefore, maxt≥0 gu(t) is achieved at a unique t = tu so that gu (t) = 0 and tuu ∈ 𝓝ϵ.

(A2) For ∀u Sϵ+ , it follows that

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

From (g2), the Sobolev embeddings and 4 < q < 6, it is easy to obtain

tuζtu3R3|u|4dx+Cζtuq1R3|u|qdx+tu5R3|u|6dxC1ζtu3+C2Cζtuq1+C3tu5,

which implies 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 u ∈ 𝓦 such that unu in Hϵ. Using the proof of [29, Lemma 3.1(ii)], it follows that Jϵ(tnun) → –∞.

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 may obtain

0<(121θ)(11K)Jϵ(vn)tn20

as n → ∞, which yields a contradiction. This completes the proof of (A2).

(A3) We first show 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) it follows that

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 > 2 and u ≠ 0. Thus, mϵ1(u)=uuϵSϵ+ is well defined and continuous. From

mϵ1(mϵ(u))=mϵ1(tuu)=tuutuuϵ=u,uSϵ+,

we know 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, it follows that

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 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. Thus, by (V1), (V2) and the Sobolev embedding, for any t ∈ [2, 6], there exists 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. From (g2), (g3) and (g5), for each t > 0, it follows that

R3G(ϵx,t2|un|2)dxΛϵ(F(t2|un|2)+t6|un|66)dx+t2KΛϵcV(ϵx)|un|2dxC1t4Λϵ|un|4dx+C2tqΛϵ|un|qdx+t66Λϵ|un|6dx+t2Kunϵ2C3t4dist(un,Sϵ+)4+C4tqdist(un,Sϵ+)q+C5t6dist(un,Sϵ+)6+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, we have

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

which implies that

lim infn{12mϵ(un)ϵ2+14R3(|x|1|mϵ(un)|2)|mϵ(un)|2dx}lim infnJϵ(mϵ(un))K22Kt2,t>0.

Since t > 0 is arbitrary, we can show that ∥mϵ(un)∥ϵ → ∞ and Jϵ(mϵ(un)) → ∞ as n → ∞.□

At this point we define the function

Ψ^ϵ:Hϵ+R,

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

From Lemma 3.2, arguing as in [35, Corollary 10] we may obtain the following lemma.

Lemma 3.3

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

  1. Ψ̂ϵC1( 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 [35], we have the variational characterization of the infimum of Jϵ over 𝓝ϵ:

cϵ=infuNϵJϵ(u)=infuHϵ+supt>0Jϵ(tu)=infuSϵ+supt>0Jϵ(tu). (3.3)

Lemma 3.4

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

Proof

Assume that {un} ⊂ Hϵ be a (PS)c sequence for Jϵ, that is, Jϵ(un) → c > 0 and Jϵ (un) → 0. From (g4), (g5) and 4 < θ < 6, it follows that

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)unϵ2+(1θ12)ΛϵcG(ϵx,|un|2)dx(121θ)unϵ2+(1θ12)1KR3V(ϵx)|un|2dx(121θ)(11K)unϵ2.

Since K > 2, from the above inequalities we know that {un} is bounded in Hϵ.□

The following lemma provides a range of levels in which the functional Jϵ verifies the Palais-Smale condition.

Lemma 3.5

The functional Jϵ satisfies the (PS)c condition at any level c(0,13S32), where S is the best constant for the Sobolev inequality

S(R3|v|6dx)1/3R3(|v|2+|v|2)dx,forvH1(R3,R).

Proof

Let (un)nHϵ be a (PS)c for Jϵ. By Lemma 3.4, (un)n 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 → +∞.

  1. We show that for any given ζ > 0, for R large enough,

    lim supnBRc(0)(|Aϵun|2+Vϵ(x)|un|2)dxζ. (3.4)

    Let R > 0 such that ΛϵBR/2(0) and 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)n 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)n in Hϵ, we obtain

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

    and so (3.4) holds.

