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# On semilinear elliptic equations involving critical Sobolev exponents and sign-changing weight function

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Volume 7, Number 2, March 2008 pp. 383–405

ON SEMILINEAR ELLIPTIC EQUATIONS INVOLVING CRITICAL SOBOLEV EXPONENTS

AND SIGN-CHANGING WEIGHT FUNCTION

Tsung-fang Wu Department of Applied Mathematics

National University of Kaohsiung, Kaohsiung 811, Taiwan (Communicated by Manuel del Pino)

Abstract. In this paper, we study the decomposition of the Nehari manifold via the combination of concave and convex nonlinearities. Furthermore, we use this result to prove that the semilinear elliptic equation with a sign-changing weight function has at least two positive solutions.

1. Introduction. In this paper, we consider the multiplicity results of positive solutions of the following semilinear elliptic equation:

½

−∆u = λf (x) |u|q−2u + |u|2∗−2u in Ω, u ∈ H1

0(Ω) ,

(Eλ,f) where Ω is a bounded domain in RN(N ≥ 3) , 1 < q < 2 < 2 = 2N

N −2, λ > 0 and f : Ω → R is a continuous function with f+(x) = max {f (x) , 0} 6≡ 0 in Ω (f is

possibly sign-changing on Ω). Associated with the equation (Eλ,f) , we consider the energy functional Jλ in H01(Ω) Jλ(u) = 1 2 Z Ω |∇u|2dx −λ q Z Ω f (x) |u|qdx − 1 2 Z Ω |u|2∗dx.

It is well known that the solutions of equation (Eλ,f) are the critical points of the energy functional Jλ in H1

0(Ω)(see Rabinowitz [15]).

It is known that the number of positive solutions of equation (Eλ,f) is affected by the concave and convex nonlinearities. This issue has been the focus of a great deal of research in recent years. If the weight function f (x) ≡ 1, the authors Ambrosetti–Brezis–Cerami [3] have investigated the following equation

½

−∆u = λ |u|q−2u + |u|p−2u in Ω, u ∈ H1

0(Ω) ,

(Eλ,1) where 1 < q < 2 < p ≤ 2∗. They found that there exists λ

0 > 0 such that the

equation (Eλ,1) admits at least two positive solutions for λ ∈ (0, λ0) , has a positive

solution for λ = λ0 and no positive solution exists for λ > λ0. Moreover, they

proved that when the domain Ω is starshaped and p = 2∗, one has the following result: given a sequence {λn} ⊂ R+ with λn & 0 as n → ∞, then there exists a

2000 Mathematics Subject Classification. Primary 35J65; Secondary 35J20.

Key words and phrases. Critical Sobolev exponent, Nehari manifold, concave-convex

nonlinearities.

Partially supported by the National Science Council of Republic of China.

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sequence {uλn} of positive solutions of equation (Eλn,f) such that kuλnk∞ → ∞ as n → ∞. Actually, Adimurthi–Pacella–Yadava [1], Damascelli–Grossi–Pacella [7], Ouyang–Shi [14] and Tang [18] proved that if 1 < q < 2 < p < 2then there exists a λ0> 0 such that the equation (Eλ,1) in unit ball BN(0; 1) has exactly two positive

solutions for all λ ∈ (0, λ0), has exactly one positive solution for λ = λ0 and no

positive solution exists for all λ > λ0.

For more general results, Ambrosetti–Azorezo–Peral [2], de Figueiredo-Grossez-Ubilla [10,11] and Wu [21] considered the following general subcritical problem:

½

−∆u = λa (x) |u|q−2u + b (x) |u|p−2u in Ω, 0 ≤ u ∈ H1

0(Ω) ,

(Eλ,a,b) where 1 < q < 2 < p ≤ 2∗ and a, b satisfy some integrability conditions or change sign in Ω. They proved that the equation (Eλ,a,b) has at least two positive solutions if λ is sufficiently small.

The main purpose of this paper is to generalize the partial results of Ambrosetti– Brezis–Cerami [3], using the decomposition of the Nehari manifold as λ varies. First, we prove that the equation (Eλ,f) has at least two positive solutions for λ sufficiently small.

Theorem 1.1. There exists λ0 > 0 such that for λ ∈ (0, λ0) , the equation (Eλ,f)

has at least two positive solutions.

As for the asymptotic behavior of the solutions obtained in Theorem1.1as λ → 0, we have the following results.

Theorem 1.2. Assume that a sequence {λn} ⊂ R satisfies λn> 0 and λn→ 0 as n → ∞.

Then there exists a subsequence {λn} and two sequences {u(j)n (x)} (j = 1, 2) of positive solutions of equation (Eλn,f) such that

(i) ku(1)n kH1→ 0 as n → ∞;

(ii) there exist two sequences {xn} ⊂ Ω, {Rn} ⊂ R+ and a positive solution v0

D1,2¡RN¢ of critical problem −∆u = |u|2∗−2u in RN, such that Rn→ ∞ as n → ∞ and ° ° °u(2)n (x) − R N −2 2 n v0(Rn(x − xn)) ° ° ° D1,2(RN)→ 0 as n → ∞.

This paper is organized as follows. In Section 2, we give some notation and preliminaries. In Section 3, we establish the existence of a local minimum for Jλon Mλ(Ω). In Section 4, we prove that the equation (Eλ,f) has at least two positive solutions for λ sufficiently small. In Section 5, we prove Theorem1.2.

2. Notation and preliminaries. We define the Palais–Smale (PS) sequences, (PS)–values, and (PS)–conditions in H1

0(Ω) for Jλ as follows:

Definition 2.1. (i) For β ∈ R, a sequence {un} is a (PS)β–sequence in H01(Ω) for

if Jλ(un) = β + o(1) and Jλ0(un) = o(1) strongly in H−1(Ω) as n → ∞; (ii) β ∈ R is a (PS)–value in H1

0(Ω) for Jλ if there exists a (PS)β–sequence in

H1

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(iii) Jλ satisfies the (PS)β–condition in H01(Ω) if every (PS)β–sequence in H01(Ω)

for Jλ contains a convergent subsequence.

Throughout this section, we denote by S the best Sobolev constant for the imbed-ding of H1 0(Ω) into L2 (Ω), i.e., S = inf u∈H1 0(Ω)\{0} R Ω|∇u| 2 dx ³R Ω|u| 2 dx ´2/2∗ > 0. In particular, kukL2∗ ≤ S− 1 2kuk

H1 for all u ∈ H01(Ω) \ {0} . Now, we consider the

Nehari minimization problem: for λ ≥ 0,

αλ(Ω) = inf {Jλ(u) | u ∈ Mλ(Ω)} , where Mλ(Ω) =©u ∈ H1

0(Ω) \ {0} | hJλ0 (u) , ui = 0 ª

. Define ψλ(u) = hJλ0 (u) , ui = kuk2H1− λ

Z Ω f (x) |u|qdx − Z Ω |u|2∗dx. Then for u ∈ Mλ(Ω) , hψ0 λ(u) , ui = 2 kuk 2 H1− λq Z Ω f (x) |u|qdx − 2∗ Z Ω |u|2∗dx.

