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Multiple positive solutions for a class of

concave-convex elliptic problems in R

N

involving sign-changing weight

Tsung-fang Wu

Department of Applied Mathematics,

National University of Kaohsiung, Kaohsiung 811, Taiwan

Abstract

In this paper, we study the multiplicity of positive solutions for the following concave-convex elliptic equation:

         −∆u + u = fλ(x) uq−1+ gµ(x) up−1 in RN, u ≥ 0 in RN, u ∈ H1 RN , where 1 < q < 2 < p < 2∗  2∗= N −22N if N ≥ 3, 2∗ = ∞ if N = 1, 2  and the parameters λ, µ ≥ 0. We assume that fλ(x) = λf+(x)+f−(x) is sign-changing and

gµ(x) = a (x) + µb (x) , where the functions f±, a and b satisfy suitable conditions.

Key words: Semilinear elliptic equations, Sign-changing weight, Multiple positive solutions

1 Introduction

In this paper, we consider the multiplicity results for positive solutions of the following concave-convex elliptic equation:

             −∆u + u = fλ(x) uq−1+ gµ(x) up−1 in RN, u ≥ 0 in RN, u ∈ H1 RN  , (Efλ,gµ)

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where 1 < q < 2 < p < 2∗ 2∗ = 2N

N −2 if N ≥ 3, 2

= ∞ if N = 1, 2

and the parameters λ, µ ≥ 0. We assume that fλ(x) = λf+(x) + f−(x) and gµ(x) = a (x) + µb (x) where the functions f±, a and b satisfy the following conditions: (D1) f ∈ Lq∗

RN



(q∗ = p−qp ) with f±(x) = ± max {±f (x) , 0} 6≡ 0 and there exists a positive number rf− such that

f−(x) ≥ −c expb 

−rf−|x|



for somec > 0 and for all x ∈ Rb

N;

(D2) a, b ∈ CRNand there are positive numbers r

a, rbwith rb < min

n

rf−, ra, q

o

such that

1 ≥ a (x) ≥ 1 − c0exp (−ra|x|) for some c0 < 1 and for all x ∈ RN and

b (x) ≥ d0exp (−rb|x|) for some d0 > 0 and for all x ∈ RN; (D3) b (x) → 0 and a (x) → 1 as |x| → ∞.

Elliptic problems in bounded domains involving concave and convex terms have been studied extensively since Ambrosetti-Brezis-Cerami [3] considered the following equation:

             −∆u = λuq−1+ up−1 in Ω, u > 0 in Ω, u ∈ H1 0(Ω) , (Eλ)

where 1 < q < 2 < p ≤ 2∗, λ > 0 and Ω is a bounded domain in RN. They found that there exists λ0 > 0 such that the equation (Eλ) admits at least two positive solutions for λ ∈ (0, λ0) , a positive solution for λ = λ0 and no positive solution exists for λ > λ0 (see also Ambrosetti-Azorezo-Peral [2] for more references therein). Actually, Adimurthi-Pacella-Yadava [5], Damascelli-Grossi-Pacella [13], Ouyang-Shi [22] and Tang [25] proved that there exists λ0 > 0 such that there have exactly two positive solutions of equation (Eλ) in the unit ball BN(0; 1) for λ ∈ (0, λ

0), exactly one positive solution for λ = λ0 and no positive solution exists for λ > λ0. Generalizations of the result of equation (Eλ) (involving sign-changing weight) were done by Brown-Wu [9,10], de Figueiredo-Gossez-Ubilla [16] and Wu [29,30]. However, little has been done for this type of problem in RN. We are only aware of the works [12,17,21,28] which studied existence of solutions for some related concave-convex elliptic problem in RN (not involving sign-changing weight). Furthermore, we do not know of any results for concave-convex elliptic problems in RN involving sign-changing weight functions. In this paper, we will study this topic. The following theorems are our main results.

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Theorem 1.1 Suppose that the functions f±, a and b satisfy the conditions (D1) − (D3) . Let Λ0 = (2 − q)2−q  p−2 kf+kLq∗ p−2  S p p−q p−q , where Sp is a best Sobolev constant for the imbedding of H1

RN  into Lp RN  . Then

(i) for each λ > 0 and µ > 0 with λp−2(1 + µ kbk)2−q <q2p−2Λ0, equation



Efλ,gµ



has at least two positive solutions;

(ii) there exist positive numbers λ0, µ0 with λp−20 (1 + µ0kbk∞) 2−q

<q2p−2Λ0 such that for λ ∈ (0, λ0) and µ ∈ (0, µ0) , equation



Efλ,gµ



has at least three positive solutions.

Note that the positive numbers λ0, µ0 are independent of f−. Therefore, if kf−kLq∗ is sufficiently small, we have the following result.

Theorem 1.2 If in addition to the condition (D1) − (D3) , we still have (D4) a (x) ≤ 1 on RN with a strict inequality on a set of positive measure; (D5) ra> 2,

then there exist positive numbers λe0 ≤ λ0, e

µ0 ≤ µ0 and ν0 such that for λ ∈

 0,λe0  , µ ∈ (0,µe0) and kf−kLq∗ < ν0, equation  Efλ,gµ 

has at least four positive solutions.

Among other interesting similar problems, Adachi-Tanaka [4] has been in-vestigated the following non-homogenous elliptic equation:

             −∆u + u = a (x) up−1 + h (x) in RN, u > 0 in RN, u ∈ H1RN, (Ea,h)

where h (x) ∈ H−1RN\ {0} is nonnegative and a (x) ∈ C RN  which sat-isfy a (x)  1 = lim |x|→∞a (x) and

a (x) ≥ 1 − c0exp (− (2 + δ) |x|) for some c0 < 1, δ > 0 and for all x ∈ RN. Using the equationEa,0



does not admit any ground state solution and Bahri-Li’s minimax argument [6], they proved that the equationEa,h



has at least four positive solutions under the assumption khkH−1 is sufficiently small.

In the following sections, we proceed to prove Theorems 1.1, 1.2. We use the variational methods to find positive solutions of equationEfλ,gµ



. Asso-ciated with the equation Efλ,gµ



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H1 RN  Jfλ,gµ(u) = 1 2kuk 2 H1 − 1 q Z RN fλ|u|qdx − 1 p Z RN gµ|u|pdx, where kukH1 =  R RN|∇u| 2

+ u2dx1/2 is the standard norm in H1 RN



. It is well known that the solutions of equation Efλ,gµ



are the critical points of the energy functional Jfλ,gµ in H

1 RN



(see Rabinowitz [23]).

This paper is organized as follows. In section 2, we give some notations and preliminaries. In section 3, we establish the existence of a local minimum for Jfλ,gµ. In section 4, we give an estimate of energy. In section 5, we discussion

some concentration behavior in the Nehari manifold. In sections 6, 7, we prove Theorems 1.1, 1.2.

2 Notations and Preliminaries

Throughout this paper, we denote by Sp the best Sobolev constant for the embedding of H1 RN  into Lp RN  is given by Sp = inf u∈H1(RN)\{0} kuk2H1 (R RN|u| p dx)2/p > 0. In particular, Z RN |u|pdx 1p ≤ S −1 2

p kukH1 for all u ∈ H

1

RN



\ {0} .

First, we define the Palais–Smale (simply by (PS)) sequences, (PS)–values, and (PS)–conditions in H1RNfor Jfλ,gµ as follows.

Definition 2.1 (i) For β ∈ R, a sequence {un} is a (PS)β–sequence in H1



RN



for Jfλ,gµ if Jfλ,gµ(un) = β + o(1) and J

0 fλ,gµ(un) = o(1) strongly in H −1 RN  as n → ∞. (ii) β ∈ R is a (PS)–value in H1RN 

for Jfλ,gµ if there exists a (PS)β–

sequence in H1 RN



for Jfλ,gµ.

(iii) Jfλ,gµ satisfies the (PS)β–condition in H

1 RN  if every (PS)β–sequence in H1 RN 

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As the energy functional Jfλ,gµ is not bounded below on H

1 RN



, it is useful to consider the functional on the Nehari manifold

Nfλ,gµ = n u ∈ H1RN\ {0} | DJf0 λ,gµ(u) , u E = 0o. Thus, u ∈ Nfλ,gµ if and only if

kuk2H1 − Z RN fλ|u|qdx − Z RN gµ|u|pdx = 0.

