# The Choquard Equation with Weighted Terms and Sobolev-Hardy Exponent

Authors: Yanbin Sang; Xiaorong Luo; Yongqing Wang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We study a nonlinear Choquard equation with weighted terms and critical Sobolev-Hardy exponent. We apply variational methods and Lusternik-Schnirelmann category to prove the multiple positive solutions for this problem.


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## Body

## 1. Introduction

The main goal of this paper is to consider the multiplicity of positive solution for the following nonlinear Choquard equation:(1)-Δu=gx∫Ωu2μ∗x-yμdyu2μ∗-2u+fλxuq-2u,inΩ,u=0,on∂Ω,where Ω⊂RN is a bounded open set with smooth boundary. N>3, 0<μ<N, 1<q<2, 2μ∗=2N-μ/N-2, and λ>0 is a parameter. There are two continuous weight functions, fλ,g satisfying the following conditions:(f1)fλ=λf++f-,(f±=±max{±f,0}≢0).(g1)g∈C(Ω¯) and g+=max{g,0}≢0.(f2) there exist two positive constants β0,ρ0 such that B(0,2ρ0)⊂Ω and fλ(x)≥β0 for every x∈B(0,2ρ0).(g2)g+∞=g(0)=maxx∈Ω¯g(x), ∀x∈B(0,2ρ0),g(x)>0, there exists β(β>N-μ) such that(2)gx-g0=oxβ,x⟶0.In recent years, much attention has been paid to nonlinear Choquard equation involving the Hardy-Littlewood-Sobolev inequality, which generalizes and complements the classical elliptic boundary value problems and Schrödinger-Poisson system [1–21]. In particularly, more and more authors have studied the critical problems [22–36]. In the case that g(x)=fλ(x)=1, Gao and Yang [22] considered the existence and multiplicity of (1) with upper critical exponent 2μ∗=2N-μ/N-2. The authors focus on the case that how the sublinear perturbation term has influence on the multiplicity of (1). When 2μ∗>2,g(x)=1,fλ(x)=0, Xiang [33] showed the uniqueness and nondegeneracy solutions for ground states of Choquard equation. Furthermore, by means of the tool of Nehari manifold, Zhang et al. [36] established the existence theorem of ground states for generalized Choquard equation when the nonlinear term is concave-convex. On the other hand, there are a great deal of results on the existence of elliptic boundary value problems with sign-changing weights [37–45]. We should point out that Hsu and Lin [38] showed the existence of positive solutions for elliptic equations with concave-convex nonlinearities and sign-changing weights. What is more, Wu [39] obtained three positive solutions discussed in [38] by using of the method of Nehari manifold combining with the Lusternik-Schnirelmann category. In view of the same method, Chen and Wu [41] obtained the existence of positive solutions for a class of critical semilinear problem. Chen et al. [42] established multiplicity theorems for Kirchhoff type equation with sign-changing weight functions. Goyal and Sreenadh [45] used fibering map approach to study p-fractional Laplacian equation with sign-changing weight function.Motivated by above results, we use variational approach to analyze the existence and multiplicity of positive solutions for (1). To the best of our knowledge, there is no result studying Choquard equation with upper critical exponent and two sign-changing weight functions.Set(3)λ∗=2-q22μ∗-qg+∞-12-q/22μ∗-q22μ∗-222μ∗-qf+∞-1Ωq-2∗/2∗CN,μq/22μ∗SH,L2μ∗2-q/22μ∗+q/2.Our main results are the following theorems.Theorem 1.
Assume that(f1),(g1) hold, for every λ∈(0,λ∗), the problem (1) has at least one positive solution in H01(Ω).Theorem 2.
Assume that(f1),(g1),(f2),(g2) hold; there exists λ∗>0, such that, for every λ∈(0,λ∗), the problem (1) has at least two positive solutions in H01(Ω).Furthermore, we will utilize the following condition.(G) There are a nonempty closed setG={x∈Ω¯∣g(x)=maxy∈Ω¯g(y)≡1}⊂Ω and β>N-μ such that (4)gz-gx=Ox-zβ,holdsuniformlyforz∈Ginthelimitx⟶z.Remark 3.
LetGr={x∈RN∣dist(x,G)<r} for r>0. According to (G), we can assume that there exist two positive solutions D and r1 such that (5)gx>0,∀x∈Gr1⊂Ω,gz-gx≤Dx-zβ∀x∈BNz,r1,∀z∈G.Theorem 4.
Assume that(G) holds; for every δ∈(0,r1), there exists λδ>0 such that for λ∈(0,λδ), (1) has at least catMδ(G)+1 positive solutions.Definition 5.
u ∈ H 0 1 ( Ω ) is called a positive weak solution of (1), if (6)∫Ω∇u∇φ=∫Ω∫Ωgxux2μ∗uy2μ∗-2uyφxx-yμdxdy+∫Ωfλxuq-1φ,foreveryφ∈H01Ω. The energy functional associated with the problem (1) is given by (7)Iλ,μu=12u2-122μ∗∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-∫Ωfλxuq, where u2=∫Ω∇u2dx is the norm in H01(Ω).


## 2. Preliminaries

Proposition 6 (Hardy-Littlewood-Sobolev inequality [22]).
Lett,r>1 and 0<μ<N with 1/t+μ/N+1/r=2,f∈Lt(RN) and h∈Lr(RN). There exists a constant C(t,N,μ,r) independent of f,h, such that (8)∫RN∫RNfxhyx-yμdxdy≤Ct,N,μ,rfthr. If t=r=2N/2N-μ, then(9)Ct,N,μ,r=CN,μ=πμ/2ΓN/2-μ/2ΓN-μ/2ΓN/2ΓN-1+μ/N. For u∈D1,2(RN), by Proposition 6, we have (10)∫RN∫RNux2μ∗uy2μ∗x-yμdxdy1/2μ∗≤CN,μ1/2μ∗u2∗2. The best constant SH,L is defined as(11)SH,L≔infu∈D1,2RN∖0⁡∫RN∇u2dx∫RN∫RNux2μ∗uy2μ∗/x-yμdxdy1/2μ∗.Proposition 7 (see [22]).
The constantSH,L is attained if and only if (12)u=Caa2+x-b2N-2/2, where C>0,b∈RN,a∈(0,∞); therefore (13)SH,L=SCN,μ1/2μ∗, where(14)S=infu∈H01Ω∖0⁡∫Ω∇u2∫Ωu2∗2/2∗,u∈H01Ω∖0.For ε>0(15)Ux=NN-2ε2N-2/4ε2+x2N-2/2 is a solution of the problem(16)-Δu=∫RNu2μ∗x-yμdyu2μ∗-2u,inRN, with ∫RN|∇U|2dx=∫RN∫RN|U(x)|2μ∗|U(y)|2μ∗/|x-y|μdxdy = SH,L2N-μ/N-μ+2.Definition 8.
(i) Forρ∈R, if Iλ,μ(un)=ρ+o(1),Iλ,μ′(un)=o(1) strongly in H-1(Ω) as n→∞, then the sequence {un} is a (PS)ρ sequence in H01(Ω);
(ii)Iλ,μ satisfies the (PS)ρ condition H01(Ω) if every (PS)ρ sequence in H01(Ω) for Iλ,μ has a convergent subsequence.
ThoughIλ,μ is not bounded below on H01(Ω), we can construct the following Nehari manifold: (17)Mλ,μ=u∈H01Ω∖0∣Iλ,μ′u,u=0=u∈H01Ω∖0∣u2-∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-∫Ωfλxuq=0.Define the fiber map Φλ,μ:R+→R by Φλ,μ(t)=Iλ,μ(tu).Lemma 9.