Similarly to the method used in Tarantello[17], we split Mλ(Ω) into three parts: M+λ(Ω) = {u ∈ Mλ(Ω) | hψλ0 (u) , ui > 0} ,

M0

λ(Ω) = {u ∈ Mλ(Ω) | hψλ0 (u) , ui = 0} , Mλ(Ω) = {u ∈ Mλ(Ω) | hψλ0 (u) , ui < 0} . Then, we have the following results.

Lemma 2.2. There exists λ1> 0 such that for each λ ∈ (0, λ1) we have M0λ(Ω) = ∅.

Proof. We consider the following two cases.

Case (I): u ∈ Mλ(Ω) and Rf (x) |u|qdx = 0. We have kuk2H1 Z Ω |u|2∗dx = 0. Thus, hψ0λ(u) , ui = 2 kuk2H1− 2∗ Z Ω |u|2∗dx = (2 − 2∗) kuk2H1 < 0 and so u /∈ M0 λ(Ω) .

Case (II): u ∈ Mλ(Ω) and Rf (x) |u|qdx 6= 0. Suppose that M0

λ(Ω) 6= ∅ for all λ > 0. If u ∈ M0λ(Ω), then we have 0 = hψ0 λ(u) , ui = 2 kuk 2 H1− λq Z Ω f (x) |u|qdx − 2∗ Z Ω |u|2∗dx = (2 − q) kuk2H1− (2∗− q) Z Ω |u|2∗dx. Thus, kuk2H1= 2− q 2 − q Z Ω |u|2∗dx (2)

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and λ Z Ω f (x) |u|qdx = kuk2H1 Z Ω |u|2∗dx = 2 − 2 2 − q Z Ω |u|2∗dx. (3) Moreover, µ 2− 2 2− qkuk2H1 = kuk2H1 Z Ω |u|2∗dx = λ Z Ω f (x) |u|qdx ≤ λ kf kLq∗S− q 2kukq H1, where q∗= 2 2− q. This implies kukH1 · λ µ 2− q 2− 2kf kLq∗S− q 2 ¸ 1 2−q . (4)

Let Iλ: Mλ(Ω) → R be given by

Iλ(u) = K (2∗, q) kuk 2N +4 N −2 H1 R Ω|u| 2 dx   N −2 4 − λ Z Ω f (x) |u|qdx, where K (2∗, q) =³2−q 2−q ´N +2 4 ³2−2 2−q ´

. Then Iλ(u) = 0 for all u ∈ M0

λ(Ω) . Indeed, by (2) and (3), Iλ(u) = K (2∗, q) kuk 2N +4 N −2 H1 R Ω|u| 2 dx   N −2 4 − λ Z Ω f (x) |u|qdx = µ 2 − q 2− qN +2 4 µ2− 2 2 − q ¶     ³ 2−q 2−q ´N +2 N −2³R Ω|u| 2 dx ´N +2 N −2 R Ω|u| 2 dx     N −2 4 2 − 2 2 − q Z Ω |u|2∗dx = 0. (5)

However, by (4) , the H¨older and Sobolev inequalities,

Iλ(u) = K (2∗, q) kuk 2N +4 N −2 H1 R Ω|u| 2 dx   N −2 4 − λ Z Ω f (x) |u|qdx ≥ K (2∗, q) kuk 2N +4 N −2 H1 R Ω|u| 2 dx   N −2 4 − λ kf kLq∗kuk q L2∗ ≥ kukqL2∗ Ã K (2∗, q) S q+N 2 kukq−1H1 − λ kf kLq∗ ! ≥ kukqL2∗ ( K (2∗, q) Sq+N 2 λ 1−q 2−q ·µ 2− q 2− 2kf kLq∗S− q 2 ¸1−q 2−q −λ kf kLq∗} .

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This implies that there exists λ1> 0 such that for λ ∈ (0, λ1) , we have Iλ(u) > 0 for all u ∈ M0

λ(Ω), this contradicts (5). Thus, we can conclude that M0λ(Ω) = ∅ for all λ ∈ (0, λ1).

Lemma 2.3. If u ∈ M+λ(Ω) , then Rf (x) |u|qdx > 0. Proof. Since kuk2H1− λ Z Ω f (x) |u|qdx − Z Ω |u|2∗dx = 0 and kuk2H1 > 2− q 2 − q Z Ω |u|2∗dx. We have λ Z Ω f (x) |u|qdx = kuk2H1 Z Ω |u|2∗dx > 2 − 2 2 − q Z Ω |u|2∗dx > 0. This completes the proof.

By Lemma2.2, for λ ∈ (0, λ1) we write Mλ(Ω) = M+λ (Ω) ∪ M−λ (Ω) and define α+ λ (Ω) = inf u∈M+ λ(Ω) Jλ(u) ; α−λ (Ω) = inf u∈M− λ(Ω) Jλ(u) .

The following lemma shows that the minimizers on Mλ(Ω) are “usually” critical points for Jλ.

Lemma 2.4. For λ ∈ (0, λ1). If u0 is a local minimizer for Jλ on Mλ(Ω) , then J0

λ(u0) = 0 in H−1(Ω) .

Proof. If u0 is a local minimizer for Jλ on Mλ(Ω) , then u0 is a solution of the

optimization problem

minimize Jλ(u) subject to ψλ(u) = 0.

Hence, by the theory of Lagrange multipliers, there exists θ ∈ R such that J0 λ(u0) = θψ0λ(u0) in H−1(Ω) . This implies hJ0 λ(u0) , u0iH1 = θ hψλ0 (u0) , u0iH1. (6) Since u0∈ Mλ(Ω) , that is ku0k2H1− λ R Ωf (x) |u0| qdx −R Ω|u0| 2 dx = 0. We have hψ0 λ(u0) , u0iH1 = (2 − q) ku0k2H1− (2∗− q) Z Ω |u0|2 dx. Moreover, hψ0

λ(u0) , u0iH1 6= 0 and so by (6) , θ = 0. This completes the proof.

For each u ∈ H1 0(Ω) \ {0} , we write tmax= Ã (2 − q) kuk2H1 (2− q)R Ω|u| 2 dx ! 1 2∗−2 > 0. Then, we have the following lemma.

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Lemma 2.5. Let q∗= 2 2−q and λ2= ³ 2−2 2−q ´ ³ 2−q 2−q ´2−q 2∗−2 S2∗−q2∗−2kf k−1 Lq∗. Then for each u ∈ H1 0(Ω) \ {0} and λ ∈ (0, λ2), we have

(i) there is a unique t−= t(u) > t

max> 0 such that t−u ∈ M−λ (Ω) and Jλ(t−u) =

maxt≥tmaxJλ(tu) ;

(ii) t−(u) is a continuous function for nonzero u; (iii) M−λ(Ω) = n u ∈ H1 0(Ω) \ {0} | kuk1H1t− ³ u kukH1 ´ = 1 o ;

(iv) if Rf (x) |u|qdx > 0, then there is a unique 0 < t+= t+(u) < t

max such that

t+u ∈ M+

λ (Ω) and Jλ(t+u) = min0≤t≤t−Jλ(tu) .

Proof. (i) Fix u ∈ H1

0(Ω) \ {0}. Let s (t) = t2−qkuk2 H1− t2 −qZ Ω |u|2∗dx for t ≥ 0.