Note that Nfλ,gµ contains every non-zero solution of equation



Efλ,gµ



. Fur-thermore, we have the following results.

Lemma 2.2 The energy functional Jfλ,gµ is coercive and bounded below on

Nfλ,gµ.

Proof. If u ∈ Nfλ,gµ, then, by the H¨older and Sobolev inequalities,

Jfλ,gµ(u) = 1 2 − 1 p ! kuk2H1 − 1 q − 1 p ! Z RN (λf++ f−) |u|qdx ≥ 1 2 − 1 p ! kuk2H1 − 1 q − 1 p ! Z RN λf+|u|qdx ≥ p − 2 2p ! kuk2H1 − λ p − q pq ! kf+kLq∗S −q2 p kukqH1. (2.1)

Thus, Jfλ,gµ is coercive and bounded below on Nfλ,gµ.

The Nehari manifold Nfλ,gµ is closely linked to the behavior of the function

of the form hu : t → Jfλ,gµ(tu) for t > 0. Such maps are known as fibering

maps and were introduced by Dr´abek-Pohozaev in [14] and are also discussed in Brown-Zhang [11] and Brown-Wu [9,10]. If u ∈ H1

RN  , we have hu(t) = t2 2 kuk 2 H1 − tq q Z RN fλ|u|qdx − tp p Z RN gµ|u|pdx; h0u(t) = t kuk2H1 − t q−1Z RN fλ|u|qdx − tp−1 Z RN gµ|u|pdx; h00u(t) = kuk2H1 − (q − 1) tq−2 Z RN fλ|u|qdx − (p − 1) tp−2 Z RN gµ|u|pdx. It is easy to see that

th0u(t) = ktuk2H1− Z RN fλ|tu| q dx − Z RN gµ|tu| p dx

and so, for u ∈ H1 RN



\ {0} and t > 0, h0

u(t) = 0 if and only if tu ∈ Nfλ,gµ,

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In particular, h0u(1) = 0 if and only if u ∈ Nfλ,gµ. Thus, it is natural to

split Nfλ,gµ into three parts corresponding to local minima, local maxima and

points of inflection. Accordingly, we define

N+f λ,gµ= n u ∈ Nfλ,gµ | h 00 u(1) > 0 o ; N0f λ,gµ= n u ∈ Nfλ,gµ | h 00 u(1) = 0 o ; N−f λ,gµ= n u ∈ Nfλ,gµ | h 00 u(1) < 0 o .

We now derive some basic properties of N+f

λ,gµ, N

0

fλ,gµ and N

− fλ,gµ.

Lemma 2.3 Suppose that u0 is a local minimizer for Jfλ,gµ on Nfλ,gµ and that

u0 ∈ N/ 0fλ,gµ. Then J 0 fλ,gµ(u0) = 0 in H −1 RN  .

Proof. The proof is essentially the same as that in Brown-Zhang [11, The-orem 2.3] (or see Binding-Dr´abek-Huang [7]).

For each u ∈ Nfλ,gµ, we have

h00u(1) = kuk2H1 − (q − 1) Z RN fλ|u| q dx − (p − 1) Z RN gµ|u| p dx = (2 − p) kuk2H1 − (q − p) Z RN fλ|u| q dx (2.2) = (2 − q) kuk2H1 − (p − q) Z RN gµ|u| p dx. (2.3)

Then we have the following result. Lemma 2.4 (i) For any u ∈ N+f

λ,gµ∪ N 0 fλ,gµ, we have R RNfλ|u| q dx > 0. (ii) For any u ∈ N−f

λ,gµ, we have

R

RNgµ|u| p

dx > 0.

Proof. The results now follows immediately from (2.2) and (2.3) . Let Λ0 = (2 − q)2−q p − 2 kf+kLq∗ !p−2 Sp p − q !p−q . Then we have the following result.

Lemma 2.5 For each λ > 0 and µ ≥ 0 with λp−2(1 + µ kbk ∞)

2−q

< Λ0, we have N0

fλ,gµ = ∅.

Proof. Suppose the contrary. Then there exist λ > 0 and µ ≥ 0 with λp−2(1 + µ kbk)2−q < Λ0

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such that N0

fλ,gµ 6= ∅. Then, for u ∈ N

0

fλ,gµ, by (2.2) and the H¨older and

Sobolev inequalities we have kuk2H1 = p − q p − 2 Z RN fλ|u|qdx ≤ λS −q 2 p p − q p − 2kf+kLq∗ kuk q H1 and so kuk2H1 ≤ S q q−2 p " λ kf+kLq∗ p − q p − 2 #2−q2 . Similarly, using (2.3) and the Sobolev inequality we have

2 − q p − qkuk 2 H1 = Z RN (a + µb) |u|pdx ≤ (1 + µ kbk) S −p 2 p kukpH1, which implies kuk2H1 ≥ S p p−2 p " 2 − q (1 + µ kbk) (p − q) # 2 p−2 for all µ ≥ 0.

Hence, we must have

λp−2(1 + µ kbk)2−q ≥ (2 − q)2−q p − 2 kf+kLq∗ !p−2 Sp p − q !p−q = Λ0

which is a contradiction. This completes the proof.

In order to get a better understanding of the Nehari manifold and fibering maps, we consider the function mu : R+ → R defined by

mu(t) = t2−qkuk2H1 − t

p−qZ RN

gµ|u|pdx for t > 0. (2.4) Clearly tu ∈ Nfλ,gµ if and only if mu(t) =

R RNfλ|u| q dx. Moreover, m0u(t) = (2 − q)t1−qkuk2H1 − (p − q)tp−q−1 Z RN gµ|u|pdx (2.5) and so it is easy to see that, if tu ∈ Nfλ,gµ, then t

q−1m0 u(t) = h 00 u(t). Hence, tu ∈ N+f λ,gµ( or N − fλ,gµ) if and only if m 0 u(t) > 0( or < 0). Suppose u ∈ H1 RN 

\ {0} . Then, by (2.5), mu has a unique critical point at t = tmax,µ(u) where

tmax,µ(u) = (2 − q) kuk2H1 (p − q)R RNgµ|u| p dx !p−21 > 0 (2.6)

and clearly mu is strictly increasing on (0, tmax,µ(u)) and strictly decreasing on (tmax,µ(u) , ∞) with limt→∞mu(t) = −∞. Moreover, if λp−2(1 + µ kbk∞)

2−q < Λ0, then

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mu(tmax,µ(u)) =   2 − q p − q !2−qp−2 − 2 − q p − q !p−qp−2  kuk 2(p−q) p−2 H1 (R RNgµ|u| p dx)2−qp−2 = kukqH1 p − 2 p − q ! 2 − q p − q !2−q p−2 kukp H1 R RN gµ|u| p dx !2−q p−2 ≥ p − 2 λ kf+kLq∗ Sp p − q !p−qp−2 2 − q 1 + µ kbk !2−qp−2 Z RN fλ|u|qdx > Z RN fλ|u|qdx.

Thus, we have the following lemma. Lemma 2.6 For each u ∈ H1

RN



\ {0} we have the following. (i) If R

RN fλ|u| q

dx ≤ 0, then there is a unique t− = t−(u) > tmax,µ(u) such that t−u ∈ N−f λ,gµ and hu is increasing on (0, t −) and decreasing on (t, ∞). Moreover, Jfλ,gµ  t−u= sup t≥0 Jfλ,gµ(tu) . (2.7) (ii) If R RNfλ|u| q

dx > 0, then there are unique

0 < t+ = t+(u) < tmax,µ(u) < t− = t−(u) such that t+u ∈ N+

fλ,gµ, t

u ∈ N

fλ,gµ, hu is decreasing on (0, t

+), increasing on (t+, t) and decreasing on (t, ∞). Moreover,

Jfλ,gµ  t+u= inf 0≤t≤tmax,µ(u) Jfλ,gµ(tu) ; Jfλ,gµ  t−u= sup t≥t+ Jfλ,gµ(tu) . (2.8)

(iii) t−(u) is a continuous function for u ∈ H1 RN  \ {0} . (iv) N−f λ,gµ = n u ∈ H1RN | 1 kukH1t − u kukH1  = 1o. Proof. Fix u ∈ H1 RN  \ {0} . (i) Suppose R RNfλ|u| q dx ≤ 0. Then mu(t) = R RNfλ|u| q

dx has a unique so-lution t− > tmax,µ(u) such that m0u(t

) < 0 and h0 u(t −) = 0. Hence, by tq−1m0 u(t) = h 00

u(t), hu has a unique critical point at t = t− and h00u(t

) < 0. Thus, t−u ∈ N−f λ,gµ and (2.7) holds. (ii) SupposeR RNfλ|u| q

dx > 0. Since mu(tmax,µ(u)) >RRNfλ|u|qdx, the equa-tion mu(t) =

R

RNfλ|u| q

dx has exactly two solutions t+ < t

max,µ(u) < t− such that m0u(t+) > 0 and m0u(t−) < 0. Hence, there are exactly two multi-ples of u lying in Nfλ,gµ, that is, t

+u ∈ N+

fλ,gµ and t

u ∈ N

fλ,gµ. Thus, by

tq−1m0

u(t) = h00u(t), huhas critical points at t = t+ and t = t−with h00u(t+) > 0 and h00u(t−) < 0. Thus, hu is decreasing on (0, t+) , increasing on (t+, t−) and decreasing on (t+, ∞) . Therefore, (2.8) must hold.

