Multiplicity and concentration behaviour of solutions for a fractional Choquard equation with critical growth

Zhipeng Yang; Fukun Zhao

doi:10.1515/anona-2020-0151

Open Access Published by De Gruyter December 1, 2020

Multiplicity and concentration behaviour of solutions for a fractional Choquard equation with critical growth

Zhipeng Yang and Fukun Zhao

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2020-0151

Abstract

In this paper, we study the singularly perturbed fractional Choquard equation

ε2s(−Δ)su+V(x)u=εμ−3(∫R3|u(y)|2μ,s∗+F(u(y))|x−y|μdy)(|u|2μ,s∗−2u+12μ,s∗f(u))inR3,

where ε > 0 is a small parameter, (−△)^s denotes the fractional Laplacian of order s ∈ (0, 1), 0 < μ < 3, 2μ,s⋆=6−μ3−2s is the critical exponent in the sense of Hardy-Littlewood-Sobolev inequality and fractional Laplace operator. F is the primitive of f which is a continuous subcritical term. Under a local condition imposed on the potential V, we investigate the relation between the number of positive solutions and the topology of the set where the potential attains its minimum values. In the proofs we apply variational methods, penalization techniques and Ljusternik-Schnirelmann theory.

Keywords: Variational method; fractional Choquard equation; critical growth

MSC 2010: 35Q40; 35J50; 58E05

1 Introduction and the main results

In the present paper we are interested in the existence, multiplicity and concentration behavior of the semi-classical solutions of the singularly perturbed nonlocal elliptic equation

(1.1) ε2s(−Δ)su+V(x)u=εμ−N(∫R3G(u(y))|x−y|μdy)g(u)inRN,

where ε > 0 is a small parameter, 0 < μ < N, V, g = G' are real continuous functions on ℝ^N and the fractional Laplacian (−△)^s is defined by

(−Δ)sΨ(x)=CN,sP.V.∫RNΨ(x)−Ψ(y)|x−y|N+2sdy,Ψ∈S(RN),

P.V. stands for the Cauchy principal value, C_N_,s is a normalized constant, S(ℝ^N) is the Schwartz space of rapidly decaying functions, s ∈ (0, 1). As ε goes to zero in (1.1), the existence and asymptotic behavior of the solutions of the singularly perturbed equation (1.1) is known as the semi-classical problem. It was used to describe the transition between Quantum Mechanics and Classical Mechanics.

Our motivation to study (1.1) mainly comes from the fact that solutions u(x) of (1.1) correspond to standing wave solutions Ψ(x,t)=e−iEt/εu(x) of the following time-dependent fractional Schrödinger equation

(1.2) iε∂Ψ∂t=ε2s(−Δ)sΨ+(V(x)+E)Ψ−(K(x)⋆|G(Ψ)|)g(Ψ)(x,t)∈ℝN×ℝ

where i is the imaginary unit, ε is related to the Planck constant. Equations of the type (1.2) was introduced by Laskin (see [25, 26]) and come from an expansion of the Feynman path integral from Brownian-like to Lévy-like quantum mechanical paths. It also appeared in several areas such as optimization, finance, phase transitions, stratified materials, crystal dislocation, flame propagation, conservation laws, materials science and water waves (see [11]).

When s = 1, the equation (1.1) turns out to be the Choquard equation

(1.3) −ε2Δu+V(x)u=εμ−N(∫RNG(u(y))|x−y|μdy)g(u)inRN,

The existence, multiplicity and concentration of solutions for (1.3) has been widely investigated. On one hand, some people have studied the classical problem, namely ε = 1 in (1.3). When V = 1 and G(u)=|u|qq, (1.3) covers in particular the Choquard-Pekar equation

(1.4) −Δu+u=(∫RN1|χ|μ⋆|u|qdy)|u|q−2u in ℝN.

The case N = 3, q = 2 and μ = 1 came from Pekar [38] in 1954 to describe the quantum mechanics of a polaron at rest. In 1976 Choquard used (1.4) to describe an electron trapped in its own hole, in a certain approximation to Hartree-Fock theory of one component plasma [27]. In this context (1.4) is also known as the nonlinear Schrödinger-Newton equation. By using critical point theory, Lions [29] obtained the existence of infinitely many radialy symmetric solutions in H¹(ℝ^N) and Ackermann [1] prove the existence of infinitely many geometrically distinct weak solutions for a general case. For the properties of the ground state solutions, Ma and Zhao [30] proved that every positive solution is radially symmetric and monotone decreasing about some point for the generalized Choquard equation (1.4) with q ≥ 2. Later, Moroz and Van Schaftingen [32, 33] eliminated this restriction and showed the regularity, positivity and radial symmetry of the ground states for the optimal range of parameters, and also derived that these solutions decay asymptotically at infinity.

On the other hand, some people have focused on the semiclassical problem, namely, ε → 0 in (1.3). The question of the existence of semiclassical solutions for the non-local problem (1.3) has been posed in [5]. Note that if v is a solution of (1.3) for x₀ ∈ ℝ^N, then u = v(εx + x₀) verifies

(1.5) −Δu+V(εx+x0)u=(∫RNG(u(y))|x−y|μdy)g(u)inRN,

which means some convergence of the family of solutions to a solution u₀ of the limit problem

(1.6) −Δu+V(x0)u=(∫RNG(u(y))|x−y|μdy)g(u)inRN.

For this case when N = 3, μ = 1 and G(u) = |u|², Wei and Winter [49] constructed families of solutions by a Lyapunov-Schmidt-type reduction when infx∈ℝNV>0. This method of construction depends on the existence, uniqueness and non-degeneracy up to translations of the positive solution of the limiting equation (1.6), which is a difficult problem that has only been fully solved in the case when N = 3, μ = 1 and G(u) = |u|². Moroz and Van Schaftingen [34] used variational methods to develop a novel non-local penalization technique to show that equation (1.3) with G(u) = |u|^q has a family of solutions concentrated at the local minimum of V, with V satisfying some additional assumptions at infinity. In addition, Alves and Yang [4] investigated the multiplicity and concentration behaviour of solutions for a quasi-linear Choquard equation via the penalization method. Very recently, in an interesting paper, Alves et al. [2] study (1.3) with a critical growth, they consider the critical problem with both linear potential and nonlinear potential, and showed the existence, multiplicity and concentration behavior of solutions when the linear potential has a global minimum or maximum.

On the contrary, the results about fractional Choquard equation (1.1) are relatively few. Recently, d’Avenia, Siciliano and Squassina [17] studied the existence, regularity and asymptotic of the solutions for the following fractional Choquard equation

(1.7) (−Δ)su+ωu=(∫RN|u(y)|q|x−y|μdy)|u|q−2uinRN,

where ω>0,2N−μN<q<2N−μN−2s. Shen, Gao and Yang [42] obtain the existence of ground states for (1.7) with general nonlinearities by using variational methods. Chen and Liu [14] studied (1.7) with nonconstant linear potential and proved the existence of ground states without any symmetry property. For critical problem, Wang and Xiang [47] obtain the existence of infinitely many nontrivial solutions and the Brezis-Nirenberg type results can be founded in [36]. For the critical Choquard equations in the sense of Hardy-Littlewood-Sobolev, Cassani and Zhang [12] developed a robust method to get the existence of ground states and qualitative properties of solutions, where they do not require the nonlinearity to enjoy monotonicity nor Ambrosetti-Rabinowitz-type conditions. For other existence results we refer to [6, 8, 23, 24, 31, 48, 52] and the references therein.

It seems that the only works concerning the concentration behavior of solutions are due to [13, 51]. Assuming the global condition on V:

0<infx∈RNV(x)<liminf|x|→∞V(x)=V∞,

which was firstly introduced by Rabinowitz [39] in the study of the nonlinear Schrödinger equations. By using the method of Nehari manifold developed by Szulkin and Weth [46], authors in [13, 51] obtained the multiplicity and concentration of positive solutions for the following fractional Choquard equation

(1.8) ε2s(−Δ)su+V(x)u=εμ−3(∫R3|u(y)|2μ,s∗+F(u(y))|x−y|μdy)(|u|2μ,s∗−2u+12μ,s∗f(u))inR3,

where ε > 0, 0 < μ < 3, F is the primitive function of f.

Different to [13, 51], in this paper, we are devote to establishing the existence and concentration of positive solutions for the fractional Choquard equation (1.8) when the potential function satisfies the following local conditions [18]:

V1V∈CR3,R and 0<infx∈R3V(x).

(V₂)There is a bounded open domain Ω such that

V0:=infΩV(x)<min∂ΩV(x).

Without loss of generality, we may assume that M={x∈Ω:V(x)=V0}≠∅andV(0)=minx∈R3V(x)=V0.

To go on studying the problem (1.8), the following Hardy-Littlewood-Sobolev inequality [28] is the starting point.

Lemma 1.1

Let t, r > 1 and 0 < μ < 3 with

1t+μ3+1r=2,

f ∈ L^t(ℝ³) and h ∈ L^r(ℝ³). There exists a sharp constant C(t, μ, r), independent of f , h such that

(1.9) ∫R3∫R3f(x)h(y)|x−y|μdydx≤C(t,μ,r)|f|t|h|r.

In particular, if t=r=66−μ, then

C(t,μ,r)=C(μ)=πμ2Γ(32−μ2)Γ(3−μ2)(Γ()32Γ(3))μ3−1.

In this case there is equality in (1.9) if and only if f ≡ Ch and

h(x)=A(a2+|x−b|2)6−μ2

for some A ∈ ℂ, a ∈ ℝ\{0} and b ∈ ℝ³.

Notice that, by the Hardy-Littlewood-Sobolev inequality, the integral

∫R3∫R3|u(x)|q|u(y)|q|x−y|μdydx

ie well defined if u^q ∈ L^t(ℝ³) satisfies 2t+μ3=2. Therefore, for u ∈ H^s(ℝ³) we will require that t⋅q∈[2,2s⋆], where 2s⋆=63−2s is fractional critical Sobolev exponent for dimension 3. Then we have

6−μ3≤q≤6−μ3−2s.

Thus, 6−μ3 is called the lower critical exponent and 2μ,s⋆:=6−μ3−2s is the upper critical exponent in the sense of Hardy-Littlewood-Sobolev inequality and the fractional Laplace operator.

For the nonlinearity term, we assume that the continuous function f vanishes in (−∞, 0) and satisfies:

(f₁) |f(u)|≤c(|u|q1−1+|u|q2−1) for some c > 0 and 6−μ3<q1≤q2<2μ,s⋆.

(f₂) The function u ⟼ f (u) is increasing in (0,∞).

(f₃) (i)

lim|u|→+∞F(u)|u|2μ,s∗−1=+∞fors∈(34,1);

(ii)

lim|u|→+∞F(u)|u|2μ,s∗−2s3−2s(log|u|)12=+∞fors=34;

(iii)

lim|u|→+∞F(u)|u|2μ,s∗−2s3−2s=+∞fors∈(0,34).

Note that there is no (AR) type assumption on f. Then it is difficult to show that the functional satisfies the (PS) condition even for the autonomous case, which is necessary to use Ljusternik-Schnirelmann category theory. We shall investigate the (PS) sequence carefully and restore the compactness for (PS) sequence via some compactness Lemmas.

In order to describe the multiplicity, we first recall that, if Y is a closed subset of a topological space X, the Ljusternik-Schnirelmann category cat_XY is the least number of closed and contractible sets in X which cover Y. Then we state our main result as follows.

Theorem 1.1

If 0 < μ < 2s, assume that V satisfies (V₁) and (V₂) and the function f satisfies (f₁)−(f₃). Then for any δ > 0 such that

Mδ={x∈R3:dist(x,M)≤δ}⊂Ω,

there exists ε_δ > 0 such that the problem (1.8) has at least catMδ(M) positive solutions for any ε ∈ (0, ε_δ). Moreover, if u_ε denotes one of these positive solutions and η_ε ∈ ℝ³ its global maximum, then

limε→0Vηε=V0.

Remark 1.1

Here, we make a few observations about the restriction on the parameter 0 < μ < 2s. In order to adapt the penalization method introduced by del Pino and Felmer in [18], we will proposed some control conditions on the non-local term (1|x|μ(|u|2μ,s∗+F(u))), which need some regularity (see Lemma 2.7 and 2.8), where we introduce the assumption 0 < μ < 2s.

We shall use the method of Nehari manifold, concentration compactness principle and category theory to prove the main results. There are some difficulties in proving our theorems. The first difficulty is that the non-linearity f is only continuous, we can not use standard arguments on the Nehari manifold. To overcome the nondifferentiability of the Nehari manifold, we shall use some variants of critical point theorems from Szulkin and Weth [46]. The second one is the lack of compactness of the embedding of H^s(ℝ³) into the space L2s*(ℝ3). We shall borrow the idea in [2, 12] to deal with the difficulties brought by the critical exponent. However, we require some new estimates, which are complicated because of the appearance of fractional Laplacian and the convolution-type nonlinearity. Moreover, the potential V satisfies (V₁) and (V₂) instead of the global condition. Since we have no information on the potential V at infinity, we adapt the truncation trick explored in [18]. It consists in making a suitable modification on the nonlinearity, solving a modified problem and then check that, for ε small enough, the solutions of the modified problem are indeed solutions of the original one. It is worthwhile to remark that in the arguments developed in [18], one of the key points is the existence of estimates involving the L^∞-norm of the modified problem. But for the critical nonlocal problem (1.8), this kind of estimates are more delicate.

This paper is organized as follows. In section 2, besides describing the functional setting to study problem (1.8), we give some preliminary Lemmas which will be used later. In section 3, influenced by the work [18] and [45], we introduce a modified functional and show it satisfies the Palais-Smale condition. In section 4, we study the autonomous problem associated. This study allows us to show that the modified problem has multiple solutions. Finally, we show the critical point of the modified functional which satisfies the original problem, and investigate its concentration behavior, which completes the proof Theorem 1.1.

2 Variational settings and preliminary results

Throughout this paper, we denote | · |_r the usual norm of the space L^r(ℝ³), 1 ≤ r < 1, B_r(x) denotes the open ball with center at x and radius r, C or C_i(i = 1, 2, · · · ) denote some positive constants may change from line to line. *and →mean the weak and strong convergence. Let E be a Hilbert space, the Fréchet derivative of a functional Φ at u, Φ'(u), is an element of the dual space E^* and we shall denote Φ'(u) evaluated at v ∈ E by 〈Φ'(u), v〉.

