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# Multiple positive solutions for Dirichlet problems involving concave and convex nonlinearities

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Nonlinear Analysis ( ) –

www.elsevier.com/locate/na

## and convex nonlinearities

### Tsung-fang Wu

Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung 811, Taiwan Received 4 September 2007; accepted 25 October 2007

Abstract

In this paper, we study the multiplicity of positive solutions for the following Dirichlet problems involving concave and convex nonlinearities

(

−∆u + u =λf (x) |u|q−2u + h(x) |u|p−2u in A; u ∈ H01(A) ,

where 1 ≤ q< 2 < p < 2∗(2∗ = 2N

N −2if N ≥ 3, 2

= ∞if N = 2), λ > 0, A = Θ × R is an infinite strip in RN, Θ is a bounded domain in RN −1and f(x) , h (x) satisfy suitable conditions.

c

Keywords:Multiple positive solutions; Nehari manifold; Concave–convex nonlinearities

1. Introduction

In this paper, we consider the multiplicity of positive solutions for the following semilinear elliptic equation:    −∆u + u =λf (x) uq−1+h(x) up−1 in A, u> 0 in A, u ∈ H01(A) , (Eλf,h) where 1 ≤ q< 2 < p < 2∗(2∗= 2N N −2if N ≥ 3, 2 ∗ = ∞if N = 2), λ > 0, A = Θ × R is an infinite strip in RN

and Θ is a bounded domain in RN −1. Let x = x0, xN ∈ RN −1× R, we assume that f(x) and h (x) satisfy

( f 1) f ∈ LH(A), where LH(A) = L

2

2−q (A) if 1 < q < 2 and L

H(A) = H−1(A) if q = 1;

( f 2) f (x) ≥ 0 for all x ∈ A, and h (x) ∈ C (A) satisfies (h1) h (x) → 1 as |xN| → ∞;

Tel.: +886 7 591 9519; fax: +886 7 591 9344. E-mail address:tfwu@nuk.edu.tw.

doi:10.1016/j.na.2007.10.056

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(h2) there exist δ > θ1and 0< C0< 1 such that h(x) ≥ 1 − C0exp  −2 √ 1 +δ |xN|  for all x = x0, xN ∈ A,

whereθ1is the first eigenvalue of the Dirichlet problem −∆φ = θφ in Θ, φ = 0 on ∂Θ.

Similar problems have been the focus of a great deal of research in recent years. Authors Chabrowski–Bezzera do ´O [10] and Goncalves–Miyagaki [15] have investigated the following equation:

     −∆u + a(x) u = λf (x) uq−1+h(x) up−1 in RN, u > 0 in RN, u ∈ H1RN , (Ea,λf)

where 1< q < 2 < p < 2∗. They found that some existence and multiplicity results can be summarized as follows. In [15], it was assumed the following conditions:

(a0) a (x) ≥ a0> 0, x ∈ RN; (a∞) a (x) → ∞ as |x| → ∞; ( f0) f ≥ 0 and f ∈ L 2∗ 2∗−q RN ∩ L∞ RN.

Then they proved that there existsλ0> 0 such that the Eq.(Ea,λf)has at least two positive solutions for allλ ∈ 0, λ0.

In [10], it was assumed the following conditions:

(a1) a (x) is positive, locally H¨older continuous and bounded in RN;

( fc) f is a positive constant.

Then they proved that there existsλ > 0 such that the Eq.(Ea,λf)has at least one positive solution for allλ ∈ 0, λ . Among other interesting results, Ambrosetti–Brezis–Cerami [4] has investigated the following equation:

   −∆u =λuq−1+up−1 in Ω, u > 0 in Ω, u ∈ H01(Ω) , (Eλ) where 1 < q < 2 < p ≤ 2∗(2∗ = 2N N −2if N ≥ 3, 2

= ∞if N = 1, 2), λ > 0 and Ω is a bounded domain in

RN. They found that there existsλ0> 0 such that the Eq.(Eλ)admits at least two positive solutions forλ ∈ (0, λ0),

a positive solution forλ = λ0and no positive solution exists forλ > λ0. Actually, Adimurthi–Pacella–Yadava [5],

Damascelli–Grossi–Pacella [12], Ouyang–Shi [17] and Tang [20] proved that there existsλ0> 0 such that there are

exactly two positive solutions of Eq.(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 Eq. (Eλ)were done by

Ambrosetti–Azorezo–Peral [3], Brown–Wu [7], de Figueiredo–Gossez–Ubilla [14] and Wu [25].

As the first main result in this paper, we want to show that Eq.(Eλf,h)in the infinite strip A has at least two positive solutions, where the infinite strip A is different from that of above domains. Our result is as follows.

Theorem 1.1. If the conditions( f 1), ( f 2) and (h1), (h2) hold, then there exists Λ0 > 0 such that for λ ∈ (0, Λ0),

the Eq.(Eλf,h)possesses at least two positive solutions.

Under the assumption that λ > 0, Eq. (Eλf,h) can be regarded as a perturbation equation of the following homogeneous equation:    −∆u + u = h(x) up−1 in A, u > 0 in A, u ∈ H01(A) . (E0,h)

When h ≡ 1, it is known that the equation E0,1 has a ground state positive solutionw0being axially symmetric in

xN, and for y0∈ R, w0 x0, xN−y0 is also a ground state solution of equation E0,1 (see Lien–Tzeng–Wang [16]

and Chen–Chen–Wang [11]). Then the function h(x) satisfies the conditions (h1) , (h2) and h (x) ≤ 1 on A with a strict inequality on a set of positive measure. Then for the Eq.(E0,h), we can see that the Mountain Pass value is equal

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to the first level of breaking down of Palais–Smale condition (see Chabrowski [9, p. 38]) and we cannot get a positive solution through the Mountain Pass Theorem (i.e. Eq.(E0,h)does not admit any ground state solution).

From the above situation and a similar idea in Adachi–Tanaka [1,2] who considered the following equation:      −∆u + u = a(x) up−1+h(x) in RN, u> 0 in RN, u ∈ H1RN , (Ea,h)

where a(x)  1 = lim|x |→∞a(x) and h (x) ∈ H−1 RN \ {0} is nonnegative. Using the equation Ea,0 does not

admit any ground state solution and they proved that the Eq.(Ea,h)has at least three positive solutions under the assumption that khkH−1 is sufficiently small. Furthermore, under the additional condition

a(x) − 1 ≥ −C (− (2 + δ) |x|) for all x ∈ RN

for someδ > 0, C > 0, then the Eq.(Ea,h)has at least four positive solutions. Thus, if h(x) satisfies the conditions

(h1) , (h2) and h (x) ≤ 1 on A with a strict inequality on a set of positive measures, then the existence of more than two positive solutions for Eq.(Eλf,h)is expected and so we have the following results.

Theorem 1.2. If in addition to the conditions( f 1), ( f 2) and (h1), (h2), we still have: (h3) h (x) ≤ 1 on A with a strict inequality on a set of positive measure,

then there existsΛ> 0 such that for λ ∈ (0, Λ), the Eq.(Eλf,h)possesses at least three positive solutions. Theorem 1.3. If in addition to the conditions( f 1), ( f 2) and (h1)–(h3), we still have:

( f 3) f x0, x

N = f x0, −xN ;

(h4) h x0, x

N = h x0, −xN ;

(h5) h (x) ≥ 22− p2 for all x ∈ A,

then there existsΛ∗> 0 such that for λ ∈ (0, Λ∗), the Eq.(Eλf,h)possesses at least two nonaxially symmetric positive

solutions and two axially symmetric positive solutions.

In the following sections, we give the proofs ofTheorems 1.1–1.3. We use the variational methods to find positive solutions of Eq.(Eλf,h). Associated with Eq. (Eλf,h), we define the energy functional Jλf,h in H01(A) for given λ ≥ 0, f (x) and h (x) , Jλf,h(u) = 1 2 Z A |∇u|2+u2dx − λ q Z A f |u|qdx − 1 p Z A h |u|pdx.

