Multiple positive solutions of semilinear elliptic equations involving concave and convex nonlinearities in ℝ N

Boundary Value Problems20122012:24

DOI: 10.1186/1687-2770-2012-24

Received: 13 July 2011

Accepted: 24 February 2012

Published: 24 February 2012

Abstract

In this article, we investigate the effect of the coefficient f(z) of the sub-critical nonlinearity. For sufficiently large λ > 0, there are at least k + 1 positive solutions of the semilinear elliptic equations

- Δ v + λ v = f ( z ) v p - 1 + h ( z ) v q - 1 in N ; v H 1 ( N ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equa_HTML.gif

where 1 ≤ q < 2 < p < 2* = 2N/(N - 2) for N ≥ 3.

AMS (MOS) subject classification: 35J20; 35J25; 35J65.

Keywords

semilinear elliptic equations concave and convex positive solutions

1 Introduction

For N ≥ 3, 1 ≤ q < 2 < p < 2* = 2N/(N - 2), we consider the semilinear elliptic equations
- Δ v + λ v = f ( z ) v p - 1 + h ( z ) v q - 1 in N ; v H 1 ( N ) , ( E λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equb_HTML.gif

where λ > 0.

Let f and h satisfy the following conditions:

(f 1) f is a positive continuous function in ℝ N and lim|z| → ∞f(z) = f > 0.

(f 2) there exist k points a1, a2,..., a k in ℝ N such that
f ( a i ) = f max = max z N f ( z ) for 1 i k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equc_HTML.gif

and f < fmax.

(h 1) h L p p - q ( N ) L ( N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq1_HTML.gif and h 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq2_HTML.gif.

Semilinear elliptic problems involving concave-convex nonlinearities in a bounded domain
- Δ u = c h ( z ) u q - 2 u + u p - 2 u in Ω ; u H 0 1 ( Ω ) , ( E c ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equd_HTML.gif

have been studied by Ambrosetti et al. [1] (h ≡ 1, and 1 < q < 2 < p ≤ 2* = 2N/(N- 2)) and Wu [2] h C ( Ω ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq3_HTML.gif and changes sign, 1 < q < 2 < p < 2*). They proved that this equation has at least two positive solutions for sufficiently small c > 0. More general results of Equation (E c ) were done by Ambrosetti et al. [3], Brown and Zhang [4], and de Figueiredo et al. [5].

In this article, we consider the existence and multiplicity of positive solutions of Equation (E λ ) in ℝ N . For the case q = λ = 1 and f(z) ≡ 1 for all z ∈ ℝ N , suppose that h is nonnegative, small, and exponential decay, Zhu [6] showed that Equation (E λ ) admits at least two positive solutions in ℝ N . Without the condition of exponential decay, Cao and Zhou [7] and Hirano [8] proved that Equation (E λ ) admits at least two positive solutions in ℝ N . For the case q = λ = 1, by using the idea of category and Bahri-Li's minimax argument, Adachi and Tanaka [9] asserted that Equation (E λ ) admits at least four positive solutions in ℝ N , where f(z) ≢ 1, f(z) ≥ 1 - C exp((-(2 + δ) |z|) for some C, δ > 0, and sufficiently small h H - 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq4_HTML.gif. Similarly, in Hsu and Lin [10], they have studied that there are at least four positive solutions of the general case -Δu + u = f(z)vp-1+ λh(z) vq-1in ℝ N for sufficiently small λ > 0.

By the change of variables
ε = λ - 1 2 and u ( z )  =  ε 2 p - 2 v ( ε z ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Eque_HTML.gif
Equation (E λ ) is transformed to
- Δ u + u = f ( ε z ) u p - 1 + ε 2 ( p - q ) p - 2 h ( ε z ) u q - 1 in N ; u H 1 ( N ) , ( E ε ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equf_HTML.gif
Associated with Equation (E ε ), we consider the C1-functional J ε , for uH1 (ℝ N ),
J ε ( u ) = 1 2 u H 2 - 1 p N f ( ε z ) u + p d z - 1 q N ε 2 ( p - q ) p - 2 h ( ε z ) u + q d z , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equg_HTML.gif

where u H 2 = N Δ u 2 + u 2 d z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq5_HTML.gif is the norm in H1 (ℝ N ) and u+ = max{u, 0} ≥ 0. We know that the nonnegative weak solutions of Equation (E ε ) are equivalent to the critical points of J ε . This article is organized as follows. First of all, we use the argument of Tarantello [11] to divide the Nehari manifold M ε into the two parts M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq6_HTML.gif and M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq7_HTML.gif. Next, we prove that the existence of a positive ground state solution u 0 M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq8_HTML.gif of Equation (E ε ). Finally, in Section 4, we show that the condition (f 2) affects the number of positive solutions of Equation (E ε ), that is, there are at least k critical points u 1 , . . . , u k M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq9_HTML.gif of J ε such that J ε ( u i ) = β ε i ( ( PS ) - value ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq10_HTML.gif for 1 ≤ ik.

