Multiplicity of positive solutions for semilinear
elliptic equations in R
N ∗
Tsung-fang Wu
Department of Applied Mathematics
National University of Kaohsiung, Kaohsiung 811, Taiwan
e-mail: [email protected]
Abstract
In this paper, we study the multiplicity of positive solutions for the following semilinear elliptic equation:
−∆u + λu = f (x) up−1+ h (x) uq−1 in RN, u > 0 in RN, u ∈ H1¡RN¢, (Eλ) where 1 ≤ q < 2 < p < 2∗ (2∗ = 2N N −2 if N ≥ 3 and 2∗ = ∞ if N = 1, 2),
λ > 0, h ∈ L2−q2 ¡RN¢\ {0} is nonnegative and f ∈ C¡RN¢. We will show
how the shape of the graph of f (x) affects the number of positive solutions.
1
Introduction
In this paper, we study the multiplicity of positive solutions for the following semilinear elliptic equation:
−∆u + λu = f (x) up−1+ h (x) uq−1 in RN, u > 0 in RN, u ∈ H1¡RN¢, (Eλ) where 1 ≤ q < 2 < p < 2∗ (2∗ = 2N N −2 if N ≥ 3 and 2∗ = ∞ if N = 1, 2),
λ > 0, f ∈ C¡RN¢ and h ∈ L2−q2 ¡RN¢\ {0} is nonnegative. Associated with
equation (Eλ) , we consider the energy functional:
Jλ(u) = 1 2 Z RN |∇u|2+ λu2− 1 p Z RN f (x) |u|p −1 q Z RN h (x) |u|q.
It is well known that the functional Jλ ∈ C1
¡
H1¡RN¢, R¢ and the solutions of
equation (Eλ) are the critical points of the energy functional Jλ in H1
¡ RN¢.
Under the assumption h 6≡ 0, our equation (Eλ) can be regarded as a
pertur-bation problem of the following semilinear elliptic equation: −∆u + λu = f (x) up−1 in RN, u > 0 in RN, u ∈ H1¡RN¢. (1)
It is known that the existence of positive solutions of equation (1) is affected by the shape of the graph of f (x) . This has been the focus of a great deal of research by several authors (see Bahri-Lions [7], Lions [19], Li [18], Bahri-Li [6], Cao [9] and Cao-Noussair [10], etc.). Furthermore, if f is a positive constant, then equation (1) has a unique positive solution (see Kwong [17]).
When q = 1, there have been some progresses as follows: Zhu [26] and Hirano [14] were mainly concerned with the following equation:
−∆u + λu = up−1+ h (x) in RN, u > 0 in RN, u ∈ H1¡RN¢, (2)
where h ∈ L2¡RN¢\ {0} is nonnegative, and succeeded to find the equation (2)
has at least two positive solutions under khkL2 is sufficiently small and h(x) decays
faster than exp (−c |x|) for some c > 0. Generalizations of the result of [26] and [14] were done by Cao-Zhou [11], Jeanjean [15] and Adachi-Tanaka [1, 2]. In [2], showed the existence of at least four positive solutions of the equation:
−∆u + λu = f (x) up−1+ h (x) in RN, u > 0 in RN, u ∈ H1¡RN¢,
under the assumptions 0 < f (x) ≤ f∞ = lim
|x|→∞f (x) , h ∈ H−1
¡
RN¢\ {0}
is nonnegative and khkH−1 is sufficiently small. In [1], [11] and [15], general
equations −∆u + λu = g (x, u) + h (x) in RN, u > 0 in RN, u ∈ H1¡RN¢
were studied, where g satisfies some suitable conditions and h ∈ H−1¡RN¢\ {0}
is nonnegative, and the existence of at least two positive solutions when khkH−1
sufficiently small was proved.
The main purpose of this paper is using the shape of the graph of f (x) to prove the multiplicity of positive solutions for equation (Eλ) . Moreover, we extend
q ∈ [1, 2) and without assuming khk
the following assumptions:
(Q1) f ∈ C¡RN¢ and f ≥ 0 in RN;
(Q2) f (x) → f∞ > 0 as |x| → ∞;
(Q3) There exist some points x1, x2, . . . , xk in RN such that f (xi) are strict
maxima and satisfy
f∞< f¡xi¢= fmax ≡ max
©
f (x) | x ∈ RNª for all i = 1, 2, . . . , k. Then we have the following result.
Theorem 1.1 Assume conditions (Q1) − (Q3) hold. Then there exists λ0 > 0 such that for λ > λ0, the equation (Eλ) has at least k + 1 positive solutions.
For the other similarly problems, Ambrosetti-Brezis-Cerami [4] has been in-vestigated the following equation:
−∆u = up−1+ λuq−1 in Ω, u > 0 in Ω, u ∈ H1 0 (Ω) , (3) where 1 < q < 2 < p ≤ 2N
N −2 (N ≥ 3) and Ω is a bounded domain in RN. They
found that there exists λ0 > 0 such that the equation (3) admits at least two
positive solutions for λ ∈ (0, λ0) , a positive solution for λ = λ0 and no positive
solution exists for λ > λ0. Actually, Adimurthi-Pacella-Yadava [5],
Damascelli-Grossi-Pacella [12], Ouyang-Shi [21] and Tang [23] 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 (3)
were done by Ambrosetti-Azorezo-Peral [3] and de Figueiredo-Gossez-Ubilla [13] and Wu [25].
This paper is organized as follows. In section 2, we give some notations and preliminaries. In section 3, we prove the existence of a local minimum. In section 4, we prove Theorem 1.1.
2
Notations and Preliminaries
By the change of variables η = √1
λ, v (x) = η 2/(p−2)u (ηx) , the equation (E λ) is transformed to −∆v + v = fηvp−1+ η 2(p−q) p−2 h ηvq−1 in RN, v > 0 in RN, v ∈ H1¡RN¢, (4)
where fη = f (ηx) and hη = h (ηx) .
