# Multiple Positive Solutions and Estimates of Extremal Values for a Nonlocal Problem with Critical Sobolev Exponent and Concave-Convex Nonlinearities

Authors: Zhigao Shi; Xiaotao Qian
Journal: Journal of Function Spaces
 (2022)

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## Abstract

We are concerned with the following nonlocal problem involving critical Sobolev exponent−a−b∫Ω∇u2dxΔu=λuq−2u+δu2u,x∈Ω,u=0,x∈∂Ω, where Ω is a smooth bounded domain in ℝ4, a,b>0, 1<q<2, δ, and λ are positive parameters. We prove the existence of two positive solutions and obtain uniform estimates of extremal values for the problem. Moreover, the blow-up and the asymptotic behavior of these solutions are also discussed when b↘0 and δ↘0. In the proofs, we apply variational methods.


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## Body

## 1. Introduction and Main Results

In this paper, we study a new class of Kirchhoff type problem with critical exponent and concave-convex nonlinearities(1)−a−b∫Ω∇u2dxΔu=λuq−2u+δu2u,x∈Ω,u=0,x∈∂Ω,Pb,δ,where Ω is a smooth bounded domain in ℝ4 (2∗=4 is the critical exponent in dimension four), a,b>0, 1<q<2, δ, and λ are positive parameters.We callPb,δ a Kirchhoff type problem since the presence of the term ∫Ω∇u2dx, which means that Pb,δ is no longer a pointwise identity. Such nonlocal problem arises in various models concerning physical and biological systems, see, e.g., [1–3]. Among others, Kirchhoff [2] built a model defined by the equation
(2)ρ∂2u∂t2−P0h+E2L∫0L∂u∂x2dx∂2u∂x2=0,where u=ux,t represents the lateral displacement, ρ denotes the mass density, P0 is the initial tension, h denotes the area of the cross-section, E denotes the Young modulus of the material, and L is the length of the string. This equation is an extension of the classical D’Alembert wave equation for free vibrations of elastic strings.Different from the traditional Kirchhoff type problem, the sign of nonlocal term included inPb,δ is negative, which causes some interesting difficulties. In the past few years, much attention has been paid to the existence, multiplicity, and the behaviour of solutions for this kind of nonlocal problem but without critical growth. In particular, Yin and Liu [4] were concerned with the following problem
(3)−a−b∫Ω∇u2dxΔu=up−2u,x∈Ω,u=0,x∈∂Ω,where 1<p<2∗ and Ω is a bounded domain in ℝN with N≥1 and succeeded to find the problem (3) admits at least two nontrivial solutions. In [5, 6], sign-changing solutions to (3) were further obtained. When N=3 and the nonlinear term has an indefinite potential, Lei et al. [7] and Qian and Chao [8] established the existence of positive solution of (3) for 1<p<2 and 3<p<6, respectively. For the singular nonlinearity, two positive solutions to (3) with N=3 were proved in [9]. In our previous work [10], we obtained two positive solutions of Pb,δ with δ=0, as well as their blow-up and asymptotic behavior when b↘0. For more related results, we refer the interested readers to [11–15] and the references therein.In 1994, Ambrosetti et al. [16] first studied the following critical local problem involving concave-convex nonlinearities
(4)−Δu=λuq−2u+u2∗−1u,x∈Ω,u=0,x∈∂Ω,where 1<q<2 and Ω⊂ℝN is a smooth bounded domain. The authors proved that there exists λ0>0 such that the problem (4) has two positive solutions for λ∈0,λ0 and no positive solutions for λ>λ0. Since then, many scholars have considered problems with critical exponent and concave-convex nonlinearities, see, e.g., [7, 16–21]. Also, the problem (4) of traditional Kirchhoff type is studied in [22–26] and the reference therein. An interesting question now is whether the same existence results as in [16] occur to the nonlocal problem Pb,δ with critical exponent. For λ=0 and δ=1, Wang et al. [27] proved the existence of two positive solutions of Pb,δ with an additional