It is well-known that the solutions of Eq. (Eλf,h) are the critical points of the energy functional Jλf,h (see Rabinowitz [19]).

This paper is organized as follows. In Section2, we give some notations and preliminaries. In Sections3–5, we complete the proofs of ourTheorems 1.1–1.3.

2. Notations and preliminaries

Throughout this section, we denote by Sp the best Sobolev constant for the imbedding of H01(A) in Lp(A). In

particular, kukLp ≤ SpkukH1 for all u ∈ H01(A). First, we define the Palais–Smale (simply by (PS)) sequences,

(PS)-values, and (PS)-conditions in H01(A) for Jλf,has follows.

Definition 2.1. (i) Forβ ∈ R, a sequence {un} is a(PS)β-sequence in H01(A) for Jλf,hif Jλf,h(un) = β + o(1) and

Jλf,h0 (un) = o(1) strongly in H−1(A) as n → ∞;

(ii) β ∈ R is a (PS)-value in H01(A) for Jλf,hif there exists a(PS)β-sequence in H01(A) for Jλf,h;

(iii) Jλf,hsatisfies the(PS)β-condition in H01(A) if every (PS)β-sequence in H01(A) for Jλf,hcontains a convergent subsequence.

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As the energy functional Jλf,his not bounded below on H01(A), it is useful to consider the functional on the Nehari manifold

Mλ,h(A) =nu ∈ H01(A) \ {0} |DJλf,h0 (u) , uE=0o . Thus, u ∈ Mλ,h(A) if and only if

D Jλf,h0 (u) , uE= kuk2H1 −λ Z A f |u|qdx − Z A h |u|pdx = 0. (2.1)

Note that Mλ,h(A) contains every nonzero solution of Eq.(Eλf,h). Furthermore, we have the following results. Lemma 2.2. The energy functional Jλf,his coercive and bounded below onMλ,h(A) .

Proof. If u ∈ Mλ,h(A), then by the H¨older inequality and the condition ( f 1) Jλf,h(u) ≥ p −2 2 p kuk 2 H1−  p − q q p  λZ A | f |2/(2−q)dx (2−q)/2Z A u2dx q/2 ≥ p −2 2 p kuk 2 H1−  p − q q p  λ k f kLH kuk q H1. (2.2)

Thus, Jλf,his coercive and bounded below on Mλ,h(A). 

Next, we consider the Nehari minimization problem: forλ ≥ 0, αλ,h(A) = inf Jλf,h(u) | u ∈ Mλ,h(A) .

Note thatα0,1(A) > 0 (see Willem [23, Theorem 4.2]). Define

ψλ(u) = D Jλf,h0 (u) , uE= kuk2H1 −λ Z A f |u|qdx − Z A h |u|pdx. Then for u ∈ Mλ,h(A) ,

ψ0 λ(u) , u = 2 kuk2H1−qλ Z A f |u|qdx − p Z A h |u|pdx =(2 − q) kuk2 H1−(p − q) Z A h |u|pdx (2.3) =(2 − p) kuk2 H1−(q − p) λ Z A f |u|qdx. (2.4)

Now, we split Mλ,h(A) into three parts:

M+λ,h(A) = u ∈ Mλ,h(A) | ψλ0 (u) , u > 0 ; M0λ,h(A) = u ∈ Mλ,h(A) | ψλ0 (u) , u = 0 ; M−λ,h(A) = u ∈ Mλ,h(A) | ψλ0 (u) , u < 0 . Then, we have the following results.

Lemma 2.3. Suppose that u0is a local minimizer for Jλf,honMλ,h(A) and that u06∈M0λ,h(A). Then Jλf,h0 (u0) = 0

in H−1.

Proof. Our proof is almost the same as in Brown–Zhang [8, Theorem 2.3]. 

Lemma 2.4. There exists a positive numberλ1such thatM0λ,h(A) = ∅ for all λ ∈ (0, λ1).

Proof. Our proof is almost the same as in Wu [25,Lemma 2]. 

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ByLemma 2.4, forλ ∈ (0, λ1) we write Mλ,h(A) = M+λ,h(A) ∪ M−λ,h(A) and define

α+

λ,h(A) = inf u∈M+λ,h(A)

Jλf,h(u) ; αλ,h− (A) = inf

u∈M−λ,h(A)

Jλf,h(u). Then we have the following result.

Theorem 2.5. There exists a positive number Λ0≤λ1such that if λ ∈ (0, Λ0), then

(i) αλ,h+ (A) < 0;

(ii) αλ,h− (A) > d0for some d0=d0 p, q, Sp, λ, f, h > 0.

In particular,αλ,h(A) = αλ,h+ (A) . Proof. (i) Let u ∈ M+λ,h(A) . By(2.4)

p −2 p − qkuk 2 H1 < λ Z A f |u|qdx (2.5) and so Jλf,h(u) = p −2 2 p kuk 2 H1−  p − q pq  λZ A f |u|qdx < −(p − 2) (2 − q) 2q p kuk 2 H1 < 0. Thus,α+λ,h(A) < 0.

(ii) Let u ∈ M−λ,h(A). By(2.3)and the Sobolev imbedding theorem 2 − q p − qkuk 2 H1 < Z A h |u|pdx ≤ Sppkhk∞kuk p H1 and so kukH1 > 2 − q (p − q) Sp pkhk∞ !p−21

for all u ∈ M−λ,h(A) . (2.6) By(2.2)in the proof ofLemma 2.2

Jλf,h(u) ≥ kukq H1  p − 2 2 p kuk 2−q H1 −  p − q q p  λ k f kLH  > 2 − q (p − q) Sp pkhk∞ !p−2q   p −2 2 p 2 − q (p − q) Sp pkhk∞ !2−qp−2 − p − q q p  λ k f kLH  . Thus, there exists a positive number Λ0≤λ1such that ifλ ∈ (0, Λ0), then

Jλf,h(u) > d0 for all u ∈ M−λ,h(A) ,

for some d0=d0 p, q, Sp, λ, f, h > 0. This completes the proof. 

For each u ∈ H01(A) \ {0}, we write tmax= (2 − q) kuk2 H1 (p − q) RAh |u|pdx !p−21 > 0. Then, we have the following lemma.

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Lemma 2.6. Let Λ0> 0 as inTheorem2.5. Then for eachλ ∈ (0, Λ0) and u ∈ H01(A) \ {0}, we have:

(i) there is a unique t−(u) > t

maxsuch that t−(u) u ∈ M−λ,h(A) and

Jλf,h t−(u) u = sup

t ≥0

Jλf,h(tu) .

Furthermore, if RA f |u|qdx> 0, then there is a unique 0 < t+(u) < tmaxsuch that t+(u) u ∈ M+λ,h(A) and

Jλf,h t+(u) u = inf

0≤t ≤t−(u)Jλf,h(tu) ;

(ii) t−(u) is a continuous function for nonzero u; (iii) M−λ,h(A) =nu ∈ H01(A) \ {0} | kuk1

H 1

t−kuku

H 1

 =1o;

(iv) there exists a bijective C0.1map m−fromU to M−λ,h(A) where U = {u ∈ H01(A) | kukH1 =1}. Furthermore,

M−λ,h(A) is path-connected.