For u ∈ H1¡RN¢, c ∈ R, nonnegative bounded function a ∈ C¡RN¢ and
nonnegative function b ∈ L2−q2 ¡RN¢ define
Ia,b(u) = 1 2kuk 2 H1 − 1 p Z RN a |u|p− η2(p−q)p−2 Z RN b |u|q; Ma,b(c) = © u ∈ H1¡RN¢\ {0} | I0 a,b(u) , u ® = cª;
αa,b(c) = inf {Ia,b(u) | u ∈ Ma,b(c)}
where kukH1 =
¡R
RN|∇u|
2+ u2¢1/2is a standard norm in H1¡RN¢and I0
a,bdenote
the Fr´echet derivative of Ia,b. We will write Ma,b(0) and αa,b(0) as Ma,b and αa,b,
respectively. It is well known that the functional Ia,b ∈ C1
¡
H1¡RN¢, R¢ and the
solutions of equation (4) are the critical points of the energy functional Ifη,hη (see
Rabinowitz [22]). Moreover, we have the following result.
Lemma 2.1 Suppose a is a continuous bounded and nonnegative function on RN. Then α
a,0(c) = 2c for c > 0 and
αa,0 ≤ αa,0(c) + αa,0(−c) −
p − 2
2p |c| for all c ∈ R.
Proof. See Cao-Noussair [10, Lemma 2.2]. ¤ Define ψη(u) = D I0 fη,hη(u) , u E = kuk2H1 − Z RN fη|u|p − η 2(p−q) p−2 Z RN hη|u|q. Then for u ∈ Mfη,hη, ψ0 η(u) , u ® = 2 kuk2H1 − p Z RN fη|u|p− η 2(p−q) p−2 q Z RN hη|u|q = (2 − q) kuk2H1 − (p − q) Z RN fη|u|p.
Similar to the method used in Tarantello [24], we split Mfη,hη into three parts:
M+fη,hη = ½ u ∈ Mfη,hη | (2 − q) kuk 2 H1 − (p − q) Z RN fη|u|p > 0 ¾ , M0fη,hη = ½ u ∈ Mfη,hη | (2 − q) kuk 2 H1 − (p − q) Z RN fη|u|p = 0 ¾ , M−fη,hη = ½ u ∈ Mfη,hη | (2 − q) kuk 2 H1 − (p − q) Z RN fη|u|p < 0 ¾ .
Lemma 2.2 There exists η1 > 0 such that M0fη,hη = ∅ for all η ∈ (0, η1) .
Proof. Assume the contrary, that is M0
fη,hη 6= ∅ for all η > 0. Then for u ∈
M0 fη,hη, we have kuk2H1 = p − q 2 − q Z RN fη|u|p (5) and η2(p−q)p−2 Z RN hη|u|q = kuk2H1 − Z RN fη|u|p = p − 2 2 − q Z RN fη|u|p. (6) Moreover, µ p − 2 p − q ¶ kuk2H1 = kuk2H1 − Z RN fη|u|p ≤ η 2(p−q) p−2 kh ηk L2−q2 kuk q H1 = η2(p−q)p−2 − (2−q)N 2 khk L2−q2 kuk q H1, this implies kukH1 ≤ · η2(p−q)p−2 − (2−q)N 2 µ p − q p − 2 ¶ khk L2−q2 ¸ 1 2−q . (7) Let Kη : Mfη,hη → R be given by Kη(u) = C (p, q) Ã kuk2(p−1)H1 R RNfη|u| p ! 1 p−2 − η2(p−q)p−2 Z RN hη|u|q, where C (p, q) = ³ 2−q p−q ´p−1 p−2 ³p−2 2−q ´
. Then Kη(u) = 0 for all η > 0 and u ∈ M0fη,hη.
Indeed, from (5) and (6) it follows that for u ∈ M0
fη,hη, we have Kη(u) = µ 2 − q p − q ¶p−1 p−2µp − 2 2 − q ¶ ³ p−q 2−q ´p−1¡R RNfη|u| p¢p−1 R RNfη|u| p 1 p−2 (8) −p − 2 2 − q Z RN fη|u|p = 0.
However, by (7) , the H¨older and Sobolev inequalities and µ kukpH1 R RNfmax|u| p ¶ 1 p−2 > µ 1 fmaxSp ¶ 1 p−2 for all u ∈ Mfη,hη,
where S is the best Sobolev constant, we have Kη(u) ≥ C (p, q) Ã kuk2(p−1)H1 R RNfη|u|p ! 1 p−2 − η2(p−q)p−2 − (2−q)N 2 khk L2−q2 kuk q H1 > kukqH1 Ã C (p, q) µ 1 fmaxSp ¶ 1 p−2 kuk1−qH1 − η 2(p−q) p−2 − (2−q)N 2 khk L2−q2 ! ≥ kukqH1 " C (p, q) µ 1 fmaxSp ¶ 1 p−2 ³ η2(p−q)p−2 − (2−q)N 2 ´1−q 2−q µ p − q p − 2khkL2−q2 ¶1−q 2−q −η2(p−q)p−2 − (2−q)N 2 khk L2−q2 i for all u ∈ M0 fη,hη. Since 1−q 2−q ≤ 0 and 2(p−q) p−2 − (2−q)N
2 > 0 (see Appendix A), there
exists η1 > 0 such that for each η ∈ (0, η1) and u ∈ M0fη,hη we have Kη(u) > 0,
this contradicts to (8). We can conclude that M0
fη,hη = ∅ for all η ∈ (0, η1) . ¤
By Lemma 2.2, for η ∈ (0, η1) we write Mfη,hη = M
+ fη,hη∪ M − fη,hη and define α± fη,hη = inf u∈M± fη,hη Ifη,hη(u) .
The following lemma shows that the minimizers on Mfη,hη are ”usually” critical
points for Ifη,hη.
Lemma 2.3 For η ∈ (0, η1) . If u0 is a local minimizer for Ifη,hη on Mfη,hη, then
I0
fη,hη(u0) = 0 in H
−1¡RN¢.
Proof. Similar to the proof of lemma 4 in Wu [25]. ¤ For each u ∈ H1¡RN¢\ {0} , we define
tmax = Ã kuk2H1 (p − 1)RRN fη|u| p ! 1 p−2 > 0.