2.1 The functional space setting

Firstly, fractional Sobolev spaces are the convenient setting for our problem, so we will give some skrtchs of the fractional order Sobolev spaces and the complete introduction can be found in [19]. We recall that, for any s ∈ (0, 1), the fractional Sobolev space H ^s(ℝ³) = W^s^,2(ℝ³) is defined as follows:

Hs(R3)={u∈L2(R3):∫R3(|ξ|2s|F(u)|2+|F(u)|2)dξ<∞},

whose norm is defined as

∥u∥Hs(R3)2=∫R3(|ξ|2s|F(u)|2+|F(u)|2)dξ,

where F denotes the Fourier transform. We also define the homogeneous fractional Sobolev space D^s^,2(ℝ³) as the completion of C0∞R3 with respect to the norm

∥u∥Ds,2(R3):=(∬R3×R3|u(x)−u(y)|2|x−y|3+2sdxdy)12=[u]Hs(R3).

The embedding Ds,2(ℝ3)→L2s*(ℝ3) is continuous and for any s ∈ (0, 1), there exists a best constant S_s > 0 such that

SS:=infu∈Ds,2(ℝ3)‖u‖Ds,2(ℝ3)2|u|2s⋆2

According to [16], S_s is attained by

(2.1) u0(x)=C(bb2+|x−a|2)3−2s2,x∈ℝ3,

where C ∈ ℝ, b > 0 and a ∈ ℝ³ are fixed parameters. We use S_H_,L to denote the best constant defined by

(2.2) SH,L:=infu∈Ds,2(R3)∖{0}∫R3|(−Δ)s2u|2dx(∫R3∫R3|u(y)|2μ,s∗|u(x)|2μ,s∗|x−y|μdydx)12μ,s∗.

The fractional Laplacian, (−Δ)^su, of a smooth function u : ℝ³ → ℝ, is defined by

F((−Δ)su)(ξ)=|ξ|2sF(u)(ξ), ξ∈ℝ3.

Also (−Δ)^su can be equivalently represented [19] as

(−Δ)su(x)=−12C(s)∫R3u(x+y)+u(x−y)−2u(x)|y|3+2sdy,∀x∈R3

where

C(s)=∫R31−cosξ1|ξ|3+2sdξ−1,ξ=ξ1,ξ2,ξ3.

Also, by the Plancherel formular in Fourier analysis, we have

[u]Hs(ℝ3)2=2C(s)|(−Δ)s2u|22.

For convenience, we will omit the normalization constant in the following. As a consequence, the norms on H^s(ℝ³) defined below

u→∫R3|u|2dx+∬R3×R3|u(x)−u(y)|2|x−y|3+2sdxdy12;u→∫R3|ξ|2s|F(u)|2+|F(u)|2dξ12;u→∫R3|u|2dx+(−Δ)s2u2212.

are equivalent.

Making the change of variable x ⟼ εx, we can rewrite the equation (1.8) as the following equivalent form

(2.3) (−Δ)su+V(εx)u=(∫R3|u(y)|2μ,s∗+F(u(y))|x−y|μdy)(|u|2μ,s∗−2u+12μ,s∗f(u))inR3,

If u is a solution of the equation (2.3), then v(x):=u(xε) is a solution of the equation (1.8). Thus, to study the equation (1.8), it suffices to study the equation (2.3). In view of the presence of potential V(x), we introduce the subspace

Hε={u∈HS(ℝ3):∫R3V(εx)u2dx<+∞},

which is a Hilbert space equipped with the inner product

(u,v)Hε=∫R3(−Δ)s2u(−Δ)s2vdx+∫R3V(εx)uvdx,

and the norm

∥u∥Hε2=∫R3(−Δ)s2u2dx+∫R3V(εx)u2dx.

We denote ‖ · ‖_Hε by ‖ · ‖_ε in the sequel for convenience.

For the reader’s convenience, we review some useful result for this class of fractional Sobolev spaces:

Lemma 2.1

[19] Let 0 < s < 1, then there exists a constant C = C(s) > 0, such that

|u|2s⋆2≤C[u]Hs(ℝ3)2

for every u ∈ H^s(ℝ³). Moreover, the embedding H^s(ℝ³) ↪ L^r(ℝ³) is continuous for any r∈[2,2S⋆] and is locally compact whenever r∈[2,2S⋆).

Lemma 2.2

[40] If {u_n} is bounded in H^s(ℝ³) and for some R > 0 we have

limn→∞supy∈R3∫BR(y)un2dx=0,

then u_n → 0 in L^r(ℝ³) for any 2<r<2s⋆.

Lemma 2.3

[37] Let u∈Ds,2(ℝ3),φ∈C0∞(ℝ3) and for each r>0,φr(x)=φ(xr). Then

uφr→0 in Ds,2(ℝ3) as r→0.

If, in addition, φ ≡ 1 in a neighbourhood of the origin, then

uφr→u in Ds,2(ℝ3) as r→+∞.

2.2 Preliminary lemmas

Set G(u)=|u|2μ,s⋆+F(u),g(u)=dG(u)du. In the sequel, set r0=66−μ, then 2<r0q1≤r0q2<2s⋆. So for any u ∈ H^s(ℝ³), we have

(2.4) |F(u)|r0≤C(|u|q1r0q1+|u|q2r0q2)

and

(2.5) |G(u)|r0≤C(|u|q1r0q1+|u|q2r0q2+|u|2s*2μ,s*).

Therefore, the Hardy-Littlewood-Sobolev inequality implies that

(2.6) |∫R3∫R3G(u(y))G(u(x))|x−y|μdydx|≤C|G(u)|r02≤C(|u|q1r02q1+|u|q2r02q2+|u|2s∗22μ,s∗)

and

(2.7) |∫R3∫R3G(u(y))g(u(x))u(x)|x−y|μdydx|≤C(|u|q1r02q1+|u|q2r02q2+|u|2s∗22μ,s∗).

It is clear that problem (2.3) is the Euler-Lagrange equations of the functional I : H_ε → ℝ defined by

(2.8) I(u)=12∥u∥ε2−122μ,S⋆∫R3∫R3G(u(y))G(u(x))|x−y|μdydx.

From (2.6) we know that I(u) is well defined on H_ε and belongs to C¹, with its derivative given by

(2.9) 〈I′(u),v〉=∫R3((−Δ)s2u(−Δ)s2v+V(εx)uv)dx−12μ,s*∫R3∫R3G(u(y))g(u(x))v(x)|x−y|μdydx

for all u, v ∈ H_ε. Hence the critical points of I in H_ε are weak solutions of problem (2.3). In the following, we will consider critical points of I using variational methods.

Firstly, we give the following Lemma, whose simple proof is omit.

Lemma 2.4

If (f₁) and (f₂) are satisfied, then

(2.10) 0<F(u)<f(u)u,0<G(u)<g(u)u,∀u≠0.

In addition, (f₂) and (2.10) imply

(2.11) F(u)uandG(u)uareincreasingon(0,+∞).

For the derivative of the functional I we have the following Lemma.

Lemma 2.5

Let (V₁) and (f₁) hold, then

(i) I' maps bounded sets in H^s(ℝ³) into bounded sets in (H^s(ℝ³))^*.

(ii) I' is weakly sequentially continuous. Namely, if u_n * u in H^s(ℝ³), then I'(u_n) * I'(u) in (H^s(ℝ³))^*.

Proof. (i). Let {u_n} be a bounded sequence in H^s(ℝ³). For any v ∈ H^s(ℝ³), from the Hardy-Littlewood-Sobolev inequality and (2.5) it follows that

(2.12) |∫R3∫R3G(un(y))g(un(x))v(x)|x−y|μdydx|≤C|G(un)|r0(|un|q1r0q1−1|v|q1r0+|un|q2r0q2−1|v|q2r0+|un|2s∗2μ,s∗−1|v|2s∗)≤C∥v∥ε.

Then |〈I′(un),v〉|≤C‖v‖ε. Hence, {I0(u_n)} is bounded in (H^s(ℝ³))^*.

(ii). Assume that u_n * u in H^s(ℝ³). For any v∈C0∞(ℝ3) with support Ω, by Lemma 2.1 we may assume that u_n(x) → u(x) a.e. in ℝ³ and u_n → u in Lp(Ω),p<2s⋆. We first check that if u_n ⇀ u in H^s(ℝ³), then

(2.13) ∫R3∫R3|un(y)|2μ,s*|un(x)|2*,s−2−μun(x)v(x)|x−y|μdydx→∫R3∫R3|u(y)|2μ,s*|u(x)|2μ,8−2−μu(x)v(x)|x−y|μdydx,

as n → ∞. By the Hardy-Littlewood-Sobolev inequality, we have

|h⋆|x−y|−μ|6μ≤C|h|r0, for all h∈Lr0(ℝ3),

and it is a linear bounded operator from Lr0(ℝ3) to L6μ(ℝ3). Choosing hn(y):=|un(y)|2μ,s⋆∈Lr0(ℝ3), we have

||un(y)|2μ,s⋆⋆|x−y|−μ|6μ≤C|un|2s⋆≤C.

Therefore, by Hölder inequality we can prove the sequence

(|un(y)|2μ,s⋆⋆|x−y|−μ)|un(x)|2μ,s⋆−2−μun(x)

is bounded in L2s⋆2s−1(ℝ3). Then, by duality we have

∫R3|un(y)|2μ,s∗|un(x)|2μ,s∗−2−μun(x)|x−y|μdy⇀∫R3|u(y)|2μ,s∗|u(x)|2μ,s∗−2−μu(x)|x−y|μdy,inL2s∗2s∗−1(R3)

as n → ∞. Then (2.13) follows for every v∈HS(ℝ3)⊂L2s⋆(ℝ3).

By (f₁) we have |f(u)|≤C(1+|u|q2−1) for all u ∈ ℝ⁺, and then f (u_n) → f(u) in Lq2rq2−1(Ω). Then Hardy-Littlewood-Sobolev inequality implies that

(2.14) ∫R3∫R3G(un(y))(f(un(x))−f(u(x)))v(x)|x−y|μdydx≤C|G(un)|r|f(un)−f(u)|G2rq2−1,Ω|v|q2r→0

Since F(u_n) is bounded in Lr0(ℝ3). we may assume that F(u_n) ⇀ F(u) in Lr0(ℝ3). Consequently,

∫R3F(un(y))|x−y|μdy→∫R3F(u(y))|x−y|μdy in L6μ(ℝ3).

Moreover, as (2.12) we get ∫R3|f(u(x))v(x)|rdx<∞, we infer

∫R3∫R3F(un(y))|x−y|μf(u(x))v(x)dydx→∫R3F(u(y))|x−y|μf(u(x))v(x)dydxinL6μ(ℝ3),

which combining with (2.14) we have

∫R3∫R3F(un(y))|x−y|μf(un(x))v(x)dydx→∫R3F(u(y))|x−y|μf(u(x))v(x)dydxinL6μ(ℝ3)

Notice that

∫R3∫R3G(u(y))g(u(x))v(x)|x−y|μdydx=∫R3(∫R3|u(y)|2μ,s∗+F(u(y))|x−y|μdy)(|u(x)|2μ,s∗−2−μu(x)+12μ,s∗f(u(x)))v(x)dx,

from our argument above it is then easy to prove

∫R3∫R3G(un(y))g(un(x))v(x)|x−y|μdydx→∫R3G(u(y))G(u(x))v(x)|x−y|μdydxinL6μ(ℝ3).

Hence, for any v∈C0∞(ℝ3),

(2.15) 〈I′(un),v〉→〈I′(u),v〉.

Since {I0(u_n)} is bounded in (HS(ℝ3))⋆ and Cn∞(ℝ3) is dense in H^s(ℝ³), we conclude that (2.15) holds for any v ∈ H^s(ℝ³), and so I′(un)→I′(u)in(HS(ℝ3))⋆.

2.3 Regularity of solutions and Pohožaev identity

The assumption (f₁) is too weak for the standard bootstrap method as in [4, 15, 32]. Therefore, in order to prove regularity of solutions of (2.3) we shall rely on a nonlocal version of the Brezis-Kato estimate. Note that a special case of the regularity result of Brezis and Kato [10, Theorem 2.3] states that if u ∈ H¹(ℝ^N) is a solution of the linear elliptic equation

−Δu+u=Vu in ℝN,

and V∈L∞(ℝN)+LN2(ℝN), then u∈Lp(ℝN) for every p ≥ 1. Similar to [33, 42], we extent this result to the fractional Choquard equation with critical growth. For the convenience of readers, we give here a short proof. We first have the following useful inequality.

Lemma 2.6

[33] Let p, q, r, t ∈ [1, +∞) and λ ∈ [0, 2] such that

1+N−μN−1p−1t=λq+2−λr.

If θ ∈ (0, 2) satisfies

min(q,r)(N−μN−1p)<θ<max(q,r)(1−1p)

and

min(q,r)(N−μN−1t)<2−θ<max(q,r)(1−1t),

then for every H∈LpRN,K∈LtRNandu∈LqRN∩LrRN,

∫RN∫RNH|u(y)|θK|u(x)|2−θ|x−y|μdydx≤C(∫RN|H|p)1p(∫RN|K|t)1t(∫RN|u|q)λq(∫RN|u|r)2−λr.

Applying Lemma 2.6,we have the following result, which is a nonlocal counterpart of the estimate [10, Lemma 2.1]: If V∈L∞(ℝN)+LN2(ℝN). then for every ε > 0, there exists C_ε such that

∫RNV|u|2dx≤ε2∫RN|∇u|2dx+Cε∫RN|u|2dx.

Lemma 2.7

Let N ≥ 2s, μ ∈ (0, 2s) and θ ∈ (0, 2). If H,K∈L2NN−μRN+L2NN+2s−μRNandN−μN<θ<2−N−μN, then for every ε > 0, there exists C_ε_, ∈ ℝ such that for every u ∈ H^s(ℝ^N),

∫RN∫RNH|u(y)|θK|u(x)|2−θ|x−y|μdydx≤ε2∫RN|(−Δ)s2u|2dx+Cε,θ∫RN|u|2dx.

Proof. Since 0 < μ < 2s, we may assume that H = H^* + H_* and K = K^* + K_* with H⋆,K⋆∈L2NN−μ(ℝ3) and H⋆,K⋆∈L2NN+2s−μ(ℝN). Applying Lemma 2.6 with q=r=2s⋆,p=t=2NN+2s−μ and λ = 0, we have

∫RN∫RNH∗|u(y)|θK∗|u(x)|2−θ|x−y|μdydx≤C|H∗|2NN+2s−μ|K∗|2NN+2s−μ|u|2s∗2,

where we use |θ−1|<μN−2s. Taking p=t=2NN−μ,q=r=2 and λ = 2, we have |θ−1|<μN and

∫RN∫RNH∗|u(y)|θK∗|u(x)|2−θ|x−y|μdydx≤C|H∗|2NN−μ|K∗|2NN−μ|u|22.

