# Existence and Multiplicity of Positive Solutions for Schrödinger-Kirchhoff-Poisson System with Singularity

Authors: Mengjun Mu; Huiqin Lu
Journal: Journal of Function Spaces
 (2017)

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## Abstract

We study a singular Schrödinger-Kirchhoff-Poisson system by the variational methods and the Nehari manifold and we prove the existence, uniqueness, and multiplicity of positive solutions of the problem under different conditions.


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## Body

## 1. Introduction

In the present paper, we consider the following singular Schrödinger-Kirchhoff-Poisson system:(1)-a+b∫Ω∇u2dxΔu+εηϕu=λfxu-β,inΩ,-Δϕ=ηu2,inΩ,u>0,inΩ,u=ϕ=0,on∂Ω,where Ω⊂R3 is a smooth bounded domain with boundary ∂Ω, a>0,  b≥0,  η≥0, and λ>0 are four parameters and ε=±1,  β∈(0,1) is a constant, and f(x)∈C1(Ω¯) is a nontrivial nonnegative function.Whena=1,  b=0,  η=1, and f(x)≡1, system (1) reduces to the following singular Schrödinger-Poisson system:(2)-Δu+εϕu=μu-β,inΩ,-Δϕ=u2,inΩ,u>0,inΩ,u=ϕ=0,on∂Ω, which has been studied in [1]. By using the variational method and the Nehari manifold, the existence, uniqueness, and multiplicity of solutions for system (2) have been obtained.Whena=1,  b=0, and η=0, system (1) reduces to the following singular semilinear elliptic problem:(3)Δu+λfxu-β=0,inΩ,u=0,on∂Ω;the existence and uniqueness of positive solution have been studied in [2, 3] for λ=1.Whenη=0, the system (1) reduces to the following singular Kirchhoff type problem:(4)-a+b∫Ω∇u2dxΔu=λfxu-β,inΩ,u=0,on∂Ω.It is well known that the following Kichhoff type problem:(5)-a+b∫Ω∇u2dxΔu=hx,u,inΩ,u=0,on∂Ω, where Ω⊂R3 is a smooth bounded domain and h:Ω¯×R→R is a continuous function, has been extensively studied; see [4–7] and so forth. Problem (5) is called nonlocal because of the presence of the term ∫Ω∇u2dx which implies, when b≠0, that the equation in (5) is no longer a pointwise identity.Recently, the singular Kirchhoff type problems have been considered (see [8–11]). In [8], by using the Nehari manifold methods, Liu and Sun proved that problem (5) with h(x)=f(x)u-β+λgx(up/xs) has at least two positive solutions for λ>0 small enough. In [9], by using the variational methods, Lei et al. obtained that problem (5) with h(x)=u5+μu-β has at least two positive solutions for μ>0 small enough. The common characteristic of [8–10] is that the nonlinear terms contain both singular and superlinear. The superlinear term up  (p≥3) can overcome the difficulties which are caused by the Kirchhoff type perturbation uH01(Ω)4.Motivated by the above references, especially by [8–10], we study the singular Schrödinger-Kirchhoff-Poisson system (1). Let H01(Ω) be the Sobolev space equipped with the inner product and norm(6)u,v=∫Ω∇u·∇v,u=u,u1/2. As usual, us=(∫Ω∇us)1/s,  s∈[1,+∞), denotes the norm of the Lebesgue space Ls(Ω). Let S>0 be the usual Sobolev constant defined by (7)infu∈H01Ω∖0⁡∫Ω∇u2dx∫Ωu6dx1/3; that is, (8)u62≤S-1u2,∀u∈H01Ω.By the Hölder inequality and (8), we have(9)∫Ωfxu1-β≤f∞u41-βΩ3+β/4≤f∞u61-βΩ5+β/6≤f∞S-1-β/2u1-βΩ5+β/6,where |Ω| denotes the Lebesgue measure of the domain Ω.Before we state the main results about system (1), we first recall the following well-known facts (see [1]).Lemma 1.
Letη≥0; then for every u∈H01(Ω), there exists a unique ϕu∈H01(Ω) solution of (10)-Δϕ=ηu2,inΩ,u=0,on∂Ω.Moreover(i)
ϕu2=η∫Ωϕuu2;(ii)
ϕu≥0. Moreover ϕu>0 when u≠0 and η>0;(iii)
for eacht≠0,  ϕtu=t2ϕu;(iv)
(11)∫Ωϕuu2≤ηS-1u12/54≤ηS-1u44Ω2/3≤ηS-3u4Ω,u∈H01Ω;(v)
assume thatun⇀u; then ϕun→ϕu in H01(Ω) and ∫Ωϕununv→∫Ωϕuuv for any v∈H01(Ω);(vi)
ϕu∈Wloc2,3(Ω)∩C0(Ω¯);(vii)
foru,v∈H01(Ω),  η∫Ω(ϕuu-ϕvv)(u-v)≥(1/2)ϕu-ϕv2.By Lemma1, we easily see that system (1) can be converted into a binonlocal type problem of the singular Schrödinger-kirchhoff (12)-a+b∫Ω∇u2dxΔu+εηϕuu=λfxu-β,inΩ,u>0,inΩ,u=0,on∂Ω.We define the functional(13)Jλu=a2u2+b4u4+εη4∫Ωϕuu2-λ1-β∫Ωfxu1-β,u∈H01Ω.For any u,v∈H01(Ω), by (9),(14)∫Ωfxu1-β-v1-β≤f∞∫Ωu-v1-β≤f∞u-v41-βΩ3+β/4≤f∞S-1-β/2u-v1-βΩ5+β/6.Then, by Lemma 1, Jλ(u) is well defined and continuous on H01(Ω).In general, a functionu is called a solution of (12), if u∈H01(Ω),  u(x)>0,  ∀x∈Ω, and(15)a+b∫Ω∇u2∫Ω∇u·∇v+εη∫Ωϕuuv-λ∫Ωfxu-βv=0,∀v∈H01Ω.Moreover, (u,ϕu)∈H01(Ω)×H01(Ω) is a solution of system (1) if and only if u∈H01(Ω) is a solution of (12).As far as we know, the singular Schrödinger-Kirchhoff-Poisson system has not been considered up to now, and the study on the existence, uniqueness, and multiplicity of solutions for system (1) is meaningful in mathematics. We emphasize that the combined effects of the two nonlocal terms it contains cause some mathematical difficulties which make the study of such class of problems particularly interesting.We consider system (1) in two cases: ε=1 and ε=-1. By using the variational method, the Nehari manifold, and Ekeland’s variational principle, we obtain the existence and uniqueness of positive solution for ε=1 and the multiplicity of positive solutions for ε=-1. To prove the multiplicity of positive solutions, it is necessary that Nλ±≠∅, where Nλ± are defined in Section 3.2. But the method of getting Nλ±≠∅ in [1] cannot apply to our problem. In order to show that Nλ±≠∅, we introduce a special nonempty set. Let (16)η0=b1/2S3/2Ω-1/2,λ∗=2a3+βa1+β3+β1+β/2f∞-1S1-β/2Ω-5+β/6η2S-3Ω-b1+β/2.Our main results can be described as follows.Theorem 2.
Assumea>0,  b≥0,  ε=1, and β∈(0,1), then system (1) has a unique positive solution for all η≥0 and λ>0.Theorem 3.
Assumea>0,  b≥0,  ε=-1, and β∈(0,1), then(i)
When0≤η≤η0, system (1) has at least one positive solution for all λ>0;(ii)
Whenη>η0, system (1) has at least two positive solutions for all 0<λ<λ∗.Remark 4.
Whena=1,  b=0,  η=1, and f(x)≡1, we easily see that λ∗=μ∗, where μ∗ is as in Theorem 1.2 in [1]. Consequently, our results generalize and improve those of  [1].