Proof. The proofs of case (i)–(iii) are almost the same as in Wu [25,Lemma 5]. (iv) For t ≥ 0 andv ∈ U: Then by part (i), there is a unique t−(v) > t

maxsuch that t−(v) v ∈ M−λ,h(A). That is

D Jλf,h0 t−(v) v , t−(v) vE=t−(v)2−t−(v)qλ Z A f |v|qdx −t−(v)p Z A h |v|pdx = 0. Consider the function g(t, u) : R+×U → R defined by

g(t, u) =DJλf,h0 (tu) , tuE=t2−tqλ Z A f |u|qdx − tp Z A h |u|pdx. Note that g t−(v) , v =DJλf,h0 t−(v) v , t−(v) vE=0. Thus,

∂g ∂t (t, u) ( t−(v),v) =2t−(v) − q t−(v)q−1λ Z A f |v|qdx − pt−(v)p−1 Z A h |v|pdx =t−(v)−1  (2 − q) t−(v) v 2 H1−(p − q) Z A h t−(v) v pdx  =t−(v)−1ψλ0 t−(v) v , t−(v) v < 0.

By the implicit function theorem, there exist a neighborhood W ofv in U and unique function T ∈ C0,1such that T :W → R+, T(v) = t−(v) ,

g(T (u) , u) = 0 for all u ∈ W.

Therefore, for eachv ∈ U, there exist T : U → R+ and m− : U → M−λ,h(A) with T, m− C0,1 such that

T(v) = t−(v) , m−(v) = t−(v) v. Clearly, T and m− are injective. For each u ∈ M−λ,h(A), write u = t−(v) v, where t−(v) = kukH1 andv = kuku

H 1

∈U. Since m−(v) = u is surjective and U is path-connected, this implies that

M−λ,h(A) is path-connected.  For c> 0, we define J0,ch(u) = 1 2 Z A |∇u|2+u2dx − 1 p Z A ch |u|pdx, M0,ch(A) = n u ∈ H01(A) \ {0} | J00,ch(u) , u = 0o .

Note that for each u ∈ M−λ,h(A) there is a unique s (u) > 0 such that s (u) u ∈ M0,h(A). Furthermore, we have the

following results.

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Lemma 2.7. For each u ∈ M−λ,h(A), we have: (i) there is a unique sc(u) > 0 such that sc(u) u ∈ M

0,ch(A) and max t ≥0 J0,ch(tu) = J0,ch s c(u) u = 1 2 − 1 p  c −2 p−2 kuk p H1 R Ah |u|pdx !p−22 ; (ii) forλ ∈ (0, 1) , Jλf,h(u) ≥ (1 − λ) p p−2 J 0,h(s (u) u) − λ (2 − q) 2q k f k 2 2−q LH and Jλf,h(u) ≤ (1 + λ) p p−2 J 0,h(s (u) u) + λ (2 − q) 2q k f k 2 2−q LH .

Proof. (i) Our proof is almost the same as in Brown–Zhang [8, Theorem 3.1].

(ii) For each u ∈ M−λ,h(A), let c = 1/ (1 − λ), sc =sc(u) > 0 and s (u) > 0 such that scu ∈M

0,ch(A) and

s(u) u ∈ M0,h(A). By the H¨older and Young inequalities,

Z A f scu q dx ≤ k f kL H scu q H1 ≤ 2 − q 2 k f k 2 2−q LH + q 2 scu 2 H1.

Then by part (i), sup s≥0 Jλf,h(su) ≥ Jλf,h scu ≥ (1 − λ) 2 scu 2 H1 − λ (2 − q) 2q k f k 2 2−q LH − 1 p Z A h scu p dx =(1 − λµ) J0,ch scu − λ (2 − q) 2q k f k 2 2−q LH =c p 2− p p −2 2 p kukp H1 R Ah |u|pdx !p−22 −λ (2 − q) 2q k f k 2 2−q LH =(1 − λ) p p−2 J 0,h(s (u) u) − λ (2 − q) 2q k f k 2 2−q LH . (2.7) ByLemma 2.6(i), sup s≥0 Jλf,h(su) = Jλf,h(u) . Thus, Jλf,h(u) ≥ (1 − λ) p p−2 J 0,h(s (u) u) − λ (2 − q) 2q k f k 2 2−q LH .

Similar to the argument in(2.7), for s ≥ 0 Jλf,h(su) ≤ (1 + λ) 2 ksuk 2 H1+ λ (2 − q) 2q k f k 2 2−q LH − 1 p Z A h |su|pdx ≤ (1 + λ) p p−2 J 0,h(s (u) u) +λ (2 − q) 2q k f k 2 2−q LH and so Jλf,h(u) ≤ (1 + λ) p p−2 J 0,h(s (u) u) + λ (2 − q) 2q k f k 2 2−q LH .

This completes the proof. 

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Letξ ∈ C∞([0, ∞)) such that 0 ≤ ξ ≤ 1 and ξ(t) =0 for t ∈ [01 for t ∈ [2, 1],, ∞) and let ξn(z) = ξ  2|z| n  . (2.8)

Then we have the following results.

Lemma 2.8. Let {un} be a(PS)β-sequence in H01(A) for Jλf,hsatisfying un* 0 weakly in H01(A) and let vn=ξnun.

Then there exists a subsequence {un} such that

(i) kun−vnkH1 =o(1) as n → ∞; (ii) RAh|un|pdx = R Ah|vn|pdx + o(1) = RA|vn|pdx + o(1) ; (iii) kvnk2H1 = R A|vn|pdx + o(1) ; (iv) J0,1(vn) = J0,h(vn) + o (1) = β + o (1) .

Furthermore, if kunk> c for some c > 0, then β ≥ α0,1(A) .

Proof. (i)–(ii) Since un* 0 weakly in H01(A), there exists a subsequence {un} such that un→0 strongly in Lrloc(A)

for 1< r < 2∗, and

un →0 a.e. in A, (2.9)

or there exists a subsequence {un} such that

Z A(n)|un |qdx → 0 and Z A(n)|un |pdx → 0, (2.10)

where A(n) =  x0, xN ∈ A : |xN|< n . Moreover, by(2.9),(2.10), the Egorov theorem and the H¨older inequality,

Z A f |un|qdx = Z A fξnr|un|qdx + o(1) = o (1) for r ≥ 0. (2.11)

By the fact that

kun−vnk2H1 = kunk2H1 + kvnk2H1−2hun, vniH1,

it suffices to show that hun, vniH1 = kunk2H1 +o(1) = kvnk2H1+o(1). Since

hun, vniH1 = Z Aξn h |∇un|2+u2n i dx + Z A un∇un∇ξndx, |∇ξn| ≤ c n and {un} is a(PS)β-sequence in H 1

0(A) for Jλf,h, it follows that

Z Aξ r nun∇un∇ξndx = o(1) for r ≥ 0. (2.12) Hence hun, vniH1 = Z A ξn h |∇un|2+u2n i dx + o(1) . (2.13) Similarly, we have: kvnk2 H1 = Z Aξ 2 n h |∇un|2+u2n i dx + o(1) . (2.14)

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Since {ξr

nun}is bounded in H01(A), we have:

o(1) =DJλf,h0 (un), ξnrun E = Z A (ξr n|∇un|2+rξnr −1un∇ξn∇un+ξnru2n)dx − λ Z A fξnr|un|qdx − Z A hξr n|un|pdx.

From(2.11)and(2.12), we can conclude that Z Aξ r n(|∇un|2+u2n)dx = Z A hξnr|un|pdx + o(1). (2.15) Since h x0, xN → 1 as |xN| → ∞, by(2.10)we have: Z A h|un|pdx = Z A hξnr|un|pdx + o(1) = Z Aξ r n|un|pdx + o(1). (2.16) By(2.13)–(2.16), hun, vniH1 = kunk2H1+o(1) = kvnk2H1 +o(1) and Z A h|un|pdx = Z A h|vn|pdx + o(1) = Z A |vn|p+o(1) .

Moreover, kun−vnkH1 = o(1) as n → ∞. The results of (iii) and (iv) are from(2.11)and the results of parts (i),

(ii). 