Let
S = sup u H 1 ( N ) u H = 1 u L p , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equh_HTML.gif
then
u L p S u H for any u H 1 ( N ) \ { 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ1_HTML.gif
(1.1)
For the semilinear elliptic equations
- Δ u + u = f ( ε z ) u p - 1 in N ; u H 1 ( N ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ2_HTML.gif
(E0)
we define the energy functional I ε ( u ) = 1 2 u H 2 - 1 p N f ( ε z ) u + p d z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq11_HTML.gif, and
γ ε = inf u N ε I ε ( u ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equi_HTML.gif

where N ε = {uH1 (ℝ N ) \ {0} | u+ ≢ 0 and I ε ( u ) , u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq12_HTML.gif}. Note that

(i) if ff, we define I ( u ) = 1 2 u H 2 - 1 p N f u + p d z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq13_HTML.gif and
γ = inf u N I ( u ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equj_HTML.gif

where N = {uH1 (ℝ N ) \ {0} | u+ ≢ 0 and I ( u ) , u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq14_HTML.gif};

(ii) if ffmax, we define I max ( u ) = 1 2 u H 2 - 1 p N f max u + p d z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq15_HTML.gif and
γ max = inf u N max I max ( u ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equk_HTML.gif

where Nmax = {uH1 (ℝ N ) \ {0} | u+ ≢ 0 and I max ( u ) , u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq16_HTML.gif}.

Lemma 1.1
γ max = p - 2 2 p ( f max S p ) - 2 / ( p - 2 ) > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equl_HTML.gif

Proof. It is similar to Theorems 4.12 and 4.13 in Wang [[12], p. 31].

Our main results are as follows.

(I) Let Λ = ε2(p-q)/(p-2). Under assumptions (f 1) and (h 1), if
0 < Λ < Λ 0 = ( p - 2 ) 2 - q f max 2 - q p - 2 ( p - q ) S 2 q - p p - 2 h # - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equm_HTML.gif

where ∥h# is the norm in L p p - q ( N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq17_HTML.gif, then Equation (E ε ) admits at least a positive ground state solution. (See Theorem 3.4)

(II) Under assumptions (f 1) - (f 2) and (h 1), if λ is sufficiently large, then Equation (E λ ) admits at least k + 1 positive solutions. (See Theorem 4.8)

2 The Nehari manifold

First of all, we define the Palais-Smale (denoted by (PS)) sequences and (PS)-conditions in H1(ℝ N ) for some functional J.

Definition 2.1 (i) For β ∈ ℝ, a sequence {u n } is a (PS) β -sequence in H1(ℝ N ) for J if J(u n ) = β + o n (1) and J'(u n ) = o n (1) strongly in H-1 (ℝ N ) as n → ∞, where H-1 (ℝ N ) is the dual space of H1(ℝ N );

(ii) J satisfies the (PS) β -condition in H1(ℝN) if every (PS) β -sequence in H1(ℝ N ) for J contains a convergent subsequence.

Next, since J ε is not bounded from below in H1 (ℝ N ), we consider the Nehari manifold
M ε = u H 1 ( N ) \ { 0 } u + 0 and J ε ( u ) , u = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ3_HTML.gif
(2.1)
where
J ε ( u ) , u = u H 2 - N f ( ε z ) u + p d z - N ε 2 ( p - q ) p - 2 h ( ε z ) u + q d z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equn_HTML.gif

Note that M ε contains all nonnegative solutions of Equation (E ε ). From the lemma below, we have that J ε is bounded from below on M ε .

Lemma 2.2 The energy functional J ε is coercive and bounded from below on M ε .

Proof. For uM ε , by (2.1), the Hölder inequality p 1 = p p - q , p 2 = p q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq18_HTML.gif and the Sobolev embedding theorem (1.1), we get
J ε ( u ) = 1 2 - 1 p u H 2 - 1 q - 1 p N ε 2 ( p - q ) p - 2 h ( ε z ) u + q d z u H q p p - 2 2 u H 2 - q - p - q q ε 2 ( p - q ) p - 2 h # S q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equo_HTML.gif

Hence, we have that J ε is coercive and bounded from below on M ε .

Define
ψ ε ( u ) = J ε ( u ) , u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equp_HTML.gif
Then for uM ε , we get
ψ ε ( u ) , u = 2 u H 2 - p N f ( ε z ) u + p d z - q N ε 2 ( p - q ) p - 2 h ( ε z ) u + q d z = ( p - q ) N ε 2 ( p - q ) p - 2 h ( ε z ) u + q d z - ( p - 2 ) u H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ4_HTML.gif
(2.2)
= ( 2 - q ) u H 2 - ( p - q ) N f ( ε z ) u + p d z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ5_HTML.gif
(2.3)
We apply the method in Tarantello [11], let
M ε + = { u M ε ψ ε ( u ) , u > 0 } ; M ε 0 = { u M ε ψ ε ( u ) , u = 0 } ; M ε - = { u M ε ψ ε ( u ) , u < 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equq_HTML.gif

Lemma 2.3 Under assumptions (f 1) and (h 1), if 0 < Λ (= ε2(p-q)/(p- 2)) < Λ0, then M ε 0 = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq19_HTML.gif.

Proof. See Hsu and Lin [[10], Lemma 5].

Lemma 2.4 Suppose that u is a local minimizer for J ε on M ε and u M ε 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq20_HTML.gif. Then J ε ( u ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq21_HTML.gif in H-1 (ℝ N ).

Proof. See Brown and Zhang [[4], Theorem 2.3].

Lemma 2.5 We have the following inequalities.