We have s(0) = 0, s(t) → −∞ as t → ∞, s (t) is concave and achieves its maximum at tmax. Moreover, s(tmax) = Ã (2 − q) kuk2H1 (2−q)R Ω|u| 2 dx !2−q 2∗−2 kuk2H1 Ã (2 − q) kuk2H1 (2− q)R Ω|u| 2 dx !2∗−q 2∗−2Z Ω |u|2∗dx = kukqH1   µ 2 − q 2− q2−q 2∗−2 µ 2 − q 2− q2∗−q 2∗−2   Ã kuk2H1 R Ω|u| 2 dx !2−q 2∗−2 ≥ kukqH1 µ 2− 2 2− q ¶ µ 2 − q 2− q2−q 2∗−2 SN (2−q)4 , or s (tmax) ≥ kukqH1 µ 2− 2 2− q ¶ µ 2 − q 2− q2−q 2∗−2 SN (2−q)4 . (7)

Case (I) :Rf (x) |u|qdx ≤ 0. Then there is a unique t−> t

maxsuch that s (t−) =

R Ωf (x) |u| q dx and s0(t) < 0. Now, (2 − q)°°t−u°°2 H1− (2∗− q) Z Ω ¯ ¯t−u¯¯2dx = ¡t−¢1+q · (2 − q)¡t−¢1−qkuk2 H1− (2∗− q) ¡ t−¢2∗−q−1 Z Ω ¯ ¯t−u¯¯2dx ¸ = ¡t−¢1+qs0¡t¢< 0 and ­ J0 λ ¡ t−u¢, tu® = ¡t¢2kuk2 H1 ¡ t−¢qλ Z Ω f (x) |u|qdx −¡t−¢2 Z Ω |u|2∗dx = ¡t−¢q · s¡t−¢− λ Z Ω f (x) |u|qdx ¸ = 0. Thus, t−u ∈ M

λ(Ω) . Moreover, for t > tmax, we have (2 − q) ktuk2H1− (2∗− q) Z Ω |tu|2∗dx < 0, d2 dt2Jλ(tu) < 0

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and d dtJλ(tu) = t kuk 2 H1− tq−1λ Z Ω f (x) |u|qdx − t2−1Z Ω |u|2∗dx = 0 for t = t−. This implies Jλ(t−u) = max

t≥tmaxJλ(tu) .

Case (II) :Rf (x) |u|qdx > 0. By (7) and s (0) = 0 < λ Z Ω f (x) |u|qdx ≤ λ kf kLq∗S −q 2 kukq H1 < kukqH1 µ 2− 2 2− q ¶ µ 2 − q 2− q2−q 2∗−2 SN (2−q)4 ≤ s (tmax) for λ ∈ (0, λ2),

there are unique t+ and t such that 0 < t+ < t

max< t−, s¡t= λ Z Ω f (x) |u|qdx = s¡t−¢ and s0¡t> 0 > s0¡t¢. We have t±u ∈ M±

λ(Ω) , Jλ(t−u) ≥ Jλ(tu) ≥ Jλ(t+u) for each t ∈ [t+, t−] and Jλ(t+u) ≤ Jλ(tu) for each t ∈ [0, t+] . Thus,

¡ t−u¢= max t≥tmax Jλ(tu) and Jλ ¡ t+u¢= min 0≤t≤t−Jλ(tu) .

(ii) By the uniqueness of t−(u) and the external property of t(u) , we see that t−(u) is a continuous function of u 6= 0.

(iii) For u ∈ M−

λ (Ω), let v = kukuH1. By part (i), there is a unique t−(v) > 0 such that t−(v) v ∈ M λ (Ω) or t− ³ u kukH1 ´ 1 kukH1u ∈ M λ (Ω) . Since u ∈ M−λ (Ω) , we have t−³ u kukH1 ´ 1

kukH1 = 1, and this implies M λ (Ω) ⊂ ½ u ∈ H1 0(Ω) \ {0} | 1 kukH1 t− µ u kukH1 ¶ = 1 ¾ . Conversely, let u ∈ H1

0(Ω) \ {0} such that kuk1H1t− ³ u kukH1 ´ = 1. Then t− µ u kukH1 ¶ u kukH1 ∈ M−λ (Ω) . Thus, Mλ (Ω) = ½ u ∈ H1 0(Ω) \ {0} | 1 kukH1 t− µ u kukH1 ¶ = 1 ¾ . (iv) By case (II) of part (i).

3. Existence of a local minimum. First, we establish the existence of positive solutions for the equation:

½

−∆u = λf (x) |u|q−2u in Ω, 0 ≤ u ∈ H1

0(Ω) .

(8) Associated with equation (8) , we consider the energy functional

Kλ(u) = 1 2 Z Ω |∇u|2dx −λ q Z Ω f (x) |u|qdx

(8)

and the minimization problem

βλ(Ω) = inf {Kλ(u) | u ∈ Nλ(Ω)} , where Nλ(Ω) =©u ∈ H1

0(Ω) \ {0} | hKλ0 (u) , ui = 0 ª

. Then we have the following result.

Theorem 3.1. For each λ > 0, the equation (8) there exists a positive solution vλ such that Iλ(vλ) = βλ(Ω) < 0.

Proof. First, we need to show that Iλ is bounded below on Nλ(Ω) and βλ(Ω) < 0. Then for u ∈ Nλ(Ω) , kuk2H1 = λ Z Ω f (x) |u|qdx ≤ λ kf kLq∗S− q 2kukq H1. where q∗= 2 2−q. This implies kukH1 ³ λ kf kLq∗S− q 2 ´ 1 2−q . (9) Hence Kλ(u) = µ 1 2 1 qkuk2H1 µ 1 2 1 q ¶ ³ λ kf kLq∗S− q 2 ´ 1 2−q

for all u ∈ Nλ(Ω) and βλ(Ω) < 0. Let {vn} be a minimizing sequence for Iλ on Nλ(Ω) , then by (9) and the compact imbedding theorem, there exist a subsequence {vn} and vλin H01(Ω) such that

vn* vλ weakly in H01(Ω)

and

vn→ vλ strongly in Lq(Ω). (10)

First, we claim thatRf (x) |vλ|qdx > 0. If not, by (10) we can conclude that Z Ω f (x) |vλ|qdx = 0 and Z Ω f (x) |vn|qdx → 0 as n → ∞. Thus, Z|∇vn|2dx = o (1) and Kλ(vn) = 1 2 Z Ω |∇vn|2dx −λ q Z Ω f (x) |vn|qdx = 0 as n → ∞,

this contradicts Kλ(vn) → βλ(Ω) < 0 as n → ∞. Thus, Rf (x) |vλ|qdx > 0. In particular, vλ 6≡ 0. Now, we prove that vn → vλ strongly in H01(Ω). Suppose

otherwise, then kvλkH1 < lim inf

n→∞ kvnkH1 and so kvλk2H1− λ Z Ω f (x) |vλ|qdx < lim infn→∞ µ kvnk2H1− λ Z Ω f (x) |vn|qdx= 0. ByRf (x) |vλ|qdx > 0, there is a unique t06= 1 such that t0vλ∈ Nλ(Ω) . Thus,

t0vn* t0 weakly in H01(Ω).

Moreover,

(9)

which is a contradiction. Hence vn → vλ strongly in H01(Ω). This implies vλ Nλ(Ω) and

Kλ(vn) → Kλ(vλ) = βλ(Ω) as n → ∞.