    From the Sobolev embedding and (3.4), we have that for any ζ > 0, there exists R = R(ζ) > 0 such that,

    unurunuLr(BR(0))+unuLr(BRc(0))ζ+C(unHϵ(BRc(0))+uHϵ(BRc(0)))C1ζ

    where r ∈ [2, 6) and n large enough. From this, we can obtain that

    unuinLr(R3,C),for anyr[2,6). (3.5)

    By (2), since ϕ|u| : L12/5(ℝ3, ℝ) → D1,2(ℝ3, ℝ) is continuous, from (3.5) we can get

    ϕ|un|ϕ|u|inD1,2(R3,R), (3.6)
    R3ϕ|un||un|2dxR3ϕ|u||u|2dx. (3.7)

    Using the boundedness of sequence (un)n and the Sobolev embedding again, for any φ Cc (ℝ3, ℂ), we have

    ReR3(AϵunAϵφ¯dx+Vϵ(x)unφ¯)dxReR3(AϵuAϵφ¯dx+Vϵ(x)uφ¯)dx, (3.8)
    ReR3g(ϵx,|un|2)unφ¯dxReR3g(ϵx,|u|2)uφ¯dx. (3.9)

    By (3.7)-(3.9), the Hölder inequality and the Sobolev embeddings, we have

    ReR3ϕ|un|unφ¯dxReR3ϕ|u|uφ¯dx=ReR3(ϕ|un|unϕ|u|u)φ¯dx=ReR3ϕ|un|(unu)φ¯dx+ReR3(ϕ|un|ϕ|u|)uφ¯dxCϕ|un|D1,2(R3,R)unuL12/5(R3,C)φL12/5((R3,C))+C(ϕ|un|ϕ|u|)D1,2(R3,R)uL12/5(R3,C)φL12/5(R3,C)0,asn. (3.10)

    By (3.8)-(3.10) and Jϵ (un) → 0, we have Jϵ (u) = 0 and

    uϵ2+R3(|x|1|u|2)|u|2dx=R3g(ϵx,|u|2)|u|2dx.
  2. limnR3g(ϵx,|un|2)|un|2dx=R3g(ϵx,|u|2)|u|2dx. (3.11)

    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+.

    By (g5) and (3.4), for any ζ > 0, there exists R > 0 large enough, 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.

    Thus,

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

    Now, we show that

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

    From the definition of g, we have that

    g(ϵx,|un|2)|un|2f(|un|2)|un|2+|un|6+V0K|un|2,for anyxR3Λϵ. (3.12)

    Since BR(0) ∩ (ℝ3Λϵ) is bounded, from the above estimate, (f1), (f2), the Sobolev embedding and the Lebesgue Dominated Convergence Theorem, we can infer

    limnBR(0)(R3Λϵ)g(ϵx,|un|2)|un|2=BR(0)(R3Λϵ)g(ϵx,|u|2)|u|2dx. (3.13)

    If we can prove that

    limnBR(0)Λϵg(ϵx,|un|2)|un|2=BR(0)g(ϵx,|u|2)|u|2dx, (3.14)

    from (3.13) and (3.14), it yields (3.11). Now, in order to show that (3.14) holds, we only need to prove the following limit holds

    limnΛϵ|un|6dx=Λϵ|u|6dx. (3.15)

    Using the boundedness of (un)n in Hϵ and the diamagnetic inequality (2.1), we may assume that

    ||un||2μand|un|6ν (3.16)

    in the sense of measures. Moreover, by the diamagnetic inequality (2.1) and (3.4), (un)n is a tight sequence in H1(ℝ3, ℝ), thus, using the concentration-compactness principle in [37], we can find an at most countable index I, sequences (xi) ⊂ ℝ3, (μi), (νi) ⊂ (0, ∞) such that

    μ||u||2dx+iIμiδxi,ν=|u|6+iIνiδxiandSνi1/3μi (3.17)

    for any iI, where δxi is the Dirac mass at the point xi. Let us show that (xi)iIΛϵ = ∅. Assume, by contradiction, that xiΛϵ for some iI. For any ρ > 0, we define ψρ(x)=ψ(xxiρ) where ψ C0 (ℝ3, [0, 1]) such that ψ = 1 in B1, ψ = 0 in ℝ3B2 and ∥∇ψL(ℝ3,ℝ) ≤ 2. We suppose that ρ > 0 such that supp(ψρ) ⊂ Λϵ. Since (ψρun) is bounded in Hϵ, we can see that Jϵ (un)[ψρun] = on(1), that is

    ReR3AϵunAϵ(ψρun)¯dx+R3Vϵ(x)|un|2ψρdx+R3(|x|1|un|2)|un|2ψρdx=R3g(ϵx,|un|2)|un|2ψρdx+on(1)=R3f(|un|2)|un|2ψρdx+R3|un|6ψρdx+on(1).

