Sobolev exponent and Kirchhoff-type nonlocal term

Multiplicity of positive solutions for a class of singular elliptic equations with critical

Sobolev exponent and Kirchhoff-type nonlocal term

Jiu Liu

, Ai-Jun Hou

and Jia-Feng Liao

1School of Mathematics and Statistics, Qiannan Normal University for Nationalities, Duyun, Guizhou, 558000, P. R. China

2School of Mathematics and Information, China West Normal University, Nanchong, Sichuan, 637002, P. R. China

3College of Mathematics Education, China West Normal University, Nanchong, Sichuan, 637002, P. R. China

Received 11 June 2018, appeared 19 December 2018 Communicated by Patrizia Pucci

Abstract. We study a class of singular elliptic equations involving critical Sobolev expo- nent and Kirchhoff-type nonlocal term− a+bR

Ω|∇u|²dx

∆u =u⁵+g(x,u) +λu^−γ, x ∈ _Ω,u > 0, x ∈ _Ω, u = 0, x ∈ _{∂Ω, where Ω} ⊂ _R³ is a bounded domain, a,b,λ>0, 0<γ<1 andg∈C(Ω×R)satisfies some conditions. By the perturbation method, variational method and some analysis techniques, we establish a multiplicity theorem.

Keywords: singular elliptic equation, Kirchhoff-type nonlocal term, critical Sobolev exponent, positive solutions, perturbation method.

2010 Mathematics Subject Classification: 35A15, 35B09, 35B33, 35J75.

1 Introduction

In this paper, we consider the following singular elliptic equation with critical Sobolev expo- nent and Kirchhoff-type nonlocal term











−

a+b Z

Ω|∇u|²dx

∆u= u⁵+g(x,u) +λu^−γ, x∈ _Ω,

u>0, x∈ _Ω,

u=0, x∈ ∂Ω,

(1.1)

where Ω ⊂ _R³ is a bounded domain, a,b,λ > 0, 0 < γ < 1 and 2^∗ = 6 is the well-known critical Sobolev exponent. The nonlinear term g(x,s) : Ω×_R → _R satisfies the following conditions.

BCorresponding author. Email: liaojiafeng@163.com

(g₁) g∈C(_Ω×_R),g(x,s)≥0 ifs ≥0 andg(x,s) =0 ifs≤0 for allx ∈_Ω.

(g₂) _lims→0+ g(x,s)

s =_{0 and lims→+∞} ^g(x,s)

s⁵ =0 uniformly for allx∈ _Ω.

(g₃) There exists 0 < κ a such that g(x,s)s−4G(x,s) ≥ −(a−κ)λ₁s² for all x ∈ _Ωand s≥0, whereG(x,s) =Rs

0 g(x,t)dtandλ₁ >0 is the first eigenvalue of the operator−_∆.

(g₄) There exists a nonempty open set ω ⊂ _Ω with 0 ∈ ω such that lims→+_∞ ^g(x,s)

s³ = +_∞ uniformly forx∈ ω.

Because of the presence of the termbR

Ω|∇u|²dx, which implies that the equation is no longer a pointwise identity, problem (1.1) is called the nonlocal problem. This phenomenon provokes some mathematical difficulties, which makes the study of such a class of problem particularly interesting. Its physical motivation about the operator R

Ω|∇u|²dx

∆u, which appears in the Kirchhoff equation. Thus, problem (1.1) is always called Kirchhoff-type problem. The Kirchhoff equation is related to the following stationary analogue of the equation





 u_tt−

a+b

Z

Ω|∇u|²dx

∆u= f(x,u), x∈_Ω,

u=_0, x∈∂Ω,

(1.2)

proposed by Kirchhoff [14] in 1883 as an extension of the classical D’Alembert’s wave equation for free vibration of elastic strings. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. In problem (1.2),udenotes the displacement, f(x,u)the external force andbthe initial tension while ais related to the intrinsic properties of the string (such as Young’s modulus). It is worth pointing out that problem (1.2) received much attention only after the work of Lions [23] where a function analysis framework was proposed to the problem. After that, the Kirchhoff-type problem has been extensively investi- gated, for examples [1,4,9–13,15,17–22,24,25,27–37].