Since Kλ(vλ) = Kλ(|vλ|) and |vλ| ∈ Nλ(Ω) , without loss of generality, we may assume that vλ is nonnegative. Similar to the argument of proof in Lemma 2.4, is a nonnegative solution of equation (8) . Now, by Dr´abek–Kufner–Nicolosi [8, Lemma 2.1] we have vλ∈ L∞(Ω) . Then we can apply the Harnack inequality due to Trudinger [19] (or see Gilbarg–Trudinger [12]) in order to get that vλ is positive in Ω.

Lemma 3.2. (i) Let vλ be a positive solution of equation (8) as in Theorem 3.1. Then there exists tλ> 0 such that αλ(Ω) < N2t2λβλ(Ω) < 0;

(ii) Jλ is coercive and bounded below on Mλ(Ω) for all λ ∈ (0,N (2−q)+2q4 ]. Proof. (i) Let vλ∈ H1

0(Ω) be a positive solution of equation (8) as in Theorem3.1

such that Kλ(vλ) = βλ(Ω) . Then Zf (x) |vλ|qdx = kvλk2H1= µ 2q q − 2βλ(Ω) > 0. Set tλ= t+(vλ) > 0 as defined by Lemma2.5. Hence tλv

λ∈ M+λ(Ω) and Jλ(tλvλ) = µ 1 2 1 qt2λkvλk2H1+ µ 1 q 1 2t2λ∗ Z Ω |vλ|2 dx < −1 N 2 − q q t 2 λkvλk2H1 = 2 Nt 2 λβλ(Ω) . This yields, αλ(Ω) < 2 Nt 2 λβλ(Ω) < 0. (11)

(ii) For u ∈ Mλ(Ω) , we have kuk2H1 = λ

R

f (x) |u|

q

+ dxR|u|2∗dx. Then by the H¨older inequality and the Young inequality,

Jλ(u) = 1 N kuk 2 H1− λ µ 1 q 1 2 ¶ Z Ω f (x) |u|qdx 1 N kuk 2 H1− λ µ 1 q 1 2kf kLq∗S −q 2 kukq H1 · 1 N − λ µ 2− q 2 × 2∗ ¶¸ kuk2H1− λ µ 2 − q 2q∗q ¶ ³ kf kLq∗S −q 2 ´ 2 2−q = 1 N · 1 − λN (2 − q) + 2q 4 ¸ kuk2H1− λ µ 2 − q 2q∗q ¶ ³ kf kLq∗S q 2 ´ 2 2−q . Thus, Jλ is coercive on Mλ(Ω) and

Jλ(u) ≥ −λ µ 2 − q 2q∗q ¶ ³ kf kLq∗S −q 2 ´ 2 2−q (12) for all λ ∈ (0,N (2−q)+2q4 ].

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Lemma 3.3. For each u ∈ Mλ(Ω) , there exist ² > 0 and differentiable function ξ : B (0; ²) ⊂ H1

0(Ω) → R+ such that ξ (0) = 1, the function ξ (v) (u − v) ∈ Mλ(Ω) and hξ0(0) , vi =2 R Ω∇u∇vdx − qλ R Ωf (x) |u| q−2 uvdx − 2∗R Ω|u| 2−2 uvdx (2 − q)R|∇u|2dx − (2∗− q)R Ω|u| 2 dx (13) for all v ∈ H1 0(Ω) .

Proof. For u ∈ Mλ(Ω), define a function F : R × H1

0(Ω) → R given by Fu(ξ, w) = hJλ0(ξ (u − w)) , ξ (u − w)i = ξ2 Z Ω |∇ (u − w)|2dx − ξqλ Z Ω f (x) |u − w|qdx − ξ2 Z Ω |u − w|2∗dx. Then Fu(1, 0) = hJ0 λ(u) , ui = 0 and d dtFu(1, 0) = 2 Z Ω |∇u|2dx − qλ Z Ω f (x) |u|qdx − 2∗ Z Ω |u|2∗dx = (2 − q) Z Ω |∇u|2dx − (2∗− q) Z Ω |u|2∗dx 6= 0.

According to the implicit function theorem, there exists ² > 0 and a differentiable function ξ : B (0; ²) ⊂ HRN¢→ R such that ξ (0) = 1,

hξ0(0) , vi = 2 R Ω∇u∇vdx − qλ R Ωf (x) |u| q−2 uvdx − 2∗R Ω|u| 2−2 uvdx (2 − q)R|∇u|2dx − (2∗− q)R Ω|u| 2 dx and Fu(ξ (v) , v) = 0 for all v ∈ B (0; ²) which is equivalent to hJ0

λ(ξ (v) (u − v)) , ξ (v) (u − v)i = 0 for all v ∈ B (0; ²) , that is, ξ (v) (u − v) ∈ Mλ(Ω) .

Proposition 1. Let λ0= min

n

λ1, λ2,N (2−q)+2q4

o

, then for λ ∈ (0, λ0) there exists

a minimizing sequence {un} ⊂ Mλ(Ω) such that

Jλ(un) = αλ(Ω) + o (1) and Jλ0 (un) = o (1) in H−1(Ω) . i.e. {un} is a (PS)αλ(Ω)–sequence in H

1

0(Ω) for Jλ.

Proof. By Lemma 3.2(ii) and the Ekeland variational principle [9], there exists a minimizing sequence {un} ⊂ Mλ(Ω) such that

Jλ(un) < αλ(Ω) + 1

n (14)

and

Jλ(un) < Jλ(w) + 1

nkw − unkH1 for each w ∈ Mλ(Ω) . (15)

By taking n large, from Lemma3.2(i) we have Jλ(un) = 1 N kunk 2 H1 µ 1 q− 1 2λ Z Ω f (x) |un|qdx (16) < αλ(Ω) + 1 n < 2 Nt 2 λβλ(Ω) .

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This implies λ Z Ω f (x) |un|qdx > −2 q 2− q 2 Nt 2 λβλ(Ω) > 0. (17)

Consequently un6= 0 and putting (16), (17) and the H¨older inequality together, we obtain · −2∗q (2− q) N 2 λt 2 λβλ(Ω) S q 2kf k−1 Lq∗ ¸1 q < kunkH1 < · λN (2∗− q) 2q kf kLq∗S −q 2 ¸ 1 2−q . (18) Now, we will show that

kJλ0 (un)kH−1 → 0 as n → ∞.