## 2. Proof of Theorem2

Lemma 5.
For alla,λ>0,  b,η≥0, and ε=1, the functional Jλ attains the global minimizer in H01(Ω); that is, there exists u∗∈H01(Ω) such that(17)Jλu∗=mλ=infu∈H01Ω⁡Jλu<0.Proof of Lemma5.
Foru∈H01(Ω), by Lemma 1(ii) and (9),(18)Jλu≥a2u2-λ1-βS-1-β/2f∞Ω5+β/6u1-β,so, Jλ(u) is coercive and bounded from below on H01(Ω) for any λ>0. Thus mλ=infu∈H01(Ω)⁡Ju(u) is well defined. For t>0 and given u∈H01(Ω)∖{0},(19)Jλtu=at22u2+b4t4u4+ηt44∫Ωϕuu2-λ1-βt1-β∫Ωfxu1-β;we can see that for t>0 small enough, Jλ(tu)<0. Therefore, we have mλ=infu∈H01(Ω)⁡Ju(u)<0.
According to the definition ofmλ, there exists a minimizing sequence {un}⊂H such that limn→∞⁡Jλ(un)=mλ<0. Since Jλ(un)=Jλ(|un|), we may assume that un≥0. From (18), it is easy to see that {un} is bounded in H01(Ω), up to a subsequence; there exists u∗∈H01(Ω) such that(20)un⇀u∗,inH01Ωun⟶u∗,inLpΩ,p∈1,6unx⟶u∗x,a.e. inΩ.Then by the weakly lower semicontinuity of the norm, Lemma 1(v) and (14), we have(21)mλ≤Jλu∗=a2u∗2+b4u∗4+η4∫Ωϕu∗u∗2-λ1-β∫Ωfxu∗1-β≤liminfn→∞⁡Jλun=mλ.So, we have Jλ(u∗)=mλ<0.Proof of Theorem2.
We divide three steps to prove Theorem2.
(1) We showu∗>0 in Ω.
From Lemma5, u∗≥0 and u∗≠0. Fix φ∈H01(Ω), φ>0, and t≥0; we have(22)0≤limt→0⁡inf⁡Jλu∗+tφ-Jλu∗t=∫Ωa+bu∗2∇u∗∇φ+ηϕu∗u∗φ-λ1-βlim⁡subt→0⁡∫Ωfxu∗+tφ1-β-u∗1-βt;that is(23)λ1-βlim⁡subt→0⁡∫Ωfxu∗+tφ1-β-u∗1-βt≤∫Ωa+bu∗2∇u∗∇φ+ηϕu∗u∗φ.Notice that(24)∫Ωfxu∗+tφ1-β-u∗1-βt=1-β∫Ωfxu∗+tφξ-βφ,where ξ(x)∈(0,1) and (u∗(x)+tφ(x)ξ(x))-βφ(x)→(u∗(x))-βφ(x); a.e x∈Ω, t→0. Since (u∗(x)+tφ(x)ξ(x))-βφ(x)≥0, by using Fatou’s lemma, we have(25)λ∫Ωfxu∗-βφ≤∫Ωa+bu∗2∇u∗∇φ+ηϕu∗u∗φ.By the idea of approximation, the above expression also holds for φ∈H01(Ω),  φ≥0; that is,(26)∫Ωa+bu∗2∇u∗∇φ+ηϕu∗u∗φ-λ∫Ωfxu∗-βφ≥0.Therefore,(27)-a+bu∗2Δu∗+ηϕu∗u∗≥0in the weak sense.Since u∗≥0 and u∗≠0, by Lemma 1(ii) and (vi), ϕu∗>0 and ϕu∗∈C0(Ω¯). Therefore, by the strong maximum principle for weak solutions, we obtain that u∗>0 a.e. in Ω.
(2) We show thatu∗ is a solution of (12); that is, we prove u∗ satisfying (15) for ε=1.
For givenδ>0, define h:[-δ,δ]→(-∞,+∞) by h(t)=Jλ(u∗+tu∗); then h attains its minimum at t=0 by Lemma 5. It implies that(28)h′0=au∗2+bu∗4+η∫Ωϕu∗u∗2-λ∫Ωfxu∗1-β=0.We take φ∈H01(Ω)∖{0},  ρ>0 and define Ψ=(u∗+ρφ)+. Let(29)Ω1=x∈Ω:u∗x+ρφx>0,Ω2=x∈Ω:u∗x+ρφx≤0.Then Ψ|Ω1=u∗+ρφ, Ψ|Ω2=0. Inserting Ψ into (26) and using (28), we can obtain that(30)0≤∫Ωa+bu∗2∇u∗∇Ψ+ηϕu∗u∗Ψ-λfxu∗-βΨ=∫Ω1a+bu∗2∇u∗∇u∗+ρφ+ηϕu∗u∗u∗+ρφ-λfxu∗-βu∗+ρφ=∫Ω∖Ω2a+bu∗2∇u∗∇u∗+ρφ+ηϕu∗u∗u∗+ρφ-λfxu∗-βu∗+ρφ=ρ∫Ωa+bu∗2∇u∗∇φ+ηϕu∗u∗φ-λfxu∗-βφ-ρ∫Ω2a+bu∗2∇u∗∇φ+ηϕu∗u∗φ-λfxu∗-βφ≤ρ∫Ωa+bu∗2∇u∗∇φ+ηϕu∗u∗φ-λfxu∗-βφ-ρ∫Ω2a+bu∗2∇u∗∇φ+ηϕu∗u∗φ.Since u∗>0 and the measure of the domain Ω2={x∈Ω:u∗(x)+ρφ(x)≤0} tends to zero as ρ→0, it follows that(31)∫Ω2a+bu∗2∇u∗∇φ+ηϕu∗u∗φ⟶0.Then dividing by ρ>0 and letting ρ→0 in (30), we see that(32)∫Ωa+bu∗2∇u∗∇φ+ηϕu∗u∗φ-λfxu∗-βφ≥0,φ∈H01Ω.This inequality also holds for -φ, so we get(33)∫Ωa+bu∗2∇u∗∇φ+ηϕu∗u∗φ-λfxu∗-βφ=0,φ∈H01Ω.Then u∗∈H01(Ω) is a solution of (12) for ε=1,  λ>0, and η≥0.
(3) We show thatu∗ is the unique solution of (12) for ε=1,  λ>0, and η≥0.
Assume thatv∗∈H01(Ω) is also a solution of (12) for ε=1,  λ>0, and η≥0. It follows from (15) that (34)a+b∫Ω∇u∗2∫Ω∇u∗·∇u∗-v∗+η∫Ωϕu∗u∗u∗-v∗-λ∫Ωfxu∗-βu∗-v∗=0.(35)a+b∫Ω∇v∗2∫Ω∇v∗·∇u∗-v∗+η∫Ωϕv∗v∗u∗-v∗-λ∫Ωfxv∗-βu∗-v∗=0Subtracting (34) and (35), we obtain that(36)au∗-v∗2+bu∗4-u∗2+v∗2∫Ω∇u∗∇v∗+v∗4+η∫Ωϕu∗u∗-ϕv∗v∗u∗-v∗=λ∫Ωfxu∗-β-v∗-βu∗-v∗.Since β∈(0,1),  u∗,v∗>0 in Ω, the following inequality holds:(37)λ∫Ωfxu∗-β-v∗-βu∗-v∗≤0.Consequently, it follows from Lemma 1(vii) and (36) that(38)u∗-v∗≤0,which implies that(39)u∗-v∗=0;that is (u∗,ϕu∗)∈H01(Ω)×H01(Ω) is the unique solution of system (1).