I λ , μ ( u ) is coercive and bounded below on Mλ,μ.Proof.
Ifu∈Mλ,μ, by (14) we have(18)Iλ,μu=12u2-122μ∗∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-∫Ωfλxuq=12-122μ∗u2-1q-122μ∗∫Ωfλuq=12-122μ∗u2-1q-122μ∗∫Ωλf++f-uq≥12-122μ∗u2-1q-122μ∗λf+∞S-q/2Ω2∗-q/2∗uq≥12-122μ∗u2-1q-122μ∗λf+∞Ω1/22∗SH,LCN,μ1/2μ∗-q/2uq,so Iλ,μ(u) is coercive and bounded below on Mλ,μ. This lemma is completed.Lemma 10.
Assume thatu1 is a local minimizer for Iλ,μ on Mλ,μ and u1∉Mλ,μ0. Then Iλ,μ′(u1)=0. Moreover, if u1 is a nontrivial nonnegative function in Ω, then u1 is a positive solution of (1).Proof.
Sinceu1 is a local minimizer for Iλ,μ on Mλ,μ, u1 is a solution of the optimization problem(19)minimizeIλ,μusubjecttoΓu=0, where (20)Γu=∫Ω∇u2-∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-∫Ωfλxuq.Hence, using Lagrange multipliers, there exists ξ∈R, such that Iλ,μ′(u1)=ξΓ′(u1), which implies(21)Iλ,μ′u1,u1=ξΓ′u1,u1. Since u1∈Mλ,μ,〈Iλ,μ′(u1),u1〉=0 and ∫Ω∇u12-∫Ω∫Ωg(x)|u1(x)|2μ∗|u1(y)|2μ∗/|x-y|μdxdy-∫Ωfλ(x)|u1|q, one has (22)Γ′u1,u1=2-q∇u12-22μ∗-q∫Ω∫Ωgxu1x2μ∗u1y2μ∗x-yμdxdy. Hence, if u1∉Mλ,μ,〈Γ′(u1),u1〉≠0, combining (21) ξ=0, then Iλ,μ(u) in H01(Ω). Using Harnack inequality, thus, u1 is a positive solution of (1) in Ω. The lemma is completed.FromΦλ,μ(t)=Iλ,μ(tu), we obtain that (23)Φλ,μt=12tu2-122μ∗∫Ω∫Ωgxtux2μ∗tuy2μ∗x-yμdxdy-∫Ωfλxtuq, for t>0,u∈H01(Ω), we get (24)Φλ,μ′t=tu2-t22μ∗-1∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-tq-1∫Ωfλxuq,Φλ,μ′′t=u2-22μ∗-1t22μ∗-2∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-q-1tq-2∫Ωfλxuq. Obviously, Φλ,μ′(1)=0; we divide Mλ,μ into the following three sets (25)Mλ,μ+=u∈Mλ,μ∣Φλ,μ′′1>0;Mλ,μ0=u∈Mλ,μ∣Φλ,μ′′1=0;Mλ,μ-=u∈Mλ,μ∣Φλ,μ′′1<0.For anyu∈Mλ,μ, one has(26)Φλ,μ′′1=u2-22μ∗-1∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-q-1∫Ωfλxuq=2-22μ∗u2-q-22μ∗∫Ωfλuq=2-qu2-22μ∗-q∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy.Therefore, we can deriveLemma 11.
(i) For everyu∈Mλ,μ+∪Mλ,μ0, then ∫Ωfλ(x)|u|q>0.
(ii) For everyu∈Mλ,μ-∪Mλ,μ0, then ∫Ω∫Ωg(x)|u(x)|2μ∗|u(y)|2μ∗/|x-y|μdxdy>0.Lemma 12.
Forλ∈(0,λ∗), then Mλ,μ0=∅.Proof.
Assuming the contrary, there isλ∈(0,λ∗) such that Mλ,μ0≠∅, then for u∈Mλ,μ0, by (26) and Proposition 6, which implies (27)u2=22μ∗-q2-q∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy≤22μ∗-q2-qu2∗2.2μ∗SH,L-2μ∗g+∞, so (28)u≥2-q22μ∗-qSH,L2μ∗g+∞-11/22μ∗-2.Similarly, coupling with (26) and (14), which infers that (29)u2=q-22μ∗2-22μ∗∫Ωfλxuq≤q-22μ∗2-22μ∗λf+∞SH,LCN,μ1/2μ∗-q/2uq=q-22μ∗2-22μ∗λΩ2∗-q/2∗f+∞SH,LCN,μ1/2μ∗-q/2uq, hence (30)u≤q-22μ∗2-22μ∗λΩ2∗-q/2∗f+∞SH,LCN,μ1/2μ∗-q/21/2-q.From the above inequalities, we deduce that (31)λ≥2-q22μ∗-qg+∞-12-q/22μ∗-q22μ∗-222μ∗-qf+∞-1Ωq-2∗/2∗CN,μq/22μ∗SH,L2μ∗2-q/22μ∗+q/2≔λ∗, which is a contradiction.Lettingu∈Φλ,μ(t), one has (32)ddtIλ,μtu=tu2-t22μ∗-1∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-tq-1∫Ωfλxuq=tq-1t2-qu2-t22μ∗-q∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy-∫Ωfλxuq. Let (33)φt=t2-qu2-t22μ∗-q∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy.Assume tu∈Mλ,μ if and only if φ(t)=∫Ωfλ(x)|u|q. So (34)φ′t=2-qt1-qu2-22μ∗-qt22μ∗-q-1∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy.Clearly, tu∈Mλ,μ, then tq-1φ′(1)=Φλ,μ′′(1). Hence, tu∈Mλ,μ+ (or Mλ,μ-) provided φ′(t)>0 (or <0). φ(t) can attain its maximum at t=tmax duo to limt→∞φ(t)=∞, where (35)tmax=2-quq22μ∗-q∫Ω∫Ωgxux2μ∗uy2μ∗/x-yμdxdy1/22μ∗-2>0,φtmax=2-quq22μ∗-q∫Ω∫Ωgxux2μ∗uy2μ∗/x-yμdxdy2-q/22μ∗-2u2-2-quq22μ∗-q∫Ω∫Ωgxux2μ∗uy2μ∗/x-yμdxdy22μ∗-q/22μ∗-2∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy=2-quq22μ∗-q∫Ω∫Ωgxux2μ∗uy2μ∗/x-yμdxdy22μ∗-q/22μ∗-2u2-2-q22μ∗-quq=2-q22μ∗-q22μ∗-q/22μ∗-222μ∗-222μ∗-quqSH,L2μ∗22μ∗-2/22μ∗-qu-22μ∗≥f+∞-12-q22μ∗-q22μ∗-q/22μ∗-222μ∗-222μ∗-qSH,L2μ∗22μ∗-2/22μ∗-qu-22μ∗∫Ωfλxuq>λ~∫Ωfλxuq,where λ~=f+∞-1(2-q/22μ∗-q)22μ∗-q/22μ∗-2(22μ∗-2/22μ∗-q)SH,L2μ∗22μ∗-2/22μ∗-qu-22μ∗.Lemma 13.
For everyu∈H01(Ω), it follows that
(i)if∫Ωfλ(x)|u|q≤0,λ>0, there exists a unique t-=t-(u)>tmax(u) such that t-u∈Mλ,μ-, and Φλ,μ is increasing on (0,t-), decreasing on (t-,∞).