inhomogeneous perturbation on the whole space ℝ4. When 2<q<2∗ and δ is replaced by a nonnegative function Qx, [28] showed how the shape of the graph of Qx affects the number of positive solutions to Pb,δ. However, there are no known existence results for Pb,δ provided λ>0 and 1<q<2.Motivated by the works described above, in the present paper, we try to prove the existence and multiplicity of positive solutions of problemPb,δ when λ∈0,T− for some T−>0 (see Theorem 1), provide uniform estimates of extremal values λ∗ for problem Pb,δ (see Theorem 2), and obtain the blow-up and asymptotic behavior of these positive solutions when b↘0 and δ↘0 (see Theorem 3).Denote byH01Ω the standard Sobolev space endowed with the standard norm ·. Let ·p be the norm of the space LsΩ. Denote by ⟶ (⇀) the strong (weak) convergence. C and Ci denote various positive constants whose exact values are not important. Let μ1 be the positive principal eigenvalue of the operator −Δ on Ω with corresponding positive principal eigenfunction e1. Denote by S the best constant in the Sobolev embedding H01Ω°L2∗Ω, namely,
(5)S=infu∈H01Ω\0u2u42>0.It is well known that the weak solutions of problemPb,δ correspond to the critical points of the following energy functional
(6)Ib,δu=a2u2−b4u4−λquqq−δ4u44.Moreover, we easily see thatIb,δ∈C1H01Ω,ℝ.Define the manifold(7)Mb,δ=u∈H01Ω:I′b,δu,u=0=u∈H01Ω:au2=bu4+λuqq+δu44,and decompose Mb,δ into three subsets as follows:
(8)Mb,δ0=u∈Mb,δ:a2−qu2−b4−qu4−δ4−qu44=0,Mb,δ+=u∈Mb,δ:a2−qu2−b4−qu4−δ4−qu44>0,Mb,δ−=u∈Mb,δ:a2−qu2−b4−qu4−δ4−qu44<0.Set(9)T1=2aSq/24−qΩ4−q/4a2−qb+δS−24−q2−q/2,T2=2qaS4−q/22−q4−qbS2+δΩ4−q/4δ2−qqq/2,T−=minT1,T2.Our main results are as follows.Theorem 1.
Assume thatλ∈0,T−, then problem Pb,δ has at least two positive solutions u∗∈Mb,δ+ and U∗∈Mb,δ− with u∗<U∗.Theorem 2.
Let(10)λ∗=supλ>0:Pb,δhasatleasttwopositivesolutions.
Then, we have(11)0<T−≤λ∗≤T+<∞,where T− is defined as above and
(12)T+=2aμ14−q2−qaμ14−qδ1/2+1.Theorem 3.
Assume thatbn and δn are two sequences satisfying bn↘0 and δn↘0 as n⟶∞. Let un and Un be the two positive solutions of Pb,δ corresponding to bn and δn obtained in Theorem 1 with un∈Mbn,δn+ and Un∈Mbn,δn−, then passing to a subsequence if necessary,
(i)
Un⟶∞ as n⟶∞(ii)
un⟶u¯ in H01Ω as n⟶∞, where u¯ is a positive ground state solution of the problem(13)−aΔu=λuq−2u,x∈Ω,P0,0,u=0,x∈∂Ω.Remark 4.
The multiplicity result ofPb,δ with δ=0 has been proved by [10]. So, our result presented in Theorem 1 can be viewed as an extension of [10] considering the subcritical case where δ=0. In particular, we provide uniform estimates of extremal values λ∗ for the problem, which are observed for the first time in the studies of such nonlocal problem like Pb,δ.Remark 5.
Comparing with [16], which considered problem Pb,δ with b=0, we in this paper investigate the nonlocal case of b≠0. Moreover, unlike [22–24, 26], where the nonlocal term is positive, here we study the case of negative sign of nonlocal term and additionally obtain a bound from above for the parameter.The plan of this paper is as follows. In Section2, we give some preliminaries. Section 3 is devoted to the Proof of Theorem 1. In Section 4, we prove Theorems 2 and 3. In the proof of our main results, we use variational methods, and they are inspired by [10, 16]. However, in the present paper, we encounter some new difficulties due to the critical growth and nonlocal term. Firstly, compared with [10], the calculations here are more delicate and difficult since we now face the critical problem Pb,δ. Secondly, to provide the bound from above for λ∗ of Pb,δ involving nonlocal term, we need to develop some techniques applied in [16] where dealt with local case. Thirdly, in order to obtain the asymptotic behavior of the solutions of Pb,δ as in the work of [10], we add the condition of δ↘0 and conduct some new analysis.