    Since Aϵ(unψρ) = iunψρ + ψρAϵun, using (g5), we have

    R3|Aϵun|2ψρdxR3f(|un|2)|un|2ψρdx+R3|un|6ψρdxReR3iun¯AϵunϕRdx+on(1). (3.18)

    Using the diamagnetic inequality (2.1) again, it follows that

    R3||un||2ψρdxR3f(|un|2)|un|2ψρdx+R3|un|6ψρdxReR3iun¯AϵunϕRdx+on(1).

    Due to the fact that f has the subcritical growth and ψρ has the compact support, we have that

    limρ0limnR3f(|un|2)|un|2ψρdx=limρ0R3f(|u|2)|u|2ψρdx=0. (3.19)

    It's also easy to show that

    limρ0limsupn|R3iun¯Aϵunψρdx|=0. (3.20)

    Then, taking into account (3.16), (3.18), (3.19) and (3.20), we can conclude that νiμi. Together with the inequality Sνi1/3 μi in (3.17), we have νiS3/2. Now, from (f3), (g4) and (g5), we have

    c=Jϵ(un)14Jϵ(un)[un]+on(1)=14unϵ2+R3(14g(ϵx,|un|2)|un|212G(ϵx,|un|2))dx+on(1)14unϵ2+Λϵc(14g(ϵx,|un|2)|un|212G(ϵx,|un|2))dx+112Λϵ|un|6dx+on(1)14(Λϵψρ||un||2dx+ΛϵcVϵ(x)|un|2)12ΛϵcG(ϵx,|un|2)dx+112Λϵ|un|6dx+on(1)14Λϵψρ||un||2dx+(1412K)ΛϵcVϵ(x)|un|2dx+112Λϵψρ|un|6dx+on(1)14Λϵψρ||un||2dx+112Λϵψρ|un|6dx+on(1).

    From the above arguments and (3.17), we have

    c14{iI:xiΛϵ}ψρ(xi)μi+112{iI:xiΛϵ}ψρ(xi)νi14μi+112νi13S3/2

    which gives a contradiction. This means that (3.15) holds.

  3. From Jϵ (un)[un] → 0, Jϵ (u) = 0, (3.7) and (3.15), we have

    limnunϵ2=uϵ2,

    and the proof is completed.□

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

Corollary 3.1

The functional Ψϵ satisfies the (PS)c condition on Sϵ+ at any level c ∈ (0, 13S32 ).

Proof

Let {un} ⊂ Sϵ+ be a (PS)c sequence for Ψϵ where c ∈ (0, 13S32 ). 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 completed.□

4 Multiple solutions for the modified problem

4.1 The autonomous problem

Now, we study the following limit problem

Δu+V0u+(|x|1|u|2)u=f(u2)u+|u|4u,in R3,Δϕ=u2,in R3,uH1(R3,R),u(x)>0,in R3. (4.1)

The solutions of problem (4.1) are the critical points of the C1-functional defined by

I0(u):=12R3(|u|2+V0u2)dx+14R3(|x|1|u|2)|u|2dx12R3F(u2)dx16R3(u+)6dx.

Let

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

and

cV0:=infuN0I0(u).

Let H0 := H1(ℝ3, ℝ) and define by H0+ the open set of H0 given by

H0+={uH0:|supp(u+)|>0},

and S0+ = S0 H0+ , where S0 be the unit sphere of H0.

As in Section 3, S0+ is a non-complete C1,1-manifold of codimension 1, modeled on H0 and contained in H0+ . Therefore, H0 = Tu S0+ ⨁ ℝu for each uTu S0+ , where Tu S0+ = {vH0: 〈u, v0 = 0}.