To our best knowledge, the pioneer work on the Kirchhoff-type problem with critical Sobolev exponent is Alves, Corrêa and Figueiredo [1], they considered the following criti- cal Sobolev exponent problem







−

M _Z

Ω|∇u|²dx

∆u=u⁵+λf(x,u), x∈ _Ω,

u=0, x∈ ∂Ω,

(1.3)

whereΩ⊂_R³,M :R⁺→_R⁺,f :Ω×_R→_Rare continuous functions,F(x,u) =Ru

0 f(x,s)ds is superquadratic at the origin and subcritical at infinity. By using the variational method, under appropriate conditions, they obtained that problem (1.3) has a positive solution for all λ > 0 large enough. After that, the Kirchhoff-type problem with critical exponent has been extensively studied, and some important and interesting results have been obtained, see [4,8–13,15–21,24,27–29,33–37].

Particularly, Lei, Liao and Tang [16] studied the following singular Kirchhoff-type problem with critical exponent











−

a+b Z

Ω|∇u|²dx

∆u=u⁵+λu^−γ, x∈ _Ω,

u>_0, x∈ _Ω,

u=0, x∈ ∂Ω,

(1.4)

using the variational method and perturbation method, they obtained two positive solutions for problem (1.4) when λ > 0 small. After that, Liu et al. generalized [16] to R⁴ with the following equation











−

a+b Z

Ω|∇u|²dx

∆u=µu³+ ^λ

|x|^βu^γ, x ∈_Ω,

u>0, x ∈_Ω,

u=_0, x ∈∂Ω,

(1.5)

whereΩ⊂_R⁴a bounded smooth domain andλ,µ>0, 0≤β<3, see [24]. When 0<γ< ¹₂ and 2(₁+γ) < β < 3, by the same methods in [16], they also got two positive solutions for problem (1.4) whenµ> bS²andλ>0 small, whereSis the best Sobolev constant inR⁴.

Based on [16] and [24], the mountain-pass level value is the most obstacle in proving the existence of the second solution of problem (1.4). This obstacle stems from the local term bR

Ω|∇u|²dx, which shows that the difference between the classic elliptic problem(that is, b = 0) and the Kirchhoff-type problem. In this paper, we give another way to overcome this obstacle. We add a supperlinear term g(x,u) in problem (1.4), that is problem (1.1).

Combining with the perturbation method and variational method, we obtain two positive solutions for problem (1.1).

For allu∈ H₀¹(_Ω), we define I(u) = ^a

2kuk²+^b

4kuk⁴− ¹ 6

Z

Ω(u⁺)⁶dx−

Z

ΩG(x,u⁺)dx− ^λ 1−γ

Z

Ω(u⁺)^1−γdx, where G(x,u) = Ru

0 g(x,s)ds and H₀¹(_Ω) is a Sobolev space equipped with the norm kuk = R

Ω|∇u|²dx¹₂

. Obviously, the energy functional I does not belong to C¹(H₀¹(_Ω),R). Note that a functionuis called a weak solution of problem (1.1) ifu∈ H¹₀(_Ω)such that

a+bkuk²

Z

Ω(∇u,∇ϕ)dx−

Z

Ω(u⁺)⁵ϕdx−

Z

Ωg(x,u⁺)ϕdx−λ Z

Ω(u⁺)^−γϕdx=0, (1.6) for all ϕ∈ H₀¹(_Ω).

LetSbe the best Sobolev constant, namely S:= inf

u∈H¹₀(_Ω)\{0}

R

Ω|∇u|²dx R

Ω|u|⁶dx¹₃. (1.7)

Our main results can be described as follows.

Theorem 1.1. Assume that a,b,λ>_{0, 0}<γ<₁and g satisfies(g₁)–(g₄)_,then there existsΛ>₀ such that problem(1.1)possesses two positive solutions for all0< λ<_Λ.

Remark 1.2. To the best of our knowledge, our result is up to date. As we known, [16] is the first paper which considered the singular Kirchhoff-type problem with critical exponent, that is, problem (1.4). However, there exists a small gap in the proof of the second positive solution, that is, the estimation ofB(tεvε)in Page 533 of [16]. Indeed, when using the inequality of (3.14) in Page 532 of [16] to estimateB(t_εv_ε), they need check _t^α

εvε is small enough for|x| ≤ε

1−γ

16 and

εsmall. However, it maybe is not true. Obviously, if |x| ≤ ε

1−γ

16 and ε → 0, for some C > 0, one has _t^α

εvε ≥ Cα^(ε+|x|2)

12

ε

14

, which does not implies that _t^α

εvε is small. So far there is no way to

correct it. In [24], the authors avoided the similar question by multiplying |x|^−β in front of the singular term _u^λγ, see problem (1.5). In here, in order to arrive at the same effect, we add a continuous subcritical functiongin the right hand side of equation (1.4).