(i) N h ( ε z ) u + q d z > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq22_HTML.gif for each u M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq23_HTML.gif;

(ii) u H < p - q p - 2 Λ h # S q 1 / ( 2 - q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq24_HTML.gif for each u M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq23_HTML.gif;

(iii) u H > 2 - q ( p - q ) f max S p 1 / ( p - 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq25_HTML.gif for each u M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq26_HTML.gif;

(iv) If 0 < Λ = ε 2 ( p - q ) / ( p - 2 ) < q Λ 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq27_HTML.gif, then J ε (u) > 0 for each u M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq26_HTML.gif.

Proof. (i) It can be proved by using (2.2).

(ii) For any u M ε + M ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq28_HTML.gif, by (2.2), we apply the Hölder inequality ( p 1 = p p - q , p 2 = p q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq29_HTML.gif to obtain that
0 < ( p - q ) N Λ h ( ε z ) u + q d z - ( p - 2 ) u H 2 ( p - q ) Λ h # S q u H q - ( p - 2 ) u H 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equr_HTML.gif
(iii) For any u M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq26_HTML.gif, by (2.3), we have that
u H 2 < p - q 2 - q N f ( ε z ) u + p d z p - q 2 - q S p u H p f max . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equs_HTML.gif
(iv) For any u M ε - M ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq30_HTML.gif, by (iii), we get that
J ε ( u ) = 1 2 - 1 p u H 2 - 1 q - 1 p N Λ h ( ε z ) u + q d z u H q p p - 2 2 u H 2 - q - p - q q Λ h # S q > 1 p 2 - q ( p - q ) f max S p q p - 2 p - 2 2 2 - q ( p - q ) f max S p 2 - q p - 2 - p - q q Λ h # S q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equt_HTML.gif

Thus, if 0 < Λ < q 2 ( p - 2 ) 2 - q f max 2 - q p - 2 ( p - q ) S 2 q - p p - 2 h # - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq31_HTML.gif, we get that J ε (u) ≥ d0 > 0 for some constant d0 = d0(ε, p, q, S, ∥h # , fmax).

For uH1 (ℝ N ) \ {0} and u + ≢ 0, let
t ̄ = t ̄ ( u ) = ( 2 - q ) u H 2 ( p - q ) N f ( ε z ) u + p d z 1 / ( p - 2 ) > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equu_HTML.gif

Lemma 2.6 For each uH1 (ℝ N )\ {0} and u + ≢ 0, we have that

(i) if N h ( ε z ) u + q d z = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq32_HTML.gif, then there exists a unique positive number t - = t - ( u ) > t ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq33_HTML.gif such that t - u M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq34_HTML.gif and J ε (t-u) = supt ≥ 0J ε (tu);

(ii) if 0 < Λ ( = ε2(p-q)/(p- 2)) < Λ0 and N h ( ε z ) u + q d z > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq35_HTML.gif, then there exist unique positive numbers t + = t + ( u ) < t ̄ < t - = t - ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq36_HTML.gif such that t + u M ε + , t - u M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq37_HTML.gif and
J ε ( t + u ) = inf 0 t t ̄ J ε ( t u ) , J ε ( t - u ) = sup t t ̄ J ε ( t u ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equv_HTML.gif

Proof. See Hsu and Lin [[10], Lemma 7].

Applying Lemma 2.3 ( M ε 0 = for 0 < Λ < Λ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq38_HTML.gif, we write M ε = M ε + M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq39_HTML.gif, where
M ε + = u M ε | ( 2 - q ) u H 2 - ( p - q ) N f ( ε z ) u + p d z > 0 , M ε - = u M ε | ( 2 - q ) u H 2 - ( p - q ) N f ( ε z ) u + p d z < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equw_HTML.gif
Define
α ε = inf u M ε J ε ( u ) ; α ε + = inf u M ε + J ε ( u ) ; α ε - = inf u M ε - J ε ( u ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equx_HTML.gif

Lemma 2.7 (i) If 0 < Λ ( = ε2(p-q)/(p- 2)) < Λ0, then α ε α ε + < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq40_HTML.gif;

(ii) If 0 < Λ < q Λ0/2, then α ε - d 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq41_HTML.gif for some constant d0 = d0 (ε, p, q, S, ∥h # , fmax).

Proof. (i) Let u M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq23_HTML.gif, by (2.2), we get
( p - 2 ) u H 2 < ( p - q ) N Λ h ( ε z ) u + q d z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equy_HTML.gif
Then
J ε ( u ) = 1 2 - 1 p u H 2 - 1 q - 1 p N Λ h ( ε z ) u + q d z < 1 2 - 1 p - 1 q - 1 p p - 2 p - q u H 2 = - ( 2 - q ) ( p - 2 ) 2 p q u H 2 < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equz_HTML.gif

By the definitions of α ε and α ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq42_HTML.gif, we deduce that α ε α ε + < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq43_HTML.gif.

(ii) See the proof of Lemma 2.5 (iv).

Applying Ekeland's variational principle and using the same argument in Cao and Zhou [7] or Tarantello [11], we have the following lemma.

Lemma 2.8 (i) There exists a ( P S ) α ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq44_HTML.gif -sequence {u n } in M ε for J ε ;

(ii) There exists a ( P S ) α ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq45_HTML.gif-sequence {u n } in M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq46_HTML.gif for J ε ;

(iii) There exists a ( P S ) α ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq47_HTML.gif-sequence {u n } in M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq48_HTML.gif for J ε .