Now, arguing as in Lemma 3.2, we have the following important property.

Lemma 4.1

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

  1. For any u H0+ , 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 u S0+ . Moreover, for each compact 𝓦 ⊂ S0+ there is C𝓦 such that tuC𝓦, for all u ∈ 𝓦;

  3. The map : H0+ → 𝓝0 given by (u) = tuu is continuous and m0 = m^0|S0+ is a homeomorphism between S0+ and 𝓝0. Moreover, m1(u)=uu0;

  4. If there is a sequence {un} ⊂ S0+ such that dist(un, S0+ ) → 0, thenm(un)∥0 → ∞ and I0(m(un)) → ∞.

We shall consider the functional defined by

Ψ^0(u)=I0(m^(u))andΨ0:=Ψ^0|S0+,

arguing as in [35, Proposition 9 and Corollary 10], the following result holds.

Lemma 4.2

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

  1. Ψ^0+ C1( H0+ , ℝ) and

    Ψ^0(u)v=m^(u)0u0I0(m^(u))[v],uH0+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

    infS0+Ψ0=infN0I0.

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

cV0=infuN0I0(u)=infuH0+{0}supt>0I0(tu)=infuS0+supt>0I0(tu). (4.2)

From [26, Lemma 2.6], we have 0 < cV0 < 13S32 .

Arguing as in [26, Lemma 2.8], the following important result holds.

Lemma 4.3

Let {un} ⊂ H0 be a (PS)c sequence for I0 with c ∈ (0, 13S32 ) such that un⇀ 0. Then, one of the following alternatives occurs:

  1. un → 0 in H0 as n → +∞;

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

    lim infnBR(yn)|un|2dxβ.

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 I0|𝓝0. Since infS0 Ψ0 = inf𝓝0 I0 = cV0 and S0 is a C1 manifold, by Ekeland’s 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), 1 ≤ r < 6 and unu a.e. in ℝ3, thus I0 (u) = 0. From [26, Lemma 2.6], we know that cV0 < 13S32 . Moreover, from Remark 4.1, we have that u ≠ 0. Now, 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))dx+112R3(u+)6dxlim infn{(121θ)un02+(141θ)R3(|x|1|un|2)|un|2dx+R3(1θf(un)un212F(un2))dx+112R3(un+)6dx}=lim infn{I0(un)1θI0(un)[un]}=cV0,

thus, u is a ground state solution. From the assumption of f, u ≥ 0. Moreover, using the standard argument, we may prove that u(x) > 0 for x ∈ ℝ3. The proof is complete.□

Note that, arguing as in [26, Proposition 3.3, Proposition 3.4 and Lemma 3.11], the ground state solution of problem decays exponentially at infinity with its gradient, and is C2(ℝ3, ℝ) ∩ L(ℝ3, ℝ). This result is very important for the proof of Lemma 4.6 later.

Lemma 4.5

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

Proof

Since (un)n ⊂ 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=infuS0+Ψ0(u).

Although S0+ is not a complete C1 manifold, we still can use the Ekeland’s variational principle [18] to the functional 𝓔0 : H → ℝ ∪ {∞} defined by 𝓔0(u) := Ψ̂0(u) if u S0+ and 𝓔0(u) := ∞ if u S0+ , where H = S0+¯ is the complete metric space equipped with the metric d(u, v) := ∥uv0. In fact, by Lemma 4.1(a4), 𝓔0C(H, ℝ ∪ {∞}), and from Lemma 4.2(b4), 𝓔0 is bounded below. Therefore, 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.□

Now, we show the relationship between cϵ and cV0.

Lemma 4.6

The numbers cϵ and cV0 satisfy the following inequality

limϵ0cϵ=cV0<13S32.