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t−(u) is a continuous function for u ∈ H1 RN



\ {0} . (iv) For u ∈ N−fλ,gµ. Let v = kuku

H1. By parts (i), (ii), there is a unique

t−(v) > 0 such that t−(v) v ∈ N−f λ,gµ or t − u kukH1  1 kukH1u ∈ N − fλ,gµ. Since u ∈ N−f λ,gµ, we have t − u kukH1  1

kukH1 = 1, and this implies

N−f λ,gµ ⊂ ( u ∈ H1RN | 1 kukH1 t− u kukH1 ! = 1 ) .

Conversely, let u ∈ H1RN such that kuk1

H1t − u kukH1  = 1. Then t− u kukH1 ! u kukH1 ∈ N−fλ,gµ. Thus, N−f λ,gµ = ( u ∈ H1RN | 1 kukH1 t− u kukH1 ! = 1 ) . This completes the proof.

Remark 2.1 (i) If λ = 0, then, by Lemma 2.6 (i) N+f

0,gµ = ∅, and so Nf0,gµ =

N−f

0,gµ for all µ ≥ 0.

(ii) If λp−2(1 + µ kbk)2−q < Λ0, then, by (2.2) , for each u ∈ N+fλ,gµ we have

kuk2H1 < p − q p − 2 Z RN fλ|u| q dx ≤ Λ1/(p−2)0 S −q 2 p p − q p − 2kf+kLq∗ kuk q H1, and so kukH1 ≤ Λ 1/(p−2) 0 S −q 2 p p − q p − 2kf+kLq∗ !1/(2−q) for all u ∈ N+f λ,gµ. (2.9)

3 Existence of a first solution

First, we remark that it follows from Lemma 2.5 that Nfλ,gµ = N

+

fλ,gµ∪ N

− fλ,gµ

for all λ > 0 and µ ≥ 0 with λp−2(1 + µ kbk ∞)

2−q

< Λ0. Furthermore, by Lemma 2.6 it follows that N+f

λ,gµ and N

fλ,gµ are non-empty and, by Lemma

2.2, we may define α+f λ,gµ = inf u∈N+fλ,gµ Jfλ,gµ(u) and α − fλ,gµ = inf u∈N−fλ,gµ Jfλ,gµ(u) .

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Theorem 3.1 We have the following: (i) α+f

λ,gµ < 0 for all λ > 0 and µ ≥ 0 with λ

p−2(1 + µ kbk ∞) 2−q < Λ0. (ii) If λp−2(1 + µ kbk ∞) 2−q

<q2p−2Λ0, then α−fλ,gµ > c0 for some c0 > 0.

In particular, for each λ > 0 and µ ≥ 0 with λp−2(1 + µ kbk)2−q <q2p−2Λ0, we have αf+

λ,gµ = infu∈Nfλ,gµJfλ,gµ(u) .

Proof. (i) Let u ∈ N+f

λ,gµ. Then, by (2.2) , kuk2H1 < p − q p − 2 Z RN fλ|u|qdx.

Hence, by (2.1) and Lemma 2.4,

Jfλ,gµ(u) = p − 2 2p kuk 2 H1 − p − q pq Z RN fλ|u|qdx < −(p − q) (2 − q) 2pq Z RN fλ|u| q dx < 0 and so α+f λ,gµ < 0. (ii) Let u ∈ N−f

λ,gµ. Then, by (2.3) and the Sobolev inequality,

2 − q p − qkuk 2 H1 < Z RN gµ|u| p dx ≤ (1 + µ kbk) S −p 2 p kukpH1, which implies kukH1 > S p 2(p−2) p 2 − q (1 + µ kbk) (p − q) !1/(p−2) for all u ∈ N−f λ,gµ. (3.1) By (2.1) and (3.1) , we have Jfλ,gµ(u) ≥ kukqH1 p − 2 2p kuk 2−q H1 − λ p − q pq ! kf+kLq∗ S −q 2 p ! > S pq 2(p−2) p 2 − q (1 + µ kbk) (p − q) !p−2q ·   p − 2 2p S p(2−q) 2(p−2) p 2 − q (1 + µ kbk) (p − q) !2−q p−2 − λ p − q pq ! kf+kLq∗ S −q2 p  . Thus, if λp−2(1 + µ kbk ∞) 2−q <q2p−2Λ0, then α−f λ,gµ > c0 for some c0 > 0.

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This completes the proof.

Now, we consider the following semilinear elliptic problem:

     −∆u + u = |u|p−2u in RN, u ∈ H1RN. (E∞)

Associated with the equation (E∞) , we consider the energy functional J∞ in H1 RN  J∞(u) = 1 2kuk 2 H1 − 1 p Z RN |u|pdx. Consider the minimizing problem:

inf u∈N∞J ∞ (u) = α∞ where N∞ =nu ∈ H1RN\ {0} | D(J∞)0(u) , uE= 0o.

It is known that equation (E∞) has a unique positive radially solution w0(x) such that J∞(w0) = α∞ (see [8,19]). Then the following proposition provides a precise description for the (PS)–sequence of Jfλ,gµ.

Proposition 3.2 (i) If {un} is a (PS)β–sequence in H1



RN



for Jfλ,gµ with

β < α+fλ,gµ + α∞, then there exists a subsequence {un} and a non-zero u0 in H1RN such that un → u0 strongly in H1



RN



and Jfλ,gµ(u0) = β.

Moreover, u0 is a solution of equation

 Efλ,gµ  . (ii) If {un} ⊂ N−fλ,gµ is a (PS)β–sequence in H 1 RN  for Jfλ,gµ with α+fλ,gµ+ α∞< β < α−fλ,gµ+ α∞,

then there exists a subsequence {un} and a non-zero u0 in H1

 RN  such that un → u0 strongly in H1  RN 

and Jfλ,gµ(u0) = β. Moreover, u0 is a solution

of equation Efλ,gµ



.

Proof. Similarly the argument in Wu [28, Proposition 4.6] (or see Adachi-Tanaka [4, Proposition 1.9]).

Theorem 3.3 For each λ > 0 and µ ≥ 0 with λp−2(1 + µ kbk)2−q < Λ0, the functional Jfλ,gµ has a minimizer u

+ λ,µ in N + fλ,gµ and it satisfies (i) Jfλ,gµ  u+λ,µ= α+f λ,gµ,

(ii) u+λ,µ is a positive solution of equation Efλ,gµ

 , (iii) u + λ,µ H1 → 0 as λ → 0.

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Proof. By the Ekeland variational principle [15] (or see Wu [29, Proposition 1]), there exists {un} ⊂ N+fλ,gµ such that it is a (PS)α+

fλ,gµ–sequence for Jfλ,gµ.

Then, by Proposition 3.2, there exist a subsequence {un} and u+λ,µ ∈ N + fλ,gµ

a non-zero solution of equation Efλ,gµ



such that un → u+λ,µ strongly in H1(RN) and Jfλ,gµ  u+λ,µ = α+fλ,gµ. Since Jfλ,gµ  u+λ,µ = Jfλ,gµ  u + λ,µ  and u + λ,µ ∈ N +

fλ,gµ, by Lemma 2.3 we may assume that u

+

λ,µ is a positive solution of equationEfλ,gµ



. Finally, by (2.2) and the H¨older and Sobolev inequalities,

u + λ,µ 2−q H1 < λ p − q p − 2kf+kLq∗S −q 2 p and so u + λ,µ H1 → 0 as λ → 0.