Similarly, with p=2NN+2s−μ,t=2NN−μ,q=2,r=2S⋆ and λ = 1, we have

∫RN∫RNH∗|u(y)|θK∗|u(x)|2−θ|x−y|μdydx≤C|H∗|2NN−+2s−μ|K∗|2NN−μ|u|2|u|2s∗

and with p=2NN−μ,t=2NN+2s−μ,q=2,r=2S⋆ and λ = 1,

∫RN∫RNH⋆|u(y)|θK⋆|u(x)|2−θ|χ−y|μdydx≤C|K⋆|2NN−2s−μ|H⋆|2NN−μ|u|2|u|2s*.

By the Sobolev inequality, we have thus prove that for every u ∈ H^s(ℝ^N),

∫RN∫RNH|u(y)|θK|u(x)|2−θ|x−y|μdydx≤C(∫RN|H⋆|2NN+2s−μ∫RN|K⋆|2NN+2s−μ)N+2s−μ2N∫RN|(−Δ)s2u|2dx +(∫RN|H*|2NN−μ∫RN|K⋆|2NN−μ)N−μ2N∫RN|u|2dx).

The conclusion follows by choosing H^* and K^* such that

C(∫RN|H⋆|2NN+2s−μ∫RN|K⋆|2NN+2s−μ)N+2s−μ2N≤ε2.

Now, we have the following result, which is a nonlocal Brezis-Kato type regularity estimate.

Lemma 2.8

Let N ≥ 2s and 0 < μ < 2s. If H,K∈L2NN−μ(ℝN)+L2NN+2s−μ(ℝN) and u ∈ H^s(ℝ^N) solves

(2.16) (−Δ)su+u=(∫RNH(u(y))u(y)|x−y|μdy)K(u(x)),

then u ∈ L^p(ℝ^N) for every p∈[2,NN−μ2NN−2s). Moreover, there exists a constant C_p independent of u such that

(∫RN|u|pdx)1p≤Cp(∫RN|u|2dx)12.

Proof. By Lemma 2.7 with θ = 1, there exists λ > 0 such that for every φ∈Hs(ℝN),

∫RN∫RN|Hφ‖Kφ||x−y|μdydx≤12∫RN|(−Δ)s2φ|2dx+λ2∫RN|φ|2dx.

Choose sequences {H_k}_k2N and {Kk}k∈ℕinL2NN−μ(ℝN) such that |H_k| ≤ |H|, |K_k| ≤ |K|, and H_k → H and K_k → K almost everywhere in ℝ^N. For each k ∈ ℕ, for φ,ψ∈HS(ℝN), the form a_k : H^s(ℝ^N) × H^s(ℝ^N) → ℝ defined by

ak(φ,ψ)=∫RN(−Δ)s2φ(−Δ)s2ψ+λφψ−∫RN∫RNHkφKkψ|x−y|μdydx

is bilinear and coercive. Therefore, applying the Lax-Milgram theorem [9, Corollary 5.8], there exists a unique solution u_k ∈ H^s(ℝ^N) satisfies

(2.17) (−Δ)suk+λuk=∫RN(Hkuk|x−y|μdy)Kkuk+(λ−1)u,

where u ∈ H^s(ℝ^N) is the given solution of (2.16). Moreover, we can prove that the sequences {u_k}_k2N converges weakly to u in H^s(ℝ^N) as k → ∞.

For μ > 0, we define the truncation u_k_,μ by

uk,μ(x)={−μ if uk(x)≤−μuk(x) if −μ<uk(x)<μμ if uk(x)≥μ

For p ≥ 2, we have |uk,μ|p−2uk,μ∈HS(ℝN), so we can take it as a test function in (2.17), we have

By Lemma 2.7 with θ = 1, there exists C > 0 such that

∫RN∫RN|Hk||uk,μ||Kk||uk,μ|p−2|uk,μ||x−y|μdydx≤2(p−1)p2∫RN|(−Δ)s2(uk,μ)p2|2dx+C∫RN||uk,μ|p2|2dx.

We have thus

2(p−1)p2∫RN|(−Δ)s2(uk,μ)p2|2dx≤C1∫RN(|uk|2+|u|2)dx+∫Ak,μ(|Hkuk||x−y|μdy)|Kkuk|,

where

Ak,μ={x∈ℝ3:|uk(x)|>μ}.

Since p<2NN−μ, by the Hardy-Littlewood-Sobolev inequality with 1r=2N−μ2N+1−1p and 1t=2N−μ2N+1p,

∫Ak,μ(1|x|μ⋆|Kk‖uk|p−1dy)|Hkuk|≤C(∫RN‖Kk‖uk|p−1|rdx)1r(∫Ak,μ|Hkuk|tdx)1t

By Hölder inequality, if u_k ∈ L^p(ℝ^N), then |Kk||uk|p−1∈Lr(ℝN),|Hk||uk|∈Lt(ℝN), thus by Lebesgue’s dominated convergence theorem we have

limμ→∞∫Ak,μ(1|x|μ⋆|Kk||uk|p−1dy)|Hkuk|=0.

In view of the Sobolev estimate, we have proved the inequality

limsupk→∞(∫RN|uk|2s*dx)22⋆≤Climsupk→∞∫RN|uk|2dx.

By iterating over p a finite number of times we cover the range p∈[2,NN−μ2NN−2s).

3 The penalized problem

In this section, we will adapt for our case an argument explored by the penalization method introduction by del Pino and Felmer [18] to overcome the lack of compactness. Let K > 2 to be determined later, and take a > 0 to be the unique number such that G(a)a=V0K where V₀ is given by (V₁). We define

G˜(u)={G(u) if u≤a,V0Ku if u>a,

and

H(x,u)=χΩ(x)G(u)+(1−χΩ(x))G˜(u),

where χ is characteristic function of set Ω. From hypotheses (f₁)−(f₃)we get that H is a Carathéodory function and satisfies the following properties:

(g₁) H(x,u)≤G(u)≤C(|u|q1+|u|q2+|u|2μ,s∗).

(g₂) The function H(x,u)u is increasing for u > 0.

(g₃)(i)

lim|u|→+∞H(x,u)|u|2⋆,s−1=+∞ for s∈(34,1);

(ii)

lim|u|→+∞H(x,u)|u|21μ,s−2s3−2s(log|u|)12=+∞ for s=34;

(iii)

lim|u|→+∞H(x,u)|u|2μ,s−23−2S=+∞ for s∈(0,34).

Moreover, in order to find positive solutions, we shall henceforth consider H(x, u) = 0 for all u ≤ 0. It is easy to check that if u is a positive solution of the equation

ε2s(−Δ)su+V(x)u=εμ−312μ,s∗∫R3H(εx,u)|x−y|μdyh(εx,u) in R3u∈Cloc0,αR3∩HsR3

such that u(x) ≤ a for all x∈ℝ3\Ω, then H (x, u) = G(u) and therefore u is also a solution of problem (1.8). In view of this argument above, we shall deal with in the following with the penalized problem

(3.1) (−Δ)su+V(εx)u=12μ,s⋆(∫R3H(εx,u)|x−y|μdy)h(εx,u) in ℝ3,

and we will look for solutions u_ε of problem (3.1) verifying

uε(x)≤a for all x∈ℝ3\Ωε,

where Ωε={x∈ℝ3:εx∈Ω}.

The energy functional associated with (3.1) is

Jε(u)=12∥u∥ε2−122μ,s∗∫R3∫R3H(εx,u(y))H(εx,u(x))|x−y|μdydx.

which is of C¹ class and whose derivative is given by

〈Jε′(u),v〉=∫R3((−Δ)s2u(−Δ)s2v+V(εx)uv)dx−12μ,s*∫R3∫R3H(εx,u(y))h(εx,u(x))v(x)|x−y|μdydx

for all u, v ∈ H_ε. Hence the critical points of J_ε in H_ε are weak solutions of problem (3.1).

Now, we denote the Nehari manifold associated to J_ε by

Nε=u∈Hε∖{0}:Jε′(u),u=0.

Obviously, N_ε contains all nontrivial critical points of I_ε. But we do not know whether N_ε is of class C¹ under our assumptions and therefore we cannot use minimax theorems directly on N_ε. To overcome this difficulty, we will adopt a technique developed in [45, 46] to show that N_ε is still a topological manifold, naturally homeomorphic to the unit sphere of H_ε, and then we can consider a new minimax characterization of the corresponding critical value for I_ε.

For this we denote by Hε+ the subset of H_ε given by

Hε+={u∈Hε:|supp(u+)∩Ωε|>0}

and Sε+=Sε∩Hε+, where S_ε is the unit sphere of H_ε.

Lemma 3.1

The set Hε+ is open in H_ε.

Proof. Suppose by contradiction there are a sequence {un}⊂Hε\Hε+and u∈Hε+ such that u_n → u in H_ε. Hence |supp(un+)∩Ωε|=0 for all n∈ℕ and un+(x)→u+(x) a.e. in x∈Ωε. So,

u+(x)=limn→∞un+(x)=0, a.e. in x∈Ωε.

But, this contradicts the fact that u∈Hε+ Therefore Hε+ is open.

From definition of Sε+ and Lemma 3.1 it follows that Sε+ is a incomplete C ^1,1-manifold of codimension 1, modeled on H_ε and contained in the open Hε+. Hence, Hε=TuSε+⊕ℝu for each u∈Sε+, where TuSε+={v∈Hε: (u, v)_ε = 0}.

In the rest of this section, we show some Lemmas related to the function J_ε and the set Hε+. First, we show the functional J_ε satisfying the Mountain Pass geometry.

Lemma 3.2

The functional J_ε satisfies the following conditions:

(i) There exist α, ρ > 0 such that J_ε(u) ≥ α with ‖u‖_ε = ρ;

(ii) There exists e ∈ H_ε satisfying ‖e‖_ε > ρ such that J_ε(e) < 0.

Proof. (i). For any u ∈ H_ε\{0}, it follows from (g₁) and the Hardy-Littlewood-Sobolev inequality that

(3.2) |∫R3∫R3H(εx,u(y))H(εx,u(x))|x−y|μdydx|≤C(|u|q1r02q1+|u|q2r02q2+|u|2s∗22μ,s∗)

Hence,

Jε(u)=12∫R3(|(−Δ)s2u|2+V(εx)u2)dx−122μ,s⋆∫R3∫R3H(εx,u(y))H(εx,u(x))|x−y|μdydx. ≥12‖u‖ε2−C1‖u‖2q1−C2‖u‖2q2−C3‖u‖22μ,s*.

Therefore, we can choose positive constants α, ρ such that

Jε(u)≥α with ‖u‖ε=ρ.

(ii). Fix a positive function u0∈Hε+ with suppu0⊂Ωε, and we set

ψ(t):=Σε(tu0‖u0‖ε)>0,

where

Σε(u)=122μ,s∗∫R3∫R3H(εx,u(y))H(εx,u(x))|x−y|μdydx.

Since H(εx, u₀) = F(u₀) and by using Lemma 2.4, we deduce that

(3.3) ψ′(t)=Σε′(tu0∥u0∥ε)u0∥u0∥ε=∫R3∫R3(F(tu0∥u0∥ε)f(tu0∥u0∥ε)u0∥u0∥ε|x−y|μ)dxdy=22μ,s∗t∫R3∫R3122μ,s∗(F(tu0∥u0∥ε)f(tu0∥u0∥ε)tu0∥u0∥ε|x−y|μ)dxdy≥22μ,s∗tψ(t).

Integrating (3.3) on [1, t‖u₀‖_ε] with t>1∥u0∥ε, we have

Σε(tu0)≥Σε(u0∥u0∥ε)∥u0∥ε22μ,s∗t22μ,s∗

Therefore, we have

Jε(tu0)=t22∥u∥ε2−Σε(tu0)≤C1t2−C2t22μ,s∗fort>1∥u0∥ε.

Taking e = tu₀ with t sufficiently large, we can see (ii) holds.

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

Lemma 3.3

Assume that the potential V satisfies (V₁)−(V₂) and the functional f satisfies (f₁)−(f₃). Then the following properties hold:

(A₁)For each u∈Sε+. let φ_u : ℝ⁺ → ℝ be given by φ_u(τ) = J_ε(τu). Then there exists a unique τ_u > 0 such that φu′(τ)>0in(0,τu)andφu′(τ)<0in(τu,∞)

(A₂)There is a σ > 0 independent on u such that τ_u > σ for all u∈Sε+. Moreover, for each compact setW⊂Sε+ there is C_W > 0 such that τ_u ≤ C_W for all u ∈ W.

(A₃)The map m^ε:Hε+→Nε given by m_ε(u) = τ_uu is continuous and mε:=m^ε|Sε+ is a homeomorphism between Sε+ and N_ε. Moreover, mε−1(u)=u∥u∥ε.

Proof. (A₁) From Lemma 3.2, it is sufficient to note that, φ_u(0) = 0, φ_u(τ) > 0 when τ > 0 is small and φ_u(τ) < 0 when τ > 0 is large. Since φ_u ∈ C¹(ℝ⁺, ℝ), there is τ_u > 0 global maximum point of φ_u and φu′(τu)=0. Thus,Jε′(τuu)(τuu)=0 and τ_uu ∈ N_ε. We see that τ_u > 0 is the unique positive number such that φu′(τu)=0. Indeed, suppose by contradiction that there are τ₁ > τ₂ > 0 with φu′(τ1)=φu′(τ2)=0. Then, for i = 1, 2 we have that

(3.4) 0=∫R3∫R31|x−y|μ(H(εx,τ1u(u))τ1h(εx,τ1u(x))u(x)−H(εx,τ2u(u))τ2h(εx,τ2u(x))u(x))dydx.

By Lemma 2.4 and (h₂) we know that h(εx,τu(x))u(x) andH(εx,τu(x))τ are positive and increasing in τ. Then (3.4) in impossible and (A₁) is proved.

(A₂) Suppose u∈Sε+, then as (2.5) we have

τu≤∫R3∫R3H(εx,u(y))h(εx,u(x))u(x)|x−y|μdydx≤C(∥τu∥2q1+∥τu∥2q2+∥τu∥22μ,s∗).

From previous inequality we obtain σ > 0 independent on u, such that τ_u > σ.

Finally, if W⊂Sε+ is compact, and suppose by contradiction that there is {u_n} ⊂ W such that τ_n := τn:=τun→∞. Since W is compact, there is u ∈ W such that u_n → u in H_ε. Then u∈W⊂Sε+. By (h₂) we obtain

0≤Jε(τnun)τn2=12−122μ,s∗∫R3∫R3H(εx,τnun(y))|x−y|μτnH(εx,τnun(x))τndydx→−∞,

which yields a contradiction. Therefore (A₂) is true.

