Electronic Journal of Qualitative Theory of Differential Equations2018, No. 100, 1–20; https://doi.org/10.14232/ejqtde.2018.1.100 www.math.u-szeged.hu/ejqtde/

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

Sobolev exponent and Kirchhoff-type nonlocal term

Jiu Liu1, Ai-Jun Hou2 and Jia-Feng LiaoB 2, 3

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 + b

∫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 and g ∈ C(Ω×R) satisfies some conditions. By the perturbationmethod, variational method and some analysis techniques, we establish a multiplicitytheorem.

Keywords: singular elliptic equation, Kirchhoff-type nonlocal term, critical Sobolevexponent, 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∫

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-knowncritical Sobolev exponent. The nonlinear term g(x, s) : Ω ×R → R satisfies the followingconditions.BCorresponding author. Email: liaojiafeng@163.com

2 J. Liu, A. J. Hou and J. F. Liao

(g1) g ∈ C(Ω×R), g(x, s) ≥ 0 if s ≥ 0 and g(x, s) = 0 if s ≤ 0 for all x ∈ Ω.

(g2) lims→0+g(x,s)

s = 0 and lims→+∞g(x,s)

s5 = 0 uniformly for all x ∈ Ω.

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

∫ s0 g(x, t)dt and λ1 > 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 for x ∈ ω.

Because of the presence of the term b∫

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

(∫Ω |∇u|2dx

)∆u, which appears in

the Kirchhoff equation. Thus, problem (1.1) is always called Kirchhoff-type problem. TheKirchhoff equation is related to the following stationary analogue of the equationutt −

(a + b

∫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 equationfor free vibration of elastic strings. Kirchhoff’s model takes into account the changes in lengthof the string produced by transverse vibrations. In problem (1.2), u denotes the displacement,f (x, u) the external force and b the initial tension while a is related to the intrinsic propertiesof the string (such as Young’s modulus). It is worth pointing out that problem (1.2) receivedmuch attention only after the work of Lions [23] where a function analysis framework wasproposed 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 criticalSobolev exponent is Alves, Corrêa and Figueiredo [1], they considered the following criti-cal Sobolev exponent problem−

[M(∫

Ω|∇u|2dx

)]∆u = u5 + λ f (x, u), x ∈ Ω,

u = 0, x ∈ ∂Ω,(1.3)

where Ω ⊂ R3, M : R+ → R+, f : Ω×R→ R are continuous functions, F(x, u) =∫ u

0 f (x, s)dsis 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 forall λ > 0 large enough. After that, the Kirchhoff-type problem with critical exponent hasbeen 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 problemwith critical exponent

−(

a + b∫

Ω|∇u|2dx

)∆u = u5 + λu−γ, x ∈ Ω,

u > 0, x ∈ Ω,

u = 0, x ∈ ∂Ω,

(1.4)

Singular elliptic equations with critical exponent and Kirchhoff term 3

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

−(

a + b∫

Ω|∇u|2dx

)∆u = µu3 + λ

|x|βuγ , x ∈ Ω,

u > 0, x ∈ Ω,

u = 0, x ∈ ∂Ω,

(1.5)

where Ω ⊂ R4 a bounded smooth domain and λ, 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 forproblem (1.4) when µ > bS2 and λ > 0 small, where S is the best Sobolev constant in R4.

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

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 overcomethis 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 positivesolutions for problem (1.1).

For all u ∈ H10(Ω), we define

I(u) =a2‖u‖2 +

b4‖u‖4 − 1

6

∫Ω(u+)6dx−

∫Ω

G(x, u+)dx− λ

1− γ

∫Ω(u+)1−γdx,

where G(x, u) =∫ u

0 g(x, s)ds and H10(Ω) is a Sobolev space equipped with the norm ‖u‖ =(∫

Ω |∇u|2dx) 1

2 . Obviously, the energy functional I does not belong to C1(H10(Ω), R). Note

that a function u is called a weak solution of problem (1.1) if u ∈ H10(Ω) such that(

a + b‖u‖2) ∫Ω(∇u,∇ϕ)dx−

∫Ω(u+)5ϕdx−

∫Ω

g(x, u+)ϕdx− λ∫

Ω(u+)−γ ϕdx = 0, (1.6)

for all ϕ ∈ H10(Ω).

Let S be the best Sobolev constant, namely

S := infu∈H1

0 (Ω)\0

∫Ω |∇u|2dx(∫Ω |u|6dx

) 13

. (1.7)

Our main results can be described as follows.

Theorem 1.1. Assume that a, b, λ > 0, 0 < γ < 1 and g satisfies (g1)–(g4), then there exists Λ > 0such that problem (1.1) possesses two positive solutions for all 0 < λ < Λ.

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

tεvεis small enough for |x| ≤ ε

1−γ16 and

ε small. However, it maybe is not true. Obviously, if |x| ≤ ε1−γ16 and ε → 0, for some C > 0,

one has αtεvε≥ Cα (ε+|x|2)

12

ε14

, which does not implies that αtεvε

is small. So far there is no way to

4 J. Liu, A. J. Hou and J. F. Liao

correct it. In [24], the authors avoided the similar question by multiplying |x|−β in front ofthe singular term λ

uγ , see problem (1.5). In here, in order to arrive at the same effect, we add acontinuous subcritical function g in the right hand side of equation (1.4).