Moreover(36)Iλ,μt-u=supt≥0⁡Iλ,μtu; (ii)if ∫Ωfλ(x)|u|q>0, then there exist unique 0<t+=t+(u)<tmax<t-=t-(u) such that t-(u)u∈Mλ,μ-,t+(u)u∈Mλ,μ+ and Φλ,μ is increasing on (t+,t-), decreasing on (t-,+∞)∪(0,t+). Moreover(37)Iλ,μt-u=supt≥tmax⁡Iλ,μtu,Iλ,μt+u=inf0≤t≤tmax⁡Iλ,μtu.Proof.
(i)Assume that∫Ωfλ(x)|u|q≤0. There exists a solution t-(t-≥tmax), with φ=∫Ωfλ(x)|u|q, then φt-′<0, and Φλ,μ′(t-)=0. Consequently, Φλ,μ has a unique critical point at t=t- and φt-′′<0. Since tq-1φ′(1)=Φλ,μ′′(1). Hence, (36) holds and t-u∈Mλ,μ-.
(ii) Assume∫Ωfλ(x)|u|q>0, then φtmax(u)>∫Ωfλ(x)|u|q. There are two solutions t-,t+,(t+<tmax<t-), satisfying φ(t)=∫Ωfλ(x)|u|q, such that φ′(t-)<0,φ′(t+)>0, which means that these two solutions of u depend on Mλ,μ, where t-u∈Mλ,μ-,t2u∈Mλ.μ+. So according to tq-1φ′(1)=Φλ,μ′′(1),Φλ,μ has critical points at t-=t+,Φλ,μ(t-)<0,Φλ,μ(t+)>0. Thus, Φλ,μ is increasing on (t+,t-), and (0,t+)∪(t+,+∞), (37) holds.
This lemma is completed.Furthermore, we have(38)Mλ,μ=Mλ,μ+∪Mλ,μ-. Define (39)αλ,μ=infu∈Mλ,μ⁡Iλ,μu,αλ,μ+=infu∈Mλ,μ+⁡Iλ,μu,αλ,μ-=infu∈Mλ,μ-⁡Iλ,μu.Theorem 14.
(i)αλ,μ+<0 for all λ∈(0,λ∗).
(ii) Ifλ∈(0,q/2λ∗), then there exists a positive number d1, such that αλ,μ->d1.
Especially,αλ,μ+=infu∈Mλ,μIλ,μ(u) for all λ∈(0,q/2λ∗).Proof.
(i) Supposing thatu∈Mλ,μ+ from (26), one has (40)u2<q-22μ∗2-22μ∗∫Ωfλxuq, by Lemma 9, we derive (41)Iλ,μu=12-122μ∗u2-1q-122μ∗∫Ωfλxuq≤12-122μ∗q-22μ∗2-22μ∗∫Ωfλxuq-1q-122μ∗∫Ωfλxuq=q-22q∫Ωfλxuq<0,Thus, αλ,μ+<0.
(ii) Suppose thatu∈Mλ,μ- from (11), (26), and (g1), we get (42)u2≤2-q22μ∗SH,L2μ∗g+∞-11/22μ∗-2,∀u∈Mλ,μ-. Moreover (43)Iλ,μu≥12-122μ∗u2-1q-122μ∗λΩ2∗-q/2∗f+∞SH,LCN,μ1/2μ∗-q/2uq=uq2μ∗-122μ∗u2-q-22μ∗-q22μ∗qλΩ2∗-q/2∗f+∞SH,LCN,μ1/2μ∗-q/2≥2-q22μ∗-qSH,L2μ∗g+∞-1q/22μ∗-22μ∗-122μ∗2-q22μ∗-qSH,L2μ∗g+∞-12-q/22μ∗-2-22μ∗-q22μ∗qλΩ2∗-q/2∗f+∞SH,LCN,μ1/2μ∗-q/2. Hence, if λ∈(0,q/2λ∗), such that (44)Iλ,μu>d1,∀u∈Mλ,μ-, where d1 depends on λ,μ,s,q,N,SH,L,|Ω|,|f+|∞,|g+|∞.
This completes the proof.Lemma 15.
Every sequence{un}∈H01(Ω) that satisfies Iλ,μ(un)=β1+o(1) with β1<αλ,μ++SH,LN/2C(N,μ)N/22μ∗ and Iλ,μ′(un) = o1inH-1(Ω) has a convergent subsequence.Proof.
For{un}, there exists u0≥0 such that(45)un⇀u0,inH01Ω,un⟶u0,inLpΩ,1≤p<2∗,unx⟶u0x,a.e.inΩ.Therefore (46)Iλ,μ′u0,u0=u02-∫RN∫RNu0x2μ∗u0y2μ∗x-yμdxdy-∫Ωfλxu0q=0,∫Ωfλxunq=∫Ωfλxu0q+o1. So, by Theorem 14, Iλ,μ(u0)≥αλ,μ+, we know that (47)∫Ω∇un2=∫Ω∇u02+∫Ω∇un-u02+o1. Thus (48)Iλ,μun=Iλ,μu0+Iλ,μun-u0+o1,o1=un-u0,Iλ,μun=un-u0,Iλ,μun-Iλ,μu0=un-u02-∫Ω∫Ωu0x2μ∗u0y2μ∗x-yμdxdy,which implies (49)4-μ22N-μun-u02=Iλ,μun-u0=Iλ,μun-Iλ,μu0+o1≤Iλ,μun-αλ,μ++o1. Furthermore (50)un-u02<SH,LN/2CN,μN/22μ∗.Lemma 16.
For everyλ∈(0,λ∗), then Iλ,μ has a minimizer u1 in Mλ,μ+, satisfying that(i)
Iλ,μ(u1)=αλ,μ+;(ii)
u1 is a positive solution of (1);(iii)
Iλ,μ(u1)→0 as λ→0.Proof.
(i)From Lemma15, there exists a minimizing sequence {un} for Iλ,μ, such that(51)Iλ,μun=β+o1,Iλ,μ′un=o1,inH01Ω. Noting that Iλ,μ is coercive and bounded on Mλ,μ in H01(Ω). Going if necessary to a subsequence, we can assume that there exists u1∈H01(Ω) such that(52)un⇀u1,inH01Ω,un⟶u1,inLpΩ,1≤p<2μ∗,unx⟶u1x,a.e.inΩ.We show that u1 is a solution of (1). un∈Mλ,μ, which derives that (53)Iλ,μun=12un2-122μ∗∫Ω∫Ωgxunx2μ∗uny2μ∗x-yμdxdy-∫Ωfλxunq=12-122μ∗un2-1q-122μ∗∫Ωfλunq≥2μ∗-122μ∗un2-22μ∗-q22μ∗qλ∫Ωf+unq, thus(54)λ∫Ωf+unq=22μ∗-1q22μ∗-qun2-22μ∗q22μ∗-qIλ,μun. By Theorem 14, (51), (52), (54), one has(55)λ∫Ωf+unq≥22μ∗q22μ∗-qIλ,μ>0, combining Fatou’s Lemma, which implies that(56)αλ,μ+≤Iλ,μu1=2μ∗-122μ∗u12-22μ∗-q22μ∗q∫Ωfλu1q≤liminfn→∞⁡2μ∗-122μ∗un2-22μ∗-q22μ∗qλ∫Ωfunq≤liminfn→∞⁡Iλ,μun=αλ,μ+. Clearly, (57)Iλ,μ+u1=αλ,μ,limn→∞⁡un2=u12.(ii) Let wn=un-u1. Using Brézis-Lieb Lemma, such that (58)wn2=un2-u12+o1, which infers that un→u1 in H01(Ω) and u1∈Mλ,μ+. On the contrary, if u1∈Mλ,μ-, from (26), (55), which gets ∫Ω∫Ωg(x)|u1(x)|2μ∗|u1(y)|2μ∗/|x-y|μdxdy>0 and ∫Ωfλ|u1|q>0. Therefore, by Lemma 13, there are unique t1 and t2 such that t1u1∈Mλ,μ-,t2u1∈Mλ,μ+. Moreover, t2<t1=1. By simple calculation, one has (59)Iλ,μ′t2u1=0,Iλ,μ′′t2u1>0. So there exists tλ∈(t2,t1) such that Iλ,μ(t2u1)<Iλ,μ(tλu1). Coupling with Lemma 13 which indicates that (60)Iλ,μt2u1<Iλ,μtλu1≤Iλ,μt1u1=Iλ,μu1, which is a contradiction. Furthermore, Iλ,μ(u1)=Iλ,μ(|u1|) and |u1|∈Mλ,μ+. From Lemma 10, we claim that u1 is a positive solution of (1). Combining the standard elliptic regularity argument and f,g∈C(Ω¯), we can get u2∈C2(Ω)∖{0}. By Harnack inequality, we have that u1>0 in Ω∖{0}; that is u1 is a positive solution of (1).