(A₃) First of all we observe that m^ε,mε andmε−1 are well defined. In fact, by (A₁), for each u∈Hε+. there exists a unique τ_u > 0 such that τ_uu ∈ N_ε, hence there is a unique m^ε(u)=τuu∈Nε. On the other hand, if u∈Nε thenu∈Hε+. Therefore, mε−1(u)=u∥u∥ε∈Sε+, is well defined and it is a continuous function. Since

mε−1(mε(u))=mε−1(tuu)=τuuτu∥u∥ε=u,∀u∈Sε+,

we conclude that m_ε is a bijection.

To prove m^ε:Hε+→Nε is continuous, let {un}⊂Hε+andu∈Hε+ be such that u_n → u in H_ε. By (A₂), there is a τ₀ > 0 up to a subsequence such that τn:=τun→τ0. Since τ_nu_n ∈ N_ε we obtain

τn2∥un∥ε2=12μ,s∗∫R3∫R3H(εx,τnun(y))h(εx,τnu(x))τnun(x)|x−y|μdydx,∀n∈N.

By Lemma 2.5 and passing to the limit as n → ∞, it follows that

τ02∥u∥ε2=12μ,s∗∫R3∫R3H(εx,τ0un(y))h(εx,τ0u(x))τ0un(x)|x−y|μdydx,

which means that τ0u∈Nεandτu=τ0. This proves m^ε(un)→m^ε(u) inHε+. So, m^ε,mε are continuous functions and (A₃) is proved. ☐

Now we define the functions

Ψ^ε:Hε+→RandΨε:Sε+→R,

by Ψ^ε(u)=Iε(m^ε(u)) and Ψε:=Ψ^ε|Sε+. The next result is a direct consequence of Lemma 3.3. The details can be seen in the relevant material from [46]. For the convenience of the reader, here we do a sketch of the proof.

Lemma 3.4

Assume that (V₁)− (V₂) and (f₁)− (f₃) are satisfied. Then: (B1)Ψ^ε∈C1(Hε+,R) and

Ψ^ε′(u)v=∥m^ε(u)∥ε∥u∥εJε′(m^ε(u))v,∀u∈Hε+and∀v∈Hε.

(B2)Ψε∈C1(Sε+,R) and

Ψε′(u)v=∥mε(u)∥εJε′(mε(u))v,∀v∈TuSε+.

(B₃)If {u_n} is a (PS)_c sequence of Ψε, then {m_ε(u_n)} is a (PS)_c sequence of J_ε. If {u_n} ⊂ N_ε is a bounded (PS)_c sequence for Jε,then{mε−1(un)} is a (PS)_c sequence of Ψε.

(B₄)u is a critical point of Ψε if and only if, m_ε(u) is a critical point of J_ε. Moreover, corresponding critical values coincide and

infSε+Ψε=infNεIε.

Proof. (B₁) Let u∈Hε+ and v ∈ H_ε. From definition of Ψ^ε andtu and the mean value theorem, we obtain

Ψ^ε(u+hv)−Ψ^ε(u)=Jε(τu+hv(u+hv))−Jε(τuu)≤Jε(τu+hv(u+hv))−Jε(τu+hvu)=Jε′(τu+hv(u+θhv))τu+hvhv,

where |h| is small enough and θ ∈ (0, 1). Similarly,

Ψ^ε(u+hv)−Ψ^ε(u)≥Jε(τu(u+hv))−Jε(τuu)=Jε′(τu(u+ςhv))τuhv,

where ζ ∈ (0, 1). Since the mapping u ⟼ τ_u is continuous according to Lemma 3.3, we see combining these two inequalities that

limh→0Ψ^ε(u+hv)−Ψ^ε(u)h=τuJε′(τuu)v=∥m^ε(u)∥ε∥u∥εJε′(m^ε(u))v.

Since J_ε ∈ C¹, it follows that the Gâteaux derivative of Ψ^ε is bounded linear in v and continuous on u. From [50] we know that Ψ^ε∈C1(Hε+,R) and

Ψ^ε′(u)v=∥m^ε(u)∥ε∥u∥εJε′(m^ε(u))v,∀u∈Hε+and∀v∈Hε.

The item (B₁) is proved.

(B₂) The item (B₂) is a direct consequence of the item (B₁).

(B₃) We first note that Hε=TuSε+⊕Ru for every u∈Sε+ and the linear projection P:Hε→TuSε+ defined by P(v + τu) = v is continuous, namely, there is C > 0 such that

(3.5) ∥v∥ε≤C∥v+τu∥ε,∀u∈Sε+,v∈TuSε+andτ∈R.

Moreover, by (B₁) we have

(3.6) ∥Ψε′∥=supv∈TuSε+,∥v∥ε=1Ψε′(u)v=∥w∥εsupv∈TuSε+,∥v∥ε=1Jε′(w)v,

where w=mε(u).Sincew∈Nε, we conclude that

(3.7) Jε′(w)u=Jε′(w)w∥w∥ε=0.

Hence, from (3.5) and (3.7) we have

∥Ψε′(u)∥≤∥w∥ε∥Jε′(w)∥≤C∥w∥εsupv∈TuSε+∖{0}Jε′(w)v∥v∥ε=C∥Ψε′(u)∥,

which showing that

(3.8) ∥Ψε′(u)∥≤∥w∥ε∥Jε′(w)∥≤C∥Ψε′(u)∥,∀u∈Sε+.

Since w ∈ N_ε, we have ‖w‖ ≥ > 0. Therefore, the inequality in (3.8) together with Jε(w)=Ψε(u) imply the item (B₃).

(B₄) It follow from (3.8) that Ψε′(u)=0 if and only if Jε′(w)=0. The remainder follows from definition of Ψε.

As in [46], using the mountain pass theorem without the (PS) condition, we get the existence of a (PS)_cε sequence {u_n} ⊂ H_ε with

cε=infu∈NεIε(u)=infu∈Hε+maxτ>0Iε(τu)=infu∈Sε+maxτ>0Iε(τu)>0.

Lemma 3.5

Suppose that (f₁) − (f₃) hold. Assume that {un}⊂Nεisa(PS)c sequence with

0<cε≤c<2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1.

Then {u_n} is bounded in H_ε. Moreover, {u_n} cannot be vanishing, namely there exist r, δ > 0 and a sequence {y_n} ⊂ ℝ³ such that

lim infn→∞∫Br(yn)|un|2dx≥δ.

Proof. We first prove the boundedness of {u_n}. Argue by contradiction we suppose that {u_n} is unbounded in H_ε. Without loss of generality, assume that ∥un∥ε→∞.Letvn:=un∥un∥ε up to a subsequence, then there exists v ∈ H_ε such that

un⇀uinHε,un→uinLlocr(R3),2≤r<2s∗,un(x)→u(x)a.e.inR3.

If v_n is vanishing, i.e.

limn→∞supy∈R3∫Br(y)vn2(x)dx=0,

then Lemma 2.2 implies that vn→0inLr0q1(R3)andLr0q2(R3). By (2.4) and (2.5) we get

(3.9) ∫R3∫R3F(τvn(x))F(τvn(y))|x−y|μdxdy→0,∫R3∫R3F(τvn(y))|x−y|μ|τvn(x)|2μ,s∗dxdy→0.

Then for sufficiently large n we have

cε+on(1)=Iε(un)≥supτ≥0Iε(τvn)≥supτ≥0(τ22−τ22μ,s∗22μ,s∗SH,L−2μ,s∗)+on(1)=2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1+on(1),

which is a contradiction. Therefore, {v_n} is non-vanishing, namely there exists y_n ∈ ℝ³ and δ > 0 such that

(3.10) ∫Br(yn)vn2(x)dx>δ.

Denote vn~(⋅)=vn(⋅+yn), then we can assume that

vn~⇀v~inHε,vn~→v~inLlocr(R3),2≤r<2s∗,vn~(x)→v~(x)a.e.inR3.

With the use of (3.10), we have v~⧸≡0. Then there exists a measure set 𝛬 such that v~(x)≠0 for x ∈ 𝛬. Let |un¯|:=|vn~|∥un∥ε. Then |un¯(x)|→+∞ for x∈Λ. By Lemma 2.4, we have

∫Λ∫Λ1|x−y|μF(un¯(y))|un¯(y)||vn~(y)|F(un¯(x))|un¯(x)||vn~(x)|dydx→+∞.

Therefor, Lemma 2.4 and Fatou Lemma imply that

limn→∞∫R3∫R31|x−y|μF(un¯(y))|un¯(y)||vn~(y)|F(un¯(x))|un¯(x)||vn~(x)|dydx=+∞.

Namely

limn→∞∫R3∫R31|x−y|μF(un(y))∥un∥εF(un(x))∥un∥εdydx=+∞.

Then,

cε∥un∥ε+on(1)=Iε(un)∥un∥ε→−∞,

which is a contradiction. Therefore, {u_n} is bounded in H^s(ℝ³).

Next we show the second conclusion. We argue by contradiction, if {u_n} is vanishing, then similar to (3.9)

we have

(3.11) cε+on(1)=Iε(un)=12∥un∥ε2−122μ,s∗∫R3∫R3|un(y)|2μ,s∗|un(x)|2μ,s∗|x−y|μdydx+on(1),

and

(3.12) 0=∥un∥ε2−∫R3∫R3|un(y)|2μ,s∗|un(x)|2μ,s∗|x−y|μdydx+on(1).

If ∥un∥ε→0, then it follows from (3.11) and (3.12) that c_ε = 0, which is impossible. Then ∥un∥ε↛0 and by virtue of (3.12) we get

∥un∥ε2≤SH,L2μ,s∗∥un∥ε22μ,s∗+on(1).

Hence,

lim infn→∞∥un∥ε2≥SH,L2μ,s∗2μ,s∗−1.

From (3.11) and (3.12) we deduce that

(3.13) cε+on(1)=Iε(un)≥2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1,

which is a contradiction. Therefore, {u_n} is non-vanishing.

Lemma 3.6

Assume (V₁) − (V₂) and (f₁) − (f₃) hold, let {u_n} be a (PS)_c sequence for J_ε with c ∈ [cε,2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1) Then, for each η > 0 there exists R = R(η) > 0 such that

lim supn→∞∫R3∖BR|(−Δ)s2u|2+V(εx)un2dx<η.

Proof. By Lemma 3.5, we can have {u_n} is bounded in H_ε. Therefore, we may assume that un⇀u in H_ε and un→uinLlocr(R3) for any r∈[2,2s∗). Fix R > 0 and let ψR∈C∞(R3) be such that ψR=0inBR2(0),ψR=1 in BRc,ψR∈[0,1] and |∇ψR|≤CR, where C is a constant independent of R. Since {unψR} is bounded we can see that

∫R3((−Δ)s2un(−Δ)s2(unψR)+V(εx)ψRun2)dx=〈Jε′(un),unψR〉+12μ,s∗∫R3(1|x|μ∗H(εx,un))h(εx,un)unψRdx=on(1)+12μ,s∗∫R3(1|x|μ∗H(εx,un))h(εx,un)unψRdx.

By Lemma 2.8, taking

H(u):=|u|2μ,s∗+F(u)u,K(u):=|u|2μ,s∗−2−μ+12μ,s∗f(u)∈L23−μ(R3)+L63+2s−μ(R3),

we have

∫R3(1|x|μ∗H(εx,un))h(εx,un)undx≤∫R3(1|x|μ∗H(un)un)K(un)undx≤ε2∫RN|(−Δ)s2u|2dx+Cε∫RN|u|2dx.

For n ≥ n₀ and ε > 0 fixed, take R > 0 big enough such that Ωε⊂BR/2. Then we have

∫R3∖BR/2(|(−Δ)s2un|2+V(εx)un2)dx≤12μ,s∗∫R3∖BR/2(1|x|μ∗H(εx,un))h(εx,un)undx+on(1)−∫R3∫R3(un(x)−un(y))(ψR(x)−ψR(y))|x−y|3+2sun(y)dxdy.

which means

12∫R3∖BR/2(|(−Δ)s2un|2+V(εx)un2)dx≤on(1)−∫R3∫R3(un(x)−un(y))(ψR(x)−ψR(y))|x−y|3+2sun(y)dxdy

Now, we note that the Hölder inequality and the boundedness of {u_n} imply that

|∫R3∫R3(un(x)−un(y))(ψR(x)−ψR(y))|x−y|3+2sun(y)dxdy|≤(∫R3∫R3|un(x)−un(y)|2|x−y|3+2sdxdy)12(∫R3∫R3|ψR(x)−ψR(y)|2|x−y|3+2sun2(y)dxdy)12≤C(∫R3∫R3|ψR(x)−ψR(y)|2|x−y|3+2sun2(y)dxdy)12.

Therefore, it is enough to prove that

limR→∞lim supn→∞∫R3∫R3|ψR(x)−ψR(y)|2|x−y|3+2sun2(y)dxdy=0

to conclude our result.

Let us note that ℝ³ × ℝ³ can be written as

R3×R3=((R3∖B2R)×(R3∖B2R))∪((R3∖B2R)×B2R)∪(B2R×R3):=X1∪X1∪X3.

Then

(3.14) ∫∫R3×R3|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy=∫∫X1|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy+∫∫X2|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy+∫∫X3|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy.

Now, we estimate each integral in (3.14). Since ψR=1inR3∖B2R, we have

(3.15) ∫∫X1|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy=0.

Let k > 4, we have

X2=(R3∖B2R)×B2R=(R3∖BkR)×B2R∪(BkR∖B2R)×B2R.

Let us note that, if (x,y)∈(R3∖BkR)×B2R, then

|x−y|≥|x|−|y|≥|x|−2R>|x|2.

Therefore, taking into account 0≤ψR≤1,|∇ψR|≤CR and applying Hölder inequality, we can see

(3.16) ∫∫X2|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy=∫R3∖BkR∫B2R|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy+∫BkR∖B2R∫B2R|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy≤25+2s∫R3∖BkR∫B2R|un(x)|2|x|3+2sdxdy+CR2∫BkR∖B2R∫B2R|un(x)|2|x−y|3+2(s−1)dxdy≤CR3∫R3∖BkR|un(x)|2|x|3+2sdxdy+CR2(kR)2(1−s)∫BkR∖B2Run2(x)dxdy≤CR3(∫R3∖BkR|un(x)|2s∗dx)22s∗(∫R3∖BkR1|x|92s+3)2s3+Ck2(1−s)R2s∫BkR∖B2Run2(x)dx≤Ck3(∫R3∖BkR|un(x)|2s∗dx)22s∗+Ck2(1−s)R2s∫BkR∖B2Run2(x)dx≤Ck3+Ck2(1−s)R2s∫BkR∖B2Run2(x)dx.