3 Existence of a ground state solution

In order to prove the existence of positive solutions, we claim that J ε satisfies the (PS) β -condition in H1(ℝ N ) for β - , γ - C 0 Λ 2 2 - q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq49_HTML.gif, where Λ = ε2(p-q)/(p- 2)and C0 is defined in the following lemma.

Lemma 3.1 Assume that h satisfies (h 1) and 0 < Λ ( = ε2(p-q)/(p- 2)) < Λ0. If {u n } is a (PS) β -sequence in H1(ℝ N ) for J ε with u n u weakly in H1 (ℝ N ), then J ε ( u ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq50_HTML.gif in H-1 (ℝ N ) and J ε ( u ) - C 0 Λ 2 2 - q - C 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq51_HTML.gif, where
C 0 = ( 2 - q ) ( p - q ) h # S q 2 2 - q / 2 p q ( p - 2 ) q 2 - q , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equaa_HTML.gif
and
C 0 = ( p - 2 ) ( 2 - q ) p p - 2 / 2 p q f max ( p - q ) 2 p - 2 S 2 p p - 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equab_HTML.gif
Proof. Since {u n } is a (PS) β -sequence in H1(ℝ N ) for J ε with u n u weakly in H1 (ℝ N ), it is easy to check that J ε ( u ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq50_HTML.gif in H -1 (ℝ N ) and u ≥ 0. Then we have J ε ( u ) , u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq52_HTML.gif, that is, N f ( ε z ) u p d z = u H 2 - N Λ h ( ε z ) u q d z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq53_HTML.gif. Hence, by the Young inequality p 1 = 2 q and p 2 = 2 2 - q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq54_HTML.gif
J ε ( u ) = 1 2 - 1 p u H 2 - 1 q - 1 p N Λ h ( ε z ) u q d z p - 2 2 p u H 2 - p - q p q Λ h # S q u H q p - 2 2 p u H 2 - p - 2 p q q u H 2 2 + p - q p - 2 Λ h # S q 2 2 - q 2 - q 2 - ( p - 2 ) ( 2 - q ) p p - 2 2 p q f max ( p - q ) 2 p - 2 S 2 p p - 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equac_HTML.gif

Lemma 3.2 Assume that f and h satisfy (f 1) and (h 1). If 0 < Λ ( = ε2(p-q)/(p- 2)) < Λ0, then J ε satisfies the (PS) β -condition in H1(ℝ N ) for β - , γ - C 0 Λ 2 2 - q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq49_HTML.gif.

Proof. Let {u n } be a (PS) β -sequence in H1(ℝ N ) for J ε such that J ε (u n ) = β + o n (1) and J ε ( u n ) = o n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq55_HTML.gif (1) in H-1(ℝ N ). Then
β + c n + d n u n H p J ε ( u n ) - 1 p J ε ( u n ) , ( u n ) = 1 2 - 1 p u n H 2 - 1 q - 1 p N ε 2 ( p - q ) p - 2 h ( ε z ) ( u n ) + q d z p - 2 2 p u n H 2 - p - q p q Λ h # S q u n H q , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equad_HTML.gif
where c n = o n (1), d n = o n (1) as n → ∞. It follows that {u n } is bounded in H1(ℝ N ). Hence, there exist a subsequence {u n } and a nonnegative uH1 (ℝ N ) such that J ε ( u ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq50_HTML.gif in H-1 (ℝ N ), u n u weakly in H1 (ℝ N ), u n u a.e. in ℝ N , u n u strongly in L l o c s N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq56_HTML.gif for any 1 ≤ s < 2*. Using the Brézis-Lieb lemma to get (3.1) and (3.2) below.
N f ( ε z ) ( u n - u ) + p d z = N f ( ε z ) ( u n ) + p d z - N f ( ε z ) u p d z + o n ( 1 ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ6_HTML.gif
(3.1)
N h ( ε z ) ( u n - u ) + q d z = N h ( ε z ) ( u n ) + q d z - N h ( ε z ) u q d z + o n ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ7_HTML.gif
(3.2)
Next, claim that
N h ( ε z ) u n - u q d z 0 as n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ8_HTML.gif
(3.3)
For any σ > 0, there exists r > 0 such that [ B N ( 0 ; r ) ] c h ( ε z ) p p - q d z < σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq57_HTML.gif. By the Hölder inequality and the Sobolev embedding theorem, we get
N h ( ε z ) u n - u q d z B N ( 0 ; r ) h ( ε z ) u n - u q d z + [ B N ( 0 ; r ) ] c h ( ε z ) u n - u q d z h # B N ( 0 ; r ) u n - u p d z q / p + S q [ B N ( 0 ; r ) ] c h ( ε z ) p p - q d z p - q p u n - u H q C σ + o n ( 1 ) . ( { u n } is bounded in H 1 ( N ) and u n u in L l o c p ( N ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equae_HTML.gif
Applying (f 1) and u n u in L l o c p ( N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq58_HTML.gif, we get that
N f ( ε z ) ( u n - u ) + p d z = N f ( u n - u ) + p d z + o n ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ9_HTML.gif
(3.4)
Let p n = u n - u. Suppose p n ↛ 0 strongly in H1 (ℝ N ). By (3.1)-(3.4), we deduce that
p n H 2 = u n H 2 - u H 2 + o n ( 1 ) = N f ( ε z ) ( u n ) + p d z - N ε 2 ( p - q ) p - 2 h ( ε z ) ( u n ) + q d z - N f ( ε z ) u p d z + N ε 2 ( p - q ) p - 2 h ( ε z ) u q d z + o n ( 1 ) = N f ( ε z ) ( u n - u ) + p d z + o n ( 1 ) = N f ( p n ) + p d z + o n ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equaf_HTML.gif
Then
I ( p n ) = 1 2 p n H 2 - 1 p N f ( p n ) + p d z = 1 2 - 1 p p n H 2 + o n ( 1 ) > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equag_HTML.gif
By Theorem 4.3 in Wang [12], there exists a sequence {s n } ⊂ ℝ+ such that s n = 1 + o n (1), {s n p n } ⊂ N and I(s n p n ) = I(p n ) + o n (1). It follows that
γ I ( s n p n ) = I ( p n ) + o n ( 1 ) = J ε ( u n ) - J ε ( u ) + o n ( 1 ) = β - J ε ( u ) + o n ( 1 ) < γ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equah_HTML.gif