Comparing with Theorem 1.1 in [28], our problem (1.1) is a singular perturbing problem of that paper. Thanks to this perturbation, we get another solution. Moreover, our condition (g₃)is more general than the following condition(g⁰₃)in [28],

(g⁰₃) There exists a constant θ ∈ (4, 6) such that g(x,s)s−θG(x,s) ≥ 0 for all x ∈ _Ω and s≥0, whereG(x,s) =Rs

0 g(x,t)dt.

We should point out that [28] generalized a part of Brézis–Nirenberg’s result in [7] to the Kirchhoff-type problem.

Our condition(g₄)is first given by [7], which is used to estimate the level of the mountain- pass value. Thanks to(g₄), we obtain the second positive solution of problem (1.1).

In view of the typical power nonlinearities, Theorem1.1allows us to ensure the following corollary.

Corollary 1.3. Assume that a,b,λ > _{0, 0} < γ < _{1, 4} < p < ₆ and g(x,u) = u^p−1, then there existsΛ˜ >0such that problem(1.1)possesses two positive solutions for all0<λ<_Λ.^˜

Remark 1.4. Wheng(x,u) =u^p−1(4< p<6), then clearlygsatisfies(g₁)–(g₄). For the proof, we can consider instead withg(x,u) = (u⁺)^p−1.

This paper is organized as following: in Section 2, we consider an auxiliary problem, and in Section 3, we give the proof of Theorem 1.1. For the convenience of writing, we denote C,C₁,C₂, . . . as various positive constants in the following.

2 The auxiliary problem

In order to overcome the difficulty of the singular term, for everye>0, we study the following perturbation problem







−

a+b Z

Ω|∇u|²dx

∆u= (u⁺)⁵+g(x,u) +λ(u⁺+e)^−γ, x∈ _Ω,

u=0, x∈ _∂Ω,

(2.1)

whereu⁺ =max{u, 0}. The energy functional corresponding to problem (2.1) is I_e(u) = ^a

2kuk²+^b

4kuk⁴− ¹ 6

Z

Ω(u⁺)⁶dx−

Z

ΩG(x,u⁺)dx− ^λ 1−γ

Z

Ω

h

(u⁺+e)^1−γ−e^1−γ i

dx,

Obviously, the energy functional Ieis of classC¹on H₀¹(_Ω). As well known that there exists a one-to-one correspondence between all solutions of problem (2.1) and the critical points of I_e onH₀¹(_Ω). We mean a functionuis called a weak solution of problem (2.1) ifu∈ H₀¹(_Ω)_such that

a+bkuk²

Z

Ω(∇u,∇ϕ)dx−

Z

Ω(u⁺)⁵ϕdx−

Z

Ωg(x,u⁺)ϕdx−λ Z

Ω

ϕ

(u⁺+e)^γdx=0, (2.2) for all ϕ∈ H₀¹(_Ω).

First, we prove that I_e satisfies the local(PS)_ccondition.

Lemma 2.1. Suppose that a,b,λ > 0, 0 < γ < 1 and g satisfies (g₁)−(g₃),then I_e satisfies the (PS)_ccondition, where c<_Θ−Dλ^1+2γ with D=D(γ,S,κ,Ω)is a positive constant and

Θ= ^abS

3

4 + ^b

3S⁶ 24 + ^aS

√b²S⁴+4aS

6 + ^b

2S⁴√

b²S⁴+4aS

24 .