Now, fix ε∈(0,12), and we note that

(3.17) ∫∫X3|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy≤∬B2R∖BεRR3|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy+∬BεRR3|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy.

Let us estimate the first integral in (3.17). First, we have

∫B2R∖BεR∫R3∩{y:|x−y|<R}|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy≤CR2s∫B2R∖BεRun2(x)dx

and

∫B2R∖BεR∫R3∩{y:|x−y|≥R}|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy≤CR2s∫B2R∖BεRun2(x)dx.

Then

(3.18) ∫B2R∖BεR∫R3|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy≤CR2s∫B2R∖BεRun2(x)dx.

By using the definition of ψR,ε∈(0,1) and ψR≤1, we have

(3.19) ∫BεR∫R3|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy=∫BεR∫R3∖BR|ψR(x)−ψR(y)|2|x−y|3+2sun2(x)dxdy≤4∫BεR∫R3∖BRun2(x)|x−y|3+2sdxdy≤C∫BεRun2(x)dx∫(12−ε)R∞1r2+2sdr=C((12−ε)R)2s∫BεRun2(x)dx

where we use the fact that if (x,y)∈BεR×(R3∖BR), then |x−y|>(12−ε)R. Taking into account (3.17)-(3.19) we deduce

(3.20) ∫∫X3un2(x)|ψR(x)−ψR(y)|2|x−y|μdxdy≤CR2s∫B2R∖BϵR|un(x)|2dx+C((1−ϵ)R)2s∫BϵRun2(x)dx.

Putting together (3.14),(3.15),(3.16) and (3.20), we can infer

(3.21) ∫∫R3×R3un2(x)|ψR(x)−ψR(y)|2|x−y|μdxdy≤Ck3+Ck2(1−s)R2s∫BkR∖B2Run2(x)dx+CR2s∫B2R∖BεRun2(x)dx+C((1−ϵ)R)2s∫BϵRun2(x)dx.

Since {u_n} is bounded in H_ε, we may assume that un→uinLloc2(R3) for some u∈Hε. Then, taking the limit as n → ∞ in (3.21), we have

lim supn→∞∫∫R3×R3un2(x)|ψR(x)−ψR(y)|2|x−y|μdxdy≤Ck3+Ck2(1−s)R2s∫BkR∖B2Ru2(x)dx+CR2s∫B2R∖BεRu2(x)dx+C((1−ϵ)R)2s∫BϵRu2(x)dx≤Ck3+Ck2(∫BkR∖B2R|u|2s∗(x)dx)22s∗+C(∫B2R∖BεR|u|2s∗(x)dx)22s∗+C(ε1−ε)2s(∫BεR|u(x)|2s∗)22s∗,

where in the last passage we use Hölder inequality.

Since u∈L2s∗(R3),k>4andε∈(0,12), we obtain

lim supR→∞∫BkR∖B2R|u(x)|2s∗dx=lim supR→∞∫B2R∖BεR|u(x)|2s∗dx=0.

Choosing ε=1k, we get

lim supR→∞lim supn→∞∫∫R3×R3un2(x)|ψR(x)−ψR(y)|2|x−y|μdxdy≤limk→∞lim supR→∞(Ck3+Ck2(∫BkR∖B2R|u(x)|2s∗dx)22s∗+C(∫B2R∖BRk|u(x)|2s∗dx)22s∗+C(1k−1)2s(∫BRk|u(x)|2s∗dx)22s∗)≤limk→∞Ck3+C(1k−1)2s(∫BRk|u(x)|2s∗dx)22s∗=0,

which complete our proof.

Lemma 3.7

Under the conditions of Lemma 3.6, the functional J_ε satisfies the (PS)_c condition for all c ∈ [cε,2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1).

Proof. Since {un} is bounded in H_ε, we may assume

un⇀uinHε,un→uinLlocr(R3),2≤r<2s∗,un(x)→u(x)a.e.inR3.

Let us prove that un→uinHε asn→∞. Setting ωn=∥un−u∥ε2, we have

(3.22) ωn=〈Jε′(un),un〉−〈Jε′(un),u〉+12μ,s∗∫R3(1|x|μ∗H(εx,un))h(εx,un)(un−u)dx+on(1).

Note that 〈Jε′(un),un〉=〈Jε′(un),u〉=on(1) so we only need to show that

(3.23) ∫R3(1|x|μ∗H(εx,un))h(εx,un)(un−u)dx=on(1).

Similar the proof in Lemma 2.7, we can see that

(3.24) 1|x|μH(εx,un)⇀1|x|μH(εx,u)inL6μ(R3).

By using Lemma 2.1, we have

(3.25) limn→∞∫BR1|x|μH(εx,un)h(εx,un)(un−u)dx=0.

By Lemma 2.1 and 3.6, for any η > 0 there exists R=R(η)>0 such that

lim supn→∞∫R3∖BR|1|x|μH(εx,un)h(εx,un)un|dx≤C1η

and

lim supn→∞∫R3∖BR|1|x|μH(εx,un)h(εx,un)u|dx≤C2η

Taking into account the above limits we can deduce that

limn→∞∫R3(1|x|μ∗H(εx,un))h(εx,un)(un−u)dx=0.

Lemma 3.8

The functional Ψε verifies the (PS)_c condition in Sε+forallc∈[cε,2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1).

Proof. Let {un}⊂Sε+ be a (PS)_c sequence for Ψε. Thus, Ψε(un)→cand∥Ψε′∥∗→0, where ∥⋅∥∗ is the norm in the dual space (TunSε+)′. It follows from Lemma 3.4-(B₃) that {mε(un)}isa(PS)c sequence for I_ε in H_ε. From Lemma 3.7 we see that there is a u∈Sε+ such that mε(un)→mε(u) in H_ε. From Lemma 3.3-(A₃), it follows that un→uinSε+.

4 The autonomous problem

Since we are interestd in giving a multiplicity result for the modified problem, we start by considering the limit problem associated to (1.8), namely, the problem

(4.1) (−Δ)su+V0u=(∫R3|u(y)|2μ,s∗+F(u(y))|x−y|μdy)(|u|2μ,s∗−2u+12μ,s∗f(u))inR3.

Set G(u)=|u|2μ,s∗+F(u),g(u)=dG(u)du, then the equation (4.1) changes into

(4.2) (−Δ)su+V0u=12μ,s∗(∫R3G(u(y))|x−y|μdy)g(u)inR3.

which has the following associated functional

I0(u)=12∫R3(|(−Δ)s2u|2+V0u2)dx−122μ,s∗∫R3∫R3G(u(y))G(u(x))|x−y|μdydx.

The functional I₀ is well defined on the Hilbert space H₀ = H^s(ℝ³) with the inner product

(u,v)0=∫R3(−Δ)s2u(−Δ)s2vdx+∫R3V0uvdx,

and the norm

∥u∥02=∫R3|(−Δ)s2u|2dx+∫R3V0u2dx.

We denote the Nehari manifold associated to I₀ by

N0={u∈H0∖{0}:〈I0′(u),u〉=0},

and by H0+ the open subset of H₀ given by

H0+={u∈H0:|supp(u+)|>0},

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

As in section 3, S0+ a incomplete C1,1. manifold of codimension 1, modeled on H₀ and contained in the open H0+. Thus, H0=TuS0+⊕Ru for each u∈S0+, where TuS0+={v∈H0:(u,v)0=0}.

Next we have the following Lemmas and the proofs follow from a similar argument used in the proofs of Lemma 3.3 and Lemma 3.4.

Lemma 4.1

Let V₀ be given in (V₁) and the functional f satisfies (f₁)−(f₃). Then the following properties hold:

(a₁)For each u∈H0+, let ϕu:R+→R be given by ϕu(τ)=I0(τu). Then there exists a unique τ_u > 0 such that ϕu′(τ)>0in(0,τu)andϕu′(τ)<0in(τu,∞).

(a₂)There is a σ > 0 independent on u such that τ_u > σ for all u∈S0+. Moreover, for each compact setW⊂S0+ there is C_W > 0 such that τu≤CW for all u ∈ W.

(a₃) The map m^:H0+→N0 given by m^(u)=τuu is continuous and m:=m^|S0+ is a homeomorphism between S0+ and N₀. Moreover, m−1(u)=u∥u∥0.

We define the applications

Ψ^0:H0+→RandΨ0:S0+→R,

byΨ^0(u)=I0(m^(u))andΨ0:=Ψ^0|S0+.

Lemma 4.2

Let V₀ be given in (V₁) and (f₁) − (f₃) are satisfied. Then: (b1)Ψ^0∈C1(H0+,R) and

Ψ^0′(u)v=∥m^(u)∥0∥u∥0I0′(m^(u))v,∀u∈H0+and∀v∈H0.

(b2)Ψ0∈C1(S0+,R) and

Ψ0′(u)v=∥m(u)∥0I0′(m(u))v,∀v∈TuS0+.

(b₃)If {u_n} is a (PS)_c sequence of Ψ0, then {m(u_n)} is a (PS)_c sequence of I₀. If fu_ng ⊂ N₀ is a bounded (PS)_c sequence for I₀, then {m⁻¹(u_n)} is a (PS)_c sequence of Ψ0.

(b₄)u is a critical point of Ψ0 if and only if, m(u) is a critical point of I₀. Moreover, corresponding critical values coincide and

infS0+Ψ0=infN0I0.

As in the previous section, we have the following variational characterization of the infimum of I₀ over N₀:

cV0=infu∈N0I0(u)=infu∈H0+maxτ>0I0(τu)=infu∈S0+maxτ>0I0(τu)>0.

The next Lemma allows us to assume that the weak limit of a (PS)_c sequence is non-trivial.

Lemma 4.3

Let {u_n} ⊂ H₀ be a (PS)_c sequence with c∈[cV0,2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1) for I₀. Then, only one of the following conclusions holds.

(i) u_n → 0 in H₀, or

(ii) There exist a sequence {y_n} ⊂ ℝ³ and constants R, β > 0 such that

lim infn→+∞∫Br(yn)un2dx≥β>0.

Proof. Suppose (ii) does not occur. Then, for any R > 0, we have

limn→+∞supy∈R3∫Br(y)un2dx=0.

Similarly to Lemma 3.5, we have {u_n} is bounded in H₀, then by Lemma 2.2, we have

un→0 in Lr(R3) for r∈(2,2s∗).

Thus, by (f₁) we have

∫R3∫R3G(u(y))g(u(x))u(x)|x−y|μdydx=on(1).

Recalling that I0′(un)un→0, we get

∥un∥02=on(1).

Therefore the conclusion follows.

From Lemma 4.3 we can see that, if u is the weak limit of a (PS)cV0 sequence {u_n} for the functional I₀, then we can assume u≠0. Otherwise we would have u_n ⇀ 0 and once it doesn’t occur u_n → 0, we conclude from Lemma 4.3 that there exist {y_n} ⊂ ℝ³ and R, β > 0 such that

lim infn→+∞∫Br(yn)un2dx≥β>0.

Then set vn(x)=un(x+yn), making a change of variable, we can prove that {v_n} ia also a (PS)cV0 sequence for the functional I₀, it is bounded in H₀ and there is v ∈ H₀ such that v_n → v in H₀ with v≠0.

Next we devote to estimating the least energy cV0. Recall that the best Sobolev constant S_s of the embedding Ds,2(R3)↪L2s∗(R3) is defined by

Ss:=infu∈Ds,2(R3)∥u∥Ds,2(R3)2|u|2s∗2

In particular, we consider the following family of functions U_ε defined as

Uε(x)=ε2s−32u¯(xεSs12s),u¯=un(x)|u0|2s∗,

for ε > 0 and x ∈ ℝ³, the minimizer of S_s (see,[41]), which satisfies

(−Δ)su=|u|2s∗−2u,x∈R3.

Then, by a simple calculation, we know

U~(x)=Ss(3−μ)(2s−3)4(2+2s−μ)C(μ)2s−32(3+2s−μ)Uε(x)

is the unique minimizer for S_H_,L that satisfies

(−Δ)su=(∫R3|u(y)|2μ,s∗|x−y|μ)|u|2μ,s∗u,inR3.

Moreover,

∫R3|(−Δ)s2U~|2dx=∫R3∫R3|U~(x)|2μ,s∗|U~(y)|2μ,s∗|x−y|μdxdy=SH,L6−μ3−μ+2s.

Let φ∈C0∞(R3, [0, 1]) and small δ > 0 be such that φ ≡ 1 in B(0) and ϕ≡0 in R3∖B2δ(0). For any ε > 0, define the best function by uε=φUε.

Similar to [22, Lemma 1.2], we can easily draw the following conclusion.

Lemma 4.4

The constant S_H_,L defined in (2.2) is achieved if and only if

u=C(bb2+|x−a|2)3−2s2,

where C > 0 is a fixed constant, a ∈ ℝ³ and b > 0 are parameters. Furthermore,

SH,L=SsC(μ)12μ,s∗.

Proof

We sketch the proof for the completeness of this paper. By the Hardy-Littlewood-Sobolev inequality,we have

SH,L≥1C(μ)12μ,s∗∫u∈Ds,2(R3)∖{0}∫R3|(−Δ)s2u|2dx|u|2s∗2=SsC(μ)12μ,s∗.

On the other hand, the equality in the Hardy-Littlewood-Sobolev (1.6) holds if and only if

f(x)=h(x)=C(bb2+|x−a|2)6−μ2,

where C > 0 is a fixed constant, a∈R3andb>0 are parameters. Thus

(∫R3∫R3|u(x)|2μ,s∗|u(y)|2μ,s∗|x−y|μdxdy)12μ,s∗=C(μ)12μ,s∗|u|2s∗2,

if and only if

u=C(bb2+|x−a|2)3−2s2.

Then, by the definition of S_H_,L, we get

(4.3) SH,L≤∫R3|(−Δ)s2u|dx(∫R3∫R3|u(y)|2μ,s∗|u(x)|2μ,s∗|x−y|μdydx)12μ,s∗=1C(μ)12μ,s∗∫R3|(−Δ)s2u|2dx|u|2s∗2

an thus we have

SH,L≤SsC(μ)12μ,s∗.

From the arguments above, we know that S_H_,L is achieved if and only if u=C(bb2+|x−a|2)3−2s2 and

SH,L=SsC(μ)12μ,s∗.

Next. repeat the proof in [22, Lemma 1.3],we have the following important information about the best constant S_H_,L.