which is a contradiction. Hence, u n u strongly in H1(ℝ N ).

Remark 3.3 By Lemma 1.1, we obtain
C 0 = 2 - q q 2 - q p - q 2 p - 2 γ max < γ max < γ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equai_HTML.gif

and γ - C 0 Λ 2 2 - q > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq59_HTML.gif for 0 < Λ < Λ0.

By Lemma 2.8 (i), there is a ( PS ) α ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq60_HTML.gif-sequence {u n } in M ε for J ε . Then we prove that Equation (E ε ) admits a positive ground state solution u0 in ℝ N .

Theorem 3.4 Under assumptions (f 1), (h 1), if 0 < Λ ( = ε2(p-q)/(p- 2)) < Λ0, then there exists at least one positive ground state solution u0 of Equation (E ε ) in N . Moreover, we have that u 0 M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq61_HTML.gif and
J ε ( u 0 ) = α ε = α ε + - C 0 Λ 2 2 - q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ10_HTML.gif
(3.5)

Proof. By Lemma 2.8 (i), there is a minimizing sequence {u n } ⊂ M ε for J ε such that J ε (u n ) = α ε + o n (1) and J ε ( u n ) = o n ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq62_HTML.gif in H-1 (ℝ N ). Since α ε < 0 < γ - C 0 Λ 2 2 - q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq63_HTML.gif, by Lemma 3.2, there exist a subsequence {u n } and u0H1 (ℝ N ) such that u n u0 strongly in H1 (ℝ N ). It is easy to see that u 0 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq64_HTML.gif is a solution of Equation (E ε ) in ℝ N and J ε (u0) = α ε . Next, we claim that u 0 M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq61_HTML.gif. On the contrary, assume that u 0 M ε - ( M ε 0 = for 0 < Λ ( = ε 2 ( p - q ) / ( p - 2 ) ) < Λ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq65_HTML.gif.

We get that
N Λ h ( ε z ) ( u 0 ) + q d z > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equaj_HTML.gif
Otherwise,
0 = N Λ h ( ε z ) ( u 0 ) + q d z = N Λ h ( ε z ) ( u n ) + q d z + o n ( 1 ) = u n H 2 - N f ( ε z ) ( u n ) + p d z + o n ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equak_HTML.gif
It follows that
α ε + o n ( 1 ) = J ε ( u n ) = 1 2 - 1 p u n H 2 + o n ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equal_HTML.gif
which contradicts to α ε < 0. By Lemma 2.6 (ii), there exist positive numbers t + < t ̄ < t - = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq66_HTML.gif such that t + u 0 M ε + , t - u 0 M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq67_HTML.gif and
J ε ( t + u 0 ) < J ε ( t - u 0 ) = J ε ( u 0 ) = α ε , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equam_HTML.gif
which is a contradiction. Hence, u 0 M ε + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq61_HTML.gif and
- C 0 Λ 2 2 - q J ε ( u 0 ) = α ε = α ε + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equan_HTML.gif

By Lemma 2.4 and the maximum principle, then u0 is a positive solution of Equation (E ε ) in ℝ N .

4 Existence of k+ 1 solutions

From now, we assume that f and h satisfy (f 1)-(f 2) and (h 1). Let wH1 (ℝ N ) be the unique, radially symmetric, and positive ground state solution of Equation (E 0) in ℝ N for f = fmax. Recall the facts (or see Bahri and Li [13], Bahri and Lions [14], Gidas et al. [15], and Kwong [16]).

(i) w L ( N ) C l o c 2 , θ ( N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq68_HTML.gif for some 0 < θ < 1 and lim z w ( z ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq69_HTML.gif;

(ii) for any ε > 0, there exist positive numbers C1, C 1 , C 2 ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq70_HTML.gif, and C 3 ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq71_HTML.gif such that for all z ∈ ℝ N
C 2 ε exp - 1 - ε z w ( z ) C 1 exp - z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equao_HTML.gif
and
w ( z ) C 3 ε exp - ( 1 - ε ) z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equap_HTML.gif
For 1 ≤ ik, we define
w ε i ( z ) = w z - a i ε , where f ( a i ) = f max . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equaq_HTML.gif

Clearly, w ε i ( z ) H 1 N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq72_HTML.gif. By Lemma 2.6 (ii), there is a unique number t ε i - > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq73_HTML.gif such that t ε i - w ε i M ε - M ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq74_HTML.gif, where 1 ≤ ik.