Comparing with Theorem 1.1 in [28], our problem (1.1) is a singular perturbing problemof that paper. Thanks to this perturbation, we get another solution. Moreover, our condition(g3) is more general than the following condition (g′3) in [28],

(g′3) There exists a constant θ ∈ (4, 6) such that g(x, s)s − θG(x, s) ≥ 0 for all x ∈ Ω ands ≥ 0, where G(x, s) =

∫ s0 g(x, t)dt.

We should point out that [28] generalized a part of Brézis–Nirenberg’s result in [7] to theKirchhoff-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, Theorem 1.1 allows us to ensure the followingcorollary.

Corollary 1.3. Assume that a, b, λ > 0, 0 < γ < 1, 4 < p < 6 and g(x, u) = up−1, then thereexists Λ > 0 such that problem (1.1) possesses two positive solutions for all 0 < λ < Λ.

Remark 1.4. When g(x, u) = up−1(4 < p < 6), then clearly g satisfies (g1)–(g4). For the proof,we can consider instead with g(x, u) = (u+)p−1.

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

2 The auxiliary problem

In order to overcome the difficulty of the singular term, for every ε > 0, we study the followingperturbation problem−

(a + b

∫Ω|∇u|2dx

)∆u = (u+)5 + g(x, u) + λ(u+ + ε)−γ, x ∈ Ω,

u = 0, x ∈ ∂Ω,(2.1)

where u+ = maxu, 0. The energy functional corresponding to problem (2.1) is

Iε(u) =a2‖u‖2 +

b4‖u‖4 − 1

6

∫Ω(u+)6dx−

∫Ω

G(x, u+)dx− λ

1− γ

∫Ω

[(u+ + ε)1−γ − ε1−γ

]dx,

Obviously, the energy functional Iε is of class C1 on H10(Ω). As well known that there exists a

one-to-one correspondence between all solutions of problem (2.1) and the critical points of Iε

on H10(Ω). We mean a function u is called a weak solution of problem (2.1) if u ∈ H1

0(Ω) suchthat(

a + b‖u‖2) ∫Ω(∇u,∇ϕ)dx−

∫Ω(u+)5ϕdx−

∫Ω

g(x, u+)ϕdx− λ∫

Ω

ϕ

(u+ + ε)γdx = 0, (2.2)

for all ϕ ∈ H10(Ω).

First, we prove that Iε satisfies the local (PS)c condition.

Singular elliptic equations with critical exponent and Kirchhoff term 5

Lemma 2.1. Suppose that a, b, λ > 0, 0 < γ < 1 and g satisfies (g1)− (g3), then Iε satisfies the(PS)c condition, where c < Θ− Dλ

21+γ with D = D(γ, S, κ, Ω) is a positive constant and

Θ =abS3

4+

b3S6

24+

aS√

b2S4 + 4aS6

+b2S4√

b2S4 + 4aS24

.

Proof. Suppose that un is a (PS)c sequence for c ∈ (0, Θ− Dλ2

1+γ ), that is,

Iε(un)→ c, I′ε(un)→ 0, (2.3)

as n→ +∞. We claim that un is bounded in H10(Ω). In fact, from (g1) and (g2), there exists

C0 > 0 such that ∣∣∣∣15 g(x, s)s− G(x, s)∣∣∣∣ ≤ 1

30|s|6 + C0. (2.4)

Note that the subadditivity of t1−γ, one has

(u+n + ε)1−γ − ε1−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)‖un‖

≥ Iε(un)−15〈I′ε(un), un〉

=3a10‖un‖2 +

b20‖un‖4 +

∫Ω

[15

g(x, u+n )u

+n − G(x, u+

n )

]dx

+130

∫Ω(u+

n )6dx− λ

1− γ

∫Ω

[(u+

n + ε)1−γ − ε1−γ]

dx− λ

5εγ

∫Ω

u−n dx

≥ 3a10‖un‖2 +

b20‖un‖4 − λ

1− γ

∫Ω(u+

n )1−γdx− λ

5εγ

∫Ω

u−n dx− C0|Ω|,

≥ 3a10‖un‖2 +

b20‖un‖4 − C‖un‖1−γ − C1‖un‖ − C0|Ω|,

(2.6)

since 0 < γ < 1, which implies that un is bounded in H10(Ω). Going if necessary to a

subsequence, still denoted by un, there exists u ∈ H10(Ω) such that

un u, weakly in H10(Ω),

un → u, strongly in Ls(Ω), 1 ≤ s < 6,

un(x)→ u(x), a.e. in Ω,

there exists k ∈ L1(Ω) such that for all n, |un(x)| ≤ k(x) a.e. in Ω.

(2.7)

For every ε > 0, since|u|

(u+n + ε)γ

≤ |u|εγ

,

by the dominated convergence theorem and (2.7), one has

limn→∞

∫Ω(u+

n + ε)−γudx =∫

Ω(u+ + ε)−γudx. (2.8)

6 J. Liu, A. J. Hou and J. F. Liao

Moreover, for every ε > 0, by (2.7), one gets∣∣∣∣ un

(u+n + ε)γ

∣∣∣∣ ≤ ∣∣∣∣ un

(u+n + ε)γ

∣∣∣∣≤ |un|

εγ

≤ 1εγ

k(x).