We need the following proposition to provide a precise description for the (PS)-sequence of Jλf,hin M−λ,h(A). Proposition 2.9. Let Λ0 > 0 as inTheorem2.5. Then for eachλ ∈ (0, Λ0) and sequence {un} ⊂ M−λ,h(A) which

satisfies

(i) Jλf,h(un) = σ + o (1) with σ < αλ,h(A) + α0,1(A) ;

(ii) Jλf,h0 (un) = o (1) in H−1(A)

has a convergent subsequence.

Proof. ByLemma 2.2(ii), there exist a subsequence {un} and u0in H01(A) such that

un* u0 weakly in H01(A).

First, we claim that u06≡0. Suppose otherwise, then byLemma 2.8, there exists a subsequence {un} such that

kunk2H1 = Z A |un|pdx + o(1) (2.17) and so lim n→∞  1 2 − 1 p  kunkLpp =σ.

Moreover, {un} ⊂M−λ,h(A) and kunkH1 > c for some c > 0. ByLemma 2.8, we getσ ≥ α0,h(A), this contradicts

σ < αλ,h(A)+α0,h(A). Thus, u0is a nontrivial solution of Eq.(Eλf,h)and Jλf,h(u0) ≥ αλ,h(A). Write un=u0+vn

withvn* 0 weakly in H01(A). By the Brezis–Lieb lemma [6],

kunkLpp = ku0+vnkLpp = ku0kLpp + kvnkpLp+o(1) .

Since {un} is a bounded sequence in H01(A) and so {vn} is also a bounded sequence in H01(A). Moreover, by

f ∈ LH(A), the Egorov theorem and the H¨older inequality, we get

Z A f |vn|qdx = Z A f |un|qdx − Z A f |u0|qdx + o(1) = o (1) .

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Hence, for n large enough, we can conclude that αλ,h(A) + α0,1(A) > Jλf,h(u0+vn) = Jλf,h(u0) + 1 2kvnk 2 H1 − 1 pkvnk p Lp +o(1) ≥ αλ,h(A) +1 2kvnk 2 H1 − 1 pkvnk p Lp +o(1) or lim n→∞  1 2kvnk 2 H1 − 1 pkvnk p Lp  < α0,1(A) . (2.18)

Also from Jλf,h0 (un) = o (1) in H−1(A), where {un} is uniformly bounded and u0 is a solution of Eq.(Eλf,h),

follows kvnk2

H1− kvnkLpp =o(1) . (2.19)

We claim that(2.18)and(2.19)can hold simultaneously only if {vn} admits a subsequencevni which converges

strongly to zero. If not, then kvnkH1is bounded away from zero, that is

kvnkH1 ≥c for some c> 0. From(2.19), it follows lim n→∞kvnk p Lp ≥ 2 p p −2α0,1(A) . By(2.18)and(2.19), α0,1(A) ≤ lim n→∞  1 2− 1 p  kvnkp Lp = lim n→∞  1 2kvnk 2 H1− 1 pkvnk p Lp  < α0,1(A)

which is a contradiction. Consequently, un→u0strongly in H01(A) and Jλf,h(u0) = σ . 

3. Proof ofTheorem 1.1

Throughout this section, assume that the conditions( f 1)–( f 2) and (h1)–(h2) hold. First, we will establish the existence of a local minimum for Jλf,hon M+λ,h(A) .

Theorem 3.1. Let Λ0 > 0 as inTheorem2.5, then for each λ ∈ (0, Λ0), the Eq.(Eλf,h)has a positive solution

umin∈M+λ,h(A) such that

(i) Jλf,h(umin) = αλ,h(A) = α+λ,h(A) ;

(ii) kuminkH1 →0 asλ → 0.

Proof. Analogously to the proof of Wu [25,Proposition 9], one can prove the Ekeland variational principle (see [13]), that there exists a sequence {un} ⊂M+λ,h(A) such that

Jλf,h(un) = αλ,h+ (A) + o (1) ,

Jλf,h0 (un) = o (1) in H−1(A) .

Then byLemma 2.2, {un} is a bounded sequence in H01(A). Thus there exists a subsequence {un} and umin∈ H01(A)

is a solution of Eq.(Eλf,h)such that un * umin weakly in H01(A),

un →umin a.e. in A.

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Moreover, by f ∈ LH(A), the Egorov theorem and the H¨older inequality, we have: Z A f |un|qdx → Z A f |umin|qdx.

Now we prove thatRA f |umin|qdx 6= 0. Suppose otherwise, then

kunk2H1 = Z A h |un|pdx + o(1) and so lim n→∞  1 2 − 1 p  Z A h |un|pdx =αλ,h(A) ,

this contradictsαλ,h(A) < 0. Thus, RA f |umin|qdx 6= 0. In particular, umin ∈ Mλ,h(A) is a nontrivial solution of

equation Eλ, f. Now we prove that un→uminstrongly in H01(A). If not, then kuminkH1 < lim inf kunkH1 and so

αλ,h(A) ≤ Jλf,h(umin) =  1 2 − 1 p  kumink2H1 −  1 q − 1 p  λZ A f |umin|qdx

< lim inf Jλf,h(un) = αλ,h+ (A) ,

which is a contradiction. Consequently, un → umin strongly in H01(A) and Jλf,h(umin) = αλ,h+ (A). Moreover, by

Theorem 2.5we have umin∈ M+λ,h(A) and Jλf,h(umin) = αλ,h(A) = α+λ,h(A). Since Jλf,h(umin) = Jλf,h(|umin|)

and |umin| ∈M+λ,h(A). ByLemma 2.3and the maximum principle, we may assume that uminis a positive solution of

Eq.(Eλf,h). Moreover, by(2.5)and the H¨older inequality kumink2−qH1 < λ  p − q p −2  k f kLH and so kuminkH1 →0 asλ → 0. 

Next, we will establish the existence of a local minimum for Jλf,hon M−λ,h(A) .We need the following result. Proposition 3.2. Let Λ0> 0 as inTheorem2.5. Then for eachλ ∈ (0, Λ0)

α−

λ,h(A) < αλ,h(A) + α0,1(A) . (3.1)

Proof. Let w0 be a positive solution of equation E0,1 such that J0,1(w0) = α0,1(A) and let wy x0, xN =

w0 x0, xN−y. Clearly, J0,1(w0) = J0,1 wy. Then

Jλf,h(umin+lwy) = 1 2 umin+lwy 2 H1− λ q Z A f umin+lwy q dx − 1 p Z A h umin+lwy p dx = Jλf,h(umin) + J0,1(lw0) + 1 p Z (1 − h) lpwp ydx −λ Z A f Z lwy 0

h(umin+s)q−1−uq−1min

i ds  dx −1 p Z A

hh(umin+lwy)p−uminp −lpwyp−puminp−1lwy

i dx ≤ Jλf,h(umin) + J0,1(lw0) + 1 p Z A (1 − h) lpwp ydx −1 p Z A

hh(umin+lwy)p−uminp −lpwyp−pu p−1 minlwy

i

dx. (3.2)

Let g2(l) = J0,1(lw0) for t ≥ 0. Then

g20(l) = l kw0k2H1−l p−1Z A(w0) pdx and g00 2(l) = kw0k2H1−(p − 1) l p−2Z Aw p 0dx.

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Moreover, there is a unique l = kw 0k2 H 1 R Aw p 0dx p−21

=1 such that g20(l) = 0 and g002(l) < 0. Thus, g2(l) has an absolute

maximum at l = 1 and sup

t ≥0

J0,1(lw0) = J0,1(w0) = α0,1(A) .

Thus,

Jλf,h(umin+lwy) ≤ Jλf,h(umin) + α0,1(A) +

1 p Z (1 − h) lpwp ydx −1 p Z A

hh(umin+lwy)p−uminp −lpwyp−pu p−1 minlwy

i dx. Since

Jλf,h(umin+lwy) → Jλf,h(umin) = αλ,h(A) < 0 as l → 0

and

Jλf,h(umin+lwy) → −∞ as l → ∞,

we can easily find 0< l1< l2such that

Jλf,h(umin+lwy) < αλ,h(A) + α0,1(A) for all l ∈ [0, l1] ∪ [l2, ∞). (3.3)

Thus, we only need to show that sup

l1≤l≤l2

Jλf,h(umin+lwy) < αλ,h(A) + α0,1(A) .