We need to prove that
lim ε 0 + J ε t ε i - w ε i γ max uniformly in i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equar_HTML.gif
Lemma 4.1 (i) There exists a number t0 > 0 such that for 0tt0 and any ε > 0, we have that
J ε t w ε i < γ max u n i f o r m l y i n i ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equas_HTML.gif
(ii) There exist positive numbers t1 and ε1 such that for any t > t1 and ε < ε1, we have that
J ε t w ε i < 0 u n i f o r m l y i n i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equat_HTML.gif
Proof. (i) Since J ε is continuous in H 1 N , { w ε i } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq75_HTML.gif is uniformly bounded in H1 (ℝ N ) for any ε > 0, and γmax > 0, there is t0 > 0 such that for 0 ≤ tt0 and any ε > 0
J ε t w ε i < γ max . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equau_HTML.gif
(ii) There is an r0 > 0 such that f (z) ≥ fmax/2 for zB N (a i ; r0) uniformly in i. Then there exists ε1 > 0 such that for ε < ε1
J ε t w ε i = t 2 2 w ε i H 2 - t p p N f ( ε z ) w ε i p d z - t q q N Λ h ( ε z ) w ε i q d z t 2 2 N w 2 w 2 d z - t p 2 p B N ( 0 ; 1 ) f max w p d z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equav_HTML.gif
Thus, there is t1 >0 such that for any t > t1 and ε < ε1
J ε t w ε i < 0 uniformly in i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equaw_HTML.gif
Lemma 4.2 Under assumptions (f 1), (f 2), and (h 1). If 0 < Λ ( = ε2(p-q)/(p- 2)) < q Λ0/ 2, then
lim ε 0 + sup t 0 J ε t w ε i γ max u n i f o r m l y i n i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equax_HTML.gif
Proof. By Lemma 4.1, we only need to show that
lim ε 0 + sup t 0 t t 1 J ε t w ε i γ max uniformly in i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equay_HTML.gif
We know that supt ≥0Imax (tw) = γmax. For t0tt1, we get
J ε t w ε i = 1 2 t w ε i H 2 - 1 p N f ( ε z ) t w ε i p d z - 1 q N Λ h ( ε z ) t w ε i q d z = t 2 2 N w z - a i ε 2 + w z - a i ε d z - t p p N f ( ε z ) w z - a i ε p d z - t q q N Λ h ( ε z ) w z - a i ε q d z = t 2 p N w 2 + w 2 d z - t p p N f max w p d z + t p p N ( f max - f ( ε z ) ) w z - a i ε p d z - t q q Λ N h ( ε z ) w z - a i ε q d z γ max + t 1 p p N ( f max - f ( ε z ) ) w z - a i ε p d z - t 0 q q Λ N h ( ε z ) w z - a i ε d z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equaz_HTML.gif
Since
N ( f max - f ( ε z ) ) w z - a i ε p d z = N f max - f ( ε z + a i ) w p d z = o ( 1 ) as ε 0 + uniformly in i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equba_HTML.gif
and
Λ N h ( ε z ) w z - a i ε q d z ε 2 ( p - q ) p - 2 h # S q w H q = o ( 1 ) as ε 0 + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbb_HTML.gif

then lim ε 0 + sup t 0 t t 1 J ε t w ε i γ max http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq76_HTML.gif, that is, lim ε 0 + sup t 0 J ε t w ε i γ max http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq77_HTML.gif uniformly in i.

Applying the results of Lemmas 2.6, 2.7(ii), and 4.2, we can deduce that
0 < d 0 α ε - γ max + o ( 1 ) as ε 0 + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbc_HTML.gif
Since γmax < γ, there exists ε0 > 0 such that
γ max < γ - C 0 Λ 2 2 - q for any ε < ε 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ11_HTML.gif
(4.1)
Choosing 0 < ρ0 < 1 such that
B ρ 0 N ( a i ) ¯ B ρ 0 N ( a j ) ¯ = for i j and 1 i , j k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbd_HTML.gif

where B ρ 0 N ( a i ) ¯ = z N | z - a i ρ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq78_HTML.gif and f(a i ) = fmax. Define K = {a i | 1 ≤ ik} and K ρ 0 / 2 = i = 1 k B ρ 0 / 2 N ( a i ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq79_HTML.gif. Suppose i = 1 k B ρ 0 N ( a i ) ¯ B r 0 N ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq80_HTML.gif for some r0 > 0.

Let Q ε : H1 (ℝ N ) \ {0} → ℝ N be given by
Q ε ( u ) = N χ ( ε z ) u p d z N u p d z , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Eqube_HTML.gif

where χ : ℝ N → ℝ N , χ (z) = z for |z| ≤ r0 and χ (z) = r0z/|z| for |z| > r0.

Lemma 4.3 There exists 0 < ε0ε0 such that if ε < ε0, then Q ε t ε i - w ε i K ρ 0 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq81_HTML.gif for each 1 ≤ ik.