## 2. Preliminaries

Lemma 6.
Letλ∈0,T1. Then, Mb,δ±≠∅ and Mb,δ0=0.Proof.
A simple calculation shows that(14)∂Ib,δϕ∂ttu=tq−1at2−qu2−bt4−qu4−λuqq−δt4−qu44.
For anyu∈H01Ω\0, t>0, set
(15)ψt=at2−qu2−bt4−qu4−δt4−qu44,t>0,ψ1t=at2−qu2−t4−qb+δS−2u4,t>0.
Since1<q<2, it is clear that limt⟶0+ψ1t=0 and limt⟶+∞ψ1t=−∞. Moreover, ψ1t is concave and achieves its maximum at the point tmax=a2−qu2/b+δS−24−qu41/2 with
(16)ψ1tmax=24−q2−q4−q2−q/2au24−qb+δS−2u42−q1/2.
By Hölder and Sobolev inequalities, forλ∈0,T1, we obtain
(17)λuqq≤λΩ4−q4S−q/2uq<ψ1tmax≤ψtmax.
From which we infer that there exist two constantst+=t+u and t−=t−u satisfying t+>tmax>t−>0 and
(18)ψt+=λuqq=ψt−,ψ′t+<0<ψ′t−.
This gives thatt+u∈Mb,δ− and t−u∈Mb,δ+.
In what follows, we prove thatMb,δ0=0. Suppose to the contrary that there is w∈Mb,δ0 with w≠0. By w∈Mb,δ0, we have
(19)a2−qw2=b4−qw4+δ4−qw44.
As a consequence, by Sobolev inequality,(20)a2−qw2≤b4−qw4+δ4−qS−2w4=b+δS−24−qw4.
Moreover, we can also infer fromw∈Mb,δ0 that −2aw2+λ4−qwqq=0 and so
(21)λwqq=2a4−qw2.
Combining (20) and (21), for λ∈0,T1, we conclude that
(22)0<24−q2−q4−q2−q/2aw24−qb+δS−2w42−q1/2−λwqq≤24−q2−q4−q2−q/2aw24−qa2−q/4−qw22−q1/2−λwqq=2a4−qw2−λwqq=0,which is absurd. The proof of Lemma 6 is completed.Lemma 7.
Assume thatλ∈0,T1, then there is a gap structure in Mb,δ:
(23)u≤Aλ<A0≤U,∀u∈Mb,δ+,U∈Mb,δ−,where
(24)A0=a2−q4−qb+δS−21/2,Aλ=λ4−qΩ4−q/42aSq/21/2−q.Proof.
In the case ofU∈Mb,δ−, using Sobolev inequality, we have
(25)a2−qU2<b4−qU4+δ4−qU44≤b+δS−24−qU4,which yields U≥A0.
In the case ofu∈Mb,δ+, it holds
(26)2au2<λ4−quqq≤λ4−qΩ4−q/4S−q/2uq,which gives that u≤Aλ. Moreover, we easily check that if λ∈0,T1, then Aλ<A0.Lemma 8.
For anyu∈Mb,δ±, there exist ρu>0 and a differential functional gρu:Bρu0⟶ℝ+ such that
(27)gρu0=1,gρuwu−w∈Mb,δ±,gρu′0,ϕ=2a−4bu2∫Ω∇u∇ϕdx−qλ∫Ωuq−2uϕdx−4δ∫Ωu2uϕdxa2−qu2−b4−qu4−δ4−qu44.Proof.
Fixu∈Mb,δ− and define F:ℝ+×H⟶ℝ by
(28)Ft,w=at2−qu−w2−bt4−qu−w4−λu−wqq−δt4−qu−w44.
Since foru∈Mb,δ−⊂Mb,δ, one has F1,0=0 and
(29)Ft1,0=a2−qu2−b4−qu4−δ4−qu44<0,then we can employ the implicit function theorem for F at the point 1,0 and derive ρ¯>0 and a differential functional g=gw>0 defined for w∈H01Ω, w<ρ¯ such that
(30)g0=1,gwu−w∈Mb,δ,∀w∈H,w<ρ¯.
In view of the continuity ofg, we may choose ρ>0 possibly smaller (ρ<ρ¯) such that for any w∈H01Ω, w<ρ, it holds
(31)gwu−w∈Mb,δ−.