(iii) By (26), we get (61)u12=q-22μ∗2-22μ∗∫Ωfλxu1q≤q-22μ∗2-22μ∗λf+∞SH,LCN,μ1/2∗-q/2u1q. Moreover (62)u12-q=q-22μ∗λf+∞2-22μ∗SH,LCN,μ1/2μ∗q/2. Obviously, Iλ,μ(u1)→0 as λ→0. This completes the proof.Lemma 17.
For everyu∈Mλ,μ, there exist ε>0 and a differentiable function ζ=ζ(w)>0 with w∈H01(Ω) and w<ε such that (63)ζ0=1,ζwu-w∈Mλ,μ,ζ′0,w=2∫Ω∇u∇w-22μ∗∫Ω∫Ωux2μ∗uy2μ∗-2uywy/x-yμdxdy-q∫Ωfλuq-2uw2-qu2-22μ∗-q∫Ω∫Ωux2μ∗uy2μ∗/x-yμdxdy.Proof.
∀ u ∈ M λ , μ; define a function F:R×H01(Ω)→R:(64)Fζ,w=Iλ,μ′ζu-w,ζu-w=ζ2∫Ω∇u-w2-22μ∗ζ22μ∗∫Ω∫Ωux2μ∗uy2μ∗x-yμdxdy-ζq∫Ωfλu-wq.F(1,0)=〈Iλ,μ′(u),u〉=0, and (65)∂F∂ζ1,0=2∫Ω∇u2-22μ∗∫Ω∫Ωux2μ∗uy2μ∗x-yμdxdy-q∫Ωfλuq=2-qu2-22μ∗-q∫Ω∫Ωux2μ∗uy2μ∗x-yμdxdy≠0. Applying the implicit function theorem, we obtain (66)ζ′0,w=2∫Ω∇u∇w-22μ∗∫Ω∫Ωux2μ∗uy2μ∗-2uywy/x-yμdxdy-q∫Ωfλuq-2uw2-qu2-22μ∗-q∫Ω∫Ωux2μ∗uy2μ∗/x-yμdxdy,Fζw,w=0,which infers that (67)Iλ,μ′ζwu-w,ζwu-w=0,w∈H01Ω, so (ζ(w)(u-w))∈Mλ,μ.


## 3. Proof of Theorem1

According to Lemma16, we can derive that the problem (1) has a positive solution.


## 4. Proof of Theorem2

Proposition 18.
(i) For everyλ∈(0,λ∗), there exists (PS)αλ,μ sequence {un}⊂Mλ,μ in H01(Ω) for Iλ,μ;
(ii)For everyλ∈(0,q/2λ∗), there exists (PS)αλ,μ- sequence {un}⊂Mλ,μ- in H01(Ω) for Iλ,μ.Proof.
By Lemma9 and Ekeland variational principle there exists a minimizing sequence {un}⊂Mλ,μ such that(68)Iλ,μun<αλ,μ+1n,Iλ,μun<Iλ,μw+1nw-un∀w∈Mλ,μ.n is large enough, from Theorem 14(i), which yields that(69)Iλ,μun=12-122μ∗un2-1q-122μ∗∫Ωfλuq<αλ,μ+1n<q-22q∫Ωfλuq, which derives that(70)λf∞S-q/2Ω2∗-q/2∗unq≥∫Ωfλunq>2qq-2β1+o1. Hence u≢0 by (70). From Lemma 15, we have(71)u>2qq-2β1+o1Ω-2μ∗-q/2μ∗Sq/21/q;(72)u<q-22μ∗2-22μ∗λΩ2∗-q/2∗SH,LCN,μ1/2μ∗-q/2f+∞1/2-q. We claim that limn→∞Iλ,μ′(un)=0. Using Lemma 17, then there exists a differentiable function ζn:B(0;ϵn)→R+ for some small enough number ϵn>0, such that ζn(w)(un-w)∈Mλ,μ. Take ϱ∈(0,ϵn), and let u∈H01(Ω)(u≢0), w=ϱIλ,μ′(un)/Iλ,μ′(un)H-1, wϱ=ζn(w)(un-w). Coupling with (68), for wϱ∈Mλ,μ, we obtain(73)1nwϱ-un≥Iλ,μun-Iλ,μwϱ=1-ζnϱIλ,μ′wϱ,un+ϱζnϱIλ,μ′wϱ,Iλ,μ′unIλ,μ′unH-1+oϱ,(74)limϱ→01ϱwϱ-un=limϱ→01wζnϱ-1un-ζnww=limϱ→01ϱζnw-1un-ζnwIλ,μ′unIλ,μ′unH-1≤ζn′0un+1.where (75)ζn′0≔ζ′0.Iλ,μ′Iλ,μ′unH-1.Dividing by ϱ>0 in (73) and passing to the limit as ϱ→0, which infers that (76)1n1+ζn′0un≥-ζn′0Iλ,μ′un,un+Iλ,μ′unH-1=Iλ,μ′unH-1. Applying (71), (72), we have (77)Iλ,μ′un≤C′n1+ζn′0. Noting that (78)ζn′0,w=2∫Ω∇un∇w-22μ∗∫Ω∫Ωunx2μ∗uny2μ∗-2unywy/x-yμdxdy-q∫Ωfλunq-2uw2-qun2-22μ∗-q∫Ω∫Ωunx2μ∗uny2μ∗/x-yμdxdy.By Lemma 17, for some constant C′′>0, we have (79)ζn′0≤C′′w2-qun2-22μ∗-q∫Ω∫Ωunx2μ∗uny2μ∗/x-yμdxdy. Now we prove that(80)2-qun2-22μ∗-q∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy>c,c>0. Arguing by contradiction, assume that there exists a subsequence {un} such that(81)2-qun2-22μ∗-q∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy=o1. From Lemma 17 and (72), there exists a constant ν>0 such that(82)∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy≥ν,fornsufficientlylarge,(83)∫Ωfλunq=un2-∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy=22μ∗-22-q∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy+o1. Set (84)Ψun≔CN,μun222μ∗-q+1/22μ∗-2-∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy2μ∗-q+2/2μ∗22μ∗-2∫Ωfλunq;ϑ=infu∈ΩCN,μu2-∫Ωfλuq,wn=un∫Ω∫Ωunx2μ∗uny2μ∗/x-yμdxdy1/22μ∗.Moreover (85)Ψun=∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy2μ∗-q+2/2μ∗22μ∗-2CN,μwn2-∫Ωfλwnq≥ϑ∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy2μ∗-q+2/2μ∗22μ∗-2≥ϑν2μ∗-q+2/2μ∗22μ∗-2>0, On the other hand, combining (82) with (83), we conclude that (86)Φun=CN,μun222μ∗-q/22μ∗-2-22μ∗-22-q∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy22μ∗-q/22μ∗-2+o1=2-q22μ∗-q22μ∗-q/22μ∗-222μ∗-q2-qun222μ∗-q/22μ∗-2-22μ∗-q2-q∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy22μ∗-q/22μ∗-2+o1=22μ∗-22-q2-q22μ∗-qun22μ∗-q/22μ∗-2-∫Ω∫Ωunx2μ∗uny2μ∗x-yμdxdy22μ∗-q/22μ∗-2=o1, which is a contradiction; thus (80) is true.