Therefore, it follows from the dominated convergence theorem that

limn→∞

∫Ω(u+

n + ε)−γundx =∫

Ω(u+ + ε)−γudx. (2.9)

From (2.7), one also has∫Ω|∇un|2dx =

∫Ω|∇wn|2dx +

∫Ω|∇u|2dx + o(1), (2.10)(∫

Ω|∇un|2dx

)2

= ‖wn‖4 + ‖u‖4 + 2‖wn‖2‖u‖2 + o(1). (2.11)

By (g2) and (2.7), one has ∫Ω

g(x, u+n )udx =

∫Ω

g(x, u+)udx + o(1), (2.12)∫Ω

g(x, u+n )undx =

∫Ω

g(x, u+)udx + o(1), (2.13)∫Ω

G(x, u+n )dx =

∫Ω

G(x, u+)dx + o(1). (2.14)

As usually, letting wn = un − u, we need prove that ‖wn‖ → 0 as n→ ∞. Let limn→∞ ‖wn‖ =l ≥ 0. If l = 0, our conclusion is true. Suppose that l > 0. By the Brézis–Lieb Lemma (see [6]),one has ∫

Ω(u+

n )6dx =

∫Ω(w+

n )6dx +

∫Ω(u+)6dx + o(1). (2.15)

From (2.3), (2.7), (2.9) and (2.13), one obtains

a‖un‖2 + b‖un‖4 −∫

Ω(u+

n )6dx−

∫Ω

g(x, u+)udx− λ∫

Ω(u+ + ε)−γudx = o(1),

consequently, it follows from (2.10), (2.11) and (2.15) that

a‖u‖2 + a‖wn‖2 + b‖u‖4 + b‖wn‖4 + 2b‖wn‖2‖u‖2

−∫

Ω(w+

n )6dx−

∫Ω(u+)6dx−

∫Ω

g(x, u+)udx− λ∫

Ω(u+ + ε)−γudx = o(1). (2.16)

It follows from (2.3), (2.8) and (2.13) that

0 = limn→∞〈I′ε(un), u〉

= a‖u‖2 + b‖u‖4 + bl2‖u‖2 −∫

Ω(u+)6dx−

∫Ω

g(x, u+)udx− λ∫

Ω(u+ + ε)−γudx.

(2.17)

Moreover, by (2.3), for any ϕ ∈ H10(Ω), one has limn→∞〈I′ε(un), ϕ〉 = 0, that is,

(a + bd)∫

Ω(∇u,∇ϕ)dx−

∫Ω(u+)5ϕdx−

∫Ω

g(x, u+)ϕdx− λ∫

Ω(u+ + ε)−γ ϕdx = 0, (2.18)

Singular elliptic equations with critical exponent and Kirchhoff term 7

where d = ‖un‖2 + o(1) is a positive constant. Particularly, choosing ϕ = u− in (2.18), one hasu− = 0. Thus, we have u ≥ 0 in Ω. On the one hand, from (2.5), (2.17) and (g3), by the Hölderinequality, Sobolev inequality and Poincaré inequality, we have

Iε(u) =a2‖u‖2 +

b4‖u‖4 − 1

6

∫Ω(u+)6dx−

∫Ω

G(x, u)dx

− λ

1− γ

∫Ω

[(u+ + ε)1−γ − ε1−γ

]dx

=a4‖u‖2 +

112

∫Ω(u+)6dx +

∫Ω

[14

g(x, u+)u− G(x, u+)

]dx

− λ

1− γ

∫Ω

[(u+ + ε)1−γ − (u+)1−γ

]dx +

λ

4

∫Ω(u+ + ε)−γudx− bl2

4‖u‖2

≥ a4‖u‖2 − λ1(a− κ)

4

∫Ω

u2dx− λ

1− γ

∫Ω

u1−γdx− bl2

4‖u‖2

≥ κ

4‖u‖2 − λ

1− γ|Ω|

5+γ6 S−

1−γ2 ‖u‖1−γ − bl2

4‖u‖2

≥ − Dλ2

1+γ − bl2

4‖u‖2,

(2.19)

where the last inequality is obtained by the Young inequality and

D =1 + γ

1− γ2−2γ1+γ (κS)−

1−γ1+γ |Ω|

5+γ3(1+γ) .

On the other hand, it follows from (2.14), (2.16) and (2.17) that

a‖wn‖2 + b‖wn‖4 + b‖wn‖2‖u‖2 −∫

Ω(w+

n )6dx = o(1), (2.20)

and

Iε(un) = Iε(u) +a2‖wn‖2 +

b4‖wn‖4 +

b2‖wn‖2‖u‖2 − 1

6

∫Ω(w+

n )6dx + o(1). (2.21)

From (1.7), one has ∫Ω(w+

n )6dx ≤

∫Ω|wn|6dx ≤ ‖wn‖6

S3 ,

consequently, it follows from (2.20) that

al2 + bl4 + bl2‖u‖2 ≤ l6

S3 ,

which implies that

l2 ≥ bS3 +√

b2S6 + 4S3(a + b‖u‖2)

2. (2.22)

8 J. Liu, A. J. Hou and J. F. Liao

Thus, from (2.20)–(2.22), we obtain

Iε(u) = limn→∞

[Iε(un)−

a2‖wn‖2 − b

4‖wn‖4 − b

2‖wn‖2‖u‖2 +

16

∫Ω(w+

n )6dx]

= c−(

a3

l2 +b

12l4 +

b3

l2‖u‖2)

≤ c−[ a(

bS3 +√

b2S6 + 4S3(a + b‖u‖2))

6

+b

48

(bS3 +

√b2S6 + 4S3(a + b‖u‖2)

)2

+b‖u‖2

6

(bS3 +

√b2S6 + 4S3(a + b‖u‖2)

) ]− bl2

4‖u‖2

≤ c−(

abS3

4+

b3S6

24+

aS√

b2S4 + 4aS6

+b2S4√

b2S4 + 4aS24

)− bl2

4‖u‖2

< − Dλ2

1+γ − bl2

4‖u‖2,

which contradicts (2.19). Hence, l ≡ 0, that is, un → u in H10(Ω) as n → ∞. Therefore, Iε

satisfies the (PS)c condition for all c < Θ−Dλ2

1+γ . This completes the proof of Lemma 2.1.