Proof. Suppose that{u_n}is a(PS)_c sequence forc∈(0,Θ−Dλ^1+2γ), that is,

Ie(un)→c, I_e⁰(un)→0, (2.3) asn→+∞. We claim that{u_n}is bounded in H₀¹(_Ω). In fact, from(g₁)_and(g₂), there exists C₀ >0 such that

1

5g(x,s)s−G(x,s)

≤ ¹

30|s|⁶+C0. (2.4)

Note that the subadditivity oft^1−γ, one has

(u⁺_n +e)^1−γ−e^1−γ ≤(u⁺_n)^1−γ. (2.5) Consequently, combining with the Sobolev inequality, it follows from (2.3) and (2.5) that

1+c+o(1)kunk

≥ Ie(un)− ¹

5hI_e⁰(un),uni

= ^3a

10kunk²+ ^b

20kunk⁴+

Z

Ω

1

5g(x,u⁺_n)u⁺_n −G(x,u⁺_n)

dx

+ ¹ 30

Z

Ω(u⁺_n)⁶dx− ^λ 1−γ

Z

Ω

h

(u⁺_n +e)^1−γ−e^1−γ i

dx− ^λ 5e^γ

Z

Ωu⁻_ndx

≥ ^3a

10ku_nk²+ ^b

20ku_nk⁴− ^λ 1−γ

Z

Ω(u⁺_n)^1−γdx− ^λ 5e^γ

Z

Ωu⁻_ndx−C₀|_Ω|,

≥ ^3a

10kunk²+ ^b

20kunk⁴−Ckunk^1−γ−C₁kunk −C₀|_Ω|,

(2.6)

since 0 < γ < 1, which implies that {u_n} is bounded in H₀¹(_Ω). Going if necessary to a subsequence, still denoted by{un}, there existsu∈ H¹₀(_Ω)such that















un*u, weakly in H₀¹(_Ω),

u_n→u, strongly in L^s(_Ω), 1≤ s<6, un(x)→u(x), a.e. in Ω,

there existsk∈ L¹(_Ω)such that for alln, |u_n(x)| ≤k(x)a.e. in Ω.

(2.7)

For everye>0, since

|u|

(u⁺_n +e)^γ ≤ |u| e^γ, by the dominated convergence theorem and (2.7), one has

nlim→_∞ Z

Ω(u⁺_n +e)^−γudx=

Z

Ω(u⁺+e)^−γudx. (2.8)

Moreover, for everye>0, by (2.7), one gets

u_n (u⁺_n +e)^γ

≤

u_n (u⁺_n +e)^γ

≤ |u_n| e^γ

≤ ¹ e^γk(x).

Therefore, it follows from the dominated convergence theorem that

nlim→_∞ Z

Ω(u⁺_n +e)^−γu_ndx=

Z

Ω(u⁺+e)^−γudx. (2.9) From (2.7), one also has

Z

Ω|∇u_n|²dx =

Z

Ω|∇w_n|²dx+

Z

Ω|∇u|²dx+o(1), (2.10) _Z

Ω|∇u_n|²dx 2

=kw_nk⁴+kuk⁴+2kw_nk²kuk²+o(1). (2.11) By(g₂)and (2.7), one has

Z

Ωg(x,u⁺_n)udx=

Z

Ωg(x,u⁺)udx+o(1), (2.12) Z

Ωg(x,u⁺_n)u_ndx=

Z

Ωg(x,u⁺)udx+o(1), (2.13) Z

ΩG(x,u⁺_n)dx=

Z

ΩG(x,u⁺)dx+o(1). (2.14) As usually, lettingwn=un−u, we need prove thatkwnk →0 asn →_{∞. Let limn}→_∞kwnk= l≥0. Ifl=0, our conclusion is true. Suppose thatl>0. By the Brézis–Lieb Lemma (see [6]),

one has _Z

Ω(u⁺_n)⁶dx=

Z

Ω(w⁺_n)⁶dx+

Z

Ω(u⁺)⁶dx+o(1). (2.15) From (2.3), (2.7), (2.9) and (2.13), one obtains

akunk²+bkunk⁴−

Z

Ω(u⁺_n)⁶dx−

Z

Ωg(x,u⁺)udx−λ Z

Ω(u⁺+e)^−γudx= o(1), consequently, it follows from (2.10), (2.11) and (2.15) that

akuk²+akwnk²+bkuk⁴+bkwnk⁴+2bkwnk²kuk²

−

Z

Ω(w⁺_n)⁶dx−

Z

Ω(u⁺)⁶dx−

Z

Ωg(x,u⁺)udx−λ Z

Ω(u⁺+e)^−γudx=o(1). (2.16) It follows from (2.3), (2.8) and (2.13) that

0= lim

n→_∞hI_e⁰(u_n),ui

=akuk²+bkuk⁴+bl²kuk²−

Z

Ω(u⁺)⁶dx−

Z

Ωg(x,u⁺)udx−λ Z

Ω(u⁺+e)^−γudx. (2.17) Moreover, by (2.3), for anyϕ∈ H₀¹(_Ω), one has limn→_∞hI_e⁰(un),ϕi=0, that is,