We recall the fact that for any 0< ε < 1 + θ1, there exist aε> 0 and bε> 0 such that

aεφ1 x0 e− √ 1+θ1+ε|xN|w 0 x0, xN ≤ bεφ1 x0 e− √ 1+θ1−ε|xN|

for all x0, xN ∈ A, where θ1 is the first eigenvalue andφ1 the corresponding first positive eigenfunction of the

Dirichlet problem −∆φ = θφ in Θ, φ = 0 on ∂Θ. (see Chen–Chen–Wang [11]). In particular, we have that there exists b0> 0 such that

w0 x0, xN ≤ b0φ1 x0 e−|xN| for all x0, xN ∈ A.

We also remark that

(i) (u + v)p−up−vp−pup−1v ≥ 0 for all (u, v) ∈ [0, ∞) × [0, ∞), (ii) for any r > 0 we can find a constant C (r) > 0 such that

(u + v)pupvppup−1v ≥ C (r) v2,

for all(u, v) ∈ [r, ∞) × [0, ∞). Thus, for A−1,1 = x0, xN ∈ A | −1 < xN < 1 is a finite strip, setting

Cλ=C  min x ∈A−1,1 umin(x)  > 0, we have: Z A

hh(umin+lwy)p−uminp −l pwp y −pu p−1 minlwy i dx ≥ Z A−1,1

hh(umin+lwy)p−uminp −lpwyp−puminp−1lwy

i dx ≥hminCλ Z A−1,1 w2 x0, x N−y dx ≥hminCλaεexp  −2p1 +θ1+ε |y| . (3.4)

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From the condition(h3), we also have: Z A(1 − h) l pwp ydx ≤ Z A C0exp  −2 √ 1 +δ |y|b0pφ1 x0 exp(−p |xN−y|) ≤Cexp−minnp, 2 √ 1 +δo|y| .

Since(3.4)holds for any 0< ε < 1 + θ1, choosing ε < δ − θ1, we can find R0> 0 such that

Z A(1 − h) l pwp ydx< Z A

hh(umin+lwy)p−uminp −l pwp

y −puminp−1lwy

i

dx (3.5)

for all |y| ≥ R0. Thus, by(3.2),(3.3)and(3.5), we obtain

sup

l≥0

Jλf,h(umin+lwy) < αλ,h(A) + α0,1(A) for all |y| ≥ R0. (3.6)

To complete the proof of Proposition 3.2, it remains to show that there exists a positive number l such that umin+lwy ∈M−λ,h(A). Let

A1=  u ∈ H01(A) \ {0} 1 kukH1 t−  u kukH1  > 1  ∪ {0}, A2=  u ∈ H01(A) \ {0} 1 kukH1 t−  u kukH1  < 1  .

Then M−λ,h(A) disconnects H01(A) into two connected components A1and A2and H01(A) \ M −

λ,h(A) = A1∪A2.

For each u ∈ M+λ,h(A), we have: 1< tmax(u) < t−(u).

Since t−(u) = kuk1

H 1

t−kuku

H 1



, then M+λ,h(A) ⊂ A1. In particular, umin ∈ A1. We claim that there exists l0 > 0

such that umin+l0wy ∈ A2. First, we find a constant c > 0 such that 0 < t−

 u

min+lwy

kumin+lwykH 1



< c for each l ≥ 0. Otherwise, there exists a sequence {ln} such that ln → ∞and t−

 u min+lnwy kumin+lnwykH 1  → ∞as n → ∞. Let vn= umin+lnwy kumin+lnwykH 1. Since t −(v

n) vn∈M−λ,h(A) ⊂ Mλ,h(A) and by the Lebesgue dominated convergence theorem,

Z Av p ndx = 1 umin+lnwy p H1 Z A umin+lnwy p dx = 1 umin+lnwy p H1 Z A  umin ln +wy p dx → R Aw p ydx wy p H1 as n → ∞. We have: Jλf,h t−(vn) vn = 1 2t −(v n) 2 −t −(v n)q q λ Z A fvnqdx − t−(v n)p p Z A hvnpdx → −∞ as n → ∞,

this contradicts Jλf,hwhich is bounded below on Mλ,h(A). Let l0= c 2−ku mink2H 1 1 2 kw0k H 1 +1. Then umin+l0wy 2 H1 = kumink2H1 +l 2 0 wy 2 H1+2l0umin, wy H1 > kumink2H1 + c 2− ku mink2H1 +2l0 Z A huminp−1wydx +λ Z A f uq−1minwydx 

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> c2> " t− umin+l0wy umin+l0wy H1 !#2 ,

that is umin+l0wy ∈ A2. Define a pathγ (s) = umin+sl0wyfor s ∈ [0, 1], then

γ (0) = umin∈ A1, γ (1) = umin+l0wy ∈ A2,

and there exists s0∈(0, 1) such that umin+s0l0wy∈M−λ,h(A). This completes the proof. 

Now, we begin to show the proof ofTheorem 1.1: Analogously to the proof of Wu [25,Proposition 9], one can prove the Ekeland variational principle (see [13]), that there exists a sequence {un} ⊂M−λ,h(A) such that

Jλf,h(un) = αλ,h− (A) + o (1) and Jλf,h0 (un) = o (1) in H−1(A) .

Sinceα−λ,h(A) < αλ,h(A) + α0,1(A), byProposition 2.9, there exist subsequences {un} and u0 ∈ M−λ,h(A) is the

nonzero solution of Eq.(Eλf,h)such that un →u0 strongly in H01(A).

Hence Jλf,h(u0) = Jλf,h(|u0|) and |u0| ∈M−λ,h(A). ByLemma 2.3and the maximum principle, we may assume

that u0is a positive solution of Eq.(Eλf,h). Combining this with the result ofTheorem 3.1the Eq.(Eλf,h)has two

positive solutions uminand u0 such that umin ∈ M+λ,h(A), u0 ∈ M−λ,h(A). Hence M+λ,h(A) ∩ M−λ,h(A) = ∅. This

implies that uminand u0are distinct. 

4. Proof ofTheorem 1.2

In this section, we will use the filtration of the submanifold M−λ,h(A) to proveTheorem 1.2. First, we consider the following elliptic equation:

−∆u + u = h(x) |u|p−2u

in A,

u ∈ H01(A) , (E0,h)

where h(x) satisfies (h1)–(h3). It is known that the Eq.(E0,h)does not admit any solutionw0such that J0,h(w0) =

α0,h(A) and α0,h(A) = α0,1(A) (see Chabrowski [9, p. 38]). Moreover, we have the following results.

Lemma 4.1. For each(PS)α0,h(A)-sequence {un}in H01(A) for J0,h, there exists a subsequence {un} such that {ξnun}

is a(PS)α0,1(A)-sequence in H01(A) for J0,1.

Proof. The Eq.(E0,h)does not admit any solutionw0such that J0,h(w0) = α0,h(A). Thus, there exists a subsequence

{un} such that

un * 0 weakly in H01(A) .

Letvn=ξnun, then byLemma 2.8, kvnk2H1 =

R

A|vn|

pdx + o(1) and J

0,1(vn) = α0,1(A)+o (1). By Wang–Wu [22,

Lemma 7], {vn} is a(PS)α0,1(A)-sequence in H01(A) for J0,1. 

Denote the upper infinite strip A+t and the lower infinite strip A −

t as follows

A+t = {(x, y) ∈ A | y > t} ;

A−t = {(x, y) ∈ A | y < t} .

For a positive numberδ, we consider the filtration of the manifold M0,h(A) as follows.