Then we have the following lemma.
Lemma 2.4 For each u ∈ H1¡RN¢\ {0} , we have
(i) there is a unique t− = t−(u) > t
max > 0 such that t−u ∈ M−fη,hη and
Ifη,hη(t
−u) = max
t≥tmaxIfη,hη(tu) ;
(ii) t−(u) is a continuous function for nonzero u;
(iii) M− fη,hη = n u ∈ H1¡RN¢\ {0} | 1 kukH1t− ³ u kukH1 ´ = 1 o ; (iv) if RRNh |u|
q > 0, then there is a unique 0 < t+ = t+(u) < t
max such that t+u ∈ M+
fη,hη and Ifη,hη(t+u) = min0≤t≤t−Ifη,hη(tu) .
3
Existence of a Local Minimum
In this section, we will establish the existence of a local minimum for Ifη,hη on
Mfη,hη. Let d =
2(p−q)
p−2 −
(2−q)N
2 > 0 (see Appendix A). Then we have the following
results.
Lemma 3.1 (i) For each u ∈ M+fη,hη, we have RRNhη|u|
q > 0 and I
fη,hη(u) < 0.
In particular, αfη,hη ≤ α
+
fη,hη < 0;
(ii) Ifη,hη is coercive and bounded below on Mfη,hη for all η ∈
µ 0, ³ p−2 p−q ´1 d ¶ .
Proof. (i) For each u ∈ M+fη,hη, (2 − q) kuk2H1 − (p − q)
R RNfη|u| p > 0 and kuk2H1 = Z RN fη|u|p+ η 2(p−q) p−2 Z RN hη|u|q, we have η2(p−q)p−2 Z RN hη|u|q= kuk2H1 − Z RN fη|u|p > (p − 2) Z RN fη|u|p > 0, and Ifη,hη(u) = µ 1 2− 1 p ¶ Z RN fη|u|p− 1 2η 2(p−q) p−2 Z RN hη|u|q < −(p − 1) (p − 2) 2p Z RN fη|u|p < 0.
(ii) For u ∈ Mfη,hη, we have kuk
2 H1 = R RN fη|u| p + η2(p−q)p−2 R RNhη|u| q. Then by
the H¨older and Young inequalities,
Ifη,hη(u) ≥ µ p − 2 2p ¶ kuk2H1 − µ p − q pq ¶ ηdkhk L2−q2 kuk q H1 ≥ · p − 2 2p − η d µ p − q 2p ¶¸ kuk2H1 − ηd µ (p − q) (2 − q) 2pq ¶ khk 2 2−q L2−q2 = 1 2p £ (p − 2) − ηd(p − q)¤kuk2 H1 − ηd µ (p − q) (2 − q) 2pq ¶ khk 2 2−q L2−q2 .
Thus, Ifη,hη is coercive and bounded below on Mfη,hη for all η ∈
µ 0, ³ p−2 p−q ´1 d ¶ . ¤
Theorem 3.2 For each positive number η < η∗ = min ½ η1, ³ p−2 p−q ´1 d ¾ the equa-tion (4) has a positive soluequa-tion uη ∈ M+fη,hη such that Ifη,hη(uλ) = α
+
fη,hη = αfη,hη
and Ifη,hη(uη) → 0 as η → 0.
Proof. Similar to the proof of proposition 9 in Wu [25], there exists a sequence
{un} ⊂ Mfη,hη such that
Ifη,hη(un) = αfη,hη + o (1) ,
I0
fη,hη(un) = o (1) in H
−1¡RN¢.
Then by Lemma 3.1 (ii), there exist a subsequence {un} and uη ∈ H1(RN) is a
solution of equation (4) such that
un* uη weakly in H1(RN) and un → uη a.e. in RN.
Moreover, by h ∈ L2−q2 ¡RN¢, the Egorov theorem and the H¨older inequality, we
have Z RN hη|un|q → Z RN hη|uη|q.
Now we prove that RRNhη|uη|q 6= 0. Suppose otherwise, then
kunk2H1 = Z RN fη|un|p+ o (1) and µ 1 2 − 1 p ¶ Z RN fη|un|p = 1 2kunk 2 H1 − 1 p Z RN fη|un|p− η 2(p−q) p−2 1 q Z RN hη|un|q+ o (1) = αfη,hη + o (1) ,
this contradicts to αfη,hη < 0. Thus,
R
RNhη|uη|
q 6= 0. In particular, u η is a
nontrivial solution of equation (4) . Now we prove that un → uη strongly in
H1(RN). If not, then ku ηkH1 < lim inf n→∞ kunkH1 and so αfη,hη ≤ Ifη,hη(uη) = µ 1 2− 1 p ¶ kuηk2H1− µ 1 q − 1 p ¶ η2(p−q)p−2 Z RN hη|uη|q < lim n→∞Ifη,hη(un) = αfη,hη,
which is a contradiction. Consequently, un → uη strongly in H1(RN) and
Ifη,hη(uη) = αfη,hη. Moreover, we have uη ∈ M
+
fη,hη. In fact, if uη ∈ M
− fη,hη,
by Lemma 2.4, there exist unique t+
0 and t−0 such that t+0uη ∈ M+fη,hη and
t−0uη ∈ M−fη,hη, we have t + 0 < t−0 = 1. Since d dtIfη,hη ¡ t+0uη ¢ = 0 and d 2 dt2Ifη,hη ¡ t+0uη ¢ > 0,
there exists ¯t ∈ (t+0, t−0] such that Ifη,hη
¡ t+0uη ¢ < Ifη,hη(¯tuη) . By Lemma 2.4, Ifη,hη ¡ t+ 0uη ¢ < Ifη,hη(¯tuη) ≤ Ifη,hη ¡ t− 0uη ¢ = Ifη,hη(uη) ,
which is a contradiction. Thus, Ifη,hη(uη) = αfη,hη = α
+
fη,hη. Since Ifη,hη(uη) =
Ifη,hη(|uη|) and |uη| ∈ M
+
fη,hη. By Lemma 2.3 and the maximum principle, we