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 thefollowing problem

−∆u = u5, ∀x ∈ R3.

Now, we estimate the level value of functional Iε and obtain the following lemma.

Lemma 2.2. Assume that a, b, λ > 0, 0 < γ < 1 and g satisfies (g1), (g2) and (g4), then there existsu0 ∈ H1

0(Ω), such that supt≥0 Iε(tu0) < Θ− Dλ2

1+γ for all 0 < λ < λ∗, where Θ and D are definedby Lemma 2.1 and the positive constant λ∗ is independent of u0 and ε.

Proof. Define a cut-off function η ∈ C∞0 (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, where ω is defined by (g4). Set uε = η(x)U(x), where U(x) is definedby (2.23). As well known (see [7, 26]), one has

‖uε‖2 = ‖U‖2 + O(ε) = S32 + O(ε), (2.24)

|uε|66 = |U|66 + O(ε3) = S32 + O(ε3), (2.25)

C2εs2 ≤

∫Ω

usεdx ≤ C3ε

s2 , 1 ≤ s < 3,

C4εs2 | ln ε| ≤

∫Ω

usεdx ≤ C5ε

s2 | ln ε|, s = 3,

C6ε6−s

2 ≤∫

Ωus

εdx ≤ C7ε6−s

2 , 3 < s < 6.

(2.26)

Singular elliptic equations with critical exponent and Kirchhoff term 9

Moreover, from [35], we also have

‖uε‖4 = S3 + O(ε);

‖uε‖6 = S92 + O(ε);

‖uε‖8 = S6 + O(ε);

‖uε‖12 = S9 + O(ε).

(2.27)

For all t ≥ 0, we define Iε(tuε) by

Iε(tuε) =at2

2‖uε‖2 +

bt4

4‖uε‖4 − t6

6

∫Ω

u6ε dx−

∫Ω

G(x, tuε)dx

− λ

1− γ

∫Ω

[(tuε + ε)1−γ − ε1−γ

]dx,

from (g2), we have

limt→+0

Iε(tuε) = 0, uniformly for all 0 < ε < ε0,

and

limt→+∞

Iε(tuε) = −∞, uniformly for all 0 < ε < ε0,

where ε0 > 0 is a small constant. Thus supt≥0 Iε(tuε) attains for some tε > 0. Using thefollowing conclusion of Step 1 in the proof of Theorem 2.3, one has Iε(tεuε) > ρ > 0. So, bythe continuity of Iε, there exist two constants t0, T0 > 0, which independent of ε, such thatt0 < tε < T0. Set Iε(tuε) = Iε,1(t)− Iε,2(t)− Iε,3(t), where

Iε,1(t) =a2

t2‖uε‖2 +b4

t4‖uε‖4 − t6

6

∫Ω

u6ε dx,

and

Iε,2(t) =∫

ΩG(x, tuε)dx,

Iε,3(t) =λ

1− γ

∫Ω

[(tuε + ε)1−γ − ε1−γ

]dx.

First, we estimate the value of Iε,1. Since I′ε,1(t) = at‖uε‖2 + bt3‖uε‖4− t5∫

Ωu6

ε dx, let I′ε,1(t) = 0,

that is,

a‖uε‖2 + bt2‖uε‖4 − t4∫

Ωu6

ε dx = 0, (2.28)

one obtains

T2ε =

b‖uε‖4 +√

b2‖uε‖8 + 4a‖uε‖2∫

Ω u6ε dx

2∫

Ω u6ε dx

.

Then I′ε,1(t) > 0 for all 0 < t < Tε and I′ε,1(t) < 0 for all t > Tε, so Iε,1(t) attains its maximum

10 J. Liu, A. J. Hou and J. F. Liao

at Tε. Thus, it follows from (2.24), (2.25), (2.27) and (2.28) that

Iε,1(t) ≤ Iε,1(Tε)

= T2ε

(a2‖uε‖2 +

b4

T2ε ‖uε‖4 − T4

ε

6

∫Ω

u6ε dx)

= T2ε

(a3‖uε‖2 +

b12

T2ε ‖uε‖4

)

=ab‖uε‖6 + a‖uε‖2

√b2‖uε‖8 + 4a‖uε‖2

∫Ω u6

ε dx

6∫

Ω u6ε dx

+b3‖uε‖12 + 2ab‖uε‖6

∫Ω u6

ε dx

24(∫

Ω u6ε dx)2

+b2‖uε‖4

√b2‖uε‖8 + 4a‖uε‖2

∫Ω u6

ε dx

24(∫

Ω u6ε dx)2

=ab‖uε‖6

4∫

Ω u6ε dx

+b3‖uε‖12

24(∫

Ω u6ε dx)2

+a‖uε‖2

√b2‖uε‖8 + 4a‖uε‖2

∫Ω u6

ε dx

6∫

Ω u6ε dx

+b2‖uε‖4

√b2‖uε‖8 + 4a‖uε‖2

∫Ω u6

ε dx

24(∫

Ω u6ε dx)2

=ab(S

92 + O(ε))

4(S32 + o(ε))

+b3(S9 + O(ε))

24(S32 + o(ε))2

+a(S

32 + O(ε))

√b2S6 + 4aS3 + O(ε)

6(S32 + o(ε))

+b2(S6 + O(ε))

√b2S6 + 4aS3 + O(ε)

24(S32 + o(ε))2

≤ abS3

4+

b3S6

24+

aS√

b2S4 + 4aS6

+b2S4√

b2S4 + 4aS24

+ C8ε

= Θ + C8ε.