M0,h(δ, A) = u ∈ M0,h(A) | J (u) ≤ α0,1(A) + δ ;

M0,h+ (δ, A) = ( u ∈ M0,h(δ, A) | Z A+ 0 h |u|pdx< p p −2α0,1(A) ) ; M0,h(δ, A) = ( u ∈ M0,h(δ, A) | Z A− 0 h |u|pdx< p p −2α0,1(A) ) .

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Then we have the following results.

Lemma 4.2. There existsδ0> 0 such that if u ∈ M0,h(δ0, A), then either

Z A+0 h |u|pdx< p p −2α0,1(A) or Z A−0 h |u|pdx< p p −2α0,1(A) . Proof. We divide the proof into two steps:

Step 1 (Existence): suppose that there exists a sequence {un} ⊂M0,h(A) such that J0,h(un) = α0,h(A) + o (1) ,

Z A+0 h |un|pdx ≥ p p −2α0,1(A) and Z A−0 h |un|pdx ≥ p p −2α0,1(A) . (4.1) Similar to the argument of Wang–Wu [22,Lemma 7], {un} is a(PS)α0,h(A)-sequence in H01(A) for J0,h. Hence the

Eq.(E0,h)does not admit any solutionw0such that J0,h(w0) = α0,h(A) . ByLemma 4.1andα0,h(A) = α0,1(A)

there exists a subsequence {un} such that {ξnun} is a (PS)α0,1(A)-sequence in H01(A) for J0,1, whereξnis as in(2.8).

Letvn=ξnun, we obtain

 J0,1(vn) = α0,1(A) + o(1),

J00,1(vn) = o(1) in H−1(A) as n → ∞

(4.2)

andvn=0 in A(1) for n > 2, where A (1) =  x0, xN ∈ A | |xN|< 1 . Moreover, vn =v+n +v − n and v± n(z) = (vn(z), for z ∈ A±0 0, for z 6∈ A±0. (4.3)

Then by(4.2)and(4.3),v±n ∈ H01(A±0) and v±n are bounded sequences. This implies J0 0,1(vn), vn± = vn± 2 H1− Z A±0n p dx = o(1) . Again using(4.2), we obtain

J00,1(vn±) = o(1) strongly in H−1(A±0) and

J0,1(vn) = J0,1 vn+ + J0,1 vn− =α0,1(A) + o (1) .

Assume that J(vn±) = c±+o(1), then

c++c−=α0,1(A) . (4.4)

Since c± are (PS)-values in H01(A±0) for J0,1, by Wang [21, Lemma 2.38], they are nonnegative. Moreover,

α0,1(A) = α0,1(A±0) > 0 (see Lien–Tzeng–Wang [16, Lemma 2.6]). Thus, by(4.4)and the definition of the Nehari

minimization problem, we may assume c+=α0,1(A+

0) = α0,1(A) and c−=0. Next, byLemma 2.8for n> 2

Z A h |un|pdx = Z A |vn|pdx + o(1) = Z A+0 vn+ p dx + Z A−0 vn− p dx + o(1) = Z A+0 vn+ p dx + Z A−0 h |un|pdx + o(1) . Thus Z A− 0 h |un|pdx = Z A h |un|pdx − Z A+ 0 v+n p dx = o(1) , which contradicts(4.1).

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Step 2 (Uniqueness): suppose that there exists u0∈ M0,h(δ0, A) such that Z A+0 h |u0|pdx< p p −2α0,1(A) and Z A−0 h |u0|pdx< p p −2α0,1(A) . Then 2 p p −2α0,1(A) ≤ Z A h |u0|pdx = Z A+ 0 h |u0|pdx + Z A− 0 h |u0|pdx < 2 p p −2α0,1(A) ,

this is a contradiction. We complete the proof ofLemma 4.2. 

By the consequence ofLemma 4.2, it is easy to prove the following result. Lemma 4.3. There existsδ0> 0 such that

(i) M0±,h(δ0, A) 6= ∅;

(ii) M0+,h(δ0, A) ∩ M0,h(δ0, A) = ∅;

(iii) M0,h(δ0, A) = M0+,h(δ0, A) ∪ M0,h(δ0, A) .

Moreover, given a 0< δ < δ0, we consider the filtration of the submanifold M−λ,h(A) as follows:

Nλ,h δ, A = n

u ∈M−λ,h(A) | Jλf,h(u) ≤ α0,1(A) + δ

o ; Nλ,h+ δ, A = ( u ∈ Nλ δ, A | Z A+0 |u|p< p 2(p − 2)α0,1(A) ) ; Nλ,h− δ, A = ( u ∈ Nλ δ, A | Z A− 0 |u|p< p 2(p − 2)α0,1(A) ) .

ByLemma 2.7(i), for each u ∈ M−λ,h(A), there is a unique s (u) > 0 such that s (u) u ∈ M0,h(A) . Moreover, we

have the following results.

Lemma 4.4. Let Λ0> 0 as inTheorem2.5. Then there existsλ2≤Λ0such that forλ ∈ (0, λ2), we have:

(i) 1< sp(u) < 2 for all u ∈ M−λ,h(A) ;

(ii) RAh |u|pdx> p−2p α0,1(A) for all u ∈ M−λ,h(A) .

Proof. For u ∈ M−λ,h(A), we have: kuk2 H1−λ Z A f |u|qdx − Z A h |u|pdx = 0 (4.5) and (2 − q) kuk2 H1 < (p − q) Z A h |u|pdx. (4.6)

ByLemma 2.7(i), there is a unique s(u) > 0 such that s (u) u ∈ M0,h(A) and so

s2(u) kuk2

H1 =sp(u)

Z

A

h |u|pdx. Then by(4.5)and the H¨older inequality

1< sp−2(u) = 1 +λ RA f |u| qdx R Ah |u|pdx ≤1 +λ k f kLHkuk q H1 R Ah |u|pdx .

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Since 0< h ≤ 1, 2 − q p − qkuk 2 H1 < Z A h |u|pdx ≤ SppkukHp1 and so kukH1 > 2 − q p − q 1 Spp !p−21 . (4.7) Thus, by(4.6)and(4.7), kukq H1 R Ah |u|pdx < p − q 2 − q kuk q−2 H1 < "  p − q 2 − q 2−qp−q Spp # 2−q p−2 . This implies 1< s (u) ≤   1 +λ k f kLH "  p − q 2 − q p−q2−q Spp # 2−q p−2    1 p−2 . (4.8)

Then there exists a positive numberλ2≤Λ0such that forλ ∈ (0, λ2) ,

1< sp(u) < 2 for all u ∈ M−λ,h(A) . Hence Z A h |u|pdx ≥ 1 sp(u)  2 p p −2  α0,1(A) .

By part (i), we can conclude that Z

A

h |u|pdx> p

p −2α0,1(A) for all u ∈ M

− λ,h(A) .

This completes the proof. 

Lemma 4.5. There exists a positive numberλ3≤min {1, λ2} such that forλ ∈ (0, λ3), we have:

(i) Nλ± δ, A 6= ∅;

(ii) Nλ+ δ, A ∩ Nλ− δ, A = ∅;

(iii) Nλ δ, A = Nλ+ δ, A ∪ Nλ− δ, A .

Proof. Forλ ∈ (0, 1): Let u ∈ Nλ δ, A, then byLemma 2.7, there is a unique s(u) > 0 such that s (u) u ∈ M0,h(A)

and J0,h(s (u) u) ≤  1 1 −λ p−2p  Jλf,h(u) +λ (2 − q) 2q k f k 2 2−q LH  ≤  1 1 −λ p−2p  α0,1(A) + δ + λ (2 − q) 2q k f k 2 2−q LH  .

Sinceδ < δ0, we can conclude that there exists a positive numberλ3≤min {1, λ2} such that forλ ∈ (0, λ3),

J0,h(s (u) u) ≤ α0,1(A) + δ0 for all u ∈ Nλ δ, A . (4.9)

ByLemma 4.3and(4.9), for each u ∈ Nλ δ, A, there is either s (u) u ∈ M0+,h(δ0, A) or s (u) u ∈ M0,h(δ0, A).