may assume that uη is a positive solution of equation (4) . Moreover, by Lemma
3.1 we have 0 > Ifη,hη(uη) ≥ −η d µ (p − q) (2 − q) 2pq ¶ khk 2 2−q L2−q2 . Since d > 0. We obtain Ifη,hη(uη) → 0 as η → 0. ¤
4
Proof of Theorem 1.1
First, we use the graph of the coefficient f to find some Palais–Smale sequences which are used to prove Theorem 1.1. For a > 0, let Ca(xi) denote the hypercube
N Π j=1 ¡ xi j− a, xij − a ¢ center at xi = (xi 1, xi2, . . . , xiN) for i = 1, 2, . . . , k. Let Ca(xi)
and ∂Ca(xi) denote the closure and the boundary of Ca(xi) respectively. By the
conditions (Q1) and (Q3) , we can choose numbers K, l > 0 such that Cl(xi) are
disjoint, f (x) < f (xi) for x ∈ ∂C
l(xi) for all i = 1, 2, . . . , k and ∪ki=1Cl(xi) ⊂
QN i=1(−K, K) . Define φη ∈ C (R, R) , gη ∈ C ¡ H1¡RN¢, RN¢ by φη(t) = 2K η t > 2Kη , t −2K η ≤ t ≤ 2Kη , −2K η t < − 2K η , gj η(u) = R RNφη(xj) |u| p R RN|u| p for j = 1, 2, . . . , N and gη(u) = ¡ g1
η(u) , gη2(u) , . . . , gNη (u)
¢
Let Ci l/η ≡ Cl/η ³ xi η ´ , Ni η = n u ∈ M− fη,hη | u ≥ 0 and gη(u) ∈ C i l/η o , ∂Ni η = n u ∈ M− fη,hη | u ≥ 0 and gη(u) ∈ ∂C i l/η o
for i = 1, 2, . . . , k. It is easy to verify that Ni
η and ∂Nηi are nonempty sets for all
i = 1, 2, . . . , k. For i = 1, 2, . . . , k, consider the minimization problems in Ni η and ∂Ni η for Ifη,hη, γi η = inf u∈Ni η Ifη,hη(u) ; eγ i η = inf u∈∂Ni η Ifη,hη(u) .
Let w be a unique positive radial solution of −∆u + u = fmaxup−1 in RN, u > 0 in RN, u ∈ H1¡RN¢
such that Ifmax,0(w) = αfmax,0. By the condition (Q3) and the routine
computa-tions, we have
αfmax,0< αf∞,0. (9)
For small η > 0 satisfying 2√η < 1, we define a function ψη ∈ C1
¡ RN, [0, 1]¢ such that ψη(x) = ( 1 |x| < 1 2√η − 1, 0 |x| > 1 2√η − 1,
and |∇ψη| ≤ 2 in RN. Let xη = 2√1η (1, 1, . . . , 1) ∈ RN and
wη(x) = t−ηw µ x − x i η + x η ¶ ψη µ x − x i η + x η ¶ , where t−
η > 0 are selected such that wη ∈ M−fη,hη. Then we have the following
results. Lemma 4.1 We have (i) η2(p−q)p−2 R RNhηwq ³ x − xi η + xη ´ ψq η ³ x − xi η + xη ´ → 0 as η → 0; (ii) t− η → 1 as η → 0.
Proof. (i) Since 2(p−q)p−2 − (2−q)N2 > 0,
0 ≤ η2(p−q)p−2 Z RN hηwq µ x − xi η + x η ¶ ψq η µ x − xi η + x η ¶ ≤ η2(p−q)p−2 − (2−q)N 2 khk L2−q2 ° ° ° °w µ x − x i η + x η ¶ ψη µ x − x i η + x η ¶°° ° ° q H1
and ° ° ° °w µ x − xi η + x η ¶ ψη µ x − xi η + x η ¶°° ° ° 2 H1 → 2p p − 2αfmax,0 as η → 0. Thus, η2(p−q)p−2 Z RN hηwq µ x −xi η + x η ¶ ψq η µ x −xi η + x η ¶ → 0 as η → 0.
(ii) Since wη ∈ M−fη,hη, we have
¡ t− η ¢2"Z RN ¯ ¯ ¯ ¯∇ µ w µ x − xi η + x η ¶ ψη µ x −xi η + x η ¶¶¯¯ ¯ ¯ 2 + µ w µ x − xi η + x η ¶ ψη µ x − xi η + x η ¶¶2# = ¡t− η ¢pZ RN fηwp µ x − x i η + x η ¶ ψp η µ x −x i η + x η ¶ +η2(p−q)p−2 ¡t− η ¢qZ RN hηwq µ x − x i η + x η ¶ ψqη µ x − x i η + x η ¶ . Since kwk2H1 = R
RNfmaxwp, from part (i) that
¡ t−η¢2kwk2H1 = ¡ t−η¢2 ° ° ° °w µ x − x i η + x η ¶ ψη µ x − x i η + x η ¶°° ° ° 2 H1 = ¡t−η¢p Z RN fηwp µ x − x i η + x η ¶ ψηp µ x − x i η + x η ¶ + o (η) = ¡t− η ¢pZ RN fη ¡ ηx + xi− ηxη¢wp+ o (η) ,
where o (η) → 0 as η → 0. Moreover, ηxη → 0 as η → 0 and
t− η > tmax= ° ° °w ³ x −xi η + xη ´ ψη ³ x − xi η + xη ´°° °2 H1 (p − 1)RRNfη ¯ ¯ ¯w ³ x − xi η + xη ´ ψη ³ x − xi η + xη ´¯¯ ¯p 1 p−2 = (p − 1)2−p1 + o (η) . Thus, t− η → 1 as η → 0. ¤ Let η∗ = min ½ η1, ³ p−2 p−q ´1 d ¾
as in Theorem 3.2. Then we have the following result.
Lemma 4.2 For each ε > 0, there exists ηε∈ (0, η∗] such that α− fη,hη ≤ γ i η < min © αfmax,0+ ε, αfη,hη + αf∞,0 ª for i = 1, 2, . . . , k and η ∈ (0, ηε) .