(2.29)

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

limε→0+

∫Ω G(x, tεuε)dx

ε= +∞. (2.30)

Let m(t) = infx∈w g(x, t), from (g1) and (g4), we have

g(x, t) ≥ m(t) ≥ 0, limt→+∞

m(t)t3 = +∞,

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

Singular elliptic equations with critical exponent and Kirchhoff term 11

M(t) ≥ µt4 for all t ≥ A, where M(t) =∫ t

0 m(s)ds. Thus, one has∫Ω G(x, tεuε)dx

ε≥ ε−1

∫|x|<δ

G(x, tεuε)dx

≥ ε−1∫|x|<δ

M(tεuε)dx

= ε−1∫ δ

0M

[tε3

14 ε

12

(ε2 + r2)12

]r2dr

= ε2∫ δε−1

0M

[tε3

14 ε−

12

(1 + r2)12

]r2dr

= ε2∫ ε−1

0M

[tε3

14 ε−

12

(1 + r2)12

]r2dr− ε2

∫ ε−1

δε−1M

[tε3

14 ε−

12

(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 hasM(t) ≤ Ct2 for all t > 0 small enough. Consequently, one gets∣∣∣∣∣ε2

∫ ε−1

δε−1M

[tε3

14 ε−

12

(1 + r2)12

]r2dr

∣∣∣∣∣ ≤ Cε−1M(tε314 ε

12 ) ≤ Cε−1M(T03

14 ε

12 ) ≤ C, (2.32)

for all ε > 0 small enough. Fixing A, there exists B > 0 such that tε314 ε−

12

(1+r2)12≥ A for all

1 < r < Bε−12 . Therefore, one obtains

lim infε→0+

ε2∫ ε−1

0M

[tε3

14 ε−

12

(1 + r2)12

]r2dr ≥ lim inf

ε→0+ε2∫ Bε−

12

1M

[tε3

14 ε−

12

(1 + r2)12

]r2dr

≥ lim infε→0+

Cµε2∫ Bε−

12

1

ε−2r2

(1 + r2)2 dr

=∫ +∞

1

ε−2r2

(1 + r2)2 dr

= +∞.

(2.33)

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

Iε,3(tε) =λ

1− γ

∫Ω

[(tεuε + ε)1−γ − ε1−γ

]dx

≥ 0.(2.34)

Thus, from (2.29), (2.30) and (2.34), there exists a large enough positive constant C9 > C8 suchthat

Iε(tuε) = Iε,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− Dλ2

1+γ ,

12 J. Liu, A. J. Hou and J. F. Liao

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

D

) 1+γ2 . Thus there exists λ∗ =( (C9−C8)ε

D

) 1+γ2 > 0, choosing u0 = uε, such that Iε(tu0) < Θ − Dλ

21+γ for all 0 < λ < λ∗.

This completes the proof of Lemma 2.2.

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

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

Proof. We divide three steps to prove Theorem 2.3.

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 Iε(u)|u∈SR ≥ ρ and

infu∈BR Iε(u) < 0 for λ ∈ (0, λ∗), where SR = u ∈ H10(Ω) : ‖u‖ = R, BR = u ∈ H1

0(Ω) :‖u‖ ≤ R. In fact, by (g1) and (g2), we infer that

|G(x, s)| ≤ aλ1

4|s|2 + C10|s|6,

for all x ∈ Ω and s ∈ R. Consequently, by the Hölder inequality and (1.7) and (2.5), we have

Iε(u) =a2‖u‖2 +

b4‖u‖4 − 1

6

∫Ω(u+)6dx−

∫Ω

G(x, u+)dx

− λ

1− γ

∫Ω

[(u+ + ε)1−γ − ε1−γ

]dx

≥ a4‖u‖2 +

b4‖u‖4 − C11‖u‖6 − λ|Ω|

5+γ6

(1− γ)S1−γ

2

‖u‖1−γ

≥ a4‖u‖2 − C11‖u‖6 − λ|Ω|

5+γ6

(1− γ)S1−γ

2

‖u‖1−γ

=‖u‖1−γ

4

(a‖u‖1+γ − C12‖u‖5+γ − Bλ

)

(2.35)

where C12 = 4C11 and B = 4|Ω|5+γ

6

(1−γ)S1−γ

2. Let

h(t) = at1+γ − C12t5+γ − Bλ, ∀t ∈ [0,+∞).

Then we have h′(t) = tγ[a(1 + γ)− C12(5 + γ)t4]. Let h′(t) = 0, one has

tmax =

[a(1 + γ)

C12(5 + γ)

] 14

, maxt≥0

h(t) = h(tmax) =4a

5 + γ

[a(1 + γ)

C12(5 + γ)

] 1+γ4

− Bλ.