Without loss of generality, we may assume s(u) u ∈ M0+,h(δ0, A). Thus,

Z A+ 0 h |u|pdx< 1 sp(u)  p p −2  α0,1(A) .

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ByLemma 4.4, Z

A+0

h |u|pdx< p

p −2α0,1(A) .

Thus, u ∈ Nλ+ δ, A . To complete the proof ofLemma 4.5, it remains to show that Nλ+ δ, A ∩ Nλ− δ, A = ∅.

Suppose otherwise, then there exists u0∈Nλ δ, A such that

Z A+0 h |u0|pdx< p 2(p − 2)α0,1(A) and Z A−0 h |u0|pdx< p 2(p − 2)α0,1(A). ByLemma 4.4(ii) p p −2α (A) < Z A h |u0|pdx ≤ Z A+ 0 h |u0|pdx + Z A− 0 h |u0|pdx< p p −2α0,1(A), which is a contradiction. 

Let Nλ± δ, A denote the closure of Nλ± δ, A, then we have the following result. Lemma 4.6. Nλ± δ, A = Nλ± δ, A .

Proof. The proofs of the cases “+” and “−” are the similar arguments. Therefore, we only need to prove the case “+”. Suppose that u0is a limit point of Nλ+ δ, A, then Jλf,h(u0) ≤ α0,1(A) + δ and

Z A+ 0 h |u0|pdx ≤ p 2(p − 2)α0,1(A) . This implies u0 ∈ Nλ δ, A. If RA+

0 h |u0

|pdx = 2(p−2)p α0,1(A), then byLemma 4.5u0 ∈ Nλ− δ, A. Thus, by

Lemma 4.4(ii) p p −2α0,1(A) < Z A h |u0|pdx ≤ Z A+ 0 h |u0|pdx + Z A− 0 h |u0|pdx< p p −2α0,1(A) , which is a contradiction. Thus, u0∈ Nλ+ δ, A and so Nλ+ δ, A = Nλ+ δ, A. 

Proposition 4.7. Let λ3 be as in Lemma 4.5. Then for λ ∈ (0, λ3), there exist minimizing sequences u±n

⊂ Nλ± δ, A such that

Jλf,hn =σ± δ + o (1) and J0

λf,h u±n = o(1) in H−1(A) ,

whereσ± δ = infnJλf,h(u) | u ∈ Nλ,h± δ, Ao .

Proof. Analogously to the proof of Wu [25,Proposition 9], one can prove the Ekeland variational principle (see [13]), that there exist minimizing sequencesu±

n ⊂ N ±

λ δ, A such that

Jλf,hn =σ± δ + o (1) and Jλf,h0 u±n = o(1) in H−1(A) . We will omit the detailed proof here. 

Next, we will establish the existence of local minimums for Jλf,hon Nλ,h± δ, A. We need the following result. Proposition 4.8. Letλ3be as inLemma4.5. Then there exists a positive numberΛ ≤λ3such that forλ ∈ 0, Λ ,

σ± δ < α

λ,h(A) + α0,1(A) ,

whereσ± δ = inf n

Jλf,h(u) | u ∈ Nλ,h± δ, Ao .

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Proof. By the proof of Proposition 3.2, for each λ ∈ (0, λ3) there exist positive numbers l0 and R0 such that

umin+l0wy ∈M−λ,h(A) and

Jλf,h(umin+l0wy) < αλ,h(A) + α0,1(A) for all |y| ≥ R0.

ByTheorem 2.5,αλ,h(A) < 0, this implies umin+l0wy ∈Nλ,h δ, A. Moreover,

Z A+0 wy p

→0 as |y| → ∞ and kuminkH1 →0 asλ → 0.

Thus, there exists a positive number Λ ≤λ3such that for eachλ ∈ 0, Λ there exist positive numbers l and R such

that Z A+0 umin+lwy p < p

2(p − 2)α0,1(A) for all y ≤ −R and Z A− 0 umin+lwy p < p

2(p − 2)α0,1(A) for all y ≥ R.

This implies umin+lwy ∈ Nλ,h+ δ, A for y ≤ −R and umin+lwy ∈ Nλ,h− δ, A for y ≥ R. This completes the

proof. 

Then we have the following result.

Theorem 4.9. Let Λ be as in Proposition 4.8. Then for eachλ ∈ (0, Λ), the Eq. (Eλf,h) has positive solutions u±0 ∈Nλ± δ, A such that Jλf,h0 =σ± δ .

Proof. ByPropositions 4.7and4.8there exist sequencesu±n ⊂ Nλ± δ, A such that Jλf,hn =σ± δ + o (1) and J0

λf,h u±n = o(1) in H −1(A) ,

whereσ± δ < αλ,h(A) + α0,1(A). Thus, byProposition 2.9andLemma 4.6there exist subsequencesu±n and

0 ∈Nλ± δ, A are nonzero solutions of Eq.(Eλf,h)such that u±n →u±0 strongly in H01(A).

Since Jλf,h0 = Jλf,h

0  and

0 ∈ N

±

λ δ, A. ByLemma 2.3and the maximum principle, we may assume

that u±0 are positive solutions of Eq.(Eλf,h). 

Now, we begin to sketch out the proof ofTheorem 1.2: By combining the results ofTheorems 3.1and4.9, the Eq. (Eλf,h)has three positive solutions umin, u+0 and u−0 such that umin ∈ M+λ,h(A), u±0 ∈ Nλ± δ, A. Hence

M+λ,h(A) ∩ M−λ,h(A) = ∅ and Nλ+ δ, A ∩ Nλ− δ, A = ∅. This implies that umin, u+0 and u−0 are distinct. 

5. Proof ofTheorem 1.3

In this section, we focus on the problem on the xN-symmetric Sobolev space Hs(A) defined as follows: let Hs(A)

be the H1-closure of the space

C0s(A) = {u ∈ C0∞(A) | u x0, xN = u x0, −xN}

and let Hs−1(A) be the dual space of Hs(A). Then Hs(A) is a closed linear subspace of H01(A). For λ ≥ 0, consider

the xN-symmetric Nehari minimization problem

αs

λ,h(A) =u∈Minfs λ,h(A)

Jλf,h(u), where Msλ,h(A) =nu ∈ Hs(A) \ {0} |

D

Jλf,h0 (u) , uE=0o. Note that some properties and results of the minimization problem are as in Section2, and the proofs are omitted here. Moreover, by the principle of symmetric criticality

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(see Palais [18]), every(PS)β-sequence in Hs(A) for Jλf,h is a(PS)β-sequence in H01(A) for Jλf,h and we have the

following results.

Lemma 5.1. Suppose that the conditions(h1), (h2) and (h4) hold. Then α0,1(A) < α0s,h(A) ≤ 2α0,1(A).

Proof. The proof is similar to that of Lemma 4.3 in Wu [24]. Thus, we omit it here. 

Proposition 5.2. Suppose that the conditions(h1), (h2) and (h4) hold. Then J0,h satisfies the(PS)αs

0,h(A)-condition

in Hs(A) if and only if αs0,h(A) < 2α0,1(A).

Proof. The proof is similar to that of Proposition 4.4 in Wu [24]. Thus, we omit it here. 