In a similar way, we can prove the case ofu∈Mb,δ+, and thus, Lemma 8 follows.Lemma 9.
Ifλ∈0,T1, then we have
(i)
The functionalIb,δ is coercive and bounded from below on Mb,δ(ii)
infMb,δ+∪Mb,δ0Ib,δ=infMb,δ+Ib,δ∈−∞,0Proof.
(i)
Foru∈Mb,δ, using Hölder’s inequality, we obtain(32)Ib,δu=Ib,δu−14I′b,δu,u=a4u2−λ1q−14uqq≥a4u2−λ1q−14Ω4−q/4S−q/2uq.
This proves the conclusion (i).(ii)
Foru∈Mb,δ+, it holds(33)Ib,δu=Ib,δu−1qI′b,δu,u=a12−1qu2+b1q−14u4+δ1q−14u44<−a2−qu2+b4−qu4+δ4−qu444q<0.
Combining this and Lemma6, we have that infMb,δ+∪Mb,δ0Ib,δ=infMb,δ+Ib,δ<0. Furthermore, we deduce from (i) that infMb,δ+∪Mb,δ0Ib,δ≠−∞. Thus, infMb,δ+∪Mb,δ0Ib,δ∈−∞,0.Lemma 10.
Ifλ∈0,T1, then Mb,δ+∪Mb,δ0 and Mb,δ− are closed.Proof.
LetUn be a sequence in Mb,δ− such that Un⟶U0 in H01Ω. Since Un⊂Mb,δ−⊂Mb,δ, we have
(34)aU02−bU04=limn⟶∞aUn2−bUn4=limn⟶∞λUnqq+δUn4=λU0qq+δU04,a2−qU02−b4−qU04−δ4−qU04=limn⟶∞a2−qUn2−b4−qUn4−δ4−qUn4≤0,namely, U0∈Mb,δ−∪Mb,δ0. For λ∈0,T1, it then follows from Lemma 7 that U0∉Mb,δ0. In turn, we obtain U0∈Mb,δ−, and so, Mb,δ− is closed for λ∈0,T1. The same argument can prove that Mb,δ0∪Mb,δ+ is closed. This completes the proof of Lemma 10.


## 3. Proof of Theorem 1

Lemma 11.
Suppose thatλ∈0,T1, then problem Pb,δ admits a positive solution u∗ with u∗∈Mb+.Proof.
By Lemmas9 and 10, we can apply Ekeland variational principle to get a minimizing sequence un⊂Mb,δ+∪Mb,δ0 such that
(35)limn⟶∞Ib,δun=infMb,δ+∪Mb,δ0Ib,δ<0,(36)Ib,δz≥Ib,δun−1nz−un,∀z∈Mb,δ+∪Mb,δ0.
SinceIb,δu=Ib,δu, we can assume that un≥0 in Ω. By Lemma 9, un is bounded in H01Ω, and so, we may assume that
(37)un⇀u∗,inH01Ω,un⟶u∗,inLsΩ,1≤s<4,un⟶u∗,a.e.inΩ.
In the following, we prove thatu∗ is a positive solution to Pb,δ. To this purpose, we divide the proof into five steps.
Step 1.u∗≠0.
If, to the contrary, we haveu∗=0. Since un∈Mb,δ+∪Mb,δ0, it follows that for n large,
(38)aun2≥4−q2−qbun4+4−q2−qδun44,and hence,
(39)Ib,δun=12aun2−14bun4−14δun44+o1>4−q22−q−14bun4+4−q22−q−14δun44+o1>0,which contradicts with (35). Therefore, u∗≠0.
Step 2. There is a positive constantC1 satisfying
(40)2aun2−λ4−qunqq<−C1.
To prove that, it suffices to check that(41)2alimsupn⟶∞un2<λ4−qu∗qq.
In view ofun∈Mb,δ+∪Mb,δ0, one has
(42)2alimsupn⟶∞un2≤λ4−qu∗qq.
Assume to the contrary that(43)2alimsupn⟶∞un2=λ4−qu∗qq.
Then, we can supposeun2⟶A>0 as n⟶∞, where A satisfies
(44)λu∗qq=2aA4−q.