## 3. Proof of Theorem3

### 3.1. The Case of0≤η≤η0

In this part, let0≤η≤η0, where η0=b1/2S3/2|Ω|-1/2. From Lemma 1(iv) and (9), we have(40)Jλu=a2u2+b4u4-η4∫Ωϕuu2-λ1-β∫Ωfxu1-β≥a2u2+b4u4-η24S-3Ωu4-λ1-βf∞S-1-β/2Ω5+β/6u1-β≥a2u2-λ1-βf∞S-1-β/2Ω5+β/6u1-β,∀u∈H01Ω.Thus, Jλ(u) is coercive and bounded from below on H01(Ω) for all 0≤η≤η0 and λ>0. So(41)αλ=infu∈H01Ω⁡Jλuis well defined. Obviously, αλ<0. Similar to the proof of Lemma 5, we easily obtain the following proposition.Proposition 6.
For alla>0,  b≥0,  ε=-1, and β∈(0,1), the functional Jλ attains the global minimizer in H01(Ω); that is, there exists u∗∈H01(Ω) such that(42)Jλu∗=αλ=infu∈H01Ω⁡Jλu<0.Proof of Theorem3(i).
The proof is similar to parts (1) and (2) of the proof in Theorem2, so we omit it.


### 3.2. The Case ofη>η0

In this part, letη>η0.  Jλ(u) is not necessarily coercive and bounded from below on H01(Ω) for η>η0. We define a Nehari manifold Nλ by(43)Nλ=u∈H01Ω:au2+bu4-η∫Ωϕuu2-λ∫Ωfxu1-βdx=0.Then the solutions of  (12) must lie in Nλ. Obviously, Nλ is a closed set in H01(Ω). In order to obtain the multiplicity, we split Nλ into the following three parts:(44)Nλ+=u∈Nλ:2au2+4bu4-4η∫Ωϕuu2-λ1-β∫Ωfxu1-βdx>0,(45)Nλ0=u∈Nλ:2au2+4bu4-4η∫Ωϕuu2-λ1-β∫Ωfxu1-βdx=0,(46)Nλ-=u∈Nλ:2au2+4bu4-4η∫Ωϕuu2-λ1-β∫Ωfxu1-βdx<0.When u∈Nλ, we have(47)2au2+4bu4-4η∫Ωϕuu2-λ1-β∫Ωfxu1-β=-2au2+3+βλ∫Ωfxu1-β=2bu4-2η∫Ωϕuu2+1+βλ∫Ωfxu1-β=1+βau2+3+βbu4-3+βη∫Ωϕuu2.Lemma 7.
For anyη>η0 there exists λ∗>0 such that Nλ±≠∅ and Nλ0={0} for λ∈(0,λ∗).Proof.
For any givenu∈H01(Ω)∖{0}, t≥0, by calculating, we can get that(48)tddtJλtu=at2u2+bt4u4-ηt4∫Ωϕuu2-λt1-β∫Ωfxu1-β=t1-βat1+βu2+bt3+βu4-ηt3+β∫Ωϕuu2-λ∫Ωfxu1-β.
Let(49)Φut=at1+βu2+bt3+βu4-ηt3+β∫Ωϕuu2,t≥0.Then,(50)Φu′t=1+βatβu2+3+βbt2+βu4-3+βηt2+β∫Ωϕuu2.
Sinceη>η0, it is easy to obtain that(51)G=u∈H01Ω∣bu4-η∫Ωϕuu2<0≠∅.
Therefore, for anyu∈G, let Φu′(t)=0. Then(52)tu=a1+βu23+βη∫Ωϕuu2-bu41/2.
Moreover,Φu′(t)>0 for all 0<t<tu and Φu′(t)<0 for all t>tu. Thus Φu is increasing for all 0<t<tu and decreasing for all t>tu. So(53)maxt∈0,+∞⁡Φut=Φutu=2a3+βa1+β3+β1+β/2u3+βη∫Ωϕuu2-bu41+β/2>0,(54)Φu0=0,limt→∞⁡Φut=-∞.Thus from Lemma 1(iv), (9), and (53), it follows that(55)Φutu-λ∫Ωfxu1-βdx≥2a3+βa1+β3+β1+β/2u3+βη∫Ωϕuu2-bu41+β/2-λf∞S-1-β/2Ω5+β/6u1-β≥2a3+βa1+β3+β1+β/21η2S-3Ω-b1+β/2-λf∞S-1-β/2Ω5+β/6u1-β>0for all 0<λ<λ∗.
From (54) and (55), we obtain that there exist unique positive numbers tu+<tu<tu- such that(56)Φutu+=λ∫Ωfxu1-βdx=Φutu-,Φu′tu+>0,Φu′tu-<0.From (48)–(50), it follows that tu+u∈Nλ+ and tu-u∈Nλ- for all 0<λ<λ∗.
Next, we prove thatNλ0={0} for all 0<λ<λ∗. Assume that there exists u0∈Nλ0 and u0≠0. Then it follows that(57)a1+βu02+b3+βu04-3+βη∫Ωϕu0u02=0,0=au02+bu04-η∫Ωϕu0u02-λ∫Ωfxu01-βdx=2a3+βu02-λ∫Ωfxu01-βdx.From (57) and (55), one has u0∈G and(58)0<2a3+βa1+β3+β1+β/2u03+βη∫Ωϕu0u02-bu041+β/2-λf∞S-1-β/2Ω5+β/6u01-β≤2a3+βa1+β3+β1+β/2u03+βη∫Ωϕu0u02-bu041+β/2-λ∫Ωfxu01-βdx=2a3+βa1+β3+β1+β/2u03+βa1+β/3+βu021+β/2-2a3+βu02=0for all 0<λ<λ∗, which implies a contradiction. Thus Nλ0={0} for all 0<λ<λ∗.Lemma 8.
N
λ
- is a closed set in H01(Ω) for λ∈(0,λ∗).Proof.
Supposeun⊂Nλ- such that un→u in H01(Ω) as n→∞. From the definition of Nλ-, one has(59)aun2+bun4-η∫Ωϕunun2-λ∫Ωfxun1-βdx=0,(60)a1+βun2+b3+βun4-η3+β∫Ωϕunun2<0.
Thus it follows fromun→u in H01(Ω) as n→∞ that(61)au2+bu4-η∫Ωϕuu2-λ∫Ωfxu1-βdx=0,a1+βu2+b3+βu4-η3+β∫Ωϕuu2≤0,so u∈Nλ0∪Nλ-. If u∈Nλ0, then u=0. However, from (60), we obtain(62)u2≥a1+β3+βη2S-3Ω-b,∀un∈Nλ-,which contracts u=0. Thus u∈Nλ- for all 0<λ<λ∗. Therefore, Nλ- is closed for all 0<λ<λ∗.Lemma 9.
Assumeη>η0 and λ∈(0,λ∗), Jλ(u) is coercive and bounded from below on Nλ and infu∈Nλ+∪Nλ0⁡Jλ(u)<0.Proof.