Proof

Let η Cc (ℝ3, [0, 1]) be a cut-off function such that η = 1 in Bρ/2 and supp(η) = BρΛ for some ρ > 0. Let us define ωϵ(x) := ηϵ(x)ω(x)eiA(0)⋅x, where ηϵ(x) = η(ϵx) for ϵ > 0, ω is a positive and radial ground state solution of problem (4.1). We observe that |ωϵ| = ηϵω and ωϵHϵ in view of Lemma 2.2. Arguing as in [16, Lemma 4.1] or [24, Lemma 4.6], we obtain

limϵ0ωϵϵ2=ωV02 (4.3)

and

limϵ0R3(|x|1|ωϵ|2)|ωϵ|2dx=R3(|x|1|ω|2)|ω|2dx. (4.4)

It is also easy to check that

limϵ0R3|ωϵ|6dx=R3|ω|6dx. (4.5)

Let tϵ > 0 be the unique number such that

Jϵ(tϵωϵ)=maxt0Jϵ(tωϵ).

Then tϵ satisfies

tϵ2ωϵϵ2+tϵ4R3(|x|1|ωϵ|2)|ωϵ|2dx=R3g(ϵx,tϵ2|ωϵ|2)tϵ2|ωϵ|2dx=R3f(tϵ2|ωϵ|2)tϵ2|ωϵ|2dx+R3tϵ6|ωϵ|6dx,

where we use supp(η) ⊂ Λ and the definition of g(x, t). Moreover, combining the facts that η = 1 in Bρ/2, u is a positive continuous function and hypothesis (f4), we have

1tϵ2ωϵϵ2+R3(|x|1|ωϵ|2)|ωϵ|2dx=1tϵ2R3f(tϵ2|ωϵ|2)|ωϵ|2dx+R3tϵ2|ωϵ|6dx1tϵ2R3f(tϵ2η2(|ϵx|)ω2(x))η2(|ϵx|)ω2(x)dz1tϵ2Bρ/(2ϵ)(0)f(tϵ2ω2(z))ω2(z)dz1tϵ2Bρ/2(0)f(tϵ2ω2(z))ω2(z)dzf(tϵ2y2)tϵ2Bρ/2(0)ω2(z)dz

for all 0 < ϵ < 1 and where y = min{ω(z) : |z| ≤ ρ/2}.

If tϵ → +∞ as ϵ → 0, by (f4), we deduce that ∫3(|x|–1 * |ωϵ|2)|ωϵ|2 dx → +∞ which contradicts (4.5).

Therefore, up to a subsequence, we may assume that tϵt0 ≥ 0 as ϵ → 0.

If tϵ → 0, using the fact that f is increasing, the Lebesgue dominated convergence theorem and relation (4.5), we obtain

ωϵϵ2+tϵ4R3(|x|1|ωϵ|2)|ωϵ|2dx=R3f(tϵ2|ωϵ|2)|ωϵ|2dx+R3tϵ4|ωϵ|6dx0, as ϵ0

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

t02R3(|ω|2+V0ω2)dx+t04R3(|x|1|ω|2)|ω|2dx=R3f(t02ω2)t02ω2dx+R3t06|ω|6dx,

so that t0ω ∈ 𝓝V0. Since ω ∈ 𝓝V0, we obtain that t0 = 1 and so, using the Lebesgue dominated convergence theorem, we get

limϵ0R3F(|tϵωϵ|2)dx=R3F(ω2)dx.

Hence

limϵ0Jϵ(tϵωϵ)=IV0(u)=cV0.

Since cϵ ≤ maxt≥0 Jϵ( ϵ) = Jϵ(tϵωϵ), we can conclude that lim supϵ→0 cϵcV0. Moreover, by (3.3), (4.2) and IV0(|u|) ≤ Jϵ(u) for any uHϵ, we have cV0cϵ. Then cV0 ≤ lim infϵ→0 cϵ. Combining with the previous arguments, we conclude that limϵ→0 cϵ = cV0 < 13S32 .□

Remark 4.2

From Lemma 4.1 and Lemma 3.5, we see that for ϵ > 0 small, problem (3.2) has a ground state solution uϵ such that Jϵ(uϵ) = cϵ and Jϵ (uϵ) = 0.

4.2 The technical results

By the Ljusternik-Schnirelmann category theory, in this subsection we prove a multiplicity result for the modified problem (3.2). We first provide some useful preliminary results.