From this, we have that forλ∈0,T1,
(45)0≤24−q2−q4−q2−q/2a4−q/2b+δS−22−q/2−λΩ4−q/4S−q/2unq≤24−q2−q4−q2−q/2aun24−qb+δS−2un42−q1/2−λunqq≤24−q2−q4−q2−q/2aun24−qaun2−λunqq2−q1/2−λunqq⟶24−q2−q4−q2−q/2aA4−q2−q/4−qaA2−q1/2−2aA4−q=0,which implies that un⟶0 in H01Ω, contradicting u∗≠0. In turn, we deduce that (40) holds.
Step 3.I′b,δun⟶0 as n⟶∞.
Let0<ρ<ρn≡ρun, gn≡gun, where ρun and gun are defined as Lemma 8 with u=un. Let wρ=ρv with v=u/u. Fix n and set zρ=gnwρun−wρ. Since zρ∈Mb,δ+, it follows from (36) that
(46)Ib,δzρ−Ib,δun≥−1nzρ−un.
By the definition of Fréchet derivative, we obtain(47)Ib,δ′un,zρ−un+ozρ−un≥−1nzρ−un.
Then,(48)Ib,δ′un,−wρ+gnwρ−1un−wρ≥−1nzρ−un+ozρ−un,and hence,
(49)−ρIb,δ′un,v+gnwρ−1Ib,δ′un,un−wρ≥−1nzρ−un+ozρ−un,which yields that
(50)Ib,δ′un,v≤1nzρ−unρ+ozρ−unρ+gnwρ−1ρI′b,δun,un−wρ.
From Step 2, Lemma8, and the boundedness of un, we also have
(51)zρ−un=gnwρ−1un−wρ−wρ≤gnwρ−1C2+ρ,limρ⟶0gnwρ−1ρ=gn′0,v≤g′n0≤C3,I′b,δun,un−wρ=I′b,δun,−wρ=−ρI′b,δun,v.
As a consequence, for fixedn, we can derive letting ρ⟶0 in (50) that
(52)I′b,δun,v≤Cn,which implies that I′b,δun⟶0 as n⟶∞.
Step 4.un⟶u∗ in H01Ω.
Setvn=un−u∗. If vn⟶0, we are done, thus assume vn⟶L>0. By I′b,δun,u∗=o1 and (37),
(53)0=au∗2−bL2+u∗2u∗2−λu∗qq−δu∗44.
Moreover, fromI′b,δun⟶0, the boundedness of un, and Brézis-Lieb lemma, we have that
(54)o1=Ib,δ′un,un=avn2+u∗2−bvn4+2vn2u∗2+u∗4−λu∗qq−δvn44−δu∗44+o1.
Combining this and (53), we get
(55)o1=avn2−bvn4−bvn2u∗2−δvn44.
It then follows from Sobolev inequality that(56)avn2−bvn4−bvn2u∗2=δvn44+o1≤δS−2vn4+o1.
Passing the limit asn⟶∞, we obtain that
(57)L2≥S2a−bu∗2bS2+δ>0.
By (53), (57), and Hölder inequality,
(58)Ib,δu∗=a2u∗2−b4u∗4−λq∫u∗qdx−δ4u∗44=a4u∗2+b4L2u∗2−λ1q−14u∗qq≥abS2u∗24bS2+δ+b4L2u∗2+aδ4bS2+δu∗2−λ1q−14Ω4−q/4S−q/2u∗q.
Forξ≔aδ/4bS2+δ and η≔λ1/q−1/4Ω4−q/4S−q/2, define
(59)ft=ξt2−ηtq.
By easy calculation, we have thatft achieves its minimum value at tmin=qη/2ξ1/2−q and
(60)ftmin=−2−q2η2/2−qq2ξq/2−q.
Therefore, we obtain(61)Ib,δu∗≥abS2u∗24bS2+δ+b4L2u∗2+ftmin=abS2u∗24bS2+δ+b4L2u∗2−2−q2λ4−q4qΩ4−q/4S−q/22/2−q2qbS2+δaδq/2−q.
Using (37), (53), and (61), we deduce that for λ∈0,T1,
(62)Ib,δun=Ib,δu∗+a4vn2−b4vn2u∗2+o1≥abS2u∗24bS2+δ+a4L2−2−q24−q4qΩ4−q/4S−q/22/2−q2qbS2+δaδq/2−qλ2/2−q+o1≥a2S24bS2+δ−2−q24−q4qΩ4−q/4S−q/22/2−q2qbS2+δaδq/2−qλ2/2−q+o1>0,which is a contradiction since limn⟶∞Ib,δun<0. This implies that vn⟶L>0 is impossible. Hence, vn⟶0; that is, un⟶u∗ in H01Ω.