Proof. First, we show that gη(wη) ∈ Cl/ηi . Since
gηj(wη) = R RNφη(xj) wp ³ x − xi η + xη ´ ψp η ³ x − xi η + xη ´ R RNwp ³ x − xi η + xη ´ ψpη ³ x − xi η + xη ´ and ψη µ x −xi η + x η ¶ = 0 if ¯ ¯ ¯ ¯xj− xi j η ¯ ¯ ¯ ¯ > √1η. By the definition of ψη, we have
gηj(wη) = R Ci l/ηφη(xj) w p³x − xi η + xη ´ ψp η ³ x − xi η + xη ´ R Ci l/ηw p ³ x − xi η + xη ´ ψpη ³ x − xi η + xη ´ provided √1 η < l
η. From the definition of φη we conclude that gηj(wη) ∈ Cl/ηi . Thus,
wη ∈ Nηi. Moreover, by Lemma 4.1 Ifη,hη(wη) = ¡ t− η ¢2 2 "Z RN ¯ ¯ ¯ ¯∇ µ w µ x − x i η + x η ¶ ψη µ x − x i η + x η ¶¶¯¯ ¯ ¯ 2 (10) + µ w µ x −xi η + x η ¶ ψη µ x −xi η + x η ¶¶2# − ¡ t− η ¢p p Z RN fηwp µ x − xi η + x η ¶ ψp η µ x − xi η + x η ¶ −η2(p−q)p−2 ¡t− η ¢qZ RN hηwq µ x − x i η + x η ¶ ψηq µ x − x i η + x η ¶ = 1 2 Z RN |∇w|2+ w2 −1 p Z RN f¡ηx + xi− ηxη¢wp+ o (η) ,
where o (η) → 0 as η → 0. Since ηxη → 0 as η → 0 and from (10) , we have
Ifη,hη(wη) = Ifmax,0(w) + o (η) = αfmax,0+ o (η) .
Therefore, for any ε > 0 there exists η2 > 0 such that γi
Moreover, αfmax,0 < αf∞,0 and αfη,hη → 0 as η → 0, then there exists η3 > 0 such
that
γηi < αfη,hη + αf∞,0 for i = 1, 2, . . . , k and η ∈ (0, η3) .
We take ηε = min {η2, η3} , this implies γi η < min © αfmax,0+ ε, αfη,hη + αf∞,0 ª for i = 1, 2, . . . , k and η ∈ (0, ηε) .
This completes the proof. ¤
Lemma 4.3 There are positive numbers δ and ηδ ∈ (0, η∗] such that for i =
1, 2, . . . , k
e
γηi > αfmax,0+ δ for all η ∈ (0, ηδ) .
Proof. Fix i ∈ {1, 2, . . . , k} . Assume the contrary there exists a sequence {ηn}
with ηn→ 0 as n → ∞ such that eγηin → c ≤ αfmax,0. Consequently, there exists a
sequence {un} ⊂ ∂Nηin such that gηn(un) ∈ ∂Cl/ηi n,
Z RN |∇un|2+ u2n= Z RN fηn|un| p+ η2(p−q)p−2 n Z RN hηn|un| q (11) and Ifηn,hηn(un) → c ≤ αfmax,0 as n → ∞.
It follows that {un} is uniformly bounded in H1
¡
RN¢. Since u
n ∈ M−fηn,hηn, we
deduce from the Sobolev imbedding theorem that kunkH1 > ν > 0 for some
constant ν and for all n. Applying the concentration-compactness principle of P. L. Lions [19] to |un|p, there are positive constants R, θ and {yn} ⊂ RN such that
Z BN(yn;R) |un|p ≥ θ for all n, where BN(y n; R) = © x ∈ RN | |x − y n| < R ª
. Let eun = un(x + yn) , then there
is a nonzero u0 ∈ H1 ¡ RN¢ such that e un * u0 in H1 ¡ RN¢, e un → u0 a.e. in RN, Z BN(0;R) |eun|p → Z BN(0;R) |u0|p ≥ θ.
Set wn = eun− u0. By Brezis-Lieb Lemma [8] we obtain
Z RN f (ηnx + ηnyn) |eun|p (12) = Z RN f (ηnx + ηnyn) |u0|p+ Z RN f (ηnx + ηnyn) |wn|p+ o (1) .
Since {un} is uniformly bounded and eun* u0, we have η 2(p−q) p−2 n Z RN hηn|un| q → 0 as n → ∞ (13) and Z RN |∇eun|2+ eu2n= Z RN |∇u0|2+ u20 + Z RN |∇wn|2+ wn2 + o (1) . (14)
Moreover, from (11) we have that Z RN |∇eun|2+ eu2n= Z RN f (ηnx + ηnyn) |eun|p+ η 2(p−q) p−2 n Z RN hηn|un| q. (15)
Combining (12) , (13) , (14) and (15) we have Z RN |∇wn|2+ w2n− Z RN f (ηnx + ηnyn) |wn|p (16) = − µZ RN |∇u0|2+ u20− Z RN f (ηnx + ηnyn) |u0|p ¶ + o (1) .
We distinguish the cases: (I) kwnkH1 → 0 and (II) kwnkH1 → c > 0.
Case (I): By condition (Q3) we can choose s > 0 such that
f (x) < fmax for x ∈ C
i
l+s\Cl−si .
We complete the proof by establishing the contradiction lim
n→∞Ifηn,hηn(un) > αfmax,0.
Choose the sequence {ηnyn} . By passing to a subsequence if necessary, we may
assume that one of the following cases occurs: (I − 1) {ηnyn} ⊂ C i l+s\Cl−si ; (I − 2) {ηnyn} ⊂ Cil−s; (I − 3) {ηnyn} ⊂ RN\Cl+si , and {ηnyn} is bounded; (I − 4) {ηnyn} is unbounded.