Lemma 4.5

For every open subset Ω⊂R3 , we have

(4.4) SH,L(Ω):=infu∈D0s,2(Ω)∖{0}∫Ω|(−Δ)s2u|2dx(∫Ω∫Ω|u(y)|2μ,s∗|u(x)|2μ,s∗|x−y|μdydx)12μ,s∗=SH,L

where SH,L(Ω) is never achieved except when Ω=R3.

Proof

It is clear that SH,L≤SH,L(Ω) by D0s,2(Ω)⊂Ds,2(R3). Let {un}⊂C0∞(R3) be a minimizing sequence for S_H_,L. We make translations and dilations for {u_n} by choosing y_n ∈ ℝ³ and τ_n > 0 such that

vn:=τn3−2s2un(τnx+yn)∈C0∞(Ω),

which satisfies

∫R3|(−Δ)s2vn|2dx=∫R3|(−Δ)s2un|2dx

and

∫Ω∫Ω|vn(y)|2μ,s∗|vn(x)|2μ,s∗|x−y|μdxdy=∫R3∫R3|un(y)|2μ,s∗|un(x)|2μ,s∗|x−y|μdxdy.

Hence SH,L(Ω)≤SH,L. Moreover, since U~(x) is the only class of functions such that the equality holds in the Hardy-Littlewood-Sobolev inequality, we know that SH,L(Ω) is never achieved except for Ω=R3.

Lemma 4.6

(4.5) ∫R3∫R3|uε(x)−uε(y)|2|x−y|3+2sdxdy≤Ss32s+O(ε3−2s),

(4.6) |uε|22=Cε2s+O(ε3−2s)if4s<3,Cε2slog(1ε)+O(ε2s)if4s=3,Cε3−2s+O(ε2s)if4s>3,

(4.7) ∫R3∫R3|uε(x)|2μ,s∗|uε(y)|2μ,s∗|x−y|μdxdy≥C(μ)32sSH,L6−μ2s−O(ε6−μ2),

In addition, if q<2μ,s∗, then there holds

(4.8) ∫Bδ∫Bδ|Uε(x)|q|Uε(y)|q|x−y|μdxdy=O(ε6−μ−q(3−2s)).

Proof

For the proof of (4.5) and (4.6), we can see that in [41]. So we only need to estimate (4.7) and (4.8). Concerning (4.7), similar to [2, Lemma 7.1], we have

(4.9) ∫R3∫R3|uε(x)|2μ,s∗|uε(y)|2μ,s∗|x−y|μdxdy≥∫Bδ∫Bδ|uε(x)|2μ,s∗|uε(y)|2μ,s∗|x−y|μdxdy=∫R3∫R3|Uε(x)|2μ,s∗|Uε(y)|2μ,s∗|x−y|μdxdy−2∫R3∖Bδ∫Bδ|Uε(x)|2μ,s∗|Uε(y)|2μ,s∗|x−y|μdxdy−∫R3∖Bδ∫R3∖Bδ|Uε(x)|2μ,s∗|Uε(y)|2μ,s∗|x−y|μdxdy:=C(μ)32sSH,L6−μ2s−2A−B,

where

A=∫R3∖Bδ∫Bδ|Uε(x)|2μ,s∗|Uε(y)|2μ,s∗|x−y|μdxdy,B=∫R3∖Bδ∫R3∖Bδ|Uε(x)|2μ,s∗|Uε(y)|2μ,s∗|x−y|μdxdy.

By direct computation, we have

(4.10) A=∫R3∖Bδ∫Bδ|Uε(x)|2μ,s∗|Uε(y)|2μ,s∗|x−y|μdxdy,=Cε6−μ∫R3∖Bδ∫Bδ1(ε2b2+x2Ss1s)6−μ21|x−y|μ1(ε2b2+y2Ss1s)6−μ2dxdy≤Cε6−μ(∫R3∖Bδ1(ε2b2+x2Ss1s)3dx)6−μ6(∫Bδ1(ε2b2+y2Ss1s)3dy)6−μ6≤Cε6−μ(∫R3∖Bδ1|x|6Ss3sdx)6−μ6(∫0Δr2(ε2b2+r2Ss1s)3dr)6−μ6=O(ε6−μ2)

and

(4.11) B=∫R3∖Bδ∫R3∖Bδ|Uε(x)|2μ,s∗|Uε(y)|2μ,s∗|x−y|μdxdy≤Cε6−μ∫R3∖Bδ∫R3∖Bδ1(ε2b2+x2Ss1s)6−μ21|x−y|μ1(ε2b2+y2Ss1s)6−μ2dxdy≤Cε6−μ(∫R3∖Bδ1(ε2b2+x2Ss1s)3dx)6−μ6(∫R3∖Bδ1(ε2b2+y2Ss1s)3dy)6−μ6≤Cε6−μ(∫R3∖Bδ1|x|6Ss3sdx)6−μ6(∫R3∖Bδ1|y|6Ss3sdy)6−μ6=O(ε6−μ)

It follows from (4.9) to (4.11) that

∫R3∫R3|uε(x)|2μ,s∗|uε(y)|2μ,s∗|x−y|μdxdy≥C(μ)32sSH,L6−μ2s−O(ε6−μ2)−O(ε6−μ)=C(μ)32sSH,L6−μ2s−O(ε6−μ2).

Then (4.7) follows. Now we prove (4.8). If q<2μ,s∗, we have

∫Bδ∫Bδ|Uε(x)|q|Uε(y)|q|x−y|μdxdy=O(ε6−μ−q(3−2s))=Cεq(3−2s)∫Bδ∫Bδ1(ε2b2+x2Ss1s)3−2s2q1|x−y|μ1(ε2b2+y2Ss1s)3−2s2qdxdy≤Cεq(3−2s)(∫Bδ1(ε2b2+x2Ss1s)3−2s2q66−μdx)6−μ6(1(ε2b2+y2Ss1s)3−2s2q66−μdy)6−μ6≤Cεq(3−2s)(∫0εr2(ε2b2+r2Ss3−2s2q66−μdr)6−μ3=O(ε6−μ−q(3−2s)

Lemma 4.7

Suppose that (f₁) − (f₃) hold. Then the number cV0 satisfies that

0<cV0<2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1.

Proof

By the definition of cV0 it suffices to prove that there exists v∈N0 such that

(4.12) I0(v)<2μ,s∗−122μ,s∗SH,L2μ,s∗2μ,s∗−1.

By Lemma 4.1, there exists τε>0 such that τεuε∈N0. We claim that for ε > 0 small enough, there exist A₁ and A₂ independent of ε such that

(4.13) 0<A1≤τε≤A2<∞.

Indeed, note that N0 is bounded away from 0, we have that τε≥A1>0 using (4.5) and (4.6). Moreover, since 〈I0′(τεuε),τεuε〉=0 from (4.7) we have that

τε22μ,s∗≤C(τε2∥uε∥02+τε2q1∥uε∥02q1+τε2q2∥uε∥02q2+τεq1+2μ,s∗∥uε∥0q1+2μ,s∗+τεq2+2μ,s∗∥uε∥0q2+2μ,s∗).

Using (4.5) and (4.6) again, there exists A2>0 such that τε≤A2. Then (4.13) holds true.

Now we estimate I0(τεuε). Note that

(4.14) I0(τεuε)≤(τε22∫R3|(−Δ)s2uε(x)|2dx−τε22μ,s∗22μ,s∗∫R3∫R3|uε(y)|2μ,s∗|uε(x)|2μ,s∗|x−y|μdydx)+(τε22∫R3V0uε2dx−122μ,s∗∫R3∫R3F(τεuε(y))F(τεuε(x))|x−y|μdydx):=I1+I2.

For I₁, we set

A:=∫R3|(−Δ)s2uε(x)|2dx,B=∫R3∫R3|uε(y)|2μ,s∗|uε(x)|2μ,s∗|x−y|μdydx,

and consider the function θ:[0,∞)→R defined by

θ(τ)=12Aτ2−τε22μ,s∗22μ,s∗B,

we have that τ0=(AB)122μ,s∗−2 is a maximum point of θ and

θ(τ0)=2μ,s∗−122μ,s∗A2μ,s∗2μ,s∗−1B11−2μ,s∗.

Combining with Lemma 4.3, (4.5) and (4.7) we have

(4.15) I1≤2μ,s∗−122μ,s∗SH,L2μ,s∗−12μ,s∗+O(ε3−2s)+O(ε6−μ2).

For I₂, given A₀ > 0, we invoke (f₃) to obtain R=R(A0)>0 such that, for x∈R3,t≥R,

(4.16) F(x,t)≥A0t2μ,s∗−1if3<4s,A0t2μ,s∗−2s3−2s(logt)12if3=4s,A0t2μ,s∗−2s3−2sif3>4s.

By (4.6) and (4.16), we need to estimate I₂ in three cases. Since the argument is similar, we only consider the case that 3 < 4s. For |x|<ε<δ, noting that φ≡1 inBδ(0), by the definition of uε and (4.13), we get a constant β > 0 such that

τεuε(x)≥A1Uε(x)≥βε2s−32.

Then we can choose ε₁ > 0 such that τεuε≥R,for|x|<ε,0<ε<ε1. It follows from (4.16) that

F(x,τεuε(x))≥A0τε2μ,s∗−1uε2μ,s∗−1,

for |x|<ε,0<ε<ε1. Then for any 0<ε<ε1, by (4.8) we get

∫Bε∫BεF(τεuε(x))F(τεuε(y))|x−y|μdydx≥A02∫Bε∫Bε|τεuε(y)|2μ,s∗−1|τεuε(x)|2μ,s∗−1|x−y|μdydx≥CA02∫Bε∫Bε|Uε(y)|2μ,s∗−1|Uε(x)|2μ,s∗−1|x−y|μdydx=A02O(ε3−2s).

Note that F(u)≥0, (4.6) and (4.13), we have

(4.17) I2≤C|uε|22−122μ,s∗∫Bε∫BεF(τεuε(x))F(τεuε(y))|x−y|μdydx≤(C−C1A02)ε3−2s.

Inserting (4.15) and (4.17) into (4.14), we get

(4.18) I0(τεuε)≤2μ,s∗−122μ,s∗SH,L2μ,s∗−12μ,s∗+(C2+C−C1A02)ε3−2s+O(ε6−μ2).

Observe that 6−μ2>3−2s for 3 < 4s, and A₀ > 0 is arbitrary, we choose large enough A₀ such that C + C₂ − C1A02<0 .Then for small ε > 0 we have v:=τεuε satisfies (4.12).

Theorem 4.1

Assume that (f₁)−(f₃) hold. Then autonomous problem (4.1) has a positive ground state solution u with I0(u)=cV0.

Proof

By Lemma 3.2 with V(x) = V₀ and the Mountain Pass Theorem without (PS) condition (cf. [50]), there exists a (PS)cV0. -sequence {un}⊂H0 of I₀ with

cV0<2μ,s∗−122μ,s∗SH,L2μ,s∗−12μ,s∗.

By Lemma 3.5 and 4.3, {u_n} is bounded in H^s(ℝ³) and non-vanishing, namely there exist r,δ>0 and a sequence {yn}⊂R3 such that

lim infn→∞∫Br(yn)|un|2dx≥δ.

Up to s subsequence, there exists u ∈ H^s(ℝ³) such that

un⇀uinHs(R3),un→uinLlocr(R3),2≤r<2s∗,un(x)→u(x)a.e.inR3.

As Lemma 2.5, we have I0′(u)=0. Since I0andI0′ are both invariant by translation, without lost of generality, we can assume that {y_n} is bounded. Note that un→uinLloc2(R3). Then u≢0.Sou∈N0. Then

cV0≤I0(u)=I0(u)−12〈I0′(u),u〉=122μ,s∗∫R3∫R3G(u(y))|x−y|μ(g(u(x))u(x)−G(u(x)))dydx≤122μ,s∗lim infn→∞∫R3∫R3G(un(y))|x−y|μ(g(un(x))un(x)−G(un(x)))dydx=lim infn→∞(I0(un)−12〈I0′(un),un〉)=cV0,

where we used Fatou Lemma and Lemma 2.4. Therefore, I0(u)=cV0, which means that u is a ground state solution for (4.1).

Next we prove that the solution u is positive, using u−=max{−u,0} as a test function in (4.1) we obtain

(4.19) ∫R3(−Δ)s2u(−Δ)s2u−dx+∫R3V0|u−|2dx=0.

On the other hand,

∫R3(−Δ)s2u(−Δ)s2u−dx=12C(s)∬R3×R3(u(x)−u(y))(u−(x)−u−(y))|x−y|3+2sdxdy≥12C(s)∫{u>0}×{u<0}(u(x)−u(y))(−u−(y))|x−y|3+2sdxdy+12C(s)∫{u<0}×{u<0}(u−(x)−u−(y))2|x−y|3+2sdxdy+12C(s)∫{u<0}×{u>0}(u(x)−u(y))u−(x)|x−y|3+2sdxdy≥0.

Thus, it follows from (4.19) that u−=0andu≥0. Rewriting the equation (4.1) in the form of

(−Δ)su+V0u=(∫R3H(u(y))u(y)|x−y|μdy)K(u)inR3,

where

H(u):=|u|2μ,s∗+F(u)u,K(u):=|u|2μ,s∗−2−μ+12μ,s∗f(u)∈L23−μ(R3)+L63+2s−μ(R3).

By Lemma 2.8, we know u∈Lp(R3) for all p∈[2,18(3−μ)(3−2s)). Using the growth assumption (f₁) and the higher integrability of u, for some C > 0 we have

(4.20) |∫R3G(u(y))|x−y|μ|∞≤C||u|2μ,s∗+|u|q1+|u|q2|33−μ≤C(|u|3(6−μ)3−μ2μ,s∗+|u|3q13−μq1+|u|3213−μq2),

which is finite since the various exponents live within the range [2,18(3−μ)(3−2s)). Thus,

(−Δ)su+V0u≤C(|u|2μ,s∗−2−μu+12μ,s∗f(u))inR3.

By the Moser iteration, similar arguments developed in Lemma 6.1 below, we can get u∈L∞(R3) and lim|x|→+∞u(x)=0 uniformly in n. Then, by regularity theory [43], there exists α∈(0,1) such that u∈Cloc0,α(R3). Therefore, if u(x0)=0 for some x0∈R3, we have that (−Δ)su(x0)=0 and by [19, Lemma 3.2], we have

(−Δ)su(x0)=−C(s)2∫R3u(x0+y)+u(x0−y)−2u(x0)|y|3+2sdy,

therefore,

∫R3u(x0+y)+u(x0−y)|y|3+2sdy=0,

yielding u≡0, a contradiction. Therefore, u is a positive solution of the equation (4.1) and the proof is completed.