Step 5.u∗ is a positive solution of problem Pb,δ and u∗∈Mb,δ+.
From (35) and Steps 3 and 4, we have that, up to a subsequence, un⟶u∗ in H01Ω with Ib,δu∗<0 and I′b,δu∗=0. Namely, u∗≥0 is a weak nontrivial solution of problem Pb,δ. Moreover, by Lemmas 6 and 10, we know u∗∈Mb,δ+. Standard elliptic regularity argument and strong maximum principle provide that u∗ is positive. Therefore, the proof of Lemma 11 is completed.Lemma 12.
Letλ∈0,T1, then problem Pb,δ has a positive solution U∗ with U∗∈Mb−.Proof.
As in the proof of Lemma11, we can prove that there exists a bounded and nonnegative sequence Un⊂Mb,δ− with the properties
(i)
limn⟶∞Ib,δUn=infMb,δ−Ib,δ(ii)
Ib,δz≥Ib,δUn−1nz−un,∀z∈Mb,δ−(iii)
Un⇀U∗inH01Ω(iv)
Un⟶U∗inLsΩ,2≤s<4(v)
Un⟶U∗a.e.inΩ
Without loss of generality, we may assume that0∈Ω. Let φx∈C0∞Ω be a cut-off function such that 0≤φ≤1 in Ω and φx≡1 near zero. Set
(63)vεx=φx81/2εε2+x2.
By [29, 30], one has for ε>0 small,
(64)vε2=S2+Oε2,vε44=S2+Oε4,vε33=Oε.
In the first place, we prove the following upper bound forinfMb,δ−Ib,δ,
(65)infMb,δ−Ib,δ≤supt>0Ib,δu∗+tvε<Ib,δu∗+a2S24bS2+δ,where u∗ is the positive solution obtained in Lemma 11. Since u∗∈Mb,δ+, it is easy to verify that a−bu∗2>0. By the fact that Ib,δ′u∗,tvε=0, we also have
(66)0=ta−bu∗2∫Ω∇u∗∇vεdx−tλ∫Ωu∗q−1vεdx−tδ∫Ωu∗3vεdx,and hence,
(67)∫Ω∇u∗∇vεdx=λ∫Ωu∗q−1vεdx+δ∫Ωu∗3vεdxa−bu∗2>0.
Letwε=u∗+Rvε with R>1. It follows from (67) that
(68)wε2=u∗2+2R∫Ω∇u∗∇vεdx+R2vε2≥u∗2+R2S2+Oε2.
Letψt be given by Lemma 6. As can be seen from the proof of Lemma 6, there exist ψtε=λwε/wεqq and ψ′tε<0, where tε=t+wε/wε. From the structure of ψ and the fact of wε/wεqq>0, we easily see that tε is uniformly bounded by a suitable constant C1>0, ∀R≥1, and ∀ε>0.
Moreover, we have from (68) that there is ε1>0 satisfying
(69)wε2≥u∗2+12R2S2,∀ε∈0,ε1.
Therefore, we may findR1≥1 such that wε>C1, ∀R≥R1, and ∀ε∈0,ε1.
Define(70)E1=u:u=0oru<t+uu,E2=u:u>t+uu.
Notice thatH01Ω−Mb,δ−=E1∪E2 and Mb,δ+⊂E1. Because u∗∈Mb,δ+ and the continuity of t+u, we have that u∗+tR1vε for t∈0,1 must intersect Mb,δ−. As a consequence,
(71)infMb,δ−Ib,δ≤supt>0Ib,δu∗+tvε.
Thus, to complete the proof of (65), it suffices to show that
(72)supt>0Ib,δu∗+tvε<Ib,δu∗+a2S24bS2+δ.