4 The estimate of energy

First, we let w0(x) be a unique radially symmetric positive solution of equa-tion (E∞) such that J∞(w0) = α∞. Then, by the result in Gidas-Ni-Nirenberg [18], for any ε > 0, there exist positive numbers Aε, B0 and Cε such that

Aεexp (− (1 + ε) |x|) ≤ w0(x) ≤ B0exp (− |x|) (4.1)

and

|∇w0(x)| ≤ Cεexp (− (1 − ε) |x|) . (4.2) Let

wl(x) = w0(x + le) , for l ∈ R and e ∈ SN −1, (4.3) where SN −1=nx ∈ RN | |x| = 1o. Then we have the following results.

Proposition 4.1 For each λ > 0 and µ > 0 with λp−2(1 + µ kbk ∞) 2−q < Λ0, we have α−f λ,gµ < α + fλ,gµ+ α ∞ .

Proof. Let u+λ,µ be a positive solution of equation Efλ,gµ



as in Theorem 3.3. Then

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Jfλ,gµ(u + λ,µ+ twl) =1 2 u + λ,µ+ twl 2 H1 − 1 q Z RN fλ u + λ,µ+ twl q dx − 1 p Z RN gµ u + λ,µ+ twl p dx ≤ Jfλ,gµ  u+λ,µ+ J∞(twl) + 1 p Z RN tpwpldx − 1 p Z RN gµtpwpldx − Z RN (λf++ f−) Z twl 0  (u+λ,µ+ η)q−1−u+λ,µq−1  dη  dx −1 p Z RN  (u+λ,µ+ twl)p −  u+λ,µp− tpwp l − p  u+λ,µp−1twl  dx ≤ α+ fλ,gµ+ J ∞ (twl) + 1 p Z RN (1 − g0) tpwpldx −µ p Z RN btpwpldx + Z RN |f−| Z twl 0 ηq−1dη  dx −1 p Z RN  (u+λ,µ+ twl)p −  u+λ,µp− tpwp l − p  u+λ,µp−1twl  dx = α+f λ,gµ+ J ∞ (twl) + tp p Z RN (1 − g0) wlpdx − µtp p Z RN bwlpdx + t q q Z RN |f−| wlqdx −1 p Z RN  (u+λ,µ+ twl)p −  u+λ,µp− tpwp l − p  u+λ,µp−1twl  dx. (4.4)

By Brown-Zhang [11] and Willem [27], we know that

J∞(twl) ≤ α∞ for all l ∈ R. (4.5) Thus, by (4.4) and (4.5), we have

Jfλ,gµ(u + λ,µ+ twl) ≤ α+ fλ,gµ+ α ∞+tp p Z RN (1 − g0) wpldx − µtp p Z RN bwlpdx +t q q Z RN |f−| wlqdx −1 p Z RN  (u+λ,µ+ twy)p−  u+λ,µp − tpwp l − p  u+λ,µp−1twl  dx. (4.6) Since Jfλ,gµ(u + λ,µ+ twl) → Jfλ,gµ(u + λ,µ) = α + fλ,gµ < 0 as t → 0 and Jfλ,gµ(u + λ,µ+ twl) → −∞ as t → ∞, we can easily find 0 < t1 < t2 such that

Jfλ,gµ(u

+

λ,µ+ twl) < αf+λ,gµ+ α

for all t ∈ [0, t1] ∪ [t2, ∞). (4.7) Thus, we only need to show that there exists l0 > 0 such that for l > l0,

sup t1≤t≤t2 Jfλ,gµ(u + λ,µ+ twl) < α+fλ,gµ+ α ∞ . (4.8)

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We also remark that

(u + v)p− up− vp− pup−1v ≥ 0 for all (u, v) ∈ [0, ∞) × [0, ∞). Thus, Z RN  (u+λ,µ+ twl)p−  u+λ,µp− tpwp l − p  u+λ,µp−1twl  dx ≥ 0. (4.9)

From the condition (D2) and (4.1)

Z RN (1 − g0) tpwpldx ≤ c0B0p Z RN

exp (−ra|x|) exp (−p |x + le|) dx ≤ c0B

p 0

Z

|x|<lexp (− min {ra, p} (|x| + |x + le|)) dx +c0B

p 0

Z

|x|≥lexp (− min {ra, p} (|x| + |x + le|)) dx ≤ c0B

p 0l

NZ

|x|<1exp (− min {ra, p} l (|x| + |x + e|)) dx +c0Bp0exp (− min {ra, p} l)

Z

|x|≥lexp (− min {ra, p} (|x + le|)) dx ≤ c0B0pl

NZ

|x|<1exp (− min {ra, p} l) dx + C0B p

0exp (− min {ra, p} l) ≤ C0B0plNexp (− min {ra, p} l) for l ≥ 1 (4.10)

and Z RN bwlpdx = Z RN b (x − le) w0p(x) dx ≥ min x∈BN(1)w p 0(x) ! Z BN(1)b (x − le) dx ≥ min x∈BN(1)w p 0(x) ! d0 Z BN(1)exp (−rb|x| − rbl |e|) dx = min x∈BN(1)w p 0(x) ! D0exp (−rbl) . (4.11)

From the condition (D1) and the same argument of inequality (4.10) , we also have Z RN |f−| wqldx ≤cBb q 0 Z RN exp−rf−|x|  exp (−q |x + le|) dx ≤cBb q 0lN exp  − minnrf−, q o l for l ≥ 1. (4.12)

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Since rb < min n rf−, ra, q o ≤ minnrf−, ra, p o and t1 ≤ t ≤ t2, by (4.6)−(4.12) , we can find l1 ≥ max {l0, 1} such that

sup t≥0 Jfλ,gµ(u + λ,µ+ twl) < α+fλ,gµ+ α ∞ for all l > l1.

To complete the proof of Proposition 4.1, it remains to show that there exists a positive number t∗ such that u+λ,µ+ t∗wl ∈ N−fλ,gµ. Let

U1= ( u ∈ H1RN 1 kukH1 t− u kukH1 ! > 1 ) ∪ {0} ; U2= ( u ∈ H1RN 1 kukH1 t− u kukH1 ! < 1 ) .

Then N−fλ,gµ separates H1RN into two connected components U1 and U2, and H1 RN  \N−f λ,gµ = U1 ∪ U2. For each u ∈ N + fλ,gµ, we have

1 < tmax,µ(u) < t−(u) . Since t−(u) = kuk1

H1t − u kukH1  , then N+f λ,gµ ⊂ U1. In particular, u + λ,µ ∈ U1. We claim that there exists t0 > 0 such that u+λ,µ + t0wl ∈ U2. First, we find a constant c > 0 such that 0 < t−

 u+λ,µ+twl ku+λ,µ+twlk H1  < c for each t ≥ 0. Suppose the contrary. Then there exists a sequence {tn} such that tn→ ∞ and t−  u+λ,µ+tnwl ku+λ,µ+tnwlk H1  → ∞ as n → ∞. Let vn = u+λ,µ+tnwl ku+λ,µ+tnwlk H1 . Since t−(vn) vn ∈ N−f

λ,gµ, by the Lebesgue dominated convergence theorem,

Z RN gµvnpdx = 1 u + λ,µ+ tnwl p H1 Z RN gµ  u+λ,µ+ tnwl p dx = 1 u+λ,µ tn + wl p H1 Z RN gµ u+λ,µ tn + wl !p dx → R RNgµw p ldx kwlkpH1 as n → ∞, we have Jfλ,gµ  t−(vn) vn  = 1 2 h t−(vn) i2 − [t −(v n)]q q Z RN fλvnqdx − [t−(vn)]p p Z RN gµvnpdx → −∞ as n → ∞,

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this contradicts the fact that Jfλ,gµ is bounded below on Nfλ,gµ. Let t0 = p − 2 2pα∞ c 2 u + λ,µ 2 H1 !12 + 1. Then u + λ,µ+ t0wl 2 H1= u + λ,µ 2 H1 + t 2 0kwlk2H1 + o (1) > u + λ,µ 2 H1 + c2− u + λ,µ 2 H1 + o (1) > c2+ o (1) >  t −   u+λ,µ+ t0wl u + λ,µ+ t0wl H1     2 + o (1) as l → ∞.