For allu∈Nλ, from (9) we have(63)Jλu=a2u2+b4u4-η4∫Ωϕuu2-λ1-β∫Ωfxu1-β=a2u2+14-au2+λ∫Ωfxu1-βdx-λ1-β∫Ωfxu1-β=a4u2-λ11-β-14∫Ωfxu1-β≥a4u2-λ11-β-14f∞S-1-β/2Ω5+β/6u1-β.
It follows from0<β<1 that Jλ(u) is coercive and bounded from below on Nλ for any λ∈(0,λ∗).
SinceNλ+∪Nλ0 and Nλ- are two nonempty closed subsets of H01(Ω) for all η>η0 and λ∈(0,λ∗), it follows that αλ+=infu∈Nλ+∪Nλ0⁡Jλ(u) and αλ-=infu∈Nλ-⁡Jλ(u) are well defined. For any given u∈Nλ+, by (47), we have(64)a1+βu2+b3+βu4-η3+β∫Ωϕuu2>0.
Consequently,(65)Jλu=a2u2+b4u4-η4∫Ωϕuu2-λ1-β∫Ωfxu1-β=-a1+β21-βu2-b3+β41-βu4+η3+β41-β∫Ωϕuu2<-a1+βu2-b3+βu4+η3+β∫Ωϕuu241-β<0.Thus, by Lemma 7, αλ+=infu∈Nλ+∪Nλ0⁡Jλ(u)<0 for all η>η0 and 0<λ<λ∗.Lemma 10.
Assume thatη>η0 and λ∈(0,λ∗), given u∈Nλ+ (resp., Nλ-), there exists ε>0 and a continuous function f=f(w)>0,  w∈H01(Ω),  w<ε, satisfying that(66)f0=1,fwu+w∈Nλ+resp.Nλ-,∀w∈H01Ω,w<ε.Proof.
Please see Lemma3.5 of [1] for the similar proof.Proof of Theorem3(ii).
By Ekeland’s variational principle, there exists a minimizing sequence{un}⊂Nλ+∪Nλ0 satisfying(i)
Jλ(un)≤infu∈Nλ+∪Nλ0⁡Jλu+1/n2,(ii)
Jλ(u)≥Jλ(un)-(1/n)u-un,  u∈Nλ+∪Nλ0.
FromJλ(|u|)=Jλ(u) and Lemma 9, we may assume that un∈Nλ+ and un≥0. It is easy to obtain that {un} is bounded in H01(Ω); we assume that un≤C0. Going if necessary to a subsequence, we can assume that(67)un⇀u0,inH01Ωun⟶u0,inLpΩ,p∈1,6unx⟶u0x,a.e. inΩ.Since {un}⊂Nλ+⊂Nλ, we get(68)aun2+bun4-η∫Ωϕunun2-λ∫Ωfxun1-βdx=0.By the weakly lower semicontinuity of the norm, Lemma 1(v), (14), and (47), we have(69)au02+bu04-η∫Ωϕu0u02-λ∫Ωfxu01-βdx≤0,-2au02+3+βλ∫Ωfxu01-βdx≥0,Jλu0≤limn→∞⁡Jλun=infu∈Nλ+∪Nλ0⁡Jλu<0.Therefore, u0(x)≥0 and u0(x)≢0.
First, we prove thatu0(x)>0 in Ω. Since {un}⊂Nλ+, we can claim that there exists a constant C1>0 such that up to a subsequence we have(70)a1+βun2+b3+βun4-η3+β∫Ωϕunun2≥C1.
In order to prove (70), it suffices to verify(71)a1+βlimn→∞¯⁡un2+b3+βlimn→∞¯⁡un4>η3+β∫Ωϕu0u02.Since un∈Nλ+, (72)a1+βun2+b3+βun4-η3+β∫Ωϕunun2>0.
It follows that(73)a1+βlimn→∞¯⁡un2+b3+βlimn→∞¯⁡un4≥η3+β∫Ωϕu0u02.By contradiction, we assume that(74)a1+βlimn→∞¯⁡un2+b3+βlimn→∞¯⁡un4=η3+β∫Ωϕu0u02.From the boundedness of {un} and (72), one has(75)a1+βlim_n→∞⁡un2+b3+βlim_n→∞⁡un4≥η3+β∫Ωϕu0u02,which combines with (74); it follows that(76)a1+βlimn→∞⁡un2+b3+βlimn→∞⁡un4=η3+β∫Ωϕu0u02.
Letlimn→∞⁡un=l; then l>0 and(77)a1+βl+b3+βl2=η3+β∫Ωϕu0u02.Since un∈Nλ, one has(78)al+bl2-η∫Ωϕu0u02-λ∫Ωfxu01-βdx=0.
Consequently, we have(79)2al3+β=λ∫Ωfxu01-βdx.
For all0<λ<λ∗, from (55), (77), and (79),(80)0<2a3+βa1+β3+β1+β/2u03+βη∫Ωϕu0u02-bu041+β/2-λf∞S-1-β/2Ω5+β/6u01-β<2a3+βa1+β3+β1+β/2u03+βη∫Ωϕu0u02-bu041+β/2-λ∫Ωfxu01-βdx=2a3+βa1+β3+β1+β/2l3+β/2a1+β/3+βl1+β/2-2al3+β=0,which is a contradiction. Thus our claim is true; that is, (70) and (71) must hold.
TakingN=21-βC0C1/(1+β), for fixed φ∈H01(Ω) and φ>0 in Ω, we apply Lemma 10 with u=un, and w=sφ,  s>0, small enough; it is easy to see that gn(s)=fn(sφ) such that gn(0)=1 and gn(un+sφ)∈Nλ+. It follows from un,  gn(s)(un+sφ)∈Nλ+⊂Nλ that(81)aun2+bun4-η∫Ωϕunun2-λ∫Ωfxun1-βdx=0,agn2sun+sφ2+bgn4sun+sφ4-ηgn4s∫Ωϕun+sφun+sφ2-λgn1-βs∫Ωfxun+sφ1-βdx=0.
By the above equalities, we have(82)0=gn2s-1aun+sφ2+aun+sφ2-un2+gn4s-1bun+sφ4+bun+sφ4-un4-gn4s-1η∫Ωϕun+sφun+sφ2-η∫Ωϕun+sφ2un+sφ2-ϕunun2dx-gn1-βs-1λ∫Ωfxun+sφ1-βdx-λ∫Ωfxun+sφ1-β-un1-βdx≤gn2s-1aun+sφ2+aun+sφ2-un2+gn4s-1bun+sφ4+bun+sφ4-un4-gn4s-1η∫Ωϕun+sφun+sφ2-η∫Ωϕun+sφ2un+sφ2-ϕunun2dx-gn1-βs-1λ∫Ωfxun+sφ1-βdx.