Applying Lemma3.3with un to obtain the functions ξn: B (0; ²n) → R+ for some

²n > 0, such that ξn(w) (un− w) ∈ Mλ(Ω) . Choose 0 < ρ < ²n. Let u ∈ H01(Ω)

with u 6≡ 0 and let wρ=kukρu

H1. We set ηρ= ξn(wρ) (un− wρ) . Since ηρ∈ Mλ(Ω) ,

we deduce from (15) that

Jλ(ηρ) − Jλ(un) ≥ −1

nkηρ− unkH1

and by the mean value theorem, we have hJ0 λ(un) , ηρ− uni + o ¡ kηρ− unkH1 ¢ ≥ −1 nkηρ− unkH1. Thus, hJ0

λ(un) , −wρi + (ξn(wρ) − 1) hJλ0 (un) , (un− wρ)i (19) ≥ −1 nkηρ− unkH1+ o ¡ kηρ− unkH1 ¢ . From ξn(wρ) (un− wρ) ∈ Mλ(Ω) and (19) that

−ρ ¿ J0 λ(un) , u kukH1 À + (ξn(wρ) − 1) hJ0

λ(un) − Jλ0 (ηρ) , (un− wρ)i ≥ −1 nkηρ− unkH1+ o ¡ kηρ− unkH1 ¢ . Thus, ¿ Jλ0 (un) , u kukH1 À kηρ− unkH1 + o¡kηρ− unkH1 ¢ ρ +(ξn(wρ) − 1) ρ hJ 0

λ(un) − Jλ0 (ηρ) , (un− wρ)i . (20) Since kηρ− unkH1≤ ρ |ξn(wρ)| + |ξn(wρ) − 1| kunkH1 and lim ρ→0 |ξn(wρ) − 1| ρ ≤ kξ 0 n(0)k .

If we let ρ → 0 in (20) for a fixed n, then by (18) we can find a constant C > 0, independent of ρ, such that

¿ J0 λ(un) , u kukH1 À C n (1 + kξ 0 n(0)k) .

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We are done once we show that kξ0

n(0)k is uniformly bounded in n. By (13) , (18) and the H¨older inequality, we have

hξn0 (0) , vi ≤ b kvkH1 ¯ ¯ ¯(2 − q)R|∇un|2dx − (2∗− q) R Ω|un| 2 dx ¯ ¯ ¯ for some b > 0. We only need to show that

¯ ¯ ¯ ¯(2 − q) Z Ω |∇un|2dx − (2∗− q) Z Ω |un|2 dx ¯ ¯ ¯ ¯ > c (21)

for some c > 0 and n large. We argue by way of contradiction. Assume that there exists a subsequence {un} , we have

(2 − q) Z Ω |∇un|2dx − (2∗− q) Z Ω |un|2 dx = o (1) . (22) Combining (22) with (18) , we can find a suitable constant d > 0 such that

Z

|un|2

dx ≥ d for n sufficiently large. (23) In addition, (22) , and the fact that un∈ Mλ(Ω) also give

λ Z Ω f (x) |un|qdx = kunk2H1 Z Ω |un|2 dx = 2 − 2 2 − q Z Ω |un|2 dx + o (1) and kunkH1 · λ µ 2− 2 2− qkf kLq∗S− q 2 ¸ 1 2−q + o (1) . (24) This implies Iλ(un) = K (2∗, q) kunk 2N +4 N −2 H1 R Ω|un| 2 dx   N −2 4 − λ Z Ω f (x) |un|qdx = µ 2 − q 2− qN +2 4 µ2− 2 2 − q ¶     ³ 2−q 2−q ´N +2 N −2³R Ω|un| 2 dx´ N +2 N −2 R Ω|un| 2 dx     N −2 4 2∗− 2 2 − q Z Ω |un|2 dx + o (1) = o (1) . (25)

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However, by (23) , (24) and λ ∈ (0, λ0) Iλ(un) = K (2∗, q) kunk 2N +4 N −2 H1 R Ω|un| 2 dx   N −2 4 − λ Z Ω f (x) |un|qdx ≥ K (2∗, q) kunk 2N +4 N −2 H1 R Ω|un| 2 dx   N −2 4 − λ kf kLq∗kunkqL2∗ ≥ kunkqL2∗   K (2∗, q) Sq+N2   kunk 2N +4 N −2 H1 kunk2 (q+1)−2q H1   N −2 4 − λ kf kLq∗    ≥ kunkqL2∗ ( K (2∗, q) Sq+N2 λ 1−q 2−q ·µ 2− q 2− 2kf kLq∗S− q 2 ¸1−q 2−q −λ kf kLq∗}

> d0 for some d0> 0 and n sufficiently large.

This contradicts (25). We get ¿ J0 λ(un) , u kukH1 À ≤C n. This shows that {un} is a (PS)αλ(Ω)–sequence for Jλ.

Final, we establish the existence of a local minimum for Jλon Mλ(Ω).

Theorem 3.4. Let λ0> 0 be as in Proposition1, then for λ ∈ (0, λ0), the functional

has a minimizer u+0 in M+λ(Ω) and it satisfies (i) Jλ¡u+0

¢

= α+λ (Ω) = αλ(Ω) ; (ii) u+

0 is a nonnegative (nontrivial) solution of equation (Eλ,f) ; (iii) Jλ¡u+0¢→ 0 as λ → 0.

Proof. Let {un} ⊂ Mλ(Ω) is a (PS)αλ(Ω)–sequence for Jλ, then by Lemma3.2and

the compact imbedding theorem, there exist a subsequence {un} and u+0 ∈ H1 0(Ω),

a solution of equation (Eλ,f) such that

un* u+0 weakly in H01(Ω)

and

un→ u+0 strongly in Lq(Ω). (26)

First, we claim that Rf (x)¯¯u+ 0

¯ ¯q

dx 6= 0. If not, by Lemma 2.3 and (26) we can

conclude that Zf (x)¯¯u+0¯¯qdx = 0 and Zf (x) |un|qdx → 0 as n → ∞. Thus, Z|∇un|2dx = Z Ω |un|2 dx + o (1)

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and 1 N Z Ω |un|2 dx = 1 2 Z Ω |∇un|2dx − λ q Z Ω f (x) |un|qdx − 1 2 Z Ω |un|2 dx + o (1) = αλ(Ω) + o (1) ,

this contradicts αλ(Ω) < 0. Thus, Rf (x)¯¯u+ 0

¯ ¯q

dx 6= 0. In particular, u+ 0 is a

nontrivial solution of equation (Eλ,f) . Moreover, αλ(Ω) ≤ Jλ ¡ u+0¢= 1 N Z Ω ¯ ¯∇u+ 0 ¯ ¯2 dx − µ 1 q− 1 2λ Z Ω f (x)¯¯u+0¯¯qdx lim n→∞Jλ(un) = αλ(Ω) . Consequently, un → u+ 0 strongly in H01(Ω) and Jλ ¡ u+ 0 ¢ = αλ(Ω) . From Lemma 2.5it follows that necessarily u+0 ∈ M+λ(Ω) and Jλ¡u+0¢= α+λ (Ω) = αλ(Ω) . Since ¡ u+ 0 ¢ = Jλ ¡¯ ¯u+ 0 ¯ ¯¢and¯¯u+ 0 ¯ ¯ ∈ M+

λ(Ω) , by Lemma2.4, we may assume that u+0 is

a nonnegative (nontrivial) solution of equation (Eλ,f) . Moreover, by Lemmas3.2, 0 > Jλ¡u+0 ¢ ≥ −λ µ 2 − q 2q∗q ¶ ³ kf kLq∗S −q 2 ´ 2 2−q . We obtain Jλ¡u+ 0 ¢ → 0 as λ → 0.

4. Proof of Theorem 1.1. First, we consider

uε(x) = ε

(N −2)/2

³

ε2+ |x|(N −2)/2

, ε > 0 and x ∈ RN

is an extremal function for the Sobolev inequality in RN. Since f is a continuous function on Ω and

f+(x) = max {f (x) , 0} 6≡ 0.