Proof. Since
Q ε t ε i - w ε i = N χ ( ε z ) w z - a i ε p d z N w z - a i ε p d z = N χ ( ε z + a i ) w z p d z N w z p d z a i a s ε 0 + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbf_HTML.gif
there exists ε0 > 0 such that
Q ε t ε i - w ε i K ρ 0 / 2 for any ε < ε 0 and each 1 i k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbg_HTML.gif

Lemma 4.4 There exists a number δ ̄ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq82_HTML.gif such that if uN ε and I ε ( u ) γ max + δ ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq83_HTML.gif, then Q ε ( u ) K ρ 0 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq84_HTML.gif for any 0 < ε < ε0.

Proof. On the contrary, there exist the sequences {ε n } ⊂ ℝ+ and { u n } N ε n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq85_HTML.gif such that ε n 0 , I ε n ( u n ) = γ max ( > 0 ) + o n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq86_HTML.gif (1) as n → ∞ and Q ε n ( u n ) K ρ 0 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq87_HTML.gif for all n ∈ ℕ. It is easy to check that {u n } is bounded in H1 (ℝ N ). Suppose u n → 0 strongly in L p (ℝ N ). Since
u n H 2 = N f ( ε n z ) ( u n ) + p d z for each n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbh_HTML.gif
and
I ε n ( u n ) = 1 2 u n H 2 - 1 p N f ( ε n z ) ( u n ) + p d z = γ max + o n ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbi_HTML.gif
then
γ max + o n ( 1 ) = I ε n ( u n ) = 1 2 - 1 p N f ( ε n z ) ( u n ) + p d z = o n ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbj_HTML.gif
which is a contradiction. Thus, u n ↛ 0 strongly in L p (ℝ N ). Applying the concentration-compactness principle (see Lions [17] or Wang [[12], Lemma 2.16]), then there exist a constant d0 > 0 and a sequence z n ̃ N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq88_HTML.gif such that
B N ( z n ̃ ; 1 ) u n ( z ) 2 d z d 0 > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ12_HTML.gif
(4.2)
Let v n ( z ) = u n z + z n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq89_HTML.gif, there are a subsequence {v n } and vH1 (ℝ N ) such that v n v weakly in H1 (ℝ N ). Using the similar computation in Lemma 2.6, there is a sequence s max n + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq90_HTML.gif such that v n ̃ = s max n v n N max http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq91_HTML.gif and
0 < γ max I max v n ̃ I ε n s max n u n I ε n ( u n ) = γ max + o n ( 1 ) as n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbk_HTML.gif

We deduce that a convergent subsequence s max n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq92_HTML.gif satisfies s max n s 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq93_HTML.gif. Then there are subsequences v n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq94_HTML.gif and H 1 N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq95_HTML.gif such that n ( = s 0 v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq96_HTML.gif weakly in H1 (ℝ N ). By (4.2), then 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq97_HTML.gif. Moreover, we can obtain that v n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq98_HTML.gif strongly in H1 (ℝ N ) and I max ( ) = γ max http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq99_HTML.gif. Now, we want to show that there exists a subsequence { z n } = ε n z n ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq100_HTML.gif such that z n z0K.

(i) Claim that the sequence {z n } is bounded in ℝ N . On the contrary, assume that |z n | → ∞, then
γ max = I max ( ) < I ( ) lim inf n 1 2 v n ̃ H 2 - 1 p N f ( ε n z + z n ) ( v n ̃ ) + p d z = lim inf n ( s max n ) 2 2 u n H 2 - ( s max n ) p p N f ( ε n z ) ( u n ) + p d z = lim inf n I ε n ( s max n u n ) lim inf n I ε n ( u n ) = γ max , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbl_HTML.gif

which is a contradiction.

(ii) Claim that z0K. On the contrary, assume that z0K, that is, f(z0) < fmax. Then using the above argument to obtain that
γ max = I max ( ) < 1 2 H 2 - 1 p N f ( z 0 ) ( ) + p d z lim inf n 1 2 v n ̃ H 2 - 1 p N f ( ε n z + z n ) ( v n ̃ ) + p d z = γ max , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbm_HTML.gif
which is a contradiction. Since v n v ≠ 0 in H1 (ℝ N ), we have that
Q ε n ( u n ) = N χ ( ε n z ) v n ( z - z n ̃ ) p d z N v n ( z - z n ̃ ) p d z = N χ ( ε n z + ε n z n ̃ ) v n p d z N v n p d z z 0 K ρ 0 / 2 as n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbn_HTML.gif

which is a contradiction.

Hence, there exists a number δ ̄ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq82_HTML.gif such that if uN ε and I ε ( u ) γ max + δ ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq83_HTML.gif, then Q ε ( u ) K ρ 0 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq84_HTML.gif for any 0 < ε < ε0.

From (4.1), choosing 0 < δ 0 < δ ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq101_HTML.gif such that
γ max + δ 0 < γ - C 0 Λ 2 2 - q for any 0 < ε < ε 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ13_HTML.gif
(4.3)
For each 1 ≤ ik, define
O ε i = { u M ε - | Q ε ( u ) - a i < ρ 0 } , O ε i = { u M ε - | Q ε ( u ) - a i = ρ 0 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbo_HTML.gif

β ε i = inf u O ε i J ε ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq102_HTML.gif and β ̃ ε i = inf u O ε i J ε ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq103_HTML.gif.

Lemma 4.5 If u M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq104_HTML.gif and J ε (u) ≤ γmax + δ0/2, then there exists a number 0 < ε ̄ < ε 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq105_HTML.gif such that Q ε ( u ) K ρ 0 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq84_HTML.gif for any 0 < ε < ε ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq106_HTML.gif.

