# Twin Positive Solutions for Schrödinger-