The next result is a compactness result on autonomous problem which we will use later.

Lemma 4.8

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

Proof

Since {un}⊂N0, it follows from Lemma 4.1-(a₃), Lemma 4.2-(b₄) and the definition of cV0 that

vn=m−1(un)=un∥un∥0∈S0+,∀n∈N,

and

Ψ0(vn)=I0(un)→cV0=infS0+Ψ0.

Although S0+ is incomplete, due to Lemma 4.1, we can still apply the Ekeland’s variational principle [21] to the functional Θ0:H→R∪{∞}, defined by Θ0(u)=Ψ0(u) if u∈S0+andΘ0(u)=∞ifu∈∂S0+, where H=S0+¯is the complete metric space equipped with the metric d(u,v):=∥u−v∥0. In fact, take ε=1k2 in Theorem 1.1 of [21], we have a subsequence {vnk}⊂{vn} such that

cV0≤Ψ(vnk)≤cV0+1k2.

From Theorem 1.1 in [21], for λ=1k, there exist a sequence {v~k}⊂S0+ such that

Θ0(v~k)≤Θ0(vnk)<cV0+1k2

and

∥vnk−v~k∥0≤1k.

In particular, for any u∈S0+ we have

Ψ0(u)>Ψ0(v~k)−1k∥u−v~k∥0.

Hence, similar the proof for Theorem 3.1 in [21], we have that there exists λk∈R such that

∥Ψ^0′(v~k)−λkg0′(v~k)∥0≤1k,

where g0(u)=∥u∥02−1. Which means that

λk=1∥g0′(v~k)∥02〈Ψ^0′(v~k),g0′(v~k)〉+ok(1),g0′(v~k)=v~k.

From Lemma 4.2-(b₁),

λk=〈Ψ^0′(v~k),v~k〉+ok(1)=τv~k〈I0′(tv~kv~k),v~k〉+ok(1)=ok(1).

Therefore, we can conclude there is a sequence {v~n}⊂S0+ such that {v~n}isa(PS)cV0 sequence for Ψ0onS0+ and

∥un−v~n∥0=on(1).

Now the remainder of the proof follows from Lemma 4.2, Theorem 4.1 and arguing as in the proof of Lemma 3.8.

5 Solutions for the penalized problem

In this section, we shall prove the existence and multiplicity of solutions. We begin showing the existence of the positive ground-state solution for the penalized problem (3.1).

Theorem 5.1

Suppose that the nonlinearity f satisfies (f₁) − (f₃) and that the potential function V satisfies assumptions (V₁) − (V₂). Then, for any ε > 0, problem (3.1) has a positive ground-state solution u_ε.

Proof

Similar to Lemma 3.2, we can prove that J_ε also satisfies the Mountain Pass geometry. Let

cε:=infu∈Hε∖{0}maxτ≥0Jε(τu)=infu∈NεJε(u).

Then, we know that there exists a (PS) sequence at c_ε, i.e.

Jε′(un)→0andJε(un)→cε.

Therefore, by Lemma 3.7, the existence of ground state solution u_ε is guaranteed. Moreover, similarly to the proof in Theorem 4.1, we know that uε(x)>0 in ℝ³.

Next, we will relate the number of positive solutions of (3.1) to the topology of the set M. For this, we consider δ > 0 such that Mδ⊂Ω and by Theorem 4.1, we can choose w∈N0 with I0(w)=cV0. Let η be a smooth nonincreasing cut-off function defined in [0,+∞) such that η(t)=1if0≤t≤δ2andη(t)=0ift≥δ. For each y ∈ 𝓜, let

Ψε,y(x)=η(|εx−y|)w(εx−yε).

Then for small ε > 0, one has Ψε,y∈Hε∖{0}forally∈M. In fact, using the change of variable z=x−yε, one has

∫R3V(εx)Ψε,y2(x)dx=∫R3V(εx)η2(|εx−y|)w2(εx−yε)dx=∫R3V(εz+y)η2(|εz|)w2(z)dz≤C∫R3w2(z)dz<+∞.

Moreover, using the change of variable x′=x−yε,z′=z−yε we have

|(−Δ)s2Ψε,y|22=12C(s)∬R3×R3|η(|εx−y|)w(εx−yε)−η(|εz−y|)w(εz−yε)|2|x−z|3+2sdxdz=12C(s)∬R3×R3|η(|εx′|)w(x′)−η(|εz′|)w(z′)|2|x′−z′|3+2sdx′dz′=|(−Δ)s2η(|εx|)w(x)|22=|(−Δ)s2ηεw|22,

where ηε(x)=η(|εx|). By Lemma 2.3, we see that ηεw∈Ds,2(R3)asε→0 and hence Ψε,y∈Ds,2(R3) for ε > 0 small. Hence Ψε,y∈Hε. Now we proof Ψε,y≠0. In fact,

∫R3Ψε,y2(x)dx=∫R3η2(|εx−y|)w2(εx−yε)dx=∫|εx−y|<δη2(|εx−y|)w2(εx−yε)dx≥∫|z|≤δ2εη2(|εz|)w2(z)dz≥∫B0(δ2ε)w2(z)dz→∫R3w2(z)dz>0

as ε → 0. Then Ψε,y≠0 for small ε > 0. Therefore, there exists unique τε>0 such that

maxτ≥0Iε(τΨε,y)=Iε(τεΨε,y)andτεΨε,y∈Nε.

We introduce the map Φε:M→Nε by setting

Φε(y)=τεΨε,y.

By construction, Φε(y) has a compact support for any y∈MandΦε is a continuous map.

Lemma 5.1

The functional Φε(y) has the following property:

limε→0Jε(Φε(y))=cV0uniformlyiny∈M.

Proof

Suppose that the result is false. Then, there exist some δ0>0,{yn}⊂M and εn→0 such that

(5.1) |Jεn(Φεn(yn))−cV0|≥δ0.

By the definition of τεn we have

(5.2) 0<τεn2∫R3|(−Δ)s2Ψεn,yn|2dx+τεn2∫R3V(εnx)Ψεn,yn2dx=12μ,s∗∫R3∫R3H(εnx,τεnΨεn,yn)h(εnx,τεnΨεn,yn)τεnΨεn,yn|x−y|μdydx

It follows from (5.2) that τεn/→0, then τεn≥τ0>0 for some τ₀ > 0. If τεn→+∞, by (f₂) and the boundedness of Ψεn,yn, we get

(5.3) (∫R3|(−Δ)s2Ψεn,yn|2dx+∫R3V(εnx)Ψεn,yn2dx)=12μ,s∗∫R3∫R3H(εnx,τεnΨεn,yn)h(εnx,τεnΨεn,yn)Ψεn,yn|x−y|μτεndydx→+∞

as n → +∞. But the left side of the above inequality is boundedness, which is impossible. Hence, 0<τ0≤ τεn≤C. Without loss of generality, we may assume that τεn→T>0.

Next we claim that T = 1. By Lemma 2.3 and Lebesgue’s theorem we have

(5.4) limn→+∞∥Ψεn,yn∥εn2=∥w∥02,limn→+∞Σε(Ψεn,yn)=Σ0(w)

Moreover, from

(5.5) τεn2∥Ψεn,yn∥εn2=12μ,s∗∫R3∫R3H(εnx,τεnΨεn,yn)h(εnx,τεnΨεn,yn)τεnΨεn,yn|x−y|μdydx

we can deduce that

(5.6) ∥w∥02=limn→∞12μ,s∗∫R3∫R3H(εnx,τεnΨεn,yn)h(εnx,τεnΨεn,yn)τεnΨεn,yn|x−y|μdydx

Taking into account that w is a ground state solution to (4.1) and using (f₂), we deduce that T = 1. It follows from (5.4), we have

(5.7) limn→+∞Jεn(Φεn(yn))=J0(w)=cV0,

which is a contradiction with (5.1). This completes the proof.

Let ρ=ρ(δ)>0 be such that Mδ⊂Bρ(0). Consider χ:R3→R3 be defined as χ(x)=xfor|x|≤ρ and χ(x)=ρx|x|for|x|≥ρ. Finally, let us consider the barycenter map βε:Nε→R3 given by

βε(u)=∫R3χ(εx)u2(x)dx∫R3u2(x)dx∈R3.

Lemma 5.2

The functional β_ε satisfies

limε→0βε(Φε(y))=yuniformlyiny∈M.

Proof

Suppose by contradiction that there exist δ0>0, {yn}⊂M and εn→0 such that

(5.8) |βεn(Φεn(yn))−yn|≥δ0.

Using the change of variables z=εnx−ynεn and the definition of β_ε, we have

βεn(Φεn(yn))=yn+∫R3(χ(εnz+y)−yn)|η(|εnz|)w(z)|2dx∫R3|η(|εnz|)w(z)|2dx.

Since {yn}⊂M⊂Bρ(0) andχ|Bρ≡id, we conclude that

|βεn(Φεn(yn))−yn|=on(1),

which contradicts (5.8) and the desired conclusion holds.

Lemma 5.3

Let εn→0and{un}⊂Nεn be such that Jεn(un)→cV0. Then, there exists a sequence {y~n}⊂R3 such that vn(x)=un(x+y~n) has a convergent subsequence in H₀. Moreover, passing to a subsequence, yn:= εny~n→y0∈M.

Proof

By Lemma 3.5, {u_n} is bounded in H₀. Note that cV0>0, and since ∥un∥εn→0 would imply Jεn(un)→ 0, we can argue as in the proof of Lemma 4.3 to obtain a sequence {y~n}⊂R3 and constants R, β > 0 such that

lim infn→+∞∫BR(y~n)un2dx≥β>0.

Define vn(x):=un(x+y~n), then {vn} is also bounded in H₀ and up to a subsequence, we have

vn⇀v≠0 in H0.

Let τn>0 be such that v~n:=τnvn∈N0andsetyn=εny~n.By{un}⊂Nεn, we have

cV0≤J0(v~n)=J0(τnun)≤Jεn(τnun)≤Jεn(un)=cV0+on(1).

Which implies that limn→+∞J0(v~n)=cV0. In virtue of v~n∈N0, we obtain {v~n} is bounded in H₀. It follows from the boundedness of {v_n} in H₀ that {τ_n} is bounded, without loss of generality, we may assume that τn→τ0≥0. If τ₀ = 0, in view of the boundedness of {v_n} in H₀, we have v~n=τnvn→0inH0. Hence J0(v~n)→0, which contradicts cV0>0. Thus, τ₀ > 0 and the weak limit of {v~n} is different from zero. Hence, up to a subsequence, we have v~n⇀τ0v:=v~≠0 in H₀ by the uniqueness of the weak limit. From Lemma 4.8, we know that v~n→v~ in H₀. Moreover, v~∈N0.

Now, we will show that {y_n} is bounded in ℝ³. Suppose that after passing to a subsequence, |yn|→+∞. Choosing R > 0 such that Ω⊂BR(0). Without loss of generality we may assume that |y_n| > 2R. Then, for all z∈BR/εn(o),

(5.9) |εnz+yn|≥|yn|−|εnz|>R.

By the change of variable x↦z+tildeyn, using the fact that V0≤V(εx) and (5.9), we have

(5.10) ∥vn∥02≤C∫R3H(εz+yn,vn)vndx≤C∫BR/εnG~(vn)vndz+C∫R3∖BR/εnG(vn)vndz.

Since G~(u)≤V0Kuandvn→v in H₀, we can see that (5.10) implies that

∥vn∥02=on(1),

that is vn→0 in H₀, which is a contradiction. Therefore, up to s subsequence, we may assume that y_n → y0∈R3. It remains to check that y0∈M. Clearly, if y0∉=Ω, then we can argue as before and we deduce that un→0 in H₀, which is impossible. Hence we only need to show that V(y₀) = V₀. Arguing by contradiction again, we suppose that V(y₀) > V₀. Then, by using v~n→v~ in H₀ and Fatou’s Lemma, we have

cV0=J0(v~)<lim infn→∞(12∫R3|(−Δ)s2v~|2dx+12∫R3V(y0)v~2dx−Σ0(v~))≤lim infn→∞(12∫R3|(−Δ)s2v~n|2dx+12∫R3V(εnx+yn)v~n2dx−Σ0(v~n))≤lim infn→∞Jεn(τnun)≤lim infn→∞Jεn(un)=cV0,

which yields a contradiction. So, y0∈M and the proof is completed.

Let h:R+→R+ be any positive function satisfying h(ε)→0+asε→0+. Define the set

N~ε={u∈Nε:Jε(u)≤cV0+h(ε)}.

Given y∈M, we conclude from Lemma 5.1 that h(ε)=supy∈M|Iε(Φε(y))−cV0|→0asε→0+.Thus,Φε(y)∈N~ε and N~ε≠∅forε>0. Moreover, we have the following Lemma.

Lemma 5.4

For any δ>0, there holds that

limε→0supu∈N~εinfy∈Mδ|βε(u)−y|=0.

Proof

Let {εn}⊂R+ be such that εn→0. By definition, there exists {un}⊂N~εn such that

infy∈Mδ|βεn(un)−y|=supu∈N~εninfy∈Mδ|βεn(u)−y|+on(1).

So, it suffices to find a sequence {yn}⊂Mδ satisfying

(5.11) limn→+∞|βεn(un)−yn|=0.

Since un∈N~εn⊂Nεn, we get

cV0≤cεn≤Jεn(un)≤cV0+h(εn).

It follows that Jεn(un)→cV0. Thus, we can invoke Lemma 5.3 to obtain a sequence {y~n}⊂R3 such that yn=εny~n∈Mδ for n large enough. Then

βεn(un)=yn+∫R3(χ(εnx+yn)−yn)|un(x+y~n)|2dx∫R3|un(x+y~n)|2dx.

For ∀x∈R3 fixed, since εnx+yn→y∈Mδ, we have that the sequence {y_n} satisfies (5.11). This completes the proof.

Next we prove our multiplicity result by presenting a relation between the topology of M the number of solutions of the modified problem (3.1), we will apply the Ljusternik-Schnirelmann abstract result in [44, 46].

Theorem 5.2

Assume that conditions (V₁) − (V₂) and (f₁)− (f₃) hold. Then, given δ > 0 there is ε^δ>0 such that for any ε∈(0,ε^δ), problem (3.1) has at least catMδ(M) positive solutions.

Proof

For y∈M set γε(y)=mε−1(Φε(y)). It follows from Lemma 3.4 and Lemma 5.1 that

(5.12) limε→0Ψε(γε(y))=limε→0Iε(Φε(y))=cV0,

uniformly in y∈M. Let

S~ε+={w∈Sε+:Ψε(w)≤cV0+h(ε)},

where h is given in the definition of N~ε. From (5.12), we know that there is a number ε^ such that S~ε+≠∅ for ε∈(0,ε^).