DenoteD+gn(0) the right upper Dini derivative of gn at zero. Then, dividing by s>0 and letting s→0+, we deduce that(83)0≤2aD+gn0un2+2a∫Ω∇un∇ϕdx+4bD+gn0un4+4bun2∫Ω∇un∇φdx-4ηD+gn0∫Ωϕunun2-4η∫Ωϕunun2-1-βλD+gn0∫Ωfxun1-βdx=D+gn02aun2+4bun4-4η∫Ωϕunun2-1-βλ∫Ωfxun1-βdx+2a∫Ω∇un∇φdx+4bun2∫Ω∇un∇φdx-4η∫Ωϕunun2=D+gn01+βaun2+3+βbun4-3+βη∫Ωϕunun2+2a∫Ω∇un∇φdx+4bun2∫Ω∇un∇φdx-4η∫Ωϕunun2.By un≤C0 and Lemma 1(iv), there exist C2,C3>0 independent of n such that(84)2a∫Ω∇un∇φdx+4bun2∫Ω∇un∇φdx-4η∫Ωϕunun2≤C2,(85)11-β1+βa∫Ω∇un∇φ+3+βbun2∫Ω∇un∇φ-3+βη∫Ωϕununφ≤C3.It follows from (70), (83), and (84) that(86)D+gn0≥-C1-1C2.
In the following part, we prove(87)D+gn0≤2φ+C3C1-1,∀n≥N.
Fixingn≥N, without loss of generality, we may assume D+gn(0)≥0. Thus, from condition (ii) and (65), we have(88)gns-1·unn+sgns·φn≥1ngnsun+sφ-un≥Jλun-Jλgnsun+sφ=1+β21-βaun+sφ2-aun2+1+β21-βagn2s-1un+sφ2+3+β41-βbun+sφ4-bun4+3+β41-βbgn4s-1un+sφ4-η3+β41-βgn4s∫Ωϕun+sφun+sφ2-ϕunun2-η3+β41-βgn4s-1∫Ωϕunun2dx.Dividing by s>0 and letting s→0+, we derive that(89)D+gn0unn+φn≥a1+β1-β∫Ω∇un∇φ+a1+β1-βD+gn0un2+b3+β1-βun2∫Ω∇un∇φ+b3+β1-βD+gn0un4-η3+β1-β∫Ωϕununϕ-η3+β1-βD+gn0∫Ωϕunun2.From (89) and (70), we have(90)φ≥φn≥D+gn01-β1+βaun2+3+βbun4-3+βη∫Ωϕunun2-1-βnun+a1+β1-β∫Ω∇un∇φdx+b3+β1-βun2∫Ω∇un∇φ-η3+β1-β∫Ωϕununφ≥D+gn01-β1-β2C1+a1+β1-β∫Ω∇un∇φdx+b3+β1-βun2∫Ω∇un∇φ-η3+β1-β∫Ωϕununφ,∀n≥N.From (85) and (90), it is easy to see that (87) holds. Thus, (86) and (87) imply that(91)D+gn0≤C,∀n≥N,where C>0 is independent of n.
From condition (ii) and (13), we can obtain that(92)1ngns-1·un+sgnsφ≥1ngnsun+sφ-un≥Jλun-Jλgnsun+sφ=-gn2s-12aun2-gn4s-14bun4+gn4s-14η∫Ωϕun2un2+λ1-βgn1-βs-1∫Ωfxun1-β+gn2s2aun2-un+sφ2+gn4s4bun4-un+sφ4+gn4s4η∫Ωϕun+sφun+sφ2-ϕunun2+λ1-βgn1-βs∫Ωfxun+sφ1-β-un1-β.Dividing by s>0 and letting s→0+, we derive that(93)1nD+gn0un+φ≥-a∫Ω∇un∇φ-bun2∫Ω∇un∇φ+η∫Ωϕununφdx+lim_s→0+⁡λ1-β∫Ωfxun+sφ1-β-un1-βsdx.Since f(x)[(un+sφ)1-β-un1-β]≥0,  ∀x∈Ω,  t>0, then by Fatou’s Lemma we obtain(94)∫Ωfxun-βφdx≤lim_s→0+⁡11-β∫Ωfxun+sφ1-β-un1-βsdx.Combining (93) and (94) we deduce(95)λ∫Ωfxun-βφdx≤1nD+gn0un+φ+a+bun2∫Ω∇un∇φ-η∫Ωϕununφdx≤C0C+φn+a+bun2∫Ω∇un∇φdx-η∫Ωϕununφdxfor all n≥N, which implies that(96)λlim_n→∞⁡∫Ωfxun-βφdx≤a+blim_n→∞⁡un2∫Ω∇u0∇φ-η∫Ωϕu0u0φdx.
Using Fatou’s Lemma, we infer that(97)λ∫Ωfxu0-βφdx≤a+blim_n→∞⁡un2∫Ω∇u0∇φ-η∫Ωϕu0u0φdx.
On one hand, sinceu0(x)≥0 a.e. in Ω, choosing φ=u0 in (97)(98)a+blim_n→∞⁡un2u02≥η∫Ωϕu0u02dx+λ∫Ωfxu01-βdx.
On the other hand, according to (67) and (68), it follows that(99)a+blim_n→∞⁡un2u02≤a+blimn→∞¯⁡un2limn→∞¯⁡un2=η∫Ωϕu0u02dx+λ∫Ωfxu01-βdx.Combining (98) and (99) yields(100)lim_n→∞⁡un2=limn→∞¯⁡un2=u02.Thus, according to (97) and (100), we have(101)a+bu02∫Ω∇u0∇φdx-η∫Ωϕu0u0φdx-λ∫Ωfxu0-βφ≥0for all φ∈H01(Ω) with φ>0. Therefore, by the idea of approximation and (101), we get(102)∫Ω∇u0∇φdx≥0,∀φ∈H01Ω,φ≥0.u0≥0 and u0≠0 in Ω. By the strong maximum principle for weak solutions, we obtain that u0>0 a.e. in Ω.
Second, we prove thatu0 is a solution of problem (12). In fact, we can repeat the proof of Theorem 2 and obtain that(103)a+bu02∫Ω∇u0∇φdx-η∫Ωϕu0u0φdx-λ∫Ωfxu0-βφdx=0,∀φ∈H01Ω.Then u0∈H01(Ω) is a positive solution of (12) for ε=-1,  0<λ<λ∗, and η>η0.
Finally, we prove thatu0∈Nλ+. According to (71) and (100), we have(104)a1+βu02+b3+βu04>η3+β∫Ωϕu0u02dx,which implies that u0∈Nλ+.
By Lemma8, we know that Nλ- is a closed set in H01(Ω) for λ∈(0,λ∗). Thus, applying Lemma 9 and Ekeland’s variational principle, we can find a bounded and nonnegative sequence {vn}⊂Nλ- and v0∈H01(Ω) such that(i)
Jλ(vn)≤infu∈Nλ-⁡Jλ(u)+1/n2,(ii)
Jλ(v)≥Jλ(vn)-(1/n)v-vn,  v∈Nλ-,(iii)
vn⇀v0 in H01(Ω),(iv)
vn→v0 in Lp(Ω),  p∈[1,6),(v)
vn(x)→v0(x),  a.e. in Ω.At this point we can repeat the proof as above and conclude that v0∈Nλ- is another positive solution of (12) for ε=-1,  0<λ<λ∗, and η>η0.