(a+bd)

Z

Ω(∇u,∇ϕ)dx−

Z

Ω(u⁺)⁵ϕdx−

Z

Ωg(x,u⁺)ϕdx−λ Z

Ω(u⁺+e)^−γϕdx=0, (2.18)

whered= ku_nk²+o(1)is a positive constant. Particularly, choosing ϕ=u⁻in (2.18), one has u⁻=0. Thus, we haveu≥0 inΩ. On the one hand, from (2.5), (2.17) and(g3), by the Hölder inequality, Sobolev inequality and Poincaré inequality, we have

I_e(u) = ^a

2kuk²+ ^b

4kuk⁴−¹ 6

Z

Ω(u⁺)⁶dx−

Z

ΩG(x,u)dx

− ^λ 1−γ

Z

Ω

h

(u⁺+e)^1−γ−e^1−γ i

dx

= ^a

4kuk²+ ¹ 12

Z

Ω(u⁺)⁶dx+

Z

Ω

1

4g(x,u⁺)u−G(x,u⁺)

dx

− ^λ 1−γ

Z

Ω

h

(u⁺+e)^1−γ−(u⁺)^1−γidx+ ^λ 4

Z

Ω(u⁺+e)^−γudx− ^bl

2

4 kuk²

≥ ^a

4kuk²− ^λ1(a−κ) 4

Z

Ωu²dx− ^λ 1−γ

Z

Ωu^1−γdx−^bl

2

4 kuk²

≥ ^κ

4kuk²− ^λ

1−γ|_Ω|^5+6γS^−1−2γkuk^1−γ− ^bl2 4 kuk²

≥ −Dλ^1+2γ − ^bl2 4 kuk²,

(2.19)

where the last inequality is obtained by the Young inequality and D= ¹+γ

1−γ2^−1+2γγ(κS)^{−11−+γγ}|_Ω|^3(51++γγ). On the other hand, it follows from (2.14), (2.16) and (2.17) that

akw_nk²+bkw_nk⁴+bkw_nk²kuk²−

Z

Ω(w⁺_n)⁶dx=o(1), (2.20) and

I_e(u_n) = I_e(u) + ^a

2kw_nk²+^b

4kw_nk⁴+ ^b

2kw_nk²kuk²− ¹ 6

Z

Ω(w⁺_n)⁶dx+o(1). (2.21) From (1.7), one has

Z

Ω(w⁺_n)⁶dx≤

Z

Ω|wn|⁶dx≤ kwnk⁶ S³ , consequently, it follows from (2.20) that

al²+bl⁴+bl²kuk²≤ ^l

6

S³, which implies that

l² ≥ ^bS3+^pb²S⁶+4S³(a+bkuk²)

2 . (2.22)

Thus, from (2.20)–(2.22), we obtain I_e(u) = lim

n→_∞

I_e(u_n)− ^a

2kw_nk²− ^b

4kw_nk⁴− ^b

2kw_nk²kuk²+ ¹ 6

Z

Ω(w⁺_n)⁶dx

= c− a

3l²+ ^b 12l⁴+ ^b

3l²kuk²

≤ c− a

bS³+^pb²S⁶+4S³(a+bkuk²) 6

+ ^b 48

bS³+

q

b²S⁶+4S³(a+bkuk²) 2

+^bkuk² 6

bS³+

q

b²S⁶+4S³(a+bkuk²) − ^bl

2

4 kuk²

≤ c− ^abS

3

4 + ^b

3S⁶ 24 +^aS

√b²S⁴+4aS

6 +^b

2S⁴√

b²S⁴+4aS 24

!

−^bl

2

4 kuk²

< −Dλ^1+2γ −^bl

2

4 kuk²,

which contradicts (2.19). Hence, l ≡ 0, that is, u_n → u in H¹₀(_Ω) as n → ∞. Therefore, I_e satisfies the(PS)_ccondition for allc<_Θ−Dλ^1+2γ. This completes the proof of Lemma2.1.

As well known, the function

U(x) = (_3ε²)¹⁴ (ε²+|x|²)¹²

, x∈_R³, (2.23)

is an extremal function for the minimum problem (1.7), that is, it is a positive solution of the following problem

−_∆u=u⁵, ∀x∈_R³.