Thus, there exists l2 ≥ l1 such that for l > l2, 1 u + λ,µ+ t0wl H1 t−   u+λ,µ+ t0wl u + λ,µ+ t0wl H1  < 1

or u+λ,µ+ t0wl ∈ U2. Define a path γl(s) = u+λ,µ+ st0wl for s ∈ [0, 1] . Then γl(0) = u+λ,µ∈ U1, γl(1) = u+λ,µ+ t0wl ∈ U2. Since kuk1 H1 t−kuku H1 

is a continuous function for non-zero u and γl([0, 1]) is connected, there exists sl ∈ (0, 1) such that u+λ,µ + slt0wl ∈ N−fλ,gµ. This

completes the proof.

Then we have the following result.

Theorem 4.2 For each λ > 0 and µ > 0 with λp−2(1 + µ kbk)2−q <q2p−2Λ0, equation Efλ,gµ



has positive solution u−λ,µ ∈ N−f

λ,gµ such that Jfλ,gµ



u−λ,µ= α−f

λ,gµ.

Proof. Analogous to the proof of Wu [30, Proposition 9], one can show that by the Ekeland variational principle (see [15]), there exist minimizing sequences {un} ⊂ N−fλ,gµ such that

Jfλ,gµ(un) = α − fλ,gµ+ o (1) and J 0 fλ,gµ(un) = o (1) in H −1 RN  .

Since α−fλ,gµ < α+fλ,gµ+ α∞, by Theorem 3.1 (ii) and Proposition 3.2 there exist a subsequence {un} and u−λ,µ∈ N

fλ,gµ a non-zero solution of equation

 Efλ,gµ  such that un→ u−λ,µ strongly in H 1 (RN).

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Since Jfλ,gµ  u−λ,µ= Jfλ,gµ  u − λ,µ  and u − λ,µ ∈ N − fλ,gµ, by Lemma 2.3, we may

assume that u−λ,µ is a positive solutions of equation Efλ,gµ



.

5 Concentration Behavior

We need the following lemmas. Lemma 5.1 We have

inf

u∈Nf0,g0Jf0,g0(u) = infu∈N∞J

(u) = α∞.

Furthermore, equation (Ef0,g0) does not admit any solution u0 such that Jf0,g0(u0) =

infu∈Nf0,g0 Jf0,g0(u) .

Proof. Let wlbe as in (4.3) . Then, by Lemma 2.6, there is a unique t−(wl) >

2−q

p−q

1/(p−2)

such that t−(wl) wl∈ Nf0,g0 for all l > 0, that is

t − (wl) wl 2 H1 = Z RN f− t − (wl) wl q dx + Z RN g0 t − (wl) wl p dx. Since kwlk 2 H1 = Z RN |wl| p dx = 2p p − 2α ∞ for all l ≥ 0, Z RN f−|wl| q dx → 0 and Z RN (1 − g0) |wl| p dx → 0 as l → ∞, we have t−(wl) → 1 as l → ∞. Thus, lim l→∞Jf0,g0  t−(wl) wl  = lim l→∞J ∞ t−(wl) wl  = α∞. Then inf

u∈Nf0,g0Jf0,g0(u) ≤ infu∈N∞J

(u) = α∞.

Let u ∈ Nf0,g0. Then, by Lemma 2.6 (i) , Jf0,g0(u) = supt≥0Jf0,g0(tu) .

More-over, there is a unique t∞> 0 such that t∞u ∈ N∞. Thus, Jf0,g0(u) ≥ Jf0,g0(t

u) ≥ J∞(t∞u) ≥ α∞ and so infu∈Nf0,g0Jf0,g0(u) ≥ α

. Therefore, inf

u∈Nf0,g0Jf0,g0(u) = infu∈N∞J

(u) = α∞.

Next, we will show that equation (Ef0,g0) does not admit any solution u0 such

that Jf0,g0(u0) = infu∈Nf0,g0Jf0,g0(u) . Suppose the contrary. Then we can

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Then, by Lemma 2.6 (i) , Jf0,g0(u0) = supt≥0Jf0,g0(tu0) . Moreover, there is a

unique tu0 > 0 such that tu0u0 ∈ N

. Thus, α∞= inf u∈Nf0,g0Jf0,g0(u) = Jf0,g0(u0) ≥ Jf0,g0(tu0u0) ≥ J∞(tu0u0) − tqu0 q Z RN f−|u0|qdx ≥ α∞− tqu0 q Z RN f−|u0|qdx. This implies R RNf−|u0| q dx = 0 and so u0 ≡ 0 in n x ∈ RN | f −(x) 6= 0 o form the condition (D1) . Therefore,

α∞= inf u∈N∞J

(u) = J∞(tu0u0) .

By the Lagrange multiplier and the maximum principle, we can assume that tu0u0 is a positive solution of (E ∞ ) . This contradicts u0 ≡ 0 in n x ∈ RN | f−(x) 6= 0 o

and completes the proof.

Lemma 5.2 Suppose that {un} is a minimizing sequence in Nf0,g0 for Jf0,g0.

Then (i) R RN f−|un| q dx = o (1) ; (ii) R RN (1 − g0) |un| p dx = o (1) . Furthermore, {un} is a (PS)α∞–sequence in H1  RN  for J∞.

Proof. For each n, there is a unique tn > 0 such that tnun∈ N∞, that is t2nkunk2H1 = tpn

Z

RN

|un|pdx.

Then, by Lemma 2.6 (i) ,

Jf0,g0(un) ≥ Jf0,g0(tnun) = J∞(tnun) + tp n p Z RN (1 − g0) |un|pdx − tq n q Z RN f−|un|qdx ≥ α∞+ t p n p Z RN (1 − g0) |un|pdx − tq n q Z RN f−|un|qdx. Since Jf0,g0(un) = α

+ o (1) from Lemma 5.1, we have tq n q Z RN f−|un|qdx = o (1) and tp n p Z RN (1 − g0) |un|pdx = o (1) .

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We will show that there exists c0 > 0 such that tn > c0 for all n. Suppose the contrary. Then we may assume tn → 0 as n → ∞. Since Jf0,g0(un) =

α∞+ o (1) , by Lemma 2.2, kunk is uniformly bounded and so ktnunkH1 → 0

or J∞(tnun) → 0 and this contradicts J∞(tnun) ≥ α∞ > 0. Thus,

Z RN f−|un| q dx = o (1) and Z RN (1 − g0) |un|pdx = o (1) , which this implies

kunk2H1 = Z RN |un|pdx + o (1) and J∞(un) = α∞+ o (1) .

Moreover, by Wang–Wu [26, Lemma 7], we have {un} is a (PS)α∞–sequence

in H1 RN



for J∞.

The following lemma is a key lemma in proving our main result.

Lemma 5.3 There exists d0 > 0 such that if u ∈ Nf0,g0 and Jf0,g0(u) ≤

α∞+ d0, then Z RN x |x|  |∇u|2 + u2dx 6= 0.

Proof. Suppose the contrary. Then there exists sequence {un} ⊂ Nf0,g0 such

that Jf0,g0(u) = α ∞+ o (1) and Z RN x |x|  |∇un|2+ u2n  dx = 0.