## 3.1. The Case of0≤η≤η0

In this part, let0≤η≤η0, where η0=b1/2S3/2|Ω|-1/2. From Lemma 1(iv) and (9), we have(40)Jλu=a2u2+b4u4-η4∫Ωϕuu2-λ1-β∫Ωfxu1-β≥a2u2+b4u4-η24S-3Ωu4-λ1-βf∞S-1-β/2Ω5+β/6u1-β≥a2u2-λ1-βf∞S-1-β/2Ω5+β/6u1-β,∀u∈H01Ω.Thus, Jλ(u) is coercive and bounded from below on H01(Ω) for all 0≤η≤η0 and λ>0. So(41)αλ=infu∈H01Ω⁡Jλuis well defined. Obviously, αλ<0. Similar to the proof of Lemma 5, we easily obtain the following proposition.Proposition 6.
For alla>0,  b≥0,  ε=-1, and β∈(0,1), the functional Jλ attains the global minimizer in H01(Ω); that is, there exists u∗∈H01(Ω) such that(42)Jλu∗=αλ=infu∈H01Ω⁡Jλu<0.Proof of Theorem3(i).
The proof is similar to parts (1) and (2) of the proof in Theorem2, so we omit it.


## 3.2. The Case ofη>η0

In this part, letη>η0.  Jλ(u) is not necessarily coercive and bounded from below on H01(Ω) for η>η0. We define a Nehari manifold Nλ by(43)Nλ=u∈H01Ω:au2+bu4-η∫Ωϕuu2-λ∫Ωfxu1-βdx=0.Then the solutions of  (12) must lie in Nλ. Obviously, Nλ is a closed set in H01(Ω). In order to obtain the multiplicity, we split Nλ into the following three parts:(44)Nλ+=u∈Nλ:2au2+4bu4-4η∫Ωϕuu2-λ1-β∫Ωfxu1-βdx>0,(45)Nλ0=u∈Nλ:2au2+4bu4-4η∫Ωϕuu2-λ1-β∫Ωfxu1-βdx=0,(46)Nλ-=u∈Nλ:2au2+4bu4-4η∫Ωϕuu2-λ1-β∫Ωfxu1-βdx<0.When u∈Nλ, we have(47)2au2+4bu4-4η∫Ωϕuu2-λ1-β∫Ωfxu1-β=-2au2+3+βλ∫Ωfxu1-β=2bu4-2η∫Ωϕuu2+1+βλ∫Ωfxu1-β=1+βau2+3+βbu4-3+βη∫Ωϕuu2.Lemma 7.
For anyη>η0 there exists λ∗>0 such that Nλ±≠∅ and Nλ0={0} for λ∈(0,λ∗).Proof.
For any givenu∈H01(Ω)∖{0}, t≥0, by calculating, we can get that(48)tddtJλtu=at2u2+bt4u4-ηt4∫Ωϕuu2-λt1-β∫Ωfxu1-β=t1-βat1+βu2+bt3+βu4-ηt3+β∫Ωϕuu2-λ∫Ωfxu1-β.
Let(49)Φut=at1+βu2+bt3+βu4-ηt3+β∫Ωϕuu2,t≥0.Then,(50)Φu′t=1+βatβu2+3+βbt2+βu4-3+βηt2+β∫Ωϕuu2.
Sinceη>η0, it is easy to obtain that(51)G=u∈H01Ω∣bu4-η∫Ωϕuu2<0≠∅.
Therefore, for anyu∈G, let Φu′(t)=0. Then(52)tu=a1+βu23+βη∫Ωϕuu2-bu41/2.
Moreover,Φu′(t)>0 for all 0<t<tu and Φu′(t)<0 for all t>tu. Thus Φu is increasing for all 0<t<tu and decreasing for all t>tu. So(53)maxt∈0,+∞⁡Φut=Φutu=2a3+βa1+β3+β1+β/2u3+βη∫Ωϕuu2-bu41+β/2>0,(54)Φu0=0,limt→∞⁡Φut=-∞.Thus from Lemma 1(iv), (9), and (53), it follows that(55)Φutu-λ∫Ωfxu1-βdx≥2a3+βa1+β3+β1+β/2u3+βη∫Ωϕuu2-bu41+β/2-λf∞S-1-β/2Ω5+β/6u1-β≥2a3+βa1+β3+β1+β/21η2S-3Ω-b1+β/2-λf∞S-1-β/2Ω5+β/6u1-β>0for all 0<λ<λ∗.
From (54) and (55), we obtain that there exist unique positive numbers tu+<tu<tu- such that(56)Φutu+=λ∫Ωfxu1-βdx=Φutu-,Φu′tu+>0,Φu′tu-<0.From (48)–(50), it follows that tu+u∈Nλ+ and tu-u∈Nλ- for all 0<λ<λ∗.
Next, we prove thatNλ0={0} for all 0<λ<λ∗. Assume that there exists u0∈Nλ0 and u0≠0. Then it follows that(57)a1+βu02+b3+βu04-3+βη∫Ωϕu0u02=0,0=au02+bu04-η∫Ωϕu0u02-λ∫Ωfxu01-βdx=2a3+βu02-λ∫Ωfxu01-βdx.From (57) and (55), one has u0∈G and(58)0<2a3+βa1+β3+β1+β/2u03+βη∫Ωϕu0u02-bu041+β/2-λf∞S-1-β/2Ω5+β/6u01-β≤2a3+βa1+β3+β1+β/2u03+βη∫Ωϕu0u02-bu041+β/2-λ∫Ωfxu01-βdx=2a3+βa1+β3+β1+β/2u03+βa1+β/3+βu021+β/2-2a3+βu02=0for all 0<λ<λ∗, which implies a contradiction. Thus Nλ0={0} for all 0<λ<λ∗.Lemma 8.
N
λ
- is a closed set in H01(Ω) for λ∈(0,λ∗).Proof.
Supposeun⊂Nλ- such that un→u in H01(Ω) as n→∞. From the definition of Nλ-, one has(59)aun2+bun4-η∫Ωϕunun2-λ∫Ωfxun1-βdx=0,(60)a1+βun2+b3+βun4-η3+β∫Ωϕunun2<0.
Thus it follows fromun→u in H01(Ω) as n→∞ that(61)au2+bu4-η∫Ωϕuu2-λ∫Ωfxu1-βdx=0,a1+βu2+b3+βu4-η3+β∫Ωϕuu2≤0,so u∈Nλ0∪Nλ-. If u∈Nλ0, then u=0. However, from (60), we obtain(62)u2≥a1+β3+βη2S-3Ω-b,∀un∈Nλ-,which contracts u=0. Thus u∈Nλ- for all 0<λ<λ∗. Therefore, Nλ- is closed for all 0<λ<λ∗.Lemma 9.
Assumeη>η0 and λ∈(0,λ∗), Jλ(u) is coercive and bounded from below on Nλ and infu∈Nλ+∪Nλ0⁡Jλ(u)<0.Proof.