Therefore, choosing λ∗ =4a

B(5+γ)

[a(1+γ)

C12(5+γ)

] 1+γ4

and R = tmax, according to (2.35), there exists

ρ > 0 such that Iε(u)|u∈SR ≥ ρ for all 0 < λ < λ∗. By (g2), for u ∈ H10(Ω) with u+ > 0 it holds

limt→0+

Iε(tu)t

= − λ

1− γlim

t→0+

1t

∫Ω[(tu+ + ε)1−γ − ε1−γ]dx

= − λ

1− γlim

t→0+

∫Ω

(1− γ)ξ−γtu+

tdx (ε < ξ < tu+ + ε)

= −λ∫

Ω

u+

εγdx (as t→ 0+, ξ → ε)

< 0.

Singular elliptic equations with critical exponent and Kirchhoff term 13

Thus there exists u ∈ H10(Ω) with ‖u‖ small enough such that Iε(u) < 0. Thus, we have

infu∈BR Iε(u) < 0. Therefore, our claim is true.Denote mε = infu∈BR Iε(u), there exists a minimizing sequence un ⊂ H1

0(Ω) such thatlimn→∞ Iε(un) = mε. Applying Corollary 4.1 in [26], there exists a subsequence of un,still denoted by un such that I′ε(un) → 0 as n → ∞. Choosing λ∗∗ =

(ΘD

) 1+γ2 such that

Θ− Dλ2

1+γ > 0 for any 0 < λ < λ∗∗. Then, taking λ∗∗ = minλ∗, λ∗∗, for any 0 < λ < λ∗∗,by Lemma 2.1, one has there exists uε ∈ H1

0(Ω) such that Iε(uε) = limn→∞ Iε(un) = mε < 0.Thus uε is nonzero solution of problem (2.1). Let u−ε = max−uε, 0, by 〈I′ε(uε), u−ε 〉 = 0, onehas u−ε = 0. Thus, uε ≥ 0. By the strong maximum principle, one has uε > 0 in Ω. Therefore,uε is a positive local minimizer solution of problem (2.1) for all 0 < λ < λ∗∗.

Step 2. We prove that there exists a positive mountain-pass type solution of problem (2.1).By (g1) and (g2), there exists Cε > 0 such that

|G(x, s)| ≤ ε|s|6 + Cε.

Consequently, for u ∈ H10(Ω)\0, one has

limt→+∞

∣∣∫Ω G(x, tu)dx

∣∣t6 ≤ lim

t→+∞

εt6∫

Ω u6dx + Cε|Ω|t6 = ε

∫Ω

u6dx.

By the arbitrary of ε, one gets

limt→+∞

∣∣∫Ω G(x, tu)dx

∣∣t6 = 0.

Thus, we have

limt→+∞

Iε(tu)t6 = −

∫Ω

u6dx.

Consequently, there exists u ∈ H10(Ω) such that ‖u‖ > R and Iε(u) < 0. Let 0 < λ < λ∗.

According to Step 1, the functional Iε satisfies the geometry of the mountain-pass lemma. Letc be defined by

c = infγ∈Γ

maxt∈[0,1]

Iε(γ(t)),

where Γ =

γ ∈ C([0, 1], H10(Ω)) : γ(0) = 0, γ(1) = u

. Obviously, one has 0 < Θ− Dλ

21+γ .

According to Lemma 2.1 and Lemma 2.2, there exists un ⊂ H10(Ω) such that

Iε(un)→ c > ρ and I′ε(un)→ 0,

then un has a convergent subsequence (still denoted by un) in H10(Ω). We assume un → vε

in H10(Ω) as n→ ∞. Then applying the mountain-pass lemma (see [3] Theorem 2.1), one gets

limn→∞

Iε(un) = Iε(vε) = c > ρ > 0 and I′ε(vε) = 0. Thus, vε is a nonzero solution of problem(2.1). Similar to uε in Step 1, by the strong maximum principle, one has vε > 0 in Ω. Therefore,vε is a positive solution of problem (2.1) with Iε(vε) > ρ > 0 for all 0 < λ < λ∗. Therefore,choosing Λ = minλ∗, λ∗∗, uε and vε are two positive solutions of problem (2.1).

14 J. Liu, A. J. Hou and J. F. Liao

Step 3. We prove that there exists a positive ground state solution of problem (2.1).Let N = u ∈ H1

0(Ω) : I′ε(u) = 0 and m0 = infu∈N

Iε(u). For all u ∈ N , by the Sobolev

inequality, it follows from (2.4) and (2.5) that

Iε(u) = Iε(u)−15〈I′ε(u), u〉

=3a10‖u‖2 +

b20‖u‖4 +

∫Ω

[15

g(x, u+)u+ − G(x, u+)

]dx

+130

∫Ω(u+)6dx− λ

1− γ

∫Ω

[(u+ + ε)1−γ − ε1−γ

]dx− λ

5εγ

∫Ω

u−dx

≥ 3a10‖u‖2 +

b20‖u‖4 − λ

1− γ

∫Ω|u|1−γdx− λ

5εγ

∫Ω|u|dx− C0|Ω|

≥ 3a10‖u‖2 +

b20‖u‖4 − C13‖u‖1−γ − C14‖u‖ − C0|Ω|,

(2.36)

since 0 < γ < 1, which implies that m0 is well defined. According to Step 1 and Step 2, onehas uε, vε ∈ N . Thus m0 = infu∈N Iε(u) ≤ Iε(uε) < 0. From (2.36), we can easy obtain thatm0 > −∞. Therefore, for the minimization problem m0, we can get a (PS)m0 sequence. ByLemma 2.1 and Lemma 2.2, there exists u ∈ H1

0(Ω) such that Iε(u) = m0 and I′ε(u) = 0.Similar to uε in Step 1, by the strong maximum principle, one obtains that u is a positiveground state solution of problem (2.1). This completes the proof of Theorem 2.3.