Moreover, by Lemma 5.2, we have {un} is a (PS)α∞–sequence in H1



RN



for J∞. It follows from Lemma 2.2 that there exist a subsequence {un} and u0 ∈ H1



RN



such that un * u0 weakly in H1



RN



. By the concentration– compactness principle (see Lions [20] or Struwe [24, Theorem 3.1]), there exist a sequence {xn} ⊂ RN, and a positive solution w0 ∈ H1

 RN  of equation (E∞) such that kun(x) − w0(x − xn)kH1 → 0 as n → ∞. (5.1)

Now we will show that |xn| → ∞ as n → ∞. Suppose the contrary. Then we may assume that {xn} is bounded and xn → x0 for some x0 ∈ RN. Thus, by (5.1)

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Z RN f−|un|qdx = Z RN f−(x) |w0(x − xn)|qdx + o (1) = Z RN f−(x + xn) |w0(x)|qdx + o (1) = Z RN f−(x + x0) |w0(x)|qdx + o (1) ,

this contradicts the result of Lemma 5.2: R

RNf−|un| q

dx = o (1) . Hence we may assume xn

|xn| → e as n → ∞, where e ∈ S

N −1. Then, by the Lebesgue dominated convergence theorem, we have

0 = Z RN x |x|  |∇un| 2 + u2ndx = Z RN x + xn |x + xn|  |∇w0| 2 + w02dx + o (1) = 2p p − 2α ∞ e + o (1) ,

which is a contradiction. This completes the proof. By (2.3) , (2.6) and Lemma 2.6, for each u ∈ N−f

λ,gµthere is a unique t

− 0 (u) > 0 such that t−0 (u) u ∈ Nf0,g0 and

t−0 (u) > tmax,0(u) =

(2 − q) kuk2H1 (p − q)R RNg0|u| p dx !p−21 > 0. Let θµ=     (p − q) (1 + µ kb/ak) 2 − q     1 + kf−kLq∗    (p − q) (1 + µ kb/ak) (2 − q) S p−q 2−q p    2−q p−2         p p−2 .

Then we have the following results.

Lemma 5.4 For each λ > 0 and µ > 0 with λp−2(1 + µ kbk)2−q <q2p−2Λ0 we have the following.

(i) ht−0 (u)ip < θµ for all u ∈ N−fλ,gµ with Jfλ,gµ(u) < α

+ fλ,gµ+ α ∞. (ii) R RNg0|u| p dx ≥ θ pq µ(p−q)α ∞ for all u ∈ N− fλ,gµ with Jfλ,gµ(u) < α + fλ,gµ+ α ∞.

Proof. (i) For u ∈ N−f

λ,gµ with Jfλ,gµ(u) < α + fλ,gµ+ α ∞, we have kuk2H1 − Z RN fλ|u|qdx − Z RN gµ|u|pdx = 0 and (2 − q) kuk2H1 < (p − q) Z RN gµ|u| p dx.

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We distinguish two cases.

Case (A): t−0 (u) < 1. Since θµ > 1 for all µ > 0, we have

h

t−0 (u)ip < 1 < θµ.

Case (B): t−0 (u) ≥ 1. Since

h t−0 (u)ip Z RN g0|u|pdx = h t−0 (u)i2kuk2H1 − h t−0 (u)iq Z RN f−|u|qdx ≤ht−0 (u)i2  kuk2H1 + Z RN |f−| |u|qdx  , we have h t−0 (u)ip−2 ≤ kuk 2 H1 + R RN|f−| |u| q dx R RNg0|u| p dx . (5.2)

Moreover, by (2.3) and the Sobolev inequality,

kuk2H1< p − q 2 − q Z RN gµ|u|pdx ≤ p − q 2 − q (1 + µ kb/ak∞) Z RN g0|u|pdx (5.3) ≤(p − q) (1 + µ kb/ak∞) (2 − q) S p 2 p kukpH1 and so kukH1 ≥   (2 − q) S p 2 p (p − q) (1 + µ kb/ak)   1 p−2 . (5.4)

Thus, by (5.2) − (5.4) and the Sobolev inequality,

h t−0 (u)ip−2 ≤ (1 + µ kb/ak) p − q 2 − q ! 1 + R RN f−|u| q dx kuk2H1 ! ≤ (1 + µ kb/ak) p − q 2 − q ! 1 + kf−kLq∗ S q 2 p kuk2−qH1   ≤(p − q) (1 + µ kb/ak∞) 2 − q     1 + kf−kLq∗    (p − q) (1 + µ kb/ak) (2 − q) S p−q 2−q p    2−q p−2     or [t−(u)]p ≤ θµ.

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α∞≤ Jf0,g0  t−0 (u) u = 1 2 − 1 q ! h t−0 (u)i2kuk2H1 + 1 q − 1 p ! h t−0 (u)ip Z RN g0|u|pdx < 1 q − 1 p ! h t−0 (u)ip Z RN g0|u|pdx,

and this implies

Z RN g0|u|pdx ≥ 1 h t−0 (u)ip pq p − q ! α∞.

By part (i) , we can conclude that

Z RN g0|u|pdx ≥ pq θµ(p − q) α∞. for all u ∈ N−f λ,gµ with Jfλ,gµ(u) < α + fλ,gµ+ α

. This completes the proof. By the proof of Proposition 4.1, there exist positive numbers t∗ and l2 such that u+λ,µ+ t∗wl ∈ N−fλ,gµ and Jfλ,gµ(u + λ,µ+ t∗wl) < α+fλ,gµ+ α ∞ for all l > l 2. Furthermore, we have the following result.

Lemma 5.5 There exist positive numbers λ0 and µ0 with

λp−20 (1 + µ0kbk∞) 2−q < q 2 p−2 Λ0

such that for every λ ∈ (0, λ0) and µ ∈ (0, µ0) , we have

Z RN x |x|  |∇u|2+ u2dx 6= 0 for all u ∈ N−f λ,gµ with Jfλ,gµ(u) < α + fλ,gµ+ α ∞. Proof. For u ∈ N−f λ,gµ with Jfλ,gµ(u) < α + fλ,gµ+ α

, by Lemma 2.6 (i) there exists t−0 (u) > 0 such that t−0 (u) u ∈ Nf0,g0. Moreover,

Jfλ,gµ(u) = sup t≥0 Jfλ,gµ(tu) ≥ Jfλ,gµ  t−0 (u) u = Jf0,g0  t−0 (u) u− λ h t−0 (u)iq q Z RN f+|u|qdx −µ h t−0 (u)ip p Z RN b |u|pdx.

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Thus, by Lemma 5.4 and the H¨older and Sobolev inequalities, Jf0,g0  t−0 (u) u ≤ Jfλ,gµ(u) + λht−0 (u)iq q Z RN f+|u| q dx +µ h t−0 (u)ip p Z RN b |u|pdx < α+fλ,gµ+ α∞+λθ q/p µ q kf+kLq∗ S −q 2 p kukqH1 + µθµkbk∞ p S −p 2 p kukpH1. Since Jfλ,gµ(u) < α + fλ,gµ + α

< α, by (2.1) in Lemma 2.2, for each λ > 0 and µ > 0 with λp−2(1 + µ kbk

∞) 2−q

< q2p−2Λ0, there exists a positive number c independent of λ, µ such that kuke H1 ≤ c for all u ∈ Ne

− fλ,gµ with Jfλ,gµ(u) < α + fλ,gµ+ α ∞. Therefore, Jf0,g0  t−0 (u) u< α+f λ,gµ+ α ∞ +λθ q/p µ q kf+kLq∗ S −q2 p ce q+µθµkbk∞ p S −p2 p ec p.

Let d0 > 0 be as in Lemma 5.3. Then there exist positive numbers λ0 and µ0 with λp−20 (1 + µ0kbk∞)

2−q

< q2p−2Λ0 such that for λ ∈ (0, λ0) and µ ∈ (0, µ0) ,

Jf0,g0



t−(u) u< α∞+ d0. (5.5) Since t−0 (u) u ∈ Nf0,g0 and t

0 (u) > 0, by Lemma 5.3 and (5.5)

Z RN x |x|  ∇  t−0 (u) u 2 +t−0 (u) u2  dx 6= 0,

and this implies

Z RN x |x|  |∇u|2+ u2dx 6= 0 for all u ∈ N−f λ,gµ with Jfλ,gµ(u) < α + fλ,gµ+ α ∞. 6 Proof of Theorem 1.1

In the following, we use an idea of Adachi-Tanaka [4]. For c ∈ R+, we denote

h Jfλ,gµ ≤ c i =nu ∈ N−f λ,gµ | u ≥ 0, Jfλ,gµ(u) ≤ c o . We then try to show for a sufficiently small σ > 0

cathJfλ,gµ ≤ α

+

fλ,gµ+ α

− σi

≥ 2. (6.1)

To prove (6.1) , we need some preliminaries. Recall the definition of Lusternik-Schnirelman category.

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Definition 6.1 (i) For a topological space X, we say a non-empty, closed subset Y ⊂ X is contractible to a point in X if and only if there exists a continuous mapping

ξ : [0, 1] × Y → X such that for some x0 ∈ X

ξ (0, x) = x for all x ∈ Y, and

ξ (1, x) = x0 for all x ∈ Y. (ii) We define

cat (X) = min {k ∈ N | there exist closed subsets Y1, ..., Yk ⊂ X such that Yj is contractible to a point in X for all j and

k ∪

j=1Yj = X



.