For allu∈Nλ, from (9) we have(63)Jλu=a2u2+b4u4-η4∫Ωϕuu2-λ1-β∫Ωfxu1-β=a2u2+14-au2+λ∫Ωfxu1-βdx-λ1-β∫Ωfxu1-β=a4u2-λ11-β-14∫Ωfxu1-β≥a4u2-λ11-β-14f∞S-1-β/2Ω5+β/6u1-β.
It follows from0<β<1 that Jλ(u) is coercive and bounded from below on Nλ for any λ∈(0,λ∗).
SinceNλ+∪Nλ0 and Nλ- are two nonempty closed subsets of H01(Ω) for all η>η0 and λ∈(0,λ∗), it follows that αλ+=infu∈Nλ+∪Nλ0⁡Jλ(u) and αλ-=infu∈Nλ-⁡Jλ(u) are well defined. For any given u∈Nλ+, by (47), we have(64)a1+βu2+b3+βu4-η3+β∫Ωϕuu2>0.
Consequently,(65)Jλu=a2u2+b4u4-η4∫Ωϕuu2-λ1-β∫Ωfxu1-β=-a1+β21-βu2-b3+β41-βu4+η3+β41-β∫Ωϕuu2<-a1+βu2-b3+βu4+η3+β∫Ωϕuu241-β<0.Thus, by Lemma 7, αλ+=infu∈Nλ+∪Nλ0⁡Jλ(u)<0 for all η>η0 and 0<λ<λ∗.Lemma 10.
Assume thatη>η0 and λ∈(0,λ∗), given u∈Nλ+ (resp., Nλ-), there exists ε>0 and a continuous function f=f(w)>0,  w∈H01(Ω),  w<ε, satisfying that(66)f0=1,fwu+w∈Nλ+resp.Nλ-,∀w∈H01Ω,w<ε.Proof.
Please see Lemma3.5 of [1] for the similar proof.Proof of Theorem3(ii).
By Ekeland’s variational principle, there exists a minimizing sequence{un}⊂Nλ+∪Nλ0 satisfying(i)
Jλ(un)≤infu∈Nλ+∪Nλ0⁡Jλu+1/n2,(ii)
Jλ(u)≥Jλ(un)-(1/n)u-un,  u∈Nλ+∪Nλ0.
FromJλ(|u|)=Jλ(u) and Lemma 9, we may assume that un∈Nλ+ and un≥0. It is easy to obtain that {un} is bounded in H01(Ω); we assume that un≤C0. Going if necessary to a subsequence, we can assume that(67)un⇀u0,inH01Ωun⟶u0,inLpΩ,p∈1,6unx⟶u0x,a.e. inΩ.Since {un}⊂Nλ+⊂Nλ, we get(68)aun2+bun4-η∫Ωϕunun2-λ∫Ωfxun1-βdx=0.By the weakly lower semicontinuity of the norm, Lemma 1(v), (14), and (47), we have(69)au02+bu04-η∫Ωϕu0u02-λ∫Ωfxu01-βdx≤0,-2au02+3+βλ∫Ωfxu01-βdx≥0,Jλu0≤limn→∞⁡Jλun=infu∈Nλ+∪Nλ0⁡Jλu<0.Therefore, u0(x)≥0 and u0(x)≢0.
First, we prove thatu0(x)>0 in Ω. Since {un}⊂Nλ+, we can claim that there exists a constant C1>0 such that up to a subsequence we have(70)a1+βun2+b3+βun4-η3+β∫Ωϕunun2≥C1.
In order to prove (70), it suffices to verify(71)a1+βlimn→∞¯⁡un2+b3+βlimn→∞¯⁡un4>η3+β∫Ωϕu0u02.Since un∈Nλ+, (72)a1+βun2+b3+βun4-η3+β∫Ωϕunun2>0.
It follows that(73)a1+βlimn→∞¯⁡un2+b3+βlimn→∞¯⁡un4≥η3+β∫Ωϕu0u02.By contradiction, we assume that(74)a1+βlimn→∞¯⁡un2+b3+βlimn→∞¯⁡un4=η3+β∫Ωϕu0u02.From the boundedness of {un} and (72), one has(75)a1+βlim_n→∞⁡un2+b3+βlim_n→∞⁡un4≥η3+β∫Ωϕu0u02,which combines with (74); it follows that(76)a1+βlimn→∞⁡un2+b3+βlimn→∞⁡un4=η3+β∫Ωϕu0u02.
Letlimn→∞⁡un=l; then l>0 and(77)a1+βl+b3+βl2=η3+β∫Ωϕu0u02.Since un∈Nλ, one has(78)al+bl2-η∫Ωϕu0u02-λ∫Ωfxu01-βdx=0.
Consequently, we have(79)2al3+β=λ∫Ωfxu01-βdx.
For all0<λ<λ∗, from (55), (77), and (79),(80)0<2a3+βa1+β3+β1+β/2u03+βη∫Ωϕu0u02-bu041+β/2-λf∞S-1-β/2Ω5+β/6u01-β<2a3+βa1+β3+β1+β/2u03+βη∫Ωϕu0u02-bu041+β/2-λ∫Ωfxu01-βdx=2a3+βa1+β3+β1+β/2l3+β/2a1+β/3+βl1+β/2-2al3+β=0,which is a contradiction. Thus our claim is true; that is, (70) and (71) must hold.
TakingN=21-βC0C1/(1+β), for fixed φ∈H01(Ω) and φ>0 in Ω, we apply Lemma 10 with u=un, and w=sφ,  s>0, small enough; it is easy to see that gn(s)=fn(sφ) such that gn(0)=1 and gn(un+sφ)∈Nλ+. It follows from un,  gn(s)(un+sφ)∈Nλ+⊂Nλ that(81)aun2+bun4-η∫Ωϕunun2-λ∫Ωfxun1-βdx=0,agn2sun+sφ2+bgn4sun+sφ4-ηgn4s∫Ωϕun+sφun+sφ2-λgn1-βs∫Ωfxun+sφ1-βdx=0.
By the above equalities, we have(82)0=gn2s-1aun+sφ2+aun+sφ2-un2+gn4s-1bun+sφ4+bun+sφ4-un4-gn4s-1η∫Ωϕun+sφun+sφ2-η∫Ωϕun+sφ2un+sφ2-ϕunun2dx-gn1-βs-1λ∫Ωfxun+sφ1-βdx-λ∫Ωfxun+sφ1-β-un1-βdx≤gn2s-1aun+sφ2+aun+sφ2-un2+gn4s-1bun+sφ4+bun+sφ4-un4-gn4s-1η∫Ωϕun+sφun+sφ2-η∫Ωϕun+sφ2un+sφ2-ϕunun2dx-gn1-βs-1λ∫Ωfxun+sφ1-βdx.