Following the method of [5], let Σ = {x ∈ Ω | f (x) > 0} be a open set of positive measure. Without loss of generality, we may assume that the set Σ is a domain. We consider the test function

wε,y(x) = ηy(x) uε,y(x) , x ∈ RN, where y ∈ Σ, uε,y(x) = uε(x − y) and ηy∈ C∞

0 (Σ) with

ηy ≥ 0 and ηy = 1 near y.

Let λ0> 0 as in Proposition 1, then for λ ∈ (0, λ0) we have the following result.

Lemma 4.1. Let u+

0 be the local minimum in Theorem3.4. Then for every l > 0

and a.e. y ∈ Σ, there exists ε0= ε0(l, y) > 0 such that

Jλ(u+ 0 + lwε,y) < αλ(Ω) + 1 NS N 2 for all ε ∈ (0, ε0) .

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Proof. Since Jλ(u+0 + lwε,y) = 1 2 ° °u+ 0 + lwε,y ° °2 H1 λ q Z Ω f (x)¯¯u+0 + lwε,y¯¯qdx 1 2 Z Ω ¯ ¯u+ 0 + lwε,y ¯ ¯2 dx = 1 2 ° °u+ 0 ° °2 H1+ l2 2 kwε,yk 2 H1+ ­ u+0, lwε,y ® H1 −λ q Z Ω f (x)¯¯u+ 0 + lwε,y ¯ ¯q dx − 1 2 Z Ω ¯ ¯u+ 0 + lwε,y ¯ ¯2 dx.(27)

A careful estimate obtained by Brezis–Nirenberg (see formulas (17) and (21) in [5]) shows that Z Ω ¯ ¯u+ 0 + lwε,y ¯ ¯2 dx = Z Ω ¯ ¯u+ 0 ¯ ¯2 dx + l2 Z Ω |wε,y|2 dx + 2∗l Z Ω ¡ u+0 ¢2−1 wε,ydx +2∗l2∗−1 Z Ω (wε,y)2∗−1u+0dx + o ³ εN −22 ´ . Also from [6] we have

kwε,yk2H1 = B + O ¡ εN −2¢ and Z Ω |wε,y|2 dx = A + O¡εN¢, where B = ku1k2H1(RN), A = R RN 1 (1+|x|2)Ndx and S = BA 2−N N . Substituting in (27)

and using the fact that u+

0 is a solution of equation (Eλ,f) , we obtain

Jλ(u+0 + lwε,y) = 1 2 ° °u+ 0 ° °2 H1+ l2 2B + l ­ u+0, wε,y ® H1 1 2 Z Ω ¯ ¯u+ 0 ¯ ¯2 dx −l 2 2∗A − l Z Ω ¡ u+0¢2∗−1wε,ydx − l2 −1Z Ω (wε,y)2∗−1u+0dx −λ q Z Ω f (x)¡u+0 + lwε,y¢qdx + o ³ εN −22 ´ = Jλ(u+ 0) + l2 2B − l2 2∗A − l 2−1Z Ω (wε,y)2∗−1u+ 0dx −λ q Z Ω f (x)¡u+ 0 + lwε,y ¢q dx +λ q Z Ω f (x)¡u+ 0 ¢q dx +λ q Z Ω f (x)¡u+ 0 ¢q−1 lwε,ydx + o ³ εN −22 ´ = Jλ(u+ 0) + l2 2B − l2 2∗A − l 2−1Z Ω (wε,y)2∗−1u+ 0dx −λ Z Ω f (x) (Z lwε,y 0 h (u+ 0 + s)q−1− ¡ u+ 0 ¢q−1i ds ) dx +o ³ εN −22 ´ . From f > 0 in Σ and wε,y≡ 0 in Σc,

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we can conclude that Jλ(u+0 + lwε,y) ≤ Jλ(u+0) +

l2 2B − l2 2∗A − l 2−1Z Ω (wε,y)2∗−1u+0dx + o ³ εN −22 ´ . Similar to the argument of Lemma 3.1 in Tarantello [17], we can conclude that for every l > 0 and a.e. y ∈ Σ, there exists ε0= ε0(l, y) > 0 such that

Jλ(u+0 + lwε,y) < αλ(Ω) + 1 NS

N

2 (28)

for all ε ∈ (0, ε0) .

The following proposition provides a precise description of the (PS)–sequence of Jλ.

Proposition 2. Each sequence {un} ⊂ M−λ (Ω) satisfying (i) Jλ(un) = σ + o (1) with σ < αλ(Ω) + 1

NS

N

2;

(ii) J0

λ(un) = o (1) in H−1(Ω) has a convergent subsequence.

Proof. By Lemma 3.2 and the compact imbedding theorem, there exist a subse-quence {un} and u0∈ H01(Ω) is a solution of equation (Eλ,f) such that

un* u0weakly in H01(Ω)

and

un→ u0 strongly in Lq(Ω). (29)

First, we claim that u06≡ 0. If not, by (29) we have

Z Ω f (x) |un|qdx → 0 as n → ∞. Thus, Z|∇un|2dx = Z Ω |un|2 dx + o (1) (30) and 1 N Z Ω |un|2 dx = 1 2 Z Ω |∇un|2dx −λ q Z Ω f (x) |un|qdx − 1 2 Z Ω |un|2 dx + o (1) = σ + o (1) , this contradicts σ < αλ(Ω) + 1 NS N

2. Thus, u0 6≡ 0 and Jλ(u0) ≥ αλ(Ω) . Write

un= u0+ vn with vn * 0 weakly in H01(Ω). By the Brezis-Lieb lemma [4], we have

Z Ω |un|2 dx = Z Ω |u0+ vn|2 dx = Z Ω |u0|2 dx + Z Ω |vn|2 dx + o (1) . Hence, for n large enough, we can conclude that

αλ(Ω) + 1 NS N 2 > Jλ(u0+ vn) = Jλ(u0) +1 2 Z Ω |∇vn|2dx − 1 2 Z Ω |vn|2 dx + o (1) ≥ αλ(Ω) +1 2 Z Ω |∇vn|2dx − 1 2 Z Ω |vn|2 dx + o (1) or 1 2 Z Ω |∇vn|2dx − 1 2 Z Ω |vn|2 dx < 1 NS N 2 + o (1) . (31)

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Also from J0

λ(un) = o (1) in H−1(Ω), {un} is uniformly bounded and u0 is a

solution of equation (Eλ,f) follows o (1) = h J0 λ(un) , uni = Z Ω |∇u0|2dx − Z Ω |u0|qdx − Z Ω |u0|2 dx + Z Ω |∇vn|2dx − Z Ω |vn|2 dx + o (1) = Z Ω |∇vn|2dx − Z Ω |vn|2 dx + o (1) , we obtain Z|∇vn|2dx − Z Ω |vn|2 dx = o (1) . (32)

We claim that (31) and (32) can hold simultaneously only if {vn} admits a subse-quence {vni} which converges strongly to zero. If not, the kvnkH1 is bounded away

from zero, that is kvnkH1 ≥ c for some c > 0. From (32) then it follows

Z

|vn|2

dx ≥ SN2 + o (1) .