When there do not exist finitely many closed subsets Y1, ..., Yk ⊂ X such that Yj is contractible to a point in X for all j and

k ∪

j=1Yj = X, we say cat (X) = ∞.

We need the following two lemmas.

Lemma 6.2 Suppose that X is a Hilbert manifold and F ∈ C1(X, R) . As-sume that there are c0 ∈ R and k ∈ N,

(i) F (x) satisfies the Palais–Smale condition for energy level c ≤ c0; (ii) cat ({x ∈ X | F (x) ≤ c0}) ≥ k.

Then F (x) has at least k critical points in {x ∈ X; F (x) ≤ c0} . Proof. See Ambrosetti [1, Theorem 2.3].

Lemma 6.3 Let X be a topological space. Suppose that there are two contin-uous maps

Φ : SN −1 → X, Ψ : X → SN −1

such that Ψ ◦ Φ is homotopic to the identity map of SN −1, that is, there exists a continuous map ζ : [0, 1] × SN −1→ SN −1 such that

ζ (0, x) = (Ψ ◦ Φ) (x) for each x ∈ SN −1, ζ (1, x) = x for each x ∈ SN −1.

Then

cat (X) ≥ 2.

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For l > l2, we define a map Φfλ,gµ : S N −1→ H1 RN  by Φfλ,gµ(e) (x) = u + λ,µ+ slt0wl for e ∈ SN −1,

where u+λ,µ+ slt0wl is as in the proof of Proposition 4.1. Then we have the following result.

Lemma 6.4 There exists a sequence {σl} ⊂ R+ with σl → 0 as l → ∞ such that Φfλ,gµ  SN −1  ⊂hJfλ,gµ ≤ α + fλ,gµ+ α ∞− σ l i .

Proof. By Proposition 4.1, for each l > l2 we have u+λ,µ+ slt0wl ∈ N−fλ,gµ

and sup t≥0 Jfλ,gµ  u+λ,µ+ twl  < α+f λ,gµ+ α ∞ uniformly in e ∈ SN −1. Since Φfλ,gµ  SN −1  is compact, Jfλ,gµ  u+λ,µ+ slt0wl  ≤ α+ fλ,gµ + α ∞− σ l, so that the conclusion holds.

From Lemma 5.5, we define Ψfλ,gµ : h Jfλ,gµ < α + fλ,gµ+ α ∞i → SN −1 by Ψfλ,gµ(u) = R RN x |x|  |∇u|2+ u2dx R RN x |x|  |∇u|2+ u2dx .

Then we have the following results.

Lemma 6.5 Let λ0, µ0 be as in Lemma 5.5. Then for each λ ∈ (0, λ0) and µ ∈ (0, µ0) , there exists l∗ ≥ l2 such that for l > l∗, the map

Ψfλ,gµ◦ Φfλ,gµ : S

N −1

→ SN −1 is homotopic to the identity.

Proof. Let Σ = nu ∈ H1 RN  \ {0} | R RN x |x|  |∇u|2+ u2dx 6= 0o. We define Ψfλ,gµ : Σ → S N −1 by Ψfλ,gµ(u) = R RN x |x|  |∇u|2+ u2dx R RN x |x|  |∇u|2+ u2dx

as an extension of Ψfλ,gµ. Since wl ∈ Σ for all e ∈ S

N −1 and for l sufficiently large, we let γ : [s1, s2] → SN −1 be a regular geodesic between Ψfλ,gµ(wl) and

Ψfλ,gµ  Φfλ,gµ(e)  such that γ (s1) = Ψfλ,gµ(wl) , γ (s2) = Ψfλ,gµ  Φfλ,gµ(e)  .

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By an argument similar to that in Lemma 5.3, there exists a positive number l∗ ≥ l2 such that for l > l∗,

w0 x + l 2 (1 − θ)e

!

∈ Σ for all e ∈ SN −1 and θ ∈ [1/2, 1) .

We define ζl(θ, e) : [0, 1] × SN −1 → SN −1 by ζl(θ, e) =              γ (2θ (s1 − s2) + s2) for θ ∈ [0, 1/2) ; Ψfλ,gµ  w0  x + 2(1−θ)l e for θ ∈ [1/2, 1) ; e for θ = 1. Then ζl(0, e) = Ψfλ,gµ  Φfλ,gµ(e) 

= Ψfλ,gµ(Φfλ(e)) and ζl(1, e) = e. By the

standard regularity, we have u+λ,µ∈ CRN. First, we claim that lim

θ→1−ζl(θ, e) = e and lim θ→1 2 −ζl(θ, e) = Ψfλ,gµ(w0(x + le)) . (a) lim θ→1−ζl(θ, e) = e : since Z RN x |x|   ∇ " w0 x + l 2 (1 − θ)e !# 2 + " w0 x + l 2 (1 − θ)e !#2 dx = Z RN x − 2(1−θ)l e x − l 2(1−θ)e  |∇ [w0(x)]|2 + [w0(x)]2  dx = 2p p − 2 ! α∞e + o(1) as θ → 1−, then lim θ→1−ζl(θ, e) = e. (b) lim θ→12− ζl(θ, e) = Ψfλ,gµ(w0(x + le)) : since Ψfλ,gµ ∈ C  Σ, SN −1, we obtain lim θ→12− ζl(θ, l) = Ψfλ,gµ(w0(x + le)) . Thus, ζl(θ, e) ∈ C  [0, 1] × SN −1, SN −1 and ζl(0, e) = Ψfλ,gµ  Φfλ,gµ(e)  for all e ∈ SN −1, ζl(1, e) = e for all e ∈ SN −1,

provided l > l∗. This completes the proof.

Lemma 6.6 For each λ ∈ (0, λ0) , µ ∈ (0, µ0) and l > l∗, functional Jfλ,gµ has

at least two critical points in hJfλ,gµ < α

+

fλ,gµ + α

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Proof. Applying Lemmas 6.3 and 6.5, we have for λ ∈ (0, λ0) , µ ∈ (0, µ0) and l > l∗, cathJfλ,gµ ≤ α + fλ,gµ+ α ∞− σ l i ≥ 2.

By Proposition 3.2, Lemma 6.2, Jfλ(u) has at least two critical points in

h

Jfλ,gµ < α

+

fλ,gµ+ α

∞i.

We can now complete the proof of Theorem 1.1: (i) by Theorems 3.3, 4.2. (ii) for λ ∈ (0, λ0) and µ ∈ (0, µ0) , from Theorem 3.3 and Lemma 6.6, equation (Efλ,gµ) has three positive solutions u

+ λ,µ, u − 1, u − 2 such that u+λ,µ ∈ N + fλ,gµand u−i ∈ N−f

λ,gµ for i = 1, 2. This completes the proof of Theorem 1.1.

7 Proof of Theorem 1.2 For c > 0, we define J0,cg0(u) = 1 2 Z RN |∇u|2+ u2dx −1 p Z RN cg0|u|pdx, N0,cg0= n u ∈ H1RN\ {0} |DJ0,cg0 0(u) , uE= 0o.

Recall that for each u ∈ H1 RN



\ {0} there exist a unique t−(u) > 0 and t0(u) > 0 such that t−(u) u ∈ N−fλ,gµ and t0(u) u ∈ N0,g0. Let

B =

n

u ∈ H1RN\ {0} | u ≥ 0 and kukH1 = 1

o

. Then we have the following result.

Lemma 7.1 For each u ∈ B we have the following. (i) There is a unique tc

0 = tc0(u) > 0 such that tc0u ∈ N0,cg0 and

sup t≥0 J0,cg0(tu) = J0,cg0(t c 0u) = p − 2 2p Z RN cg0|u| p dx p−2−2 . (ii) For ρ ∈ (0, 1) , Jfλ,gµ  t−(u) u≥ (1 − ρ) p p−2 (1 + µ kb/ak)p−22 J0,g0(t0(u) u)− 2 − q 2q (ρSp) q q−2(λ kf +kLq∗) 2 2−q and Jfλ,gµ  t−(u) u≤ (1 + ρ)p−2p J 0,g0(t0(u) u)+ 2 − q 2q (ρSp) q q−2(λ kf +kLq∗ + kf−kLq∗) 2 2−q .