DenoteD+gn(0) the right upper Dini derivative of gn at zero. Then, dividing by s>0 and letting s→0+, we deduce that(83)0≤2aD+gn0un2+2a∫Ω∇un∇ϕdx+4bD+gn0un4+4bun2∫Ω∇un∇φdx-4ηD+gn0∫Ωϕunun2-4η∫Ωϕunun2-1-βλD+gn0∫Ωfxun1-βdx=D+gn02aun2+4bun4-4η∫Ωϕunun2-1-βλ∫Ωfxun1-βdx+2a∫Ω∇un∇φdx+4bun2∫Ω∇un∇φdx-4η∫Ωϕunun2=D+gn01+βaun2+3+βbun4-3+βη∫Ωϕunun2+2a∫Ω∇un∇φdx+4bun2∫Ω∇un∇φdx-4η∫Ωϕunun2.By un≤C0 and Lemma 1(iv), there exist C2,C3>0 independent of n such that(84)2a∫Ω∇un∇φdx+4bun2∫Ω∇un∇φdx-4η∫Ωϕunun2≤C2,(85)11-β1+βa∫Ω∇un∇φ+3+βbun2∫Ω∇un∇φ-3+βη∫Ωϕununφ≤C3.It follows from (70), (83), and (84) that(86)D+gn0≥-C1-1C2.
In the following part, we prove(87)D+gn0≤2φ+C3C1-1,∀n≥N.
Fixingn≥N, without loss of generality, we may assume D+gn(0)≥0. Thus, from condition (ii) and (65), we have(88)gns-1·unn+sgns·φn≥1ngnsun+sφ-un≥Jλun-Jλgnsun+sφ=1+β21-βaun+sφ2-aun2+1+β21-βagn2s-1un+sφ2+3+β41-βbun+sφ4-bun4+3+β41-βbgn4s-1un+sφ4-η3+β41-βgn4s∫Ωϕun+sφun+sφ2-ϕunun2-η3+β41-βgn4s-1∫Ωϕunun2dx.Dividing by s>0 and letting s→0+, we derive that(89)D+gn0unn+φn≥a1+β1-β∫Ω∇un∇φ+a1+β1-βD+gn0un2+b3+β1-βun2∫Ω∇un∇φ+b3+β1-βD+gn0un4-η3+β1-β∫Ωϕununϕ-η3+β1-βD+gn0∫Ωϕunun2.From (89) and (70), we have(90)φ≥φn≥D+gn01-β1+βaun2+3+βbun4-3+βη∫Ωϕunun2-1-βnun+a1+β1-β∫Ω∇un∇φdx+b3+β1-βun2∫Ω∇un∇φ-η3+β1-β∫Ωϕununφ≥D+gn01-β1-β2C1+a1+β1-β∫Ω∇un∇φdx+b3+β1-βun2∫Ω∇un∇φ-η3+β1-β∫Ωϕununφ,∀n≥N.From (85) and (90), it is easy to see that (87) holds. Thus, (86) and (87) imply that(91)D+gn0≤C,∀n≥N,where C>0 is independent of n.
From condition (ii) and (13), we can obtain that(92)1ngns-1·un+sgnsφ≥1ngnsun+sφ-un≥Jλun-Jλgnsun+sφ=-gn2s-12aun2-gn4s-14bun4+gn4s-14η∫Ωϕun2un2+λ1-βgn1-βs-1∫Ωfxun1-β+gn2s2aun2-un+sφ2+gn4s4bun4-un+sφ4+gn4s4η∫Ωϕun+sφun+sφ2-ϕunun2+λ1-βgn1-βs∫Ωfxun+sφ1-β-un1-β.Dividing by s>0 and letting s→0+, we derive that(93)1nD+gn0un+φ≥-a∫Ω∇un∇φ-bun2∫Ω∇un∇φ+η∫Ωϕununφdx+lim_s→0+⁡λ1-β∫Ωfxun+sφ1-β-un1-βsdx.Since f(x)[(un+sφ)1-β-un1-β]≥0,  ∀x∈Ω,  t>0, then by Fatou’s Lemma we obtain(94)∫Ωfxun-βφdx≤lim_s→0+⁡11-β∫Ωfxun+sφ1-β-un1-βsdx.Combining (93) and (94) we deduce(95)λ∫Ωfxun-βφdx≤1nD+gn0un+φ+a+bun2∫Ω∇un∇φ-η∫Ωϕununφdx≤C0C+φn+a+bun2∫Ω∇un∇φdx-η∫Ωϕununφdxfor all n≥N, which implies that(96)λlim_n→∞⁡∫Ωfxun-βφdx≤a+blim_n→∞⁡un2∫Ω∇u0∇φ-η∫Ωϕu0u0φdx.
Using Fatou’s Lemma, we infer that(97)λ∫Ωfxu0-βφdx≤a+blim_n→∞⁡un2∫Ω∇u0∇φ-η∫Ωϕu0u0φdx.
On one hand, sinceu0(x)≥0 a.e. in Ω, choosing φ=u0 in (97)(98)a+blim_n→∞⁡un2u02≥η∫Ωϕu0u02dx+λ∫Ωfxu01-βdx.
On the other hand, according to (67) and (68), it follows that(99)a+blim_n→∞⁡un2u02≤a+blimn→∞¯⁡un2limn→∞¯⁡un2=η∫Ωϕu0u02dx+λ∫Ωfxu01-βdx.Combining (98) and (99) yields(100)lim_n→∞⁡un2=limn→∞¯⁡un2=u02.Thus, according to (97) and (100), we have(101)a+bu02∫Ω∇u0∇φdx-η∫Ωϕu0u0φdx-λ∫Ωfxu0-βφ≥0for all φ∈H01(Ω) with φ>0. Therefore, by the idea of approximation and (101), we get(102)∫Ω∇u0∇φdx≥0,∀φ∈H01Ω,φ≥0.u0≥0 and u0≠0 in Ω. By the strong maximum principle for weak solutions, we obtain that u0>0 a.e. in Ω.
Second, we prove thatu0 is a solution of problem (12). In fact, we can repeat the proof of Theorem 2 and obtain that(103)a+bu02∫Ω∇u0∇φdx-η∫Ωϕu0u0φdx-λ∫Ωfxu0-βφdx=0,∀φ∈H01Ω.Then u0∈H01(Ω) is a positive solution of (12) for ε=-1,  0<λ<λ∗, and η>η0.
Finally, we prove thatu0∈Nλ+. According to (71) and (100), we have(104)a1+βu02+b3+βu04>η3+β∫Ωϕu0u02dx,which implies that u0∈Nλ+.
By Lemma8, we know that Nλ- is a closed set in H01(Ω) for λ∈(0,λ∗). Thus, applying Lemma 9 and Ekeland’s variational principle, we can find a bounded and nonnegative sequence {vn}⊂Nλ- and v0∈H01(Ω) such that(i)
Jλ(vn)≤infu∈Nλ-⁡Jλ(u)+1/n2,(ii)
Jλ(v)≥Jλ(vn)-(1/n)v-vn,  v∈Nλ-,(iii)
vn⇀v0 in H01(Ω),(iv)
vn→v0 in Lp(Ω),  p∈[1,6),(v)
vn(x)→v0(x),  a.e. in Ω.At this point we can repeat the proof as above and conclude that v0∈Nλ- is another positive solution of (12) for ε=-1,  0<λ<λ∗, and η>η0.



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*Source: 5985962-2017-04-23.xml*