Now, we estimate the level value of functionalIeand obtain the following lemma.

Lemma 2.2. Assume that a,b,λ>0, 0< γ<1and g satisfies(g₁),(g₂)and(g₄),then there exists u₀ ∈ H₀¹(_Ω), such thatsup_t≥0Ie(tu₀)< _Θ−Dλ^1+2γ for all0<λ<λ^∗,whereΘand D are defined by Lemma2.1and the positive constantλ^∗ is independent of u₀ande.

Proof. Define a cut-off functionη ∈ C₀^∞(_Ω)such that 0≤ η ≤ 1, |∇η| ≤C₁. For some δ > 0, we define

η(x) =

(1, |x| ≤δ, 0, |x| ≥2δ,

and{x:|x| ≤2δ} ⊂ω, whereωis defined by(g₄). Setu_ε =η(x)U(x), whereU(x)is defined by (2.23). As well known (see [7,26]), one has

ku_εk²=kUk²+O(ε) =S³² +O(ε)_{, (2.24)}

|u_ε|⁶₆=|U|⁶₆+O(ε³) =S³² +O(ε³)_{, (2.25)}













 C₂ε^2s ≤

Z

Ωu^s_εdx≤C₃ε

s

2, 1≤s<3, C₄ε^2s|lnε| ≤

Z

Ωu^s_εdx≤C₅ε

s

2|lnε|, s=3, C₆ε

6−s

2 ≤

Z

Ωu^s_εdx≤C₇ε

6−s

2 , 3<s<6.

(2.26)

Moreover, from [35], we also have















ku_εk⁴ =S³+O(ε); kuεk⁶ =S⁹² +O(ε); kuεk⁸ =_S⁶+_O(ε)_; ku_εk¹² =S⁹+O(ε).

(2.27)

For allt ≥0, we defineIe(tuε)by Ie(tuε) = ^at

2

2 kuεk²+ ^bt

4

4 kuεk⁴−^t

6

6 Z

Ωu⁶_εdx−

Z

ΩG(x,tuε)dx

− ^λ 1−γ

Z

Ω

h

(tu_ε+e)^1−γ−e^1−γ i

dx,

from(g₂), we have

tlim→+0Ie(tuε) =0, uniformly for all 0<ε<ε₀, and

t→+lim_∞I_e(tu_ε) =−_∞, uniformly for all 0< ε<ε₀,

where ε₀ > 0 is a small constant. Thus sup_t≥0Ie(tuε) attains for some tε > 0. Using the following conclusion of Step 1in the proof of Theorem 2.3, one has Ie(tεuε) > ρ > 0. So, by the continuity of I_e, there exist two constants t₀,T₀ > 0, which independent of ε, such that t0 <tε <T0. Set Ie(tuε) =I_ε,1(t)−Iε,2(t)−Iε,3(t), where

I_ε,1(t) = ^a

2t²ku_εk²+^b

4t⁴ku_εk⁴−^t6 6

Z

Ωu⁶_εdx, and

I_ε,2(t) =

Z

ΩG(x,tu_ε)dx, I_ε,3(t) = ^λ

1−γ Z

Ω

h

(tuε+e)^1−γ−e^1−γ i

dx.

First, we estimate the value of I_ε,1. SinceI_ε,1⁰ (t) =atku_εk²+bt³ku_εk⁴−t⁵ Z

Ωu⁶_εdx, letI_ε,1⁰ (t) =0, that is,

aku_εk²+bt²ku_εk⁴−t⁴ Z

Ωu⁶_εdx=0, (2.28)

one obtains

T_ε²= ^b

kuεk⁴+^q_b²kuεk⁸+_4akuεk²R

Ωu⁶_εdx 2R

Ωu⁶_εdx .