Corollary 5.3. Suppose that the conditions(h1), (h2) and (h4), (h5) hold. Then

α0,1(A) < α0s,h(A) < 2α0,1(A). (5.1)

Proof. By Lemma 5.1, we only need to show α0s,h(A) < 2α0,1(A). By Lien–Tzeng–Wang [16] and

Chen–Chen–Wang [11], the equation E0,1 has a ground state solution w0 being axially symmetric in xN such

that J0,1(w0) = α0,1(A). Let s0> 0 with s0w0∈Ms0,h(A) . Then

s02 Z A |∇w0|2+w20dx = s0p Z A h|w0|pdx. (5.2)

Since h(x) ≥ 22− p2 for all x ∈ A and h x0, xN → 1 as |xN| → ∞, we apply(5.2)to obtain s2

0< 2. Thus, αs 0,h(A) ≤ J0,h(s0w0) =  1 2− 1 p  s02 Z A |∇w0|2+w20dx < 2α0,1(A).

This completes the proof. 

Similar toLemma 2.4, forλ ∈ (0, λ1), we write Msλ,h(A) = Msλ,h,+(A) ∪ Msλ,h,−(A) and define

αs,+

λ,h(A) = inf u∈Ms,+λ,h(A)

Jλf,h(u) ; αsλ,h,−(A) = inf

u∈Ms,−λ,h(A)

Jλf,h(u) .

Similar to the proof ofTheorem 3.1, for eachλ ∈ (0, Λ) there exists usmin ∈ Msλ,h(A) such that usmin is a positive solution of Eq.(Eλf,h)and

Jλf,h(umin) ≤ Jλf,h usmin =αsλ,h(A) = α s,+

λ,h(A) < 0.

Moreover, byLemmas 2.6and2.7 αs,− λ,h(A) ≥ (1 − λ) p p−2αs 0,h(A) − λ (2 − q) 2q k f k 2 2−q LH (5.3) and αs,− λ,h(A) ≤ (1 + λ) p p−2αs 0,h(A) + λ (2 − q) 2q k f k 2 2−q LH . (5.4)

Then we have the following results.

Lemma 5.4. Let Λ be as in Proposition4.8. Then for each positive number  < minnα0s,h(A) − α0,1(A)

o , there exists a positive numberΛ∗≤Λ such that if λ ∈ (0, Λ∗), then

σ± δ < α

0,1(A) < α0s,h(A) −  < α s,−

λ,h(A) < 2α0,1(A) − .

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Proof. By(5.1)–(5.4),

α0,1(A) < α0,hs (A) −  < αλ,hs,−(A) < 2α0,1(A) − .

Moreover, byTheorem 2.5andProposition 3.2, σ± δ < α

λ,h(A) + α0,1(A) < α0,1(A).

This completes our proof ofLemma 5.4. 

Theorem 5.5. Let Λ∗be as inLemma5.4. Then forλ ∈ (0, Λ∗), the Eq.(Eλf,h)has a positive solutioneu ∈Msλ,h,−(A)

such that Jλf,h(eu) = αλ,hs,−(A).

Proof. Analogously to the proof of Wu [25,Proposition 9], one can show that the Ekeland variational principle (see [13]) and the principle of symmetric criticality give a sequence {un} ⊂Msλ,h,−(A) which satisfies

Jλf,h(un) = αsλ,h,−(A) + o (1) and Jλf,h0 (un) = o (1) in H−1(A) . (5.5)

ByLemma 2.2(ii), there exist a subsequence {un} andeu ∈ Hs(A) such that un→eu a.e. in A

and

un*eu weakly in H

1 0(A) .

First, we claim thateu 6≡0. If not, byLemma 2.8, there exists a subsequence {un} such that

nunk2 H1 = Z A |ξnun|pdx + o(1) (5.6) and J0,1(ξnun) = αsλ,h,−(A) + o (1) ,

whereξn as in(2.8). Letvn = ξnun, then vn = 0 in A−n,n for n > 2 and there exists vn± ∈ H01 A ±

0 such that

vn = vn++v −

n. From(5.5), the fact that

v±

n is uniformly bounded and the similar argument of Lemma 2.8, we

obtain vn± 2 H1 = Z A± 0 vn± p dx + o(1) . (5.7)

Sincevn∈ Hs(A), we have v+n x 0, x N =vn− x 0, −x N, J0,1 vn+ = J0,1 vn− and αs,− λ,h(A) + o (1) = J0,1(vn) = J0,1 v+n + J0,1 v−n = 2J0,1 v+n  . This implies J0,1 vn+ = 1 2α s,− λ,h(A) + o (1) . (5.8)

By(5.7),(5.8), the definition of Nehari minimization problem andLemma 5.4, we have: 2α0,1(A) = 2α0,1 A+0 ≤α

s,−

λ,h(A) < 2α0,1(A), (5.9)

which is a contradiction. Thus,eu ∈ Hs(A) is a nontrivial solution of Eq. (Eλf,h)and Jλ(u0) ≥ αλ,hs,−(A). Write

un=eu +wn, thenwn* 0 weakly in H

1

0(A) and wn∈Hs(A). By the Brezis–Lieb lemma [6], we have:

kunkLpp = keu +wnk

p

Lp = keuk

p

Lp + kwnkLpp +o(1) .

Since {un} is a bounded sequence in H01(A) and so {wn} is also a bounded sequence in H01(A). Moreover, by

f ∈ LH(A), the Egorov theorem and the H¨older inequality, we have:

Z A f |wn|qdx = Z A f |un|qdx − Z A f |eu|qdx + o(1) = o (1) .

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Hence, for n large enough, we can conclude that 2α0,1(A) > Jλf,h(eu +wn) + o (1) = Jλf,h(eu) +1 2kwnk 2 H1− 1 pkwnk p Lp+o(1) ≥ αsλ,h(A) +1 2kwnk 2 H1− 1 pkwnk p Lp+o(1) or lim n→∞  1 2kwnk 2 H1− 1 pkwnk p Lp  ≤2α0,1(A) − αλ,hs (A) . (5.10)

Also from Jλf,h0 (un) = o (1) in H−1(A), {un} is uniformly bounded andeuis a solution of Eq.(Eλf,h)following o(1) =DJλf,h0 (un) , un

E

= kwnk2

H1− kwnkLpp +o(1) . (5.11)

We claim that(5.10)and(5.11)can hold simultaneously only if {wn} admits a subsequencewni which converges

strongly to zero. If not, then wni

H1is bounded away from zero, that is

wni

H1 ≥c for some c> 0.

From(5.11)and the similar argument of(5.9), it follows wni p Lp ≥ 4 p p −2α0,1(A) + o (1) . By(5.10)and(5.11), for n large enough

2α0,1(A) ≤  1 2 − 1 p  wni p Lp +o(1) = 1 2 wni 2 H1− 1 p wni p Lp+o(1) < 2α0,1(A),

which is a contradiction. Consequently, un → u0strongly in H01(A) and Jλf,h(eu) = αλ,hs,−(A). Hence Jλf,h(eu) =

Jλf,h(|eu|) and |eu| ∈ Msλ,h,−(A). ByLemma 2.3 and the maximum principle, we may assume thateu is a positive solution of Eq.(Eλf,h)such that Jλf,h(eu) = αλ,hs,−(A). 

Now, we begin to show the proof of Theorem 1.3: By Theorems 1.2 and 5.5, the Eq. (Eλf,h)in A has four positive solutions usmin, u+0, u0−andeu such that usmin ∈ Msλ,h(A), u±0 ∈ Nλ± δ, A andeu ∈ Msλ,h,−(A). Moreover,

Jλf,h usmin =αλ,hs (A) < 0, Jλf,h u0± =σ± δ and Jλf,h(eu) = αλ,hs,−(A). Hence Msλ,h,+(A) ∩ Nλ± δ, A = ∅, Nλ+ δ, A ∩ Nλ− δ, A = ∅

and

Jλf,h(eu) = αsλ,h,−(A) > σ± δ .

This implies that usmin, u+0, u−0 andeuare distinct. Furthermore, usminandeuare axially symmetric in xNand

Z A±0 h u±0 p dx< p 2(p − 2)α0,1(A) . ByLemma 4.4(ii), Z A h u±0 p dx> p p −2α0,1(A) .

This means that u±0 are nonaxially symmetric in xN. 

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