Let ² > 0 and R²> be such that
R |x|≥R²|eun| p R RN|eun|p ≤ ². (17)
In case (I − 1) , we may assume ηnyn → ey ∈ C i
l+s\Cl−si and f (ey) < fmax.
Conse-quently lim n→∞Ifηn,hηn(un) = lim n→∞ ½ 1 2 Z RN |∇eun|2+ eu2n− 1 p Z RN f (ηnx + ηnyn) |eun|p − η 2(p−q) p−2 n Z RN hηn|un| q ¾ = 1 2 Z RN |∇u0|2+ u20− 1 p Z RN f (ey) |u0|p > αfmax,0, which is a contradiction. In case (I − 2) , gηjn(un) = R RN φηn ³ xj + (yn)j ´ |eun|p R RN|eu| p = R |x|≤R²φηn ³ xj+ (yn)j ´ |eun|p+ R |x|≥R²φηn ³ xj + (yn)j ´ |eun|p R RN|eun| p .
In the region |xj| ≤ R², we have
xj+ (yn)j ∈ µ xi j− (l − δ) ηn − R², xi j+ (l − δ) ηn + R² ¶ ⊂ µ −2K ηn ,2K ηn ¶
for n sufficiently large. In the follows from (17) and the definition of φηn that
gj ηn(un) > µ xi j− (l − δ) ηn − R² ¶ (1 − ²) − 2K ηn ², gηjn(un) < µ xi j− (l − δ) ηn + R² ¶ (1 − ²) + 2K ηn ².
From the above inequalities it is clear that we can choose ² > 0, δ > ² sufficiently small such that
gηjn(un) ∈ µ xi j − l ηn ,x i j+ l ηn ¶
for n sufficiently large, this contradicts to gηn(un) ∈ ∂Cl/ηi n.
In case (I − 3) , we may assume that ηnyn→ ey ∈ C i
l+s as n → ∞, eyi ≥ xij+ l + s
for some i and (yn)j > xi
j+l+s/2
ηn for all n. For |xj| ≤ R² we have
xj + (yn)j >
xi
j+ l + s/2
ηn
and gj ηn(un) > µ xi j− (l − s) ηn − R² ¶ (1 − ²) − 2K ηn ²
for sufficiently small ² > 0, s < ² and n large enough. This contradicts to
gηn(un) ∈ ∂Cl/ηi n.
In case (I − 4) is excluded by assuming ηnyn→ ∞ as n → ∞, and using a similar
argument to case (I − 1) . Case (II) : Set
Z RN |∇u0|2+ u20− Z RN f (ηnx + ηnyn) |u0|p = A + o (1) , then by (16) Z RN |∇wn|2+ wn2 − Z RN f (ηnx + ηnyn) |wn|p = −A + o (1) .
Without loss of generality, we may assume that A > 0 (A < 0 can be considered similarly). We can find a sequence {tn} with tn → 1 as n → ∞ such that
vn= tnwn satisfiesZ RN |∇vn|2+ v2n− Z RN f (ηnx + ηnyn) |vn|p = −A.
Since u0 ∈ Mf (ηnx+ηnyn),0(A + o (1)) , by (12) − (14) and Lemma 2.1 we have
Ifηn,hηn(un) = 1 2 Z RN |∇u0|2 + u20− 1 p Z RN f (ηnx + ηnyn) |u0|p +1 2 Z RN |∇wn|2+ wn2 − 1 p Z RN f (ηnx + ηnyn) |wn|p+ o (1) ≥ A + o (1) 2 + 1 2 Z RN |∇vn|2+ vn2 −1 p Z RN f (ηnx + ηnyn) |vn|p + o (1) = αf (ηnx+ηnyn),0(A) + αf (ηnx+ηnyn),0(−A) + o (1) > αf (ηnx+ηnyn),0+ µ p − 2 4p ¶ A + o (1) ≥ αfmax,0+ µ p − 2 4p ¶ A + o (1) ,
which is a contradiction. If A = 0, we can find sn, tn > 0, sn→ 1 as n → ∞ such
that wn = tnwn, vn = snu0 satisfy Z RN |∇wn|2+ w2n = Z RN f (ηnx + ηnyn) |wn|p, Z RN |∇vn|2+ v2n = Z RN f (ηnx + ηnyn) |vn|p.
Hence lim n→∞Ifηn,hηn(un) = n→∞lim · 1 2 Z RN |∇vn|2 + v2n− 1 p Z RN f (ηnx + ηnyn) |vn|p +1 2 Z RN |∇wn|2+ w2n− 1 p Z RN f (ηnx + ηnyn) |wn|p ¸ > αfmax,0.
This completes the proof. ¤
Throughout this section, take η0 = min {ηε, ηδ} , ηε and ηδ are as in Lemmas
4.2, 4.3. Used the idea of Ni-Takagi [20] and Wu [25], we have the following result. Lemma 4.4 For each η ∈ (0, η0) and u ∈ Nηi, there exist ² > 0 and a
differen-tiable function t∗ : B (0; ²) ⊂ H1¡RN¢→ R+ such that t∗(0) = 1, t∗(v) (u − v) ∈
Ni
η for all v ∈ B (0; ²) and
(t∗)0(0) , v® = 2 R RN∇u∇v + uv − p R RNfη|u| p−2uv − η2(p−q)p−2 R RNhη|u| q−2uv R RN|∇u| 2+ u2− (p − 1)R RNfη|u|p for all v ∈ H1(RN). Proof. For u ∈ Ni η, define a function Fu : R × H1 ¡ RN¢→ R by Fu(t, w) = D I0 fη,hη(t (u − w)) , t (u − w) E = t2 Z RN |∇ (u − w)|2+ (u − w)2− |t|p Z RN fη|u − w|p −η2(p−q)p−2 |t|q Z RN hη|u − w|q. Then Fu(1, 0) = D I0 fη,hη(u) , u E = 0 and d dtFu(1, 0) = 2 Z RN |∇u|2+ u2− p Z RN fη|u|p− η 2(p−q) p−2 Z RN hη|u|q = Z RN |∇u|2 + u2− (p − 1) Z RN fη|u|p < 0.