For a fixed δ>0, by Lemmas 3.3, 5.1-5.2 and 5.4, we know that there exists a ε^=ε^δ>0 such that for any ε∈(0,ε^δ), the diagram

M⟶ΦεN~ε⟶mε−1S~ε+⟶mεN~ε⟶βεMδ

is well defined. From Lemma 5.2, there is a function λ(ε, y) with |λ(ε,y)|<δ2 uniformly in y∈M, for all ε∈(0,ε^), such that βε(Φε(y)):=y+λ(ε,y)for ally∈M. Define H(t,y)=y+(1−t)λ(ε,y). Then, H : [0,1]×M→Mδ is continuous. Obviously, H(0,y)=βε(Φε(y)),H(1,y)=yfor ally∈M. That is, H(t, y) is homotopy between βε∘Φε and the inclusion map id:M→Mδ. This fact and Lemma 4.3 in [7] implies that

catS~ε+γε(M)≥catMδ(M).

On the other hand, using the definition of N~ε and choosing ε^δ small if necessary, we see that I_ε satisfies the (PS) condition in N~ε recalling Lemma 3.7. By Lemma 3.4 and 3.8, we obtain that Ψε satisfies the (PS) condition inS~ε+. Therefore, the standard Ljusternik-Schnirelmann theory provides at least catS~ε+γε(M) critical points of Ψε restricted to S~ε+. Using Lemma 3.7 again, we infer that I_ε has at least catMδ(M) critical points. Using the same arguments contained in the proof Theorem 4.1, we see that the equation (3.1) has at least catMδ(M) positive solutions.

6 Proof of Theorem 1.1

In this section we will prove our main result. The idea is to show that the solutions obtained in Theorem 5.2 verify the following estimate uε(x)≤a,∀x∈Ωεcforε small enough. This fact implies that these solutions are in fact solutions of the original problem (2.3). The key ingredient is the following result, whose proof uses an adaptation of the arguments found in [20], which are related to the Moser iteration method [35].

Lemma 6.1

Let εn→0+andun∈N~εn be a solution of (3.1). Then up to a subsequence, vn=un(x+y~n) satisfies that vn∈L∞(R3) and there exists C > 0 such that

∥vn∥L∞(R3)≤C,∀n∈N,

where {y~n} is given in Lemma 5.3. Furthermore,

lim|x|→∞vn(x)=0uniformlyinn∈N.

Proof

Rewriting the equation (3.1) in the form of

(−Δ)su+V(εx)u=(∫R3H(u(y))u(y)|x−y|μdy)K(u)inR3,

where

H(u):=|u|2μ,s∗+F(u)u,K(u):=|u|2μ,s∗−2−μ+12μ,s∗f(u)∈L23−μ(R3)+L63+2s−μ(R3).

By Lemma 2.8, we know u∈Lp(R3)for allp∈[2,18(3−μ)(3−2s)). Using the growth assumption (f₁) and the higher integrability of u, for some C > 0 we have

|∫R3G(u(y))|x−y|μ|∞≤C||u|2μ,s∗+|u|q1+|u|q2|33−μ≤C(|u|3(6−μ)3−μ2μ,s∗+|u|3q13−μq1+|u|3213−μq2),

which is finite since the various exponents live within the range [2,18(3−μ)(3−2s)). Therefore, we have

(6.1) |h(εx,vn)|:=(∫R3H(εnx+εny~n,vn)|x−y|μdy)h(εnx+εny~n,vn)−V(εnx+εny~n)vn≤C(1+|vn|2s∗−1)

for n large enough.

Let T > 0, we define

H(t)=0,ift≤0,tβ,if 0<t<T,βTβ−1(t−T)+Tβ,ift≥T,

with β>1 to be determined later. Since H is Lipschitz with constant L0=βTβ−1, we have

[H(vn)]Ds,2=(∬R3×R3|H(vn(x))−H(vn(y))|2|x−y|3+2sdxdy)12≤(∬R3×R3L02|vn(x)−vn(y)|2|x−y|3+2sdxdy)12=L0[vn]Ds,2.

Therefore, H(vn)∈Ds,2(R3). Moreover, by the definition of H, we know that H is a convex function, then we have

(6.2) (−Δ)sH(vn)≤H′(vn)(−Δ)svn

in the weak sense. Thus, from H(vn)∈Ds,2(R3) and (6.1)-(6.2), we have

∥H(vn)∥2s∗2≤C∫R3|(−Δ)s2H(vn)|2dx=C∫R3H(vn)(−Δ)sH(vn)dx≤C∫R3H(vn)H′(vn)(−Δ)svndx=C∫R3H(vn)H′(vn)h(εnx,vn)dx≤C∫R3H(vn)H′(vn)dx+C∫R3H(vn)H′(vn)vn2s∗−1dx.

Using the fact that H(vn)H′(vn)≤βvn2β−1andvnH′(vn)≤βH(vn), we have

(6.3) (∫R3(H(vn))2s∗dx)22s∗=Cβ(∫R3vn2β−1dx+∫R3(H(vn))2vn2s∗−2dx),

where C is a positive constant that does not depend on β. Notice that the last integral is well defined for T in the definition of H. Indeed

∫R3(H(vn))2vn2s∗−2dx=∫vn≤T(H(vn))2vn2s∗−2dx+∫vn>T(H(vn))2vn2s∗−2dx≤T2β−2∫R3vn2s∗dx+C∫R3vn2s∗dx<∞.

We choose now β in (6.3) such that 2β−1=2s∗, and we name it β₁, that is

(6.4) β1:=2s∗+12.

Let R^>0 to be fixed later. Attending to the last integral in (6.3) and applying the Holder’s inequality with exponents γ:=2s∗2andγ′:=2s∗2s∗−2,

(6.5) ∫R3(H(vn))2vn2s∗−2dx=∫vn≤R^(H(vn))2vn2s∗−2dx+∫vn>R^(H(vn))2vn2s∗−2dx≤∫vn≤R^(H(vn))2vnR^2s∗−1dx+(∫R3(H(vn))2s∗dx)22s∗(∫vn>R^vn2s∗dx)2s∗−22s∗.

By Lemma 4.8, we know that {vn} has a convergent subsequence in H₀, therefore we can choose R^ large enough so that

(∫vn>R^vn2s∗dx)2s∗−22s∗≤12Cβ1,

where C is the constant appearing in (6.3). Therefore, we can absorb the last term in (6.5) by the left hand side of (6.3) to get

(∫R3(H(vn))2s∗dx)22s∗≤2Cβ1(∫R3vn2s∗dx+R^2s∗−1∫R3(H(vn))2vndx),

Now we use the fact that H(vn)≤vnβ1 and we take T→∞, we obtain

(∫R3vn2s∗β1dx)22s∗≤2Cβ1(∫R3vn2s∗dx+R^2s∗−1∫R3vn2s∗dx).

and therefore

(6.6) vn∈L2s∗β1(R3).

Let us suppose now β>β1. Thus, using that H(vn)≤vnβ in the right hand side of (6.3) and letting T→∞ we get

(6.7) (∫R3vn2s∗βdx)22s∗≤Cβ(∫R3vn2β−1dx+R^2s∗−1∫R3vn2β+2s∗−2dx).

Set c0:=2s∗(2s∗−1)2(β−1)andc1:=2β−1−c0. Notice that, since β>β1, then 0<c0<2s∗,c1>0. Hence, applying Young’s inequality with exponents γ:=2s∗/c0andγ′:=2s∗/2s∗−c0, we have

∫R3vn2β−1dx≤c02s∗∫R3vn2s∗dx+2s∗2s∗−c0∫R3vn2s∗c12s∗−c0dx≤∫R3vn2s∗dx+∫R3vn2β+2s∗−2dx≤C(1+∫R3vn2β+2s∗−2dx),

with C > 0 independent of β. Plugging into (6.7),

(∫R3vn2s∗βdx)22s∗≤Cβ(1+∫R3vn2β+2s∗−2dx),

with C changing from line to line, but remaining independent of β. Therefore

(6.8) (1+∫R3vn2s∗βdx)12s∗(β−1)≤(Cβ)12(β−1)(1+∫R3vn2β+2s∗−2dx)12(β−1).

Repeating this argument we will define a sequence βm,m≥1 such that

2βm+1+2s∗−2=2s∗βm.

Thus,

βm+1−1=(2s∗2)m(β1−1).

Replacing it in (6.8) one has

(1+∫R3vn2s∗βm+1dx)12s∗(βm+1−1)≤(Cβm+1)12(βm+1−1)(1+∫R3vn2s∗βmdx)12s∗(βm−1).

Defining Cm+1:=Cβm+1 and

Am:=(1+∫R3vn2s∗βmdx)12s∗(βm−1),

we conclude that there exists a constant C₀ > 0 independent of m, such that

Am≤∏k=1mCk12(βk−1)A1≤C0A1.

Thus,

(6.9) ∥vn∥∞≤C0A1<∞,

uniformly in n∈N, thanks to (6.6). Now argue as in the proof of [3, Lemma 2.6], we conclude that

un(x)→0 as |x|→∞,

uniformly in n ∈ ℕ. This finishes the proof of Lemma 6.1.

We are now ready to prove the main result of the paper.

Proof of Theorem 1.1. We fix a small δ > 0 such that Mδ⊂Ω. We first claim that there exists some ε~δ>0 such that for any ε∈(0,ε~δ) and any solution uε∈N~ε of the problem (3.1), there holds

(6.10) ∥uε∥L∞(R3∖Ωε)<a.

In order to prove the claim we argue by contradiction. So, suppose that for some sequence ε_n → 0⁺ we can obtain un∈N~εn such that Iεn′(un)=0 and

(6.11) ∥un∥L∞(R3∖Ωε)≥a.

As in the proof of Lemma 5.3, we have that Jεn(un)→cV0 and we can obtain a sequence {y~n}∈R3 such that εny~n→y0∈M.

If we take r>0 such that Br(y0)⊂B2r(y0)⊂Ω we have that

Brεn(y0εn)=1εnBr(y0)⊂Ωεn.

Moreover, for any z∈Brεn(y~n), there holds

|z−y0εn|≤||z−y~n|+|y~n−y0εn|<1εn(r+on(1))<2rεn,

for n large. For these values of n we have that Brεn(y~n)⊂Ωεn, that is, R3∖Ωεn⊂R3∖Brεn(y~n). On the other hand, it follows from Lemma 6.1 that there is R>0 such that

un(x)<afor|x|≥Rand∀n∈N,

from where it follows that

vn(x−y~n)<aforx∈BRc(y~n)andn∈N.

Thus, there exists n0∈N such that for any n≥n0and rεn>R, there holds

R3∖Ωεn⊂R3∖Brεn(y~n)⊂R3∖BR(y~n).

Then, there holds

un(x)<a∀x∈R3∖Ωεn,

which contradicts to (6.11) and the claim holds true.

Let ε^δ given by Theorem 5.2 and let εδ:=min{ε^δ,ε~δ}. We will prove the theorem for this choice of εδ. Let ε∈(0,εδ) be fixed. By using Theorem 5.2 we get catMδ(M) nontrivial solutions of problem (3.1). If u∈Hε is one of these solutions, we have that u∈N~ε, and we can use (6.10) and the definition of g to conclude that H(⋅,u)=G(u). Hence, u is also a solution of the problem (2.1). An easy calculation shows that ω(x)=u(xε) is a solution of the original problem (1.8). Then, (1.8) has at least catMδ(M) positive solutions.

Now we consider εn→0+ and take a sequence un∈Hεn of positive solutions of the problem (3.1) as above. In order to study the behavior of the maximum points of u_n, we first notice that, by the definition of H and (h₁), (h₂), there exists 0 < < a such that

(6.12) H(εnx,u)u≤V0Ku2,for allx∈R3,u≤γ.

Using a similar discussion above, we obtain R>0 and{y~n}⊂R3 such that

(6.13) ∥un∥L∞(BRc(y~n))<γ.

Up to a subsequence, we may assume that

(6.14) ∥un∥L∞(BR(y~n))≥γ.

Indeed, if this is not the case, we have ∥un∥L∞<γ, and therefore it follows from Jεn′(un)=0 and (6.12) that

(6.15) ∥un∥εn2≤V0K∫R3un2dx.

The above expression implies that ∥un∥εn→0asn→∞, which leads to a contradiction. Thus, (6.14) holds.

By using (6.13) and (6.14) we conclude that the maximum points p_n ∈ ℝ³ of u_n belongs to BR(y~n). Hence, pn=y~n+qn for some q_n ∈ B_R(0). Recalling that the associated solution of (1.8) is of the form ωn(x)=un(xε), we conclude that the maximum point ηε of vnisηε:=εny~n+εnqn. Since{qn}⊂BR(0) is bounded and εny~n→y0∈M, we obtain

limn→∞V(ηε)=V(y0)=V0.

Thus, the proof of Theorem 1.1 is completed.

Acknowledgment

We would like to thank the anonymous referee for his/her careful readings of our manuscript and the useful comments made for its improvement.

This work is supported by NSFC (11771385), China.

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Received: 2019-07-10

Accepted: 2020-09-06

Published Online: 2020-12-01

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

Multiplicity and concentration behaviour of solutions for a fractional Choquard equation with critical growth

Abstract

1 Introduction and the main results

Lemma 1.1

Theorem 1.1

Remark 1.1

2 Variational settings and preliminary results

2.1 The functional space setting

Lemma 2.1

Lemma 2.2

Lemma 2.3

2.2 Preliminary lemmas

Lemma 2.4

Lemma 2.5

2.3 Regularity of solutions and Pohožaev identity

Lemma 2.6

Lemma 2.7

Lemma 2.8

3 The penalized problem

Lemma 3.1

Lemma 3.2

Lemma 3.3

Lemma 3.4

Lemma 3.5

Lemma 3.6

Lemma 3.7

Lemma 3.8

4 The autonomous problem

Lemma 4.1

Lemma 4.2

Lemma 4.3

Lemma 4.4

Proof

Lemma 4.5

Proof

Lemma 4.6

Proof

Lemma 4.7

Proof

Theorem 4.1

Proof

Lemma 4.8

Proof

5 Solutions for the penalized problem

Theorem 5.1

Proof

Lemma 5.1

Proof

Lemma 5.2

Proof

Lemma 5.3

Proof

Lemma 5.4

Proof

Theorem 5.2

Proof

6 Proof of Theorem 1.1

Lemma 6.1

Proof

Acknowledgment

References

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