Then I_ε,1⁰ (t)> 0 for all 0< t < T_ε andI_ε,1⁰ (t)< _{0 for all}t > T_ε, so I_ε,1(t)attains its maximum

atT_ε. Thus, it follows from (2.24), (2.25), (2.27) and (2.28) that I_ε,1(t)≤ I_ε,1(T_ε)

= T_ε² a

2kuεk²+ ^b

4T_ε²kuεk⁴− ^T

4 ε

6 Z

Ωu⁶_εdx

= T_ε² a

3ku_εk²+ ^b

12T_ε²ku_εk⁴

=

abku_εk⁶+aku_εk^2qb²ku_εk⁸+4aku_εk²R

Ωu⁶_εdx 6R

Ωu⁶_εdx

+^b

3ku_εk¹²+2abku_εk⁶R

Ωu⁶_εdx 24 R

Ωu⁶_εdx2

+

b²ku_εk^4qb²ku_εk⁸+4aku_εk²R

Ωu⁶_εdx 24 R

Ωu⁶_εdx2

= ^abkuεk⁶ 4R

Ωu⁶_εdx+ ^b

3kuεk¹² 24(R

Ωu⁶_εdx)² +

aku_εk^2qb²ku_εk⁸+4aku_εk²R

Ωu⁶_εdx 6R

Ωu⁶_εdx

+^b

2kuεk^4qb²kuεk⁸+4akuεk²R

Ωu⁶_εdx 24 R

Ωu⁶_εdx2

= ^ab(S⁹² +O(ε)) 4(S³² +o(ε)) + ^b

3(S⁹+O(ε)) 24(S³² +o(ε))² + ^a(S³² +O(ε))^pb²S⁶+4aS³+O(ε)

6(S³² +o(ε)) +^b

2(S⁶+O(ε))^pb²S⁶+4aS³+O(ε) 24(S³² +o(ε))²

≤ ^abS

3

4 + ^b

3S⁶ 24 + ^aS

√

b²S⁴+4aS

6 + ^b

2S⁴√

b²S⁴+4aS 24 +C₈ε

=_Θ+C₈ε.

(2.29)

Second, we estimate the value of I_ε,2. We claim that

lim

ε→0⁺

R

ΩG(x,t_εu_ε)dx

ε = +_∞. (2.30)

Letm(t) =inf_x∈wg(x,t), from(g₁)and(g₄), we have

g(x,t)≥m(t)≥0, lim

t→+_∞

m(t)

t³ = +_∞,

for almost x ∈ ω and t > 0. Consequently, for any µ > 0, there exists A > 0 such that

M(t)≥µt⁴ for allt ≥ A, whereM(t) =Rt

0 m(s)ds. Thus, one has R

ΩG(x,t_εu_ε)dx ε ≥ε⁻¹

Z

|x|<δ

G(x,tεuε)dx

≥ε⁻¹ Z

|x|<δ

M(t_εu_ε)dx

=ε⁻¹ Z _δ

0 M

"

t_ε3¹⁴ε¹² (ε²+r²)¹²

# r²dr

=ε² Z _δε⁻¹

0 M

"

tε3¹⁴ε⁻¹² (1+r²)¹²

# r²dr

=ε² Z _ε⁻¹

0 M

"

tε3¹⁴ε⁻

1 2

(1+r²)¹²

#

r²dr−ε² Z _ε⁻¹

δε⁻¹

M

"

tε3¹⁴ε⁻

1 2

(1+r²)¹²

# r²dr.

(2.31)

Since m(t) > 0 for all t > 0, we obtain M(t) is increasing for all t > 0. From (g₂), one has M(t)≤Ct²for all t>0 small enough. Consequently, one gets

ε² Z _ε⁻¹

δε⁻¹

M

"

tε3¹⁴ε⁻

1 2

(1+r²)¹²

# r²dr

≤Cε⁻¹M(tε3¹⁴ε

12)≤Cε⁻¹M(T03¹⁴ε

12)≤C, (2.32)

for all ε > 0 small enough. Fixing A, there exists B > 0 such that ^tε3

14ε⁻

12

(₁+r²)¹2

≥ A for all 1<r <Bε⁻¹². Therefore, one obtains

lim inf

ε→0⁺ ε² Z _ε⁻¹

0 M

"

tε3¹⁴ε⁻¹² (1+r²)¹²

#

r²dr≥lim inf

ε→0⁺ ε² Z _Bε⁻

12

1 M

"

tε3¹⁴ε⁻¹² (1+r²)¹²

# r²dr

≥lim inf

ε→0⁺ Cµε² Z _Bε⁻

12

1

ε⁻²r² (1+r²)^2dr

=

Z _+∞

1

ε⁻²r² (1+r²)^2dr

= +_∞.