According to the implicit function theorem, there exist ² > 0 and a differentiable function t∗ : B (0; ²) ⊂ H1¡RN¢→ R such that t∗(0) = 1,
(t∗)0(0) , v® = 2 R RN∇u∇v + uv − p R RNfη|u|p−2uv − η 2(p−q) p−2 R RNhη|u|q−2uv R RN|∇u| 2+ u2− (p − 1)R RNfη|u| p
and Fu(t∗(v) , v) = 0 for all v ∈ B (0; ²) which is equivalent to D I0 fη,hη(t ∗(v) (u − v)) , t∗(v) (u − v)E= 0 for all v ∈ B (0; ²) . Furthermore, Z RN |∇t∗(v) (u − v)|2+ [t∗(v) (u − v)]2− (p − 1) Z RN fη|t∗(v) (u − v)|p < 0 and gη(t∗(v) (u − v)) ∈ Cl/ηi
still holds if ² is sufficiently small by the continuity of the maps gη and t∗. ¤
Proposition 4.5 For each η ∈ (0, η0) we have
α−fη,hη ≤ γηi < min©αfη,hη + αf∞,0, eγ
i η
ª
and there exists a sequence {un} ⊂ Nηi such that
Ifη,hη(un) = γ i η+ o (1) , I0 fη,hη(un) = o (1) in H −1¡RN¢ for all i = 1, 2, . . . , k.
Proof. If Niη denotes the closure of Ni
η, then first we note that, N i
η = Nηi∪ ∂Nηi
for each i = 1, 2, . . . , k. It then follows from Lemmas 4.2, 4.3 that for a positive number ε ≤ δ and taking η0 = min {ηε, ηδ} , we obtain
γηi < min©αfη,hη+ αf∞,0, eγ i η ª for i = 1, 2, . . . , k, η ∈ (0, η0) . (18) Hence γi η = inf n Ifη,hη(u) | u ∈ N i η o for i = 1, 2, . . . , k. (19) Now we fix i ∈ {1, 2, . . . , k} . Applying the Ekeland variational principle [16] there exists a minimizing sequence {un} ⊂ N
i η such that Ifη,hη(un) < γ i η + 1 n (20) and Ifη,hη(un) ≤ Ifη,hη(w) + 1 nkw − unkH1 for all w ∈ N i η. (21)
Using (18) we may assume that un∈ Nηi for n sufficiently large. Applying Lemma
4.4 with u = un we obtain the function t∗n : B (0; ²n) → R for some ²n > 0 such
that t∗ n(w) (un− w) ∈ Nηi. Let 0 < δ < ²n and u < H1 ¡ RN¢ with u 6≡ 0. We set wδ = kukδu H1 and zδ= t ∗
n(wδ) (un− wδ) . Since zδ ∈ Nηi, we deduce from (21) that
Ifη,hη(zδ) − Ifη,hη(un) ≥ −
1
n kzδ− unkH1.
By the mean value theorem, we obtain D If0η,hη(un) , zδ− un E + o (kzδ− unk) ≥ −1 nkzδ− unkH1. Therefore, D I0 fη,hη(un) , −wδ E + (t∗ n(wδ) − 1) D I0 fη,hη(un) , (un− wδ) E (22) ≥ −1 n kzδ− unkH1 + o (kzδ− unk) .
Now we observe that t∗
n(wδ) (un− wδ) ∈ Nηi and consequently we get from (22)
that −δ ¿ I0 fη,hη(un) , u kukH1 À +(t ∗ n(wδ) − 1) t∗ n(wδ) D I0 fη,hη(zδ) , t ∗ n(wδ) (un− wδ) E + (t∗ n(wδ) − 1) D I0 fη,hη(un) − I 0 fη,hη(zδ) , (un− wδ) E ≥ −1 nkzδ− unkH1 + o (kzδ− unk) .
Then we write above inequality in the following form ¿ I0 fη,hη(un) , u kukH1 À (23) ≤ kzδ− unkH1 δn + o (kzδ− unkH1) δ +(t ∗ n(wδ) − 1) δ D If0η,hη(un) − If0η,hη(zδ) , (un− wδ) E .
Since we can find a constant C > 0, independent of δ such that
kzδ− unkH1 ≤ δ + C (|t∗n(wδ) − 1|) and lim δ→0 |t∗ n(wδ) − 1| δ ≤ ° °(t∗ n)0(0) ° ° ≤ C.
For a fixed n, let δ → 0 in (23) and use the fact lim δ→0kzδ− unkH1 = 0, we obtain ¿ I0 fη,hη(un) , u kukH1 À ≤ C n. This implies Ifη,hη(un) = γ i η+ o (1) and I0 fη,hη(un) = o (1) in H −1¡RN¢. ¤ We need the following proposition to provide the precise description of the Palais–Smale sequences for Ifη,hη.
Proposition 4.6 Assume that {un} ⊂ M−fη,hη is a sequence satisfying
Ifη,hη(un) = β + o (1) ,
I0
fη,hη(un) = o (1) in H
−1¡RN¢,
where β < αfη,hη+ αf∞,0. Then there exist a subsequence {un} and u0 in H1
¡ RN¢
such that un→ u0 strongly in H1
¡
RN¢ and I
fη,hη(u0) = β.
Proof. By Lemma 3.1 (ii), there exist a subsequence {un} and u0 in H1
¡ RN¢ such that un * u0 weakly in H1 ¡ RN¢.
First, we claim that u0 6≡ 0. If not, by h ∈ L
2
2−q ¡RN¢, the Egorov theorem and
the H¨older inequality, we have
kunk2H1 = Z RN f∞|un|p+ o (1) (24) and µ 1 2− 1 p ¶ Z RN f∞|u n|p = 1 2kunk 2 H1 − 1 p Z RN fη|un|p −1 qη 2(p−q) p−2 Z RN hη|un|q+ o (1) = β + o (1) . Moreover, {un} ⊂ M−fη,hη and kunkH1 > c for some c > 0.