Using the same methods to prove (ii). The proof is completed.Lemma 19.
Assume that(f1) and (g1) hold, if {un} is a (PS)ρ sequence, for Iλ,μ with un⇀u in H01(Ω), then Iλ,μ′(u)=0 and there exists a constant C1, such that Iλ,μ(u)≥-C1λ2/2-q.Proof.
Since{un} is a (PS)ρ sequence for Iλ,μ with un⇀u in H01(Ω). Obviously, (87)Iλ,μ′u,u=0,∫Ω∫Ωux2μ∗uy2μ∗x-yμdxdy=u2-∫Ωfλuq.Thus (88)Iλ,μ=12-122μ∗u2-1q-122μ∗∫Ωfλuq. Therefore (89)Iλ,μu=12-122μ∗u2-1q-122μ∗∫Ωfλuq≥12-122μ∗u2-1q-122μ∗λ∫Ωf+uq≥12-122μ∗u2-1q-122μ∗λΩ2∗-q/2∗f+∞SH,LCN,μ1/2μ∗-q/2uq≥12-122μ∗u2-2q22μ∗-122μ∗qq/2uq2/q-2-q222μ∗-q22μ∗qq-2/222μ∗-1q/2f+∞Ω2∗-q/2∗SH,LCN,μ1/2μ∗-q/22/2-qλ2/2-q=-C1λ2/2-q, where C1=2-q/2(22μ∗-q)(22μ∗q)q-2/2/[2(2μ∗-1)]q/2|f+|∞|Ω|2∗-q/2∗SH,LC(N,μ)1/2μ∗-q/22/2-q.
The proof is completed.Lemma 20.
Assume that(f1),(g1) hold. For c∈(-∞,2μ∗-1/22μ∗|g+|∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q), Iλ,μ satisfies the (PS)c condition.Proof.
Let{un}⊂H be a (PS)c sequence, which implies (90)Iλ,μun=c+o1,Iλ,μ′un=o1. By Lemma 9, {un} is bounded in H01(Ω), so there exists a subsequence still denoted by {un}, we can suppose that there exists u∈H01(Ω) such that(91)un⇀u,inH01Ω,un⟶u,inLpΩ,1≤p<2∗,unx⟶ux,a.e.inΩ.By (f1),(g1) and Lemma 19, we know that Iλ,μ′u=0 and(92)∫Ωfλunq=∫Ωfλuq+o1. Let wn=un-u. According to Brézis-Lieb Lemma and the Vitai’s theorem, one has (93)wn2=un2-u2+o1;∫Ω∫Ωgxwnx2μ∗wny2μ∗x-yμdxdy=∫Ω∫Ωgxunx2μ∗uny2μ∗x-yμdxdy-∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy+o1. Thus(94)c+o1=12wn2-122μ∗∫Ω∫Ωgxwnx2μ∗wny2μ∗x-yμdxdy+Iλ,μu,(95)wn2-∫Ω∫Ωgxwnx2μ∗wny2μ∗x-yμdxdy=o1.Therefore, we suppose that(96)limn→∞⁡wn2=limn→∞⁡∫Ω∫Ωgxwnx2μ∗wny2μ∗x-yμdxdy=l. By the Sobolev inequality, one has (97)wn2≥Swn2μ∗2. Applying (94), we deduce that (98)l≤g+∞lSH,L-12μ∗, and, obviously, l=0, or l=|g+|∞-N-2/2+N-μSH,L2N-μ/N-μ+2. In fact, l=0. Arguing by contradiction, suppose l≥|g+|∞-N-2/2+N-μSH,L2N-μ/N-μ+2.
Coupling with (91)-(94), we have (99)c=Iλ,μwn-122μ∗Iλ,μ′wn,wn+o1=12-122μ∗wn2-1q-122μ∗∫Ωfλwnq≥2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q, which is a contradiction, that is, un→u in H01(Ω).
This completes the proof.Lemma 21.
Assume that(f1),(f2),(g1),(g2) hold, then there exist v∈H01(Ω) and λ∗>0, such that, for λ∈(0,λ∗), (100)supt≥0⁡Iλ,μtv<2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q. In particular, for every λ∈(0,λ∗), we have αλ,μ-<2μ∗-1/22μ∗|g+|∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q.Proof.
Set(101)Hu=12u2-122μ∗∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy,Ku=u2∫Ω∫Ωgxux2μ∗uy2μ∗/x-yμdxdy1/2μ∗. Based on (g2), we get that there exists δ∈(0,ρ0), for every x∈B(0,2δ) such that (102)gx-g0=oxβ. For ε>0, let η∈C0∞(Ω) be a cut function, such that (103)0≤η≤1,ηx=1,x≤δ,0,x>2δ,∇η≤C. Let (104)uεx=ηxUεx,where (105)Uεx=ε2-N/2Uxε,Ux=NN-2N-2/41+x2N-2/2.Step1. We will show that supt≥0H(tuε)≤N-μ+2/2(2N-μ)|g+|∞-N-2/2+N-μSH,L2N-μ/N+2-μ+O(εN-2).