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Proof. (i) For each u ∈ B, let K(t) = J0,cg0(tu) = 1 2t 2 1 pt pZ RN cg0|u|pdx, then K (t) → −∞ as t → ∞, K0(t) = t − tp−1R RNcg0|u| p dx and K00(t) = 1 − (p − 1) tp−2R RNcg0|u| p dx. Let tc0 = tc0(u) = Z RN cg0|u|pdx 2−p1 > 0. Then K0(tc0) = 0, tc0u ∈ N0,cg0 and K 00(tc

0) = 2 − p < 0. Thus, there is a unique tc

0 = tc0(u) > 0 such that tc0u ∈ N0,cg0 and

sup t≥0 J0,cg0(tu) = J0,cg0(t c 0u) = p − 2 2p Z RN cg0|u|pdx p−2−2 .

(ii) Let c = (1 + µ kb/ak) / (1 − ρ) . Then for each u ∈ B and ρ ∈ (0, 1) , we get Z RN fλ|tc0u| q dx ≤ λS −q 2 p kf+kLq∗ ktc0uk q H1 ≤2 − q 2  (ρSp) −q 2 λ kf +kLq∗ 2−q2 + q 2  ρq2 ktc 0uk q H1 2q =2 − q 2 (ρSp) q q−2(λ kf +kLq∗) 2 2−q + qρ 2 kt c 0uk 2 H1. (7.1)

Then, by part (i) and (7.1) ,

sup t≥0 Jfλ,gµ(tu) ≥ Jfλ,gµ(t c 0u) ≥1 − ρ 2 kt c 0uk 2 H1 − 2 − q 2q (ρSp) q q−2(λ kf +kLq∗) 2 2−q −(1 + µ kb/ak∞) p Z RN g0|tc0u| p dx = (1 − ρ) J0,cg0(t c 0u) − 2 − q 2q (ρSp) q q−2(λ kf +kLq∗) 2 2−q = (p − 2) (1 − ρ) p p−2 2p ((1 + µ kb/ak)R RNg0|u| p dx)p−22 − 2 − q 2q (ρSp) q q−2(λ kf +kLq∗) 2 2−q = (1 − ρ) p p−2 (1 + µ kb/ak)p−22 J0,g0(t0(u) u) − 2 − q 2q (ρSp) q q−2(λ kf +kLq∗) 2 2−q .

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By Lemma 2.6 and Theorem 3.1, sup t≥0 Jfλ,gµ(tu) = Jfλ,gµ  t−(u) u. Thus, Jfλ,gµ  t−(u) u≥ (1 − ρ) p p−2 (1 + µ kb/ak)p−22 J0,g0(t0(u) u)− 2 − q 2q (ρSp) q q−2(λ kf +kLq∗) 2 2−q .

Moreover, by the H¨older, Sobolev and Young inequalities,

Z RN fλ|tu| q dx ≤ (λ kf+kLq∗ + kf−kLq∗) S −q2 p ktukqH1 ≤2 − q 2 (ρSp) q q−2(λ kf +kLq∗ + kf−kLq∗) 2 2−q + qρ 2 ktuk 2 H1. Therefore, Jfλ,gµ(tu) ≤ (1 + ρ) 2 t 2 +2 − q 2q (ρSp) q q−2(λ kf +kLq∗ + kf−kLq∗) 2 2−q −1 p Z RN g0|tu| p dx ≤ (1 + ρ)p−2p J 0,g0(t0(u) u) +2 − q 2q (ρSp) q q−2(λ kf +kLq∗ + kf−kLq∗) 2 2−q and so Jfλ,gµ  t−(u) u≤ (1 + ρ)p−2p J 0,g0(t0(u) u) +2 − q 2q (ρSp) q q−2(λ kf +kLq∗ + kf−kLq∗) 2 2−q .

This completes the proof.

Since α−fλ,gµ > 0 for all λ ∈ (0, λ0) and µ ∈ (0, µ0) , we define

Ifλ,gµ(u) = sup t≥0 Jfλ,gµ(tu) = Jfλ,gµ  t−(u) u> 0, where t−(u) u ∈ N−f

λ,gµ. We observe that if λ, µ and kf−kLq∗ are sufficiently

small, Bahri-Li’s minimax argument [6] also works for Jfλ,gµ. Let

Γfλ,gµ = n γ ∈ CBN(0, l), B | γ| ∂BN(0,l) = wl/ kwlkH1 o for large l.

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Then we define βfλ,gµ = inf γ∈Γfλ,gµx∈RsupN Ifλ,gµ(γ (x)) and β0,g0 = inf γ∈Γ0,g0x∈RsupN I0,g0(γ (x)) .

By Lemma 7.1 (ii) , for 0 < ρ < 1, we have

βfλ,gµ ≥ (1 − ρ)p−2p (1 + µ kb/ak)p−22 β0,g0 − 2 − q 2q (ρSp) q q−2(λ kf +kLq∗) 2 2−q (7.2) and βfλ,gµ ≤ (1 + ρ) p p−2β 0,g0 + 2 − q 2q (ρSp) q q−2(λ kf +kLq∗ + kf−kLq∗) 2 2−q . (7.3)

We need the following results. Lemma 7.2 α∞ < β0,g0 < 2α

.

Proof. Bahri-Li [6] prove that equation (E0,g0) admits at least one positive

solution u0 and J0,g0(u0) = β0,g0 < 2α

. Moreover, by the condition (D4) , equation (E0,g0) does not have a positive ground state solution. Hence, α

< β0,g0 < 2α

.

Theorem 7.3 Let λ0, µ0 be as in Lemma 5.5. Then there exist positive num-bers λe0 ≤ λ0e0 ≤ µ0 and ν0 such that for λ ∈

 0,λe0  , µ ∈ (0,µe0) and kf−kLq∗ < ν0, we have α+f λ,gµ+ α ∞ < βfλ,gµ < α − fλ,gµ+ α ∞ . Furthermore, equation Efλ,gµ 

has a positive solution vfλ,gµ such that

Jfλ,gµ



vfλ,gµ



= βfλ,gµ.

Proof. By Lemma 7.1 (ii), we also have that for 0 < ρ < 1

α−f λ,gµ ≥ (1 − ρ)p−2p (1 + µ kb/ak)p−22 α∞− 2 − q 2q (ρSp) q q−2(λ kf +kLq∗) 2 2−q and αf− λ,gµ ≤ (1 + ρ) p p−2α+2 − q 2q (ρSp) q q−2(λ kf +kLq∗ + kf−kLq∗) 2 2−q .

For any ε > 0 there exist positive numbers λe1 ≤ λ0, e

µ1 ≤ µ0 and ν1 such that for λ ∈0,λe1  , µ ∈ (0,µe1) and kf−kLq∗ < ν1, we have α∞− ε < αf− λ,gµ < α ∞ + ε.

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Thus,

2α∞− ε < α−f

λ,gµ+ α

< 2α∞+ ε.

Applying (7.2) and (7.3) for any δ > 0 there exist positive numbers λe2

λ0,µe2 ≤ µ0 and ν2 such that for λ ∈  0,λe2  , µ ∈ (0,µe2) and kf−kLq∗ < ν2, we have β0,g0 − δ < βfλ,gµ < β0,g0 + δ.

Fix a small 0 < ε < (2α∞− β0,g0) /2, since α

< β

0,g0 < 2α

, choosing a δ > 0 such that for λ < λe0 = min

n e λ1,λe2 o , µ < µe0 = min {µe1,µe2} and kf−kLq∗ < ν0 = min {ν1, ν2} , we get α+f λ,gµ+ α ∞ < α∞ < βfλ,gµ < 2α ∞− ε < α− fλ,gµ+ α ∞ . Therefore, by Proposition 3.2, we obtain thatEfλ,gµ



has a positive solution vfλ,gµ such that Jfλ,gµ



vfλ,gµ



= βfλ,gµ.

We can now complete the proof of Theorem 1.2: for λ ∈0,λe0 

, µ ∈ (0,µe0)

and kf−kLq∗ < ν0, from Theorems 1.1, 7.3, equation (Efλ,gµ) has at least four

positive solutions.

Acknowledgements

The author is grateful for the referee’s valuable suggestions and helps.

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