3 The proof of Theorem 1.1

According to Section 2, we know that Θ, Λ, D are independent of ε. Therefore, there exist twosequences of positive solutions uεn and vεn of the auxiliary problem (2.1) with Iεn(uεn) < 0and Iεn(vεn) > 0, where εn → 0+ as n → +∞. Now, we will prove the limits of the twosequences of positive solutions are two different positive solutions of problem (1.1). Now, wegive the proof of Theorem 1.1.

Proof of Theorem 1.1. First, we prove that uεn and vεn have a convergent subsequence inH1

0(Ω), respectively. Without loss of generality, we only need prove that uεn has a conver-gent subsequence in H1

0(Ω). Similarly, we can also obtain vεn has a convergent subsequence.Since uεn is the positive solution of problem (2.1), one has

−(a + b‖uεn‖2)∆uεn = u5εn+ g(x, uεn) + λ(uεn + εn)

1−γ ≥ min

1,λ

2γ

,

which implies that

−∆uεn ≥1

a + b‖uεn‖2 min

1,λ

2γ

.

Let e be a positive weak solution of the following problem−∆u = 1, x ∈ Ω,

u = 0, x ∈ ∂Ω,

and for every Ω0 ⊂⊂ Ω, there exists e0 > 0 such that e|Ω0 ≥ e0. Therefore, by the comparisonprinciple, we get

uεn ≥min1, λ

2γ a + b‖uεn‖2 e.

Singular elliptic equations with critical exponent and Kirchhoff term 15

In particular, from e|Ω0 ≥ e0 > 0, one has

uεn |Ω0 ≥min

1, λ

2γ

a + b‖uεn‖2 e0 > 0. (3.1)

Similar to (2.6), we can easy obtain that

Θ− Dλ2

1+γ ≥ Iεn(uεn)−15〈I′εn

(uεn), uεn〉

≥ 3a10‖uεn‖2 +

b20‖uεn‖4 − λ

1− γ

∫Ω

u1−γεn dx− C0|Ω|

≥ 3a10‖uεn‖2 +

b20‖uεn‖4 − C‖uεn‖1−γ − C0|Ω|,

which implies that uεn is bounded in H10(Ω). Up to a subsequence, combining with (3.1),

there exists u∗ ∈ H10(Ω) with u∗ > 0 such that

uεn u∗ weakly in H10(Ω),

uεn → u∗ strongly in Lq(Ω)(1 ≤ q < 6),

uεn(x)→ u∗(x) a.e. in Ω.

(3.2)

Now, we shall prove that uεn → u∗ in H10(Ω) as εn → 0. By (g2) and (3.2), one has∫

Ωg(x, uεn)u∗dx =

∫Ω

g(x, u∗)u∗dx + o(1), (3.3)∫Ω

g(x, uεn)uεn dx =∫

Ωg(x, u∗)u∗dx + o(1), (3.4)∫

ΩG(x, uεn)dx =

∫Ω

G(x, u∗)dx + o(1). (3.5)

As usually, letting wεn = uεn − u∗, we need prove that ‖wεn‖→0 as n→∞. Let limn→∞ ‖wεn‖ =l ≥ 0. For every εn > 0, since

uεn

(uεn + εn)γ≤ u1−γ

εn ,

and similar to (2.3) in [22], one has∫Ω

u1−γεn dx =

∫Ω

u1−γ∗ dx + o(1),

consequently, by the dominated convergence theorem, one has

limn→∞

∫Ω(uεn + εn)

−γuεn dx =∫

Ωu1−γ∗ dx. (3.6)

Since uεn is a positive solution of problem (2.1) with ε = εn, it follows from (2.2) that(a + b‖uεn‖2)∫

Ω(∇uεn ,∇ϕ)dx−

∫Ω

u5εn

ϕdx−∫

Ωg(x, uεn)ϕdx− λ

∫Ω

ϕ

(uεn + εn)γdx = 0, (3.7)

for all ϕ ∈ H10(Ω). For any φ ∈ H1

0(Ω) ∩ C0(Ω), where C0(Ω) is the subset of C(Ω) consistingof functions with compact support in Ω, by the dominate convergence theorem, it followsfrom (3.1) with Ω0 = supp φ that

limn→∞

∫Ω

φ

(uεn + εn)γdx =

∫Ω

u−γ∗ φdx.