By (31) and (32) for n large enough 1 NS N 2 1 N Z Ω |vn|2 dx + o (1) = 1 2 Z Ω |∇vn|2dx − 1 2 Z Ω |vn|2 dx + o (1) < 1 NS N 2

which is a contradiction. Consequently, un→ u0 strongly in H01(Ω) and Jλ(u0) =

σ.

Next, we establish the existence of a local minimum for Jλ on Mλ (Ω) .

Theorem 4.2. Let λ0> 0 as in Proposition1, then for λ ∈ (0, λ0), the functional

has a minimizer u−0 in M−λ (Ω) and it satisfies (i) Jλ¡u− 0 ¢ = α− λ(Ω) < αλ(Ω) + N1S N 2;

(ii) u−0 is a nonnegative (nontrivial) solution of equation (Eλ,f) . Proof. First, we claim that α−λ (Ω) < αλ(Ω) + 1

NS N 2. Let U1 = ½ u ∈ H1 0(Ω) \ {0} ¯ ¯ ¯ ¯kuk1 H1 t− µ u kukH1 ¶ > 1 ¾ ∪ {0} , U2 = ½ u ∈ H01(Ω) \ {0} ¯ ¯ ¯ ¯kuk1 H1 t− µ u kukH1 ¶ < 1 ¾ . Then Mλ (Ω) disconnects H1

0(Ω) in two connected components U1 and U2 and

H1

0(Ω) \M−λ(Ω) = U1∪ U2. For each u ∈ M+λ(Ω) , we have 1 < tmax(u) < t−(u) .

Since t−(u) = 1 kukH1t− ³ u kukH1 ´ , then M+λ (Ω) ⊂ A1. In particular, u0∈ A1. We

claim that there exists l0> 0 such that u+0 + l0u+∈ A2. First, we find a constant

c > 0 such that 0 < t− µ u+ 0+lwε,y ku+ 0+lwε,ykH1

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a sequence {ln} such that ln → ∞ and t− µ u+ 0+lnwε,y ku+ 0+lnwε,ykH1→ ∞ as n → ∞. Let vn = u + 0+lnwε,y ku+0+lnwε,ykH1. Since t (vn) vn ∈ M

λ (Ω) ⊂ Mλ(Ω) and by the Lebesgue dominated convergence theorem,

Z Ω v2 n = 1 ° °u+ 0 + lnwε,y ° °2 H1 Z Ω ¡ u+ 0 + lnwε,y ¢2 dx = ° 1 °u+ 0 + lnwε,y ° °2 H1 Z Ω µ u+0 ln + wε,y ¶2 dx R Ω(wε,y) 2 dx kwε,yk2 H1 as n → ∞. We have ¡ t−(vn) vn¢ = 1 2 £ t−(vn)¤2[t−(vn)] q q λ Z Ω f vq ndx − [t−(vn)]2 2 Z Ω v2 n dx → −∞ as n → ∞,

this contradicts that Jλ is bounded below on Mλ(Ω). Let l+0 = ¯ ¯ ¯c2°°u+ 0 ° °2 H1 ¯ ¯ ¯ 1 2 kwε,ykH1 + 1, then ° °u+ 0 + l0wε,y ° °2 H1 = °°u+0°°2H1+ (l0) 2 kwε,yk2H1+ 2l0 ­ u+0, wε,y ® H1 > °°u+ 0 ° °2 H1+ ¯ ¯ ¯c2°°u+ 0 ° °2 H1 ¯ ¯ ¯ + 2l0 µZ Ω u+ 0wε,ydx + λ Z Ω u+ 0wε,ydx> c2> " t− Ã u+0 + l0wε,y ° °u+ 0 + l0wε,y ° ° H1 !#2 , that is u+

0 + l0wε,y ∈ U2. Now, we define

β = inf γ∈Γs∈[0,1]max Jλ(γ (s)) , where Γ = ©γ ∈ C¡[0, 1] ; H1 0(Ω) ¢ | γ(0) = u+0 and γ(1) = u+0 + l0wε,y ª . Define a path γ0(s) = u+0 + sl0wε,y for s ∈ [0, 1], then γ0 ∈ Γ and there exists s0 ∈ (0, 1)

such that γ0(s0) ∈ M−λ (Ω) , we have β ≥ α−λ(Ω) . Moreover, by Lemma4.1 α− λ (Ω) ≤ β < αλ(Ω) + 1 NS N 2.

Analogously to the proof of Proposition1, one can show that the Ekeland variational principle gives a sequence {un} ⊂ M−λ (Ω) which satisfies

Jλ(un) = α−λ (Ω) + o (1) and Jλ0 (un) = o (1) in H−1(Ω) . Since α−λ (Ω) < αλ(Ω) + 1

NS

N

2, by Proposition 2, there exist a subsequence {un}

and u−

0 ∈ H01(Ω) such that

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This implies u−

0 ∈ M−λ(Ω) and Jλ(un) → Jλ

¡

u−0¢= α−λ (Ω) as n → ∞.

Since Jλ¡u−0¢ = Jλ¡¯¯u−0¯¯¢ and ¯¯u−0¯¯ ∈ M−λ (Ω) , by Lemma 2.4, ¯¯u−0¯¯ is also a solution of equation (Eλ,f) . Without loss of generality, we may assume that u−

0 is

nonnegative (nontrivial) solution of equation (Eλ,f).

Now, we begin to show the proof of Theorem1.1: By Theorems 3.4and4.2the equation (Eλ,f) has two nonnegative (nontrivial) solutions u+0 and u−0 such that

u+

0 ∈ M+λ(Ω) and u−0 ∈ M−λ (Ω). Since M+λ (Ω) ∩ M−λ(Ω) = ∅. This implies that u+

0 and u−0 are distinct. Now, by Dr´abek–Kufner–Nicolosi [8, Lemma 2.1] we have

u+0, u−0 ∈ L∞(Ω) . Then we can apply the Harnack inequality due to Trudinger [19]

in order to get u+0 and u−0 are positive in Ω.

5. Proof of Theorem 1.2. Finally in this section, we give a proof of Theorem 1.2. We need the following lemmas.

Lemma 5.1. For each uλ ∈ M−λ (Ω) , there is a unique su> 0 such that suλuλ

M0(Ω) , su≥1 − λ kfkLq∗ µ 2− q S (2 − q)2∗−q 2∗−2   1 2∗−2 and su≤1 + λ kfkLq∗ µ 2− q S (2 − q)2∗−q 2∗−2   1 2∗−2 . Furthermore, suλ → 1 as λ → 0.

Proof. For uλ∈ M−λ(Ω) , we have Z Ω |∇uλ|2dx − λ Z Ω f (x) |uλ|qdx − Z Ω |uλ|2 dx = 0 and (2 − q) Z Ω |∇uλ|2dx < (2∗− q) Z Ω |uλ|2 dx. Thus, there is a unique suλ > 0 such that suλuλ∈ M0(Ω) and so

s2 Z Ω |∇uλ|2dx = s2 Z Ω |uλ|2 . Then by the H¨older inequality

1 − λ kf kLq∗kuλkq−2 L2∗ ≤ s2 −2 = R Ω|∇uλ| 2 dx R Ω|uλ| 2 dx = 1 + λRf (x) |uλ|qdx R Ω|uλ| 2 dx ≤ 1 + λ kf kLq∗kuλkq−2 L2∗ . Since Z|uλ|2 > 2 − q 2− q Z Ω |∇uλ|2dx ≥ S (2 − q) 2− q kuλk 2 L2∗ or kuλkL2∗ > µ S (2 − q) 2− q ¶ 1 2∗−2 .