By mean value theorem, there existsδx∈0,1 such that
(73)u∗x+tvεxq−u∗qx=qu∗x+δxtvεxq−1tvεx≥qtu∗q−1xvεx,for any x∈Ω. Using (66), (67), and (73), we obtain
(74)Ib,δu∗+tvε=a2u∗2+at∫Ω∇u∗∇vεdx+a2t2vε2−b4u∗4−bt2∫Ω∇u∗∇vεdx2−b4t4vε4−btu∗2∫Ω∇u∗∇vεdx−b2t2u∗2vε2−bt3vε2∫Ω∇u∗∇vεdx−λq∫Ωu∗+tvεqdx−δ4∫Ωu∗+tvε4dx≤Ib,δu∗+a2t2vε2−b4t4vε4−b2t2u∗2vε2−λq∫Ωu∗+tvεq−u∗q−qtu∗q−1vεdx−δ4∫Ωu∗+tvε4−u∗4−4tu∗3vεdx≤Ib,δu∗+a2t2vε2−b4t4vε4−b2t2u∗2vε2−δ4∫Ωu∗+tvε4−u∗4−4tu∗3vεdx.
To proceed, we set(75)Jv=a2v2−b4v4−b2u∗2v2−δ4∫Ωu∗+v4−u∗4−4u∗3vdx.
Recall that, forr,s≥1, it holds
(76)r+s4−r4−4r3s≥s4+C1rs3,for some C1>0. By (73) and (76), we have that
(77)Jtvε=a2t2vε2−b4t4vε4−b2t2u∗2vε2−δ4∫Ωu∗+tvε4−u∗4−4tu∗3vεdx≤a2t2vε2−b4t4vε4−b2t2u∗2vε2−δ4∫Ωtvε4+C1u∗tvε3dx=a2t2vε2−b2t2u∗2vε2−b4t4vε4−δ4t4vε44−δ4C1t3∫Ωu∗vε3dx,which implies that there exists a constant t1>0 small enough such that
(78)sup0<t<t1Ib,δu∗+tvε<a24b.
Thus, we only need to consider the case oft≥t1. By the same argument of Lemma 11 of [21], we have
(79)∫Ωu∗vε3dx=83/2εu∗0∫ℝ411+x23dx+oε.
Combining this and (64), we have for ε>0 sufficiently small,
(80)supt≥t1Jtvε≤supt>0a2t2vε2−b2t2u∗2vε2−b4t4vε4vε4−δ4t4vε44−δ4C1t13∫Ωu∗vε3dx≤avε2−bu∗2vε224bvε4+vε44−C2ε+oε=aS2−bS2u∗224bS4+δS2+Oε2−C2ε+oε<S2a−bu∗224bS2+δ<a2S24bS2+δ,where C2>0 is a positive constant independent of ε. This together with (74) implies that (65) holds.
In the second place, we claim thatU∗≠0. If, to the contrary, we have U∗≡0. Since Un∈Mb,δ−⊂Mb,δ, it follows that
(81)aUn2−bUn4−λUnqq−δUn44=0,and so, by Sobolev inequality
(82)aUn2=bUn4+δUn44+o1≤b+δS−2Un4.
Assume thatUn2⟶ι2. By Un⊂Mb,δ− and Lemma 7, we obtain that ι2>0. Taking n⟶∞ in (82), we have ι2≥aS2/bS2+δ, and thus
(83)infMb,δ−Ib,δ=limn⟶∞Ib,δUn=limn⟶∞Ib,δUn−14Ib,δ′Un,Un=limn⟶∞a4Un2−λ1q−14Unqq=a4ι2≥a2S24bS2+δ,which is a contradiction with (65). Therefore, the claim follows. At this point, we may proceed as in the proof of Lemma 11 and conclude that U∗ is a positive solution of problem Pb,δ with U∗∈Mb,δ−. This completes the proof of Lemma 12.Proof of Theorem 1.
Theorem1 is an immediate consequence of Lemmas 7, 11, and 12.


## 4. Proofs of Theorems 2 and 3

Proof of Theorem 2.
By the definition ofλ∗ and Theorem 1, we easily see that λ∗≥T−. Hence, Proof of Theorem 2 is completed if we show that λ∗≤T+. To this goal, let us define the functions
(84)hλt=tq−1δt4−q−aμ1t2−q+λ,t>0,h~λt=δt4−q−aμ1t2−q+λ,t>0.
Obviously, we have thath~t is convex and attains its minimum at the point tmin=2−qaμ1/4−qδ1/2 with
(85)h~λtmin=−2aμ14−q2−qaμ14−qδ1/2+λ.