At first, we prove the following estimates(ε→0):(106)∫Ω∫Ωgxuεx2μ∗uεy2μ∗x-yμdxdy1/2μ∗≤ε2-Ng+∞N-2/2N-μCN,μNN-2/22N-μSH,LN-2/2+OεN-2;(107)uε2=CN,μN/22μ∗SH,LN/2+OεN-2. From [22] and definition of uε, let x/i=u,(x1,x2,…,xn)=i(u1,u2⋯,un)xi=εui,dxi=εdui, which derives that (108)∫Ω∫Ωgxuεx2μ∗uεy2μ∗x-yμdxdy1/2μ∗≤CN,μN/2S2N-μ/2g+∞+Oε2N-μ. Furthermore (109)εμ-2Ng+∞CN,μN/2SH,L2N-μ/2-∫Ω∫Ωgxuεx2μ∗uεy2μ∗x-yμdxdy=∫RN∖B0,δ∫RN∖B0,δg0-gxη2μ∗x1+x/ε22N-μ/2x-yμ1+y/ε22N-μ/2+2∫RN∖B0,δ∫B0,δg0-gxη2μ∗xε2+x22N-μ/2x-yμε2+y22N-μ/2+∫B0,δ∫B0,δg0-gx1+x/ε22N-μ/2x-yμ1+y/ε22N-μ/2-∫Ω∫Ωgxuεx2μ∗uεy2μ∗x-yμdxdy=A1+2A2+A3-CN,μN/2SH,L2N-μ/2, where (110)A1=∫RN∖B0,δ∫RN∖B0,δg0-gxη2μ∗x1+x/ε22N-μ/2x-yμ1+y/ε22N-μ/2=ε2N-μNN-22N-μ/2∫RN∖B0,δ∫RN∖B0,δg0ε2+x22N-μ/2x-yμε2+y22N-μ/2dxdy≤ε2N-μNN-22N-μ/2∫RN∖B0,δ∫RN∖B0,δg0x2N-μx-yμy2N-μdxdy≤ε2N-μNN-22N-μ/2∫RN∖B0,δg0x2N-μdx∫RN∖B0,δg0y2N-μdy=NωNε2N-μNN-22N-μ/2∫δ∞g0rN-1r2N-μdr∫δ∞g0rN-1r2N-μdr=NωNε2N-μNN-22N-μ/2g+∞2δμ-Nμ-N2;A2=∫RN∖B0,δ∫B0,δg0-gxη2μ∗x1+x/ε22N-μ/2x-yμ1+y/ε22N-μ/2=ε2N-μNN-22N-μ/2∫RN∖B0,δ∫B0,δg0ε2+x22N-μ/2x-yμε2+y22N-μ/2dxdy≤ε2N-μNN-22N-μ/2∫RN∖B0,δ∫RN∖B0,δg0x2N-μx-yμy2N-μdxdy≤ε2N-μNN-22N-μ/2∫RN∖B0,δg0x2N-μdx∫B0,δg0y2N-μdy=NωNε2N-μNN-22N-μ/2∫δ∞g0rN-1r2N-μdr∫0δg0rN-1r2N-μdr=NωNε2N-μNN-22N-μ/2g+∞2δμ-Nμ-N2;A3=∫B0,δ∫B0,δg0-gx1+x/ε22N-μ/2x-yμ1+y/ε22N-μ/2=ε2N-μNN-22N-μ/2∫B0,δ∫B0,δoxβε2+x22N-μ/2x-yμε2+y22N-μ/2dxdy≤ε2N-μNN-22N-μ/2∫B0,δ∫B0,δoxβx2N-μx-yμy2N-μdxdy≤ε2N-μNN-22N-μ/2∫B0,δoxβx2N-μdx∫B0,δoxβy2N-μdy=NωNε2N-μNN-22N-μ/2∫δ∞orβrN-1r2N-μdr∫δ∞orβrN-1r2N-μdr=NωNε2N-μNN-22N-μ/2o1δβ-N+μβ-N+μ2. Hence (111)A1+2A2+A3-CN,μN/2SH,L2N-μ/2g+∞≤3NωNε2N-μNN-22N-μ/2g+∞2δμ-Nμ-N2+NωNε2N-μNN-22N-μ/2o1δβ-N+μβ-N+μ2-CN,μN/2SH,L2N-μ/2g+∞≔C3. (ωN=2πN/2/NΓ(N/2) is the volume of the unit ball B(0,1) in RN.)
Thus(112)0≤εμ-2Ng+∞CN,μN/2SH,L2N-μ/2-∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy≤C3, moreover (113)0≤1-ε2N-μg+∞-1CN,μ-N/2SH,Lμ-2N/2∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy≤ε2N-μg+∞-1CN,μ-N/2SH,Lμ-2N/2C3, which implies that(114)1-ε2N-μg+∞-1CN,μ-N/2SH,Lμ-2N/2C3≤ε2N-μg+∞-1CN,μ-N/2SH,L-μ-2N/2∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy≤1. Taking sufficient small ε such that ε2N-μ|g+|∞-1C(N,μ)-N/2SH,Lμ-2N/2C3<1, by (114), we have (115)
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≤
1
, so (116)ε2-Ng+∞N-2/2N-μCN,μNN-2/22N-μSH,LN-2/2-εN-μ+2g+∞-N-2+μ/2N-μCN,μN-N-2+μ/22N-μSH,L-N-2+μ/2C3≤∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdyN-2/2N-μ≤εN-2g+∞N-2/2N-μCN,μNN-2/22N-μSH,LN-2/2, that is, (117)∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdyN-2/2N-μ=ε2-Ng+∞N-2/2N-μCN,μNN-2/22N-μSH,LN-2/2+OεN+2-μ. From (106),(107), we conclude that (118)Kuε=uε2∫Ω∫Ωgxux2μ∗uy2μ∗/x-yμdxdyN-2/2N-μ=CN,μNN-2/22N-μSH,LN/2+OεN-2ε2-Ng+∞N-2/2N-μCN,μNN-2/22N-μSH,LN-2/2+OεN+2-μ=g+∞-N-2/2N-μCN,μNN-2/22N-μSH,LN/2+OεN-2ε2-NCN,μNN-2/22N-μSH,LN-2/2+OεN+2-μ, moreover(119)Kuε-g+∞-N-2/2N-μSH,L=g+∞-N-2/2N-μCN,μNN-2/22N-μSH,LN/2+OεN-2ε2-NCN,μNN-2/22N-μSH,LN-2/2+OεN+2-μ-SH,L=g+∞-N-2/2N-μCN,μNN-2/22N-μSH,LN/2+OεN-2-SH,Lε2-NCN,μNN-2/22N-μSH,LN-2/2+OεN+2-με2-NCN,μNN-2/22N-μSH,LN-2/2+OεN+2-μ=OεN-2. Let (120)ϕt=t22a-t22μ∗22μ∗b,a,b>0,in fact ϕ′(t)=ta-t22μ∗-1b,andt=(a/b)1/22μ∗-2≔tmax, so (121)maxt≥0⁡ϕt=N+2-μ22N-μqb1/2μ∗2N-μ/N-μ+2, which leads to (122)supt≥0⁡Htuε=N-μ+222N-μKuε2N-μ/N-μ+2. From (119), we have (123)supt≥0⁡Htuε≤N-μ+222N-μg+∞-N-2/2+N-μSH,L2N-μ/N+2-μ+OεN-2.Step2. Let ε=λ2/(N-2)(2-q), there exists λ3>0, for all λ∈(0,λ3), and we have (124)supt≥0⁡Iλ,μtuε<2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q≔cλ. We denote δ1>0 satisfying (125)2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q>0,forλ∈0,δ1. By (f2),(g2), we obtain that (126)Iλ,μtuε≤t22uε2,fort≥0,λ>0, which implies that there exists t0∈(0,1) such that (127)sup0≤t≤t0⁡Iλ,μtuε≤2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q>0,forλ∈0,δ1. Moreover, set 0<ε≤δ(N-2)(2q+1)/2, which leads to(128)supt≥t0⁡Iλ,μtuε=supt≥t0⁡Htuε-tqq∫Ωfλuεq≤2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2+OεN-2-t0qqβ0λ∫B0,δuεq. Furthermore(129)∫B0,δuεq=∫B0,δεN-2/2NN-2N-2/41+x/ε2N-2/2q=εN-2/2NN-2N-2q/4∫B0,δ11+x/ε2N-2/2q≤NN-2qN-2/4δN-22q+1/2≔C4. Combining with (128) and (129), for ε=λ2/(N-2)(2-q)∈(0,δ(N-2)(2q+1)/2), we have (130)supt≥t0⁡Iλ,μtuε≤2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2+Oλ2/2-qN-2-t0qqβ0C4λ. It is easy to find δ2>0 such that (131)Oλ2/2-q-t0qqβ0C4λ<-C1λ2/2-q,forλ∈0,δ2. Hence, set λ3=min{δ1,δ(N-2)(2q+1)/2,δ2}>0 and ε=λ2/(2-q)(N-2), if λ∈(0,λ3), then (132)supt≥0⁡Iλ,μtuε<2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q.Step3. For every λ∈(0,λ3) then αλ,μ-<2μ∗-1/22μ∗|g+|∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q.
By(f2) and (g2), we get (133)∫Ωfλuεq>0,∫Ω∫Ωgxuεx2μ∗uεy2μ∗x-yμ>0. Using Lemma 13 and the definition of αλ,μ-, we have that for every λ∈(0,λ3) there exists tε>0 such that tεuε∈Mλ,μ- and (134)αλ,μ-≤Iλ,μtεuε≤supt≥0⁡Iλ,μtuε<2μ∗-122μ∗g+∞-N-2/N-μ+2SH,L2N-μ/N-μ+2-C1λ2/2-q. The proof of (73) is valid.Theorem 22.