16 J. Liu, A. J. Hou and J. F. Liao

Consequently, taking the test function ϕ = φ ∈ H10(Ω) ∩ C0(Ω) in (3.7), and let n → ∞, we

can obtain

(a + bl2 + b‖u∗‖2)∫

Ω(∇u∗,∇φ)dx =

∫Ω

u5∗φdx +

∫Ω

g(x, u∗)φdx + λ∫

Ωu−γ∗ φdx. (3.8)

Similar to prove (4.3) in [16] holds for any φ ∈ H10(Ω), we can prove that (3.8) holds for any

φ ∈ H10(Ω) by the same way. Choosing φ = u∗ in (3.8), one has

a‖u∗‖2 + b‖u∗‖4 + bl2‖u∗‖2 −∫

Ωu6∗dx−

∫Ω

g(x, u∗)u∗dx− λ∫

Ωu1−γ∗ dx = 0. (3.9)

Choosing ϕ = uεn in (3.7), let n → ∞, by the Brézis–Lieb Lemma, it follows from (3.2), (3.4)and (3.6) that

a‖u∗‖2 + a‖wεn‖2 + b‖u∗‖4 + b‖wεn‖4 + 2b‖wεn‖2‖u‖2

−∫

Ωw6

εndx−

∫Ω

u6∗dx−

∫Ω

g(x, u∗)u∗dx− λ∫

Ωu1−γ∗ dx = o(1). (3.10)

On the one hand, from (2.5), (3.9) and (g3), by the Hölder inequality, Sobolev inequality andPoincaré inequality, we have

I(u∗) =a2‖u∗‖2 +

b4‖u∗‖4 − 1

6

∫Ω

u6∗dx−

∫Ω

G(x, u∗)dx− λ

1− γ

∫Ω

u1−γ∗ dx

=a4‖u∗‖2 +

112

∫Ω

u6∗dx +

∫Ω

[14

g(x, u∗)u− G(x, u∗)]

dx

− λ

1− γ

∫Ω

u1−γ∗ dx− bl2

4‖u∗‖2

≥ a4‖u∗‖2 − λ1(a− κ)

4

∫Ω

u2∗dx− λ

1− γ

∫Ω

u1−γ∗ dx− bl2

4‖u∗‖2

≥ κ

4‖u∗‖2 − λ

1− γ|Ω|

5+γ6 S−

1−γ2 ‖u∗‖1−γ − bl2

4‖u∗‖2

≥ − Dλ2

1+γ − bl2

4‖u∗‖2,

(3.11)

where the last inequality is obtained by the Young inequality. On the other hand, it followsfrom (3.5), (3.9) and (3.10) that

a‖wεn‖2 + b‖wεn‖4 + b‖wεn‖2‖u∗‖2 −∫

Ωw6

εndx = o(1), (3.12)

and

Iεn(uεn) = I(u∗) +a2‖wεn‖2 +

b4‖wεn‖4 +

b2‖wεn‖2‖u∗‖2 − 1

6

∫Ω

w6εn

dx + o(1). (3.13)

From (1.7), one has ∫Ω

w6εn

dx ≤ ‖wεn‖6

S3 ,

consequently, it follows from (3.12) that

al2 + bl4 + bl2‖u∗‖2 ≤ l6

S3 ,

Singular elliptic equations with critical exponent and Kirchhoff term 17

which implies that

l2 ≥ bS3 +√

b2S6 + 4S3(a + b‖u∗‖2)

2. (3.14)

Since Iεn(uεn) < Θ− Dλ2

1+γ , from (3.12)–(3.14), we obtain

I(u∗) < limn→∞

[Θ− Dλ

21+γ − a

2‖wεn‖2 − b

4‖wεn‖4 − b

2‖wεn‖2‖u∗‖2 +

16

∫Ω

w6εn

dx]

= Θ− Dλ2

1+γ −(

a3

l2 +b

12l4 +

b3

l2‖u∗‖2)

≤ Θ− Dλ2

1+γ −[ a(

bS3 +√

b2S6 + 4S3(a + b‖u∗‖2))

6

+b

48

(bS3 +

√b2S6 + 4S3(a + b‖u∗‖2)

)2

+b‖u∗‖2

6

(bS3 +

√b2S6 + 4S3(a + b‖u∗‖2)

) ]− bl2

4‖u∗‖2

≤ Θ− Dλ2

1+γ −(

abS3

4+

b3S6

24+

aS√

b2S4 + 4aS6

+b2S4√

b2S4 + 4aS24

)− bl2

4‖u∗‖2

< − Dλ2

1+γ − bl2

4‖u∗‖2,

which contradicts (3.11). Hence, l ≡ 0, that is, uεn → u∗ in H10(Ω) as n → ∞. Moreover, since

(3.8) holds any φ ∈ H10(Ω), we get that u∗ is a positive solution of problem (1.1). Similarly, for

vεn, up to a subsequence, there exists v∗ ∈ H10(Ω) with v∗ > 0 such that vεn → v∗ in H1

0(Ω)

and v∗ is a positive solution of problem (1.1).Second, we prove that u∗ and v∗ are two different positive solutions of problem (1.1).

According to Theorem 2.3, one has I(u∗) = limn→∞ Iεn(uεn) = limn→∞ mεn ≤ 0 and I(v∗) =

limn→∞ Iεn(vεn) > ρ > 0. Therefore, u∗ and v∗ are two different positive solutions of problem(1.1). This completes the proof of Theorem 1.1.

Acknowledgements

The authors express their gratitude to the reviewer for careful reading and helpful sugges-tions which led to an improvement of the original manuscript. The third author expresseshis gratitude to Prof. Chun-Yu Lei for his discussions and valuable opinions for the revisedmanuscript. This work was supported by the National Natural Science Foundation of China(11861052), Natural Science Foundation of Education of Sichuan Province (18ZA0471), Fun-damental Research Funds of China West Normal University (16E014,18B015), MeritocracyResearch Funds of China West Normal University (17YC383) and Innovative Research Teamof China West Normal University (CXTD2018-8).

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