As a consequence, we can take(86)T+=2aμ14−q2−qaμ14−qδ1/2+1,such that
(87)h~T+t≥h~T+tmin=1>0,∀t>0.
This gives that(88)hT+t≥tq−1h~T+t>0,∀t>0,namely,
(89)T+tq−1+δt3>aμ1t,∀t>0.
Assume that anyλ>0 is such that Pb,δ admits a positive solution u. On the one hand, using (89) with t=u, multiplying by e1, and integrating over Ω, we get
(90)T+∫Ωuq−1e1dx+δ∫Ωu3e1dx>aμ1∫Ωue1dx.
On the other hand, multiplyingPb,δ by e1 and integrating over Ω, there holds
(91)a−bu2∫Ω∇u∇e1dx=λ∫Ωuq−1e1dx+δ∫Ωu3e1dx>0.
Since(92)aμ1∫Ωue1dx=a∫Ω∇u∇e1dx>a−bu2∫Ω∇u∇e1dx,we infer from (90) and (91) that λ<T+. By the arbitrariness of λ and the definition of λ∗, we conclude that λ∗≤T+<∞. Proof of Theorem 2 is thus completed.Proof of Theorem 3.
Letbn and δn be two sequences satisfying bn↘0 and δn↘0 as n⟶∞, and let un and Un be the two positive solutions of Pbn,δn obtained in Theorem 1 with un∈Mbn,δn+ and Un∈Mbn,δn−.
Using Lemma7 and Un∈Mbn−, we have that
(93)limn⟶∞Un2≥limn⟶∞a2−q4−qbn+δnS−21/2=+∞,and thus, the conclusion (i) holds.
In what follows, we prove the conclusion (ii) of Theorem3. Noting that
(94)Ibn,δnun=infMbn,δn+∪Mbn,δn0Ibn,δn<0,for all n∈ℕ, we obtain from Hölder inequality that
(95)0≥Ibn,δnun−14I′bn,δnun,un≥12−14un2−λ1q−14Ω4−q/4S−q/2unq.
As a consequence of1<q<2, we have that un is bounded in H01Ω. Thus, there is a subsequence of un (still denoted by un) such that un⇀u¯ in H01Ω as n⟶∞. Furthermore, for all ϕ∈H01Ω, it holds
(96)0=limn⟶∞Ibn,δn′un,ϕ=limn⟶∞a−bnun2∫Ω∇un∇ϕdx−λ∫Ωunq−1ϕdx−δn∫Ωun3ϕdx=a∫Ω∇u0∇ϕdx−λ∫Ωu0q−1ϕdx,which provides that u¯ is a nonnegative weak solution of problem P0,0. Let I0,0u be the corresponding functional of P0,0 defined by
(97)I0,0u=a2u2−λquqq.
Since(98)aun−u¯2=Ibn,δn′un−I0,0′u¯,un−u¯+bn∫Ω∇un2dx∫Ω∇un∇un−u¯dx+λ∫Ωunq−1−u¯q−1un−u¯dx+δn∫Ωun3un−u¯dx⟶0,as n⟶∞, it follows that un⟶u¯ in H01Ω.
Definec0=infI0,0u:u∈H01Ω. It is easy to check that there exists v0∈H01Ω\0 such that c0=I0,0v0 and c0<0. As I0,0u≥Ibn,δnu for any u∈H01Ω, we easily see that infMbn,δn+∪Mbn,δn0Ibn,δn≤c0. Set cbn,δn=Ibn,δnun and suppose that limn⟶∞cbn,δn=k. We claim k=c0. Otherwise, we have k<c0, and hence, by bn⟶0, δn⟶0 as n⟶∞ and un is bounded in H01Ω; one has for large n,
(99)c0≤I0,0un=Ibn,δnun+bn4un4+δn4un44=cbn,δn+bn4un4+δn4un44≤k+c0−k2=c0+k2<c0,a contradiction. Thus, the claim follows. Then,
(100)c0=limn⟶∞Ibn,δnun=a2u¯2−λqu¯qq=I0,0u¯,which implies that u¯ is a global minimum of I0,0. This result, together with strong maximum principle proves that u¯ is a positive ground state solution of P0,0. Theorem 3 is thus proved.



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*Source: 1011342-2022-06-02.xml*