Suppose that(f1),(f2),(g1),(g2) hold, then there exists λ¯>0(λ¯=minλ3,q/2λ∗) such that for λ∈(0,λ¯), Iλ,μ has a minimizer u2∈Mλ,μ- and
(i)Iλ,μ(u2)=αλ,μ-;
(ii)u2 is a positive solution of (1) in H01(Ω).Proof.
According to Proposition18 (ii), for any λ∈(0,q/2λ∗), there exists a (PS)αλ,μ- sequence {un}⊂Mλ,μ- in H01(Ω) for Iλ,μ. In terms of Theorem 14 (ii) and Lemmas 20 and 21, for all λ∈(0,λ3), Iλ,μ satisfies αλ,μ->0 and (PS)αλ,μ- condition. From Lemma 9, Iλ,μ is coercive on Mλ,μ, so {un} is bounded. Moreover, there exists a subsequence still denoted by {un}, and u2∈Mλ,μ- such that un→u2 in H01(Ω), for λ∈(0,λ¯), Iλ,μ(u2)=αλ,μ-. At last, we apply the same arguments as those of Lemma 16 to prove fore λ∈(0,λ¯), u2 is a positive solution of (1).
The proof is completed.Proof of Theorem2.
According to Lemma16 and Theorem 22, we admit that u1∈Mλ,μ+ and u2∈Mλ,μ- are two solutions of (1). It is easy to verify that u1,u2 is different because Mλ,μ+∪Mλ,μ-=∅.
The proof is completed.


## 5. Proof of Theorem4

In this part, we will show the proof of Theorem4. At first, we consider the following problem:(135)-Δu=∫Ω∫Ωuy2μ∗x-y2dyu2μ∗-2u,inΩ,u∈H01Ω,and the associated energy functional I∞ of (135) in H01(Ω) is(136)I∞u=12u2-122μ∗∫Ω∫Ωgxux2μ∗uy2μ∗x-yμdxdy. Recalling that (137)infu∈M∞RN⁡I∞u=infu∈M∞Ω⁡I∞u=N-μ+222N-μCN,μNN-2/22N-μSH,LN/2,Ω∈RN, where (138)M∞RN=u∈D1.2RN∖0∣I∞′u,u=0;M∞Ω=u∈H01Ω∖0∣I∞′u,u=0.Lemma 23.
(139) inf u ∈ M 0 , μ ⁡ I 0 , μ u = inf u ∈ M ∞ Ω ⁡ I ∞ u = N - μ + 2 2 2 N - μ C N , μ N N - 2 / 2 2 N - μ S H , L N / 2 . Moreover, (135) has no positive solution u1 satisfying I0,μ(u1)=N-μ+2/2(2N-μ)C(N,μ)N(N-2)/2(2N-μ)SH,LN/2.Proof.
By Lemma13, for any ε>0, we can obtain that t0(uε)>tmax(u) and t0(uε)uε∈M0,μi.e.,(140)t0uε>tmaxuε=2-quε222μ∗-q∫Ω∫Ωgxt0uεuεx2μ∗t0uεuεy2μ∗/x-yμdxdy1/22μ∗-2.(141)t0uεuε2=∫Ωf-t0uεuεq+∫Ω∫Ωgxt0uεuεx2μ∗t0uεuεy2μ∗x-yμdxdy. Furthermore(142)limε→0∫Ω∫Ωgxuεx2μ∗uεy2μ∗x-yμdxdy1/2μ∗=CN,μNN-2/22N-μSH,LN-2/2,(143)limε→0⁡∫Ωf-uεq=0. Using (140)-(143), we have (144)limε→0⁡t0uε=1. Consequently (145)infu∈M0,μ⁡I0,μu≤limε→0⁡I0,μt0uεuε=N-μ+222N-μCN,μNN-2/22N-μSH,LN/2,infu∈M0,μI0,μu≤infu∈M∞ΩI∞u=N-μ+222N-μCN,μNN-2/22N-μSH,LN/2. Applying Lemma (18)(i), for every u∈M0,μ, I0,μ(u)=supt≥0I0,μ(tu). In addition, there exists a unique t3>0 such that t3u∈M∞(Ω). Hence (146)I0,μu≥I0,μt3u≥I∞t3u≥N-μ+222N-μCN,μNN-2/22N-μSH,LN/2. Furthermore, (147)infu∈M0,μI0,μu≥N-μ+222N-μCN,μNN-2/22N-μSH,LN/2,so (148)infu∈M0,μI0,μu=infu∈M∞ΩI∞u=N-μ+222N-μCN,μNN-2/22N-μSH,LN/2.Next, we assume that (135) has a solution u1 satisfying I0,μ(u1)=infu∈M0I0,μ(u). By contradiction, suppose there exists u1∈M0,μ such that I0,μ(u1)=infu∈M0,μI0,μ(u). In fact, u1 is a positive solution owing to I0,μ(u1)=I0,μ(|u1|) and |u1|∈M0,μ, from Lemmas 10 and 13, we conclude that I0,μ(u1)=supt≥0I0,μ(tu1), and there exists a unique t3>0 with t3u1∈M∞(Ω) such that (149)N-μ+222N-μCN,μNN-2/22N-μSH,LN/2=infu∈M0,μ⁡I0,μu=I0,μu1≥I0,μt3u1≥I∞t3u1+t22μ∗22μ∗∫Ω∫Ω1-gxu1x2μ∗u1y2μ∗x-yμdxdy>N-μ+222N-μCN,μNN-2/22N-μSH,LN/2+t22μ∗22μ∗∫Ω∫Ω1-gxu1x2μ∗u1y2μ∗x-yμdxdy,which indicates that ∫Ω∫Ω(1-g(x))|u1(x)|2μ∗|u1(y)|2μ∗/|x-y|μdxdy<0. That is 1-g<0, which contradicts with g≤1.
This proof is completed.Lemma 24.
Assume that{un} is a minimizing sequence for I0,μ in M0,μ, then (i) ∫Ωf-|un|q=o(1).
(ii)∫Ω∫Ω(1-g(x))|un(x)|2μ∗|un(y)|2μ∗/|x-y|μdxdy=o(1).
In addition, forI0,μ, {un} is a (PS)N-μ+2/2(2N-μ)C(N,μ)N(N-2)/2(2N-μ)SH,LN/2 sequence in H01(Ω).Proof.
For anyn, there exists tn>0 such that tnun∈M∞(Ω), which implies that (150)tn2un2=tn2μ∗∫Ωun2μ∗. Form Lemma 13(i) (151)I0,μun≥I0,μtnun≥I∞tnun-tnqq∫Ωf-unq+t22μ∗22μ∗∫Ω∫Ω1-gxunx2μ∗uny2μ∗x-yμdxdy≥N-μ+222N-μCN,μNN-2/22N-μSH,LN/2-tnqq∫Ωf-unq+t22μ∗22μ∗∫Ω∫Ω1-gxunx2μ∗uny2μ∗x-yμdxdy. By Lemma 23, we can get (152)tnqq∫Ωf-unq=o1,t22μ∗22μ∗∫Ω∫Ω1-gxunx2μ∗uny2μ∗x-yμdxdy=o1. Indeed, ∀n, there is C5>0 such that tn>C5. Arguing with contradiction, assume that tn→0 as n→∞. Noticing that (153)I0,μun=N-μ+222N-μCN,μNN-2/22N-μSH,LN/2+o1,I0,μun=N-μ+222N-μun2+o1. By Lemma 9, we know that un is bounded, thus tnun→0 or I∞(tnun)→0 which is a contradiction with I∞(tnun)≥N-μ+2/2(2N-μ)C(N,μ)N/2SH,L2N-μ/2>0, which derives (154)∫Ωf-unq=o1,∫Ω∫Ω1-gxunx2μ∗uny2μ∗x-yμdxddy=o1,and so (155)un2∫Ωun2μ∗+o1,I∞un=N-μ+222N-μCN,μNN-2/22N-μSH,LN/2+o1.Using the same method as those of (i), we can prove (ii).Define(156)M0,μk=u∈M0,μ∣I0,μu≤N-μ+222N-μCN,μNN-2/22N-μSH,LN/2+k,k>0.Let Ψ:H01(Ω)→RN be a barycentre map: (157)Ψu=∫Ωxu2μ∗∫Ωu2μ∗.Lemma 25.