Let δ > 0 such that MδΛ, ωH1(ℝ3, ℝ) is a positive ground state solution of the limit problem (4.1), and ηC(ℝ+, [0, 1]) is 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 tϵ Ψϵ,y ∈ 𝓝ϵ.

Let us define Φϵ : M → 𝓝ϵ as

Φϵ(y):=tϵΨϵ,y.

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

Moreover, arguing as in Lemma 4.1, the energy of above function has the following behavior as ϵ → 0+.

Lemma 4.7

The limit

limϵ0+Jϵ(Φϵ(y))=cV0

holds uniformly in yM.

Now we define the barycenter map.

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

Υ(x):=x,if|x|<ρ,ρx/|x|,if|x|ρ.

The barycenter map βϵ : 𝓝ϵ → ℝ3 is defined by

βϵ(u):=1u44R3Υ(ϵx)|u(x)|4dx.

Lemma 4.8

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. (4.6)

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

βϵn(Φϵn(yn))=yn+R3(Υ(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 (4.6).□

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|)nH1(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 exists a sequence {n} ⊂ ℝ3 and constants R, β > 0 such that

lim infnBR(y~n)|un|2dxβ. (4.7)

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 (4.7), we may assume that |vn| ⇀ v in H1(ℝ3, ℝ) for some v ≠ 0.

Let now tn > 0 be such that n := tn|vn| ∈ 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|)n 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 := t0v ≠ 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+BR/ϵnc(0)|vn|6dx. (4.8)

Since |vn| → v in H1(ℝ3, ℝ) and (t) ≤ V0/K, we can see that (4.8) 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 |ϵnx + yny0| < 2r so that ϵnx + 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)dx16R3|v~|6dxlim infn(12R3(|v~n|2+V(ϵnx+yn)|v~n|2)dx+14R3(|x|1|v~n|2)|v~n|2dx12R3F(|v~n|2)dx16R3|v~n|6dx)=lim infn(tn22R3(||un||2+V(ϵnz)|un|2)dz+tn44R3(|x|1|un|2)|un|2dx12R3F(|tnun|2)dz16R3|tnun|6dx)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 4.7, |Jϵ(Φϵ(y)) – cV0| → 0 as ϵ → 0+, we get that 𝓝̃ϵ ≠ ∅ for any ϵ > 0 small enough.

The relation between 𝓝̃ϵ and the barycenter map is as follows.

Lemma 4.9

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) (4.9)

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(Υ(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 nontrivial 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 4.7 and Lemma 3.3(B4), it follows that

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

Therefore, 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 4.7, Lemma 3.2(A3), Lemma 4.8, and Lemma 4.9, 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 4.8, 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. This fact implies that

catπϵ(M)(πϵ(M))catMδ(M). (4.10)

By Corollary 3.1 and the abstract category theorem [35], Ψϵ has at least catπϵ(M)(πϵ(M)) critical points on Sϵ+ . Therefore, from Lemma 3.3(B4) and (4.10), 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 shall show that the solutions uϵ obtained in Theorem 4.1 satisfy

|uϵ(x)|2a for xΛϵc

for ϵ small and prove the main result of this paper.

Arguing as in [29] or [41], the following uniform result holds.

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} ⊂ ℝ3 such that, if vn(x) := un(x + n), we have that {|vn|} is bounded in L(ℝ3, ℝ) and

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

Now it’s the position to give the proof of Theorem 1.1.

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 |vn|2a in BRc (0) and n large enough, where vn = un(⋅ + n). Hence |un|2a in BRc (n) and n large enough. Moreover, if n is so large that r/ϵn > R, then ΛϵncBr/ϵnc(y~n) BRc (n), which gives |un|2a 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 allxR3,|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 > 2, ∥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

J.J. Liu was supported by the National Nature Science Foundation of China(11701525), C. Ji was supported by Natural Science Foundation of Shanghai(20ZR1413900, 18ZR1409100).

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Received: 2020-09-10
Accepted: 2020-10-27
Published Online: 2020-11-27

© 2021 Jingjing Liu and Chao Ji, published by De Gruyter

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

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