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Therefore, su≥1 − λ kfkLq∗ µ 2− q S (2 − q)2∗−q 2∗−2   1 2∗−2 and su≤1 + λ kfkLq∗ µ 2− q S (2 − q)2∗−q 2∗−2   1 2∗−2 . This completes the proof.

For c > 0, we define J0c(u) = 1 2 Z Ω |∇u|2+ u2 1 2 Z Ω c |u|2; Mc0(Ω) = © u ∈ H01(Ω) \ {0} | ­ (J0c)0(u) , u ® = 0ª; M0(Ω) = © u ∈ H01(Ω) \ {0} | hJ00(u) , ui = 0 ª .

Note that J0 = J0c for c = 1, and for each u ∈ M−λ(Ω) there is a unique su > 0 such that suu ∈ M0(Ω) . Furthermore, we have the following results.

Lemma 5.2. For each u ∈ M−λ(Ω) , we have

(i) there is a unique sc(u) > 0 such that sc(u) u ∈ Mc

0(Ω) and max t≥0J c 0(tu) = J0c(sc(u) u) = 1 Nc 2−N 2 Ã kuk2H1 R Ω|u| 2 !N −2 2 ; (ii) for µ ∈ (0, 1) , Jλ(u) ≥ (1 − λµ) N 2 J 0(suu) −λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q and Jλ(u) ≤ (1 + λµ) N 2 J 0(suu) + λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q . Proof. (i) For each u ∈ M−λ (Ω) , let

f (s) = J0c(su) = 1 2s 2kuk2 H1 1 2∗s 2Z Ω c |u|2∗, then f (s) → −∞ as s → ∞, f0(s) = s kuk2 H1 − s2 −1R Ωc |u| 2 and f00(s) = kuk2H1− (2∗− 1) s2 −2R Ωc |u| p . Let sc(u) = Ã kuk2H1 R Ωc |u| 2 ! 1 2∗−2 > 0. Then f0(sc(u)) = 0, sc(u) u ∈ Mc

0(Ω) and

f00(sc(u)) = kuk2H1− (2∗− 1) kuk

2

H1 = (2 − 2∗) kuk

2

H1 < 0.

Thus, there is a unique sc(u) > 0 such that sc(u) u ∈ Mc

0(Ω) and max s≥0J c 0(su) = J0c(sc(u) u) = 1 Nc 2−N 2 Ã kuk2H1 R Ω|u| 2 !N −2 2 .

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(ii) For each u ∈ M−

λ(Ω) , let c = 1/ (1 − λµ), sc= sc(u) > 0 and su> 0 such that scu ∈ Mc

0(Ω) and suu ∈ M0(Ω) . For µ ∈ (0, 1) , by the H¨older inequality and the

Young inequality, Z Ω f (x) |scu|qdx ≤ kf kLq∗kscuk q L2∗ ≤ kf kLq∗S −q 2 kscukq H1 2 − q 2 ³ kf kLq∗S −q 2 µ−q2 ´ 2 2−q +q 2 ³ µq2kscukq H1 ´2 q = 2 − q 2 µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q + 2 ks cuk2 H1.

Then by part (i) , sup s≥0Jλ(su) ≥ Jλ(s cu) ≥ (1 − λµ) 2 ks cuk2 H1 1 2 Z Ω |scu|2 −λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q = (1 − λµ) J01/(1−λµ)(scu) −λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q = (1 − λµ)N2 1 N Ã kuk2H1 R Ω|u| 2 !N −2 2 −λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q = (1 − λµ)N2 J0(suu) − λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q . By Lemma2.5(i), sup s≥0Jλ(su) = Jλ(u) . Thus, Jλ(u) ≥ (1 − λµ) 2∗ 2∗−2J 0(suu) − λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q . Moreover, Jλ(su) ≤ (1 + λµ) 2 ksuk 2 H1 1 2 Z Ω |su|2 +λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q = (1 + λµ) J01/(1+λµ)(su) +λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q and so Jλ(u) ≤ (1 + λµ) N 2 J0(suu) +λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q . This completes the proof.

Now, we begin to show the proof of Theorem1.2: Suppose that {λn} is a sequence of positive number such that λn → 0 as n → ∞. Let u(1)n = u+0,nand u

(2)

n = u−0,n be

positive solutions corresponding to Theorem1.1. (i) By Theorem3.4(iii) , we can conclude that

° ° °u(1)n ° ° ° H1 → 0 as n → ∞.

(22)

(ii) By Lemmas 5.1 and 5.2, for each n, there is a unique su(2) n > 0 such that su(2) n u (2) n ∈ M0(Ω) , su(2) n = 1 + o (1) and for µ ∈ (0, 1) J0 ³ su(2) n u (2) n ´ µ 1 1 − λnµN 2 · Jλn ³ u(2) n ´ +λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q ¸ . (33) Since Jλn ³ u(2)n ´ < Jλn(u (1) n ) + 1 NS N 2. (34) Then by (33) and (34) J0 ³ su(2) n u (2) n ´ < µ 1 1 − λnµN 2 · Jλn(u (1) n ) + 1 NS N 2 +λ (2 − q) 2q µ q q−2 ³ kf kLq∗S −q 2 ´ 2 2−q ¸ . Since Jλn(u (1) n ) → 0 as n → ∞. Thus, lim sup n→∞ J0 ³ su(2) n u (2) n ´ 1 NS N 2. This implies lim n→∞J0 ³ su(2) n u (2) n ´ = 1 NS N 2.

We can conclude that {su(2)

n u (2)

n } is a minimizing sequence for J0on M0(Ω) . Since

su(2) n → 1 as n → ∞. Thus, Z Ω ¯ ¯ ¯∇u(2)n ¯ ¯ ¯2dx = Z Ω ¯ ¯ ¯u(2)n ¯ ¯ ¯2 + o (1) and J0 ³ u(2)n ´ = 1 NS N 2 + o (1) .

Similar to the method used in Wang–Wu [20, Lemma 7] (or Ekeland [9]), we have {u(2)n } is a (PS)1

NS N

2–sequence for J0 in H

1

0(Ω) . Clearly, {u(2)n } is a bounded se-quence. Then there exist a subsequence {u(2)n } and u0∈ H01(Ω) such that

u(2)

n * u0 weakly in H01(Ω) .

Since Ω is a bounded domain, we have u0 ≡ 0. Moreover, by the concentration–

compactness principle (see Lions [13] or Struwe [16, Theorem 3.1]), there exist two sequences {xn} ⊂ Ω, {Rn} ⊂ R+ and a positive solution v0∈ D1,2

¡

RN¢of critical problem

−∆u = |u|2∗−2u in RN, such that Rn → ∞ as n → ∞ and

° ° °u(2)n (x) − R N −2 2 n v0(Rn(x − xn)) ° ° ° D1,2(RN)→ 0 as n → ∞.

This completes the proof of Theorem1.2.

Corollary 1. If u−0,λ∈ M−λ(Ω) is a positive solution of corresponding to Theorem

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Received September 2005; revised July 2007.

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