(2.33)

According to (2.31)–(2.33), (2.30) is obtained. Finally, we estimate the value of I_ε,3. From (2.26), since 0< t0 <tε <T0, one has

I_ε,3(t_ε) = ^λ 1−γ

Z

Ω

h

(t_εu_ε+e)^1−γ−e^1−γ i

dx

≥0.

(2.34)

Thus, from (2.29), (2.30) and (2.34), there exists a large enough positive constantC₉ >C₈such that

Iε(tuε) = I_ε,1(t)−Iε,2(t)−Iε,3(t)

≤ Iε,1(_tε)−Iε,2(_tε)−Iε,3(_tε)

<_Θ+C₈ε−C₉ε

<_Θ−(C₉−C₈)ε

≤_Θ−Dλ^1+2γ,

provided that ε > 0 small enough and 0 < λ < ^(C9−_D^C8)ε

1+γ

2 . Thus there exists λ^∗ =

(C9−C8)ε D

¹⁺₂^γ

> 0, choosing u₀ = uε, such that Ie(tu₀) < _Θ−Dλ^1+2γ for all 0 < λ < λ^∗. This completes the proof of Lemma2.2.

Therefore, we can obtain the following conclusion for problem (2.1).

Theorem 2.3. Assume that a,b,λ>0, 0<γ<1and g satisfies(g₁)–(g₄),then there existsΛ>0 such that problem(2.1)possesses two positive solutions for all0<λ<_Λand everye>0. Moreover, one of the solutions is a positive ground state solution.

Proof. We divide three steps to prove Theorem2.3.

Step 1. We prove that there exists a positive local minimizer solution of problem(2.1).

First, we claim that there exist λ∗ > 0 and R,ρ > 0 such that Ie(u)|_u∈SR ≥ ρ and inf_u∈BRI_e(u) < 0 for λ ∈ (0,λ∗), where S_R = {u ∈ H₀¹(_Ω) : kuk = R}, B_R = {u ∈ H₀¹(_Ω) : kuk ≤R}. In fact, by (g₁)_and(g₂), we infer that

|G(x,s)| ≤ ^aλ1

4 |s|²+C₁₀|s|⁶,

for allx∈ _Ωands∈R. Consequently, by the Hölder inequality and (1.7) and (2.5), we have Ie(u) = ^a

2kuk²+ ^b

4kuk⁴−¹ 6

Z

Ω(u⁺)⁶dx−

Z

ΩG(x,u⁺)dx

− ^λ 1−γ

Z

Ω

h

(u⁺+e)^1−γ−e^1−γ i

dx

≥ ^a

4kuk²+ ^b

4kuk⁴−C₁₁kuk⁶− ^λ|_Ω|^5+6γ (1−γ)S^1−2γ

kuk^1−γ

≥ ^a

4kuk²−C₁₁kuk⁶− ^λ|_Ω|^5+6γ (1−γ)S^1−2γ

kuk^1−γ

= kuk^1−γ 4

akuk^1+γ−C₁₂kuk^5+γ−Bλ

(2.35)

whereC₁₂ =4C₁₁andB= ^4|Ω|

5+_γ 6

(1−γ)S^1−2γ

. Let

h(t) =at^1+γ−C₁₂t^5+γ−Bλ, ∀t∈[0,+_∞). Then we haveh⁰(t) =t^γ[a(1+γ)−C₁₂(5+γ)t⁴]. Leth⁰(t) =0, one has

t_max=

a(₁+γ) C₁₂(5+γ)

¹₄

, max

t≥0 h(t) =h(t_max) = ^4a 5+γ

a(₁+γ) C₁₂(5+γ)

¹⁺₄^γ

−Bλ.

Therefore, choosing λ∗ = ^4a

B(5+γ)

h a(1+γ) C12(5+γ)

i¹⁺₄^γ

and R = t_max, according to (2.35), there exists ρ>0 such that Ie(u)|_u∈SR ≥ρfor all 0<λ<λ∗. By(g₂), foru∈ H₀¹(_Ω)withu⁺ >0 it holds

tlim→0⁺

Ie(tu)

t =− ^λ

1−γ lim

t→0⁺

1 t

Z

Ω[(tu⁺+e)^1−γ−e^1−γ]dx

=− ^λ 1−γ lim

t→0⁺

Z

Ω

(1−γ)ξ^−γtu⁺

t dx (e<ξ <tu⁺+e)

=−λ Z

Ω

u⁺

e^γdx (ast→0⁺, ξ →e)

<0.