We get β ≥ αf∞,0, this contradicts to β < αfη,hη+ αf∞,0. Thus, u0 is a nontrivial
solution of equation (4) and Ifη,hη(u0) ≥ αfη,hη. We write un = u0 + vn with
vn* 0 weakly in H1
¡
RN¢. By the Brezis-Lieb lemma [8], we have
Z RN fη|un|p = Z RN fη|u0|p+ Z RN fη|vn|p+ o (1) = Z RN fη|u0|p+ Z RN f∞|v n|p+ o (1) .
Since {un} is a bounded sequence in H1
¡
RN¢ and so {v
n} is also a bounded
sequence in H1¡RN¢. Moreover, by h ∈ L2−q2 ¡RN¢, the Egorov theorem and the
H¨older inequality, we have Z RN hη|vn|q = Z RN hη|un|q− Z RN hη|u0|q+ o (1) = o (1) .
Hence, for n large enough, we can conclude that
αfη,hη + αf∞,0 > Ifη,hη(u0+ vn) = Ifη,hη(u0) + 1 2kvnk 2 H1 − 1 p Z RN f∞|v n|p+ o (1) ≥ αfη,hη + 1 2kvnk 2 H1 − 1 p Z RN f∞|vn|p+ o (1) or 1 2kvnk 2 H1 − 1 p Z RN f∞|v n|p < αf∞,0+ o (1) . (25) Also from I0 fη,hη(un) = o (1) in H −1¡RN¢, {u
n} is uniformly bounded and u0 is
a solution of equation (4) , we obtain
o (1) = D If0η,hη(un) , un E = kvnk2H1 − Z RN f∞|vn|p+ o (1) . (26)
We claim that (25) and (26) can be hold simultaneously only if {vn} admits a
subsequence {vni} which converges strongly to zero. If not, the kvnkH1 is bounded
away from zero, that is
kvnkH1 ≥ c for some c > 0. From (26), it follows Z RN f∞|v n|p ≥ µ 2p p − 2 ¶ αf∞,0+ o (1) .
By (25) and (26) , for n large enough, αf∞,0 ≤ µ 1 2 − 1 p ¶ Z RN f∞|vn|p+ o (1) = 1 2kvnk 2 H1 − 1 p Z RN f∞|v n|p+ o (1) < αf∞,0,
which is a contradiction. Therefore, un→ u0strongly in H1
¡
RN¢and I
fη,hη(u0) =
β. ¤
Now, we begin to show the proof of Theorem 1.1: By Propositions 4.5, 4.6 for each η ∈ (0, η0) , there exist a sequence {uin} ⊂ Nηi and ui0 ∈ H1
¡ RN¢\ {0} such that Ifη,hη ¡ ui n ¢ = γi η+ o (1) , I0 fη,hη ¡ ui n ¢ = o (1) in H−1¡RN¢, and ui n→ ui0 strongly in H1 ¡ RN¢.
Obviously, the function ui
0 is a solution of the equation (4) and Ifη,hη(ui0) = γηi.
It is clear that ui
0 is nonnegative, by the maximum principle ui0 is positive. Since gi
η(ui0) ∈ Cl η (x
i),
uη ∈ M+fη,hη and ui0 ∈ M−fη,hη,
where uη is a positive solution of equation (4) as in Theorem 3.2. This implies
uη and ui0 are different. Letting λ0 = η−20 , Uη(x) = λ
1 p−2u η ³√ λx ´ and Ui(x) = λp−21 ui 0 ³√ λx ´
. We obtain Uη and Ui are positive solutions of the equation (Eλ).
¤
5
Appendix A
Lemma 5.1 2(p−q)p−2 − (2−q)N2 > 0 where 1 ≤ q < 2 < p < 2∗ and N ≥ 1.
Proof. Case (I) : 1 ≤ q < 2 < p < 2∗ and N = 1. Since q < 2 < p we have
6q − 4
2 + q < 2 < p. Thus,
2 (p − q) > (p − 2) (2 − q) 2
and so
2 (p − q)
p − 2 −
(2 − q) 2 > 0. Case (II) : 1 ≤ q < 2 < p < 2∗ and N = 2. Since 4 − 4
q < 2 < p we have 4q − 4 < pq. Thus, 2 (p − q) > (p − 2) (2 − q) and so 2 (p − q) p − 2 − (2 − q) > 0.
Case (III) : 1 ≤ q < 2 < p < 2∗ and N ≥ 3. We only need to show that
p [4 − N (2 − q)] > 4q − 2N (2 − q) . (27) Since it is equivalent to 2(p−q)p−2 − (2−q)N2 > 0.
Case (III − a) : q = 1. Then (27) becomes
p (4 − N) = 4 − 2N. (28)
Clearly, (28) holds for N = 3, 4. Since
p < 2N N − 2 <
2N − 4
N − 4 for N ≥ 5,
(28) holds for N ≥ 5.
Case (III − b) : 1 < q < 2 < p < 2∗ and N = 3, 4. Since q < 2, we have
4q − 2N (2 − q) < 8 − 4N + 2qN. Moreover, q > 1 ≥ 2 − 4 N for N = 3, 4. Thus, 2q − 2N (2 − q) 4 − N (2 − q) < 2 < p for N = 3, 4. Case (III − c) : q = 2 − 4 N and N ≥ 5. Since 4q − 2N (2 − q) = 4 µ 2 − 4 N ¶ − 8 < 0 and 4 − (2 − q) N = 0, p (4 − N (2 − q)) = 0 < 4q − 2N (2 − q) .
Case (III − d) : q ∈ ¡1, 2 − 4 N ¢ and N ≥ 5. Since 2N [N (2 − q) − 4] < (N − 2) [2N (2 − q) − 4q] and N (2 − q) − 4 > 0, p < 2N N − 2 < 2N (2 − q) − 4q N (2 − q) − 4 for N ≥ 5. Case (III − e) : q ∈ ¡2 − 4 N, 2 ¢ and N ≥ 5. Since 2 [4 − N (2 − q)] > 4q − 2N (2 − q) and 4 − N (2 − q) > 0, p > 2 > 4q − 2N (2 − q) 4 − N (2 − q) .
This completes the proof. ¤
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