Proof. We use the similar computation in Lemma 2.6 to get that there is a unique positive number
s ε u = u H 2 N f ( ε z ) u + p d z 1 / ( p - 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbp_HTML.gif
such that s ε u u N ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq107_HTML.gif. We want to show that s ε u < c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq108_HTML.gif for some constant c > 0 (independent of u). First, since u M ε - M ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq109_HTML.gif,
0 < d 0 α ε - J ε ( u ) γ max + δ 0 / 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbq_HTML.gif
and J ε is coercive on M ε , then 0 < c 2 < u H 2 < c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq110_HTML.gif for some constants c1 and c2 (independent of u). Next, we claim that u L p p > c 3 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq111_HTML.gif for some constant c3 > 0 (independent of u). On the contrary, there exists a sequence { u n } M ε - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq112_HTML.gif such that
u n L p p = o n ( 1 ) as n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbr_HTML.gif
By (2.3),
2 - q p - q < N f ( ε z ) ( u n ) + p d z u n H 2 f max u n L p p c 2 = o n ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbs_HTML.gif
which is a contradiction. Thus, s ε u < c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq108_HTML.gif for some constant c > 0 (independent of u). Now, we get that
γ max + δ 0 / 2 J ε ( u ) = sup t 0 J ε ( t u ) J ε ( s ε u u ) = 1 2 s ε u u H 2 - 1 p N f ( ε z ) ( s ε u u ) + p d z - 1 q N Λ h ( ε z ) ( s ε u u ) + q d z I ε ( s ε u u ) - 1 q N Λ h ( ε z ) ( s ε u u ) + q d z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbt_HTML.gif
From the above inequality, we deduce that
I ε ( s ε u u ) γ max + δ 0 / 2 + 1 q N Λ h ( ε z ) ( s ε u u ) + q d z γ max + δ 0 / 2 + Λ h S q s ε u u H q < γ max + δ 0 / 2 + Λ c q ( c 1 ) q / 2 h S q , where Λ = ε 2 ( p - q ) / ( p - 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbu_HTML.gif
Hence, there exists 0 < ε ̄ < ε 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq105_HTML.gif such that for 0 < ε < ε ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq106_HTML.gif
I ε ( s ε u u ) γ max + δ 0 , where s ε u u N ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbv_HTML.gif
By Lemma 4.4, we obtain
Q ε ( s ε u u ) = N χ ( ε z ) s ε u u ( z ) p d z N s ε u u ( z ) p d z K ρ 0 / 2 for any 0 < ε < ε ̄ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbw_HTML.gif

or Q ε ( u ) K ρ 0 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq84_HTML.gif for any 0 < ε < ε ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq106_HTML.gif.

Applying the above lemma, we get that
β ̃ ε i γ max + δ 0 / 2 for any 0 < ε < ε ̄ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ14_HTML.gif
(4.4)
By Lemmas 4.2, 4.3, and Equation (4.3), there exists 0 < ε * ε ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq113_HTML.gif such that
β ε i J ε ( t ε i ) - w ε i γ max + δ 0 / 3 < γ - C 0 Λ 2 2 - q for any 0 < ε < ε * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ15_HTML.gif
(4.5)
Lemma 4.6 Given u O ε i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq114_HTML.gif, then there exist an η > 0 and a differentiable functional l : B(0; η) ⊂ H1(ℝ N ) → ℝ+ such that l ( 0 ) = 1 , l ( v ) ( u - v ) O ε i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq115_HTML.gif for any vB(0;η) and
l ( v ) , ϕ | ( l , v ) = ( 1 , 0 ) = ψ ε ( u ) , ϕ ψ ε ( u ) , u f o r a n y ϕ C c ( N ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ16_HTML.gif
(4.6)

where ψ ε ( u ) = J ε ( u ) , u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq116_HTML.gif.

Proof. See Cao and Zhou [7].

Lemma 4.7 For each 1 ≤ ik, there is a ( P S ) β ε i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq117_HTML.gif-sequence { u n } O ε i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq118_HTML.gif in H1(ℝ N ) for J ε .

Proof. For each 1 ≤ ik, by (4.4) and (4.5),
β ε i < β ̃ ε i for any 0 < ε < ε * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ17_HTML.gif
(4.7)
Then
β ε i = inf u O ε i O ε i J ε ( u ) for any 0 < ε < ε * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equbx_HTML.gif
Let u n i O ε i O ε i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq119_HTML.gif be a minimizing sequence for β ε i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq120_HTML.gif. Applying Ekeland's variational principle, there exists a subsequence { u n i } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq121_HTML.gif such that J ε ( u n i ) = β ε i + 1 / n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq122_HTML.gif and
J ε ( u n i ) J ε ( w ) + w - u n i H / n for all w O ε i O ε i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_Equ18_HTML.gif
(4.8)
Using (4.7), we may assume that u n i O ε i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq123_HTML.gif for sufficiently large n. By Lemma 4.6, then there exist an η n i > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq124_HTML.gif and a differentiable functional l n i : B ( 0 ; η n i ) H 1 ( N ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq125_HTML.gif such that l n i ( 0 ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-24/MediaObjects/13661_2011_Article_123_IEq126_HTML.gif, and l n i ( v ) u n i - v O ε i