Forδ∈(0,r1), there exists kδ>0 such that (158)Ψu∈Gδ,∀u∈M0,μkδ.Proof.
According to Lemma24, there exist a (PS)N-μ+2/2(2N-μ)C(N,μ)N(N-2)/2(2N-μ)SH,LN/2 sequence {un}⊂M0,μ in H01(Ω) and δ0<r1. Since Ω is bounded and using concentration compactness principle, there exist two sequences {xn}⊂Ω and {Rn}⊂R+ with x1∈Ω¯, and v1∈D1,2(RN) is a positive solution of (16), such that xn→x1 and Rn→∞ as n→∞ satisfying(159)unx-RnN-2μ/2v1Rnx-xnD1,2RN⟶0,asn⟶∞. Thus (160)Ψun=∫Ωxun2μ∗∫Ωun2μ∗dxdy=∫ΩxRnN-2μ/2v1Rnx-xn2μ∗∫ΩRnN-2μ/2v1Rnx-xn2μ∗+o1=∫Ωx/Rn+xnv1x2μ∗∫Ωv1x2μ∗+o1=x1+o1. In fact, x1∈Gδ0. We know that {un} is a minimizing sequence for M0,μ, and, combining with Lemma 24 (ii), we have (161)0=limn→∞∫Ω1-gun2μ∗=limn→∞∫Ω1-gRnN-2μ/2v1Rnx-xnN-2μ/2=limn→∞∫Ω1-gxRn+xnv1xN-2μ/2=1-gx1SN/2μ=1-gx1SH,LCN,μ1/2μ∗N/2μ, which deduce that g(x1)=maxx∈Ω¯g(x)=1 and (162)x1∈x∈Ω¯∣gx=maxy∈Ω¯⁡gy=1=G, and this contradicts with assumption. This completes the proof.Set(163)Mλ,μc=u∈Mλ,μ-∣Iλ,μu≤c.Lemma 26.
For every0<δ<r1, there exists λδ∈(0,q/2λ∗), such that, for everyλ<λδ, we have (164)Ψu∈Gδ,foreveryu∈Mλ,μcλ.Proof.
Foru∈Mλ,μ(cλ), combining (26) with Lemma 13(i), there exists a unique t3>tmax(u), such that t3u∈M0,μ. Based on Sobolev and Höder inequalities, one has(165)Iλ,μu=supt≥tmaxu⁡Iλ,μtu≥Iλ,μt3u=I0,μt3u-λt3qq∫Ωf+uq≥I0,μt3u-λt3qqf+Lq∗S-q/2uq=I0,μt3u-λt3qqf+∞SH,LCN,μ1/2μ∗-q/2uq. We assume that there exists a positive constant m such that t3≤C6. Using (26) and Sobolev inequality, we deduce that (166)u2≤22μ∗-q2-q∫Ω∫Ωgxux2μ∗uy2μ∗x-yμ≤22μ∗-q2-qCN,μ2μ∗-1u22μ∗=22μ∗-q2-qCN,μ2μ∗-1u22μ∗SH,L-2μ∗. So(167)u≥2-qSH,L2μ∗22μ∗-qCN,μ2μ∗-1.We may assume that t3≥1, which leads to (168)t322μ∗∫Ω∫Ωgxux2μ∗uy2μ∗x-yμ=t32u2-t3q∫Ωf-uq≤t32u2+∫Ωf-uq.That is(169)t3≤u2+∫Ωf-uq∫Ω∫Ωgxux2μ∗uy2μ∗/x-yμ1/22μ∗-2. Combining (167), (169), one gets(170)t3≤22μ∗-q2-qu2+∫Ωf-uqu21/22μ∗-2≤22μ∗-q2-q1+∫Ωf-uqSq/2u2-q1/22μ∗-2≤22μ∗-q2-q1+f-∞SH,LCN,μ1/2μ∗q/2u2-q1/22μ∗-2≤22μ∗-q2-q1+f-∞SH,LCN,μ1/2μ∗2-qSH,L2μ∗22μ∗-qCN,μ2∗-11/22μ∗-2≔m.From (165) and (170), (171)I0,μt3u≤Iλ,μu+λqt3qf+∞SH,LCN,μ1/2μ∗-q/2uq≤cλ+λmqqf+∞SH,LCN,μ1/2μ∗-q/2uq.Using Lemma 9, if Iλ,μ(u)≤cλ<N-μ+2/2(2N-μ)C(N,μ)N(N-2)/2(2N-μ)SH,LN/2. For every λ∈(0,q/2λ∗), there exists C6>0 such that u≤C6 for every u∈Mλ,μ(cλ).
Thus(172)I0,μt3u≤cλ+λmqqf+∞SH,LCN,μ1/2μ∗-q/2uq.Taking kδ>0, for λ<λδ, there exists λδ∈(0,q/2λ∗), and we have (173)Iλ,μu≤N-μ+222N-μCN,μNN-2/22N-μSH,LN/2+kδ,∀u∈Mλ,μcλ. Based on Lemma 25, we derive that t3u∈M0,μ(kδ) and (174)Ψu=∫Ωxt3u2μ∗∫Ωt3u2μ∗=Ψt3u∈Gδ,∀u∈Mλ,μcλ.Let Γ(z)=u1+t3uε∈Mλ,μ-, and (175)Iλ,μu1+t3uε<cλ=αλ,μ++N-μ+222N-μCN,μNN-2/22N-μSH,LN/2+kδ,so, u1+t3uε∈Mλ,μ(cλ). At the same time, we conclude that(176)Ψu1+t3uε∈Gδ.Lemma 27 (see [41]).
Let closed setsX,Y,Z with Y⊂Z and h1⊂C(X,Z),h2⊂C(Y,Z), and suppose h1∘h2 is homotopically equivalent to identity mapping id in Z, then catx(X)≥catZ(Y).Theorem 28.
For everyλ∈(0,λδ), Iλ,μ has at least catGδ(G) critical points on Mλ,μ,+(cλ)=u∈Iλ,μcλ∣u≥0.Proof.
By Lemma15, for every λ∈(0,λδ) and z∈G(177)Iλ,μΓz<cλ=αλ,μ++N-μ+222N-μCN,μNN-2/22N-μSH,LN/2,ΨΓz=∫ΩxΓz2μ∗∫ΩΓz2μ∗=∫Ωxu2+tzuε2μ∗∫Ωu2+tzuε2μ∗=∫Ωx+zu2x+z+tzηx2μ∗∫Ωu2x+z+tzηx2μ∗=z+∫Ωzu2x+z+tηxvx2μ∗∫Ωu2x+z+tzηxvx2μ∗=z+o1z⟶0. From Lemma 27, X=Mλ,μ(cλ),Y=G,Z=Gδ,h1=Ψ,h2=Γ, which implies that catMλ,μ(cλ)(Mλ,μ(cλ)>catGδ(G)). Combined with Lemma 16, (1) has at least catGδ(G)+1 solution.
This proof is completed.



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*Source: 7210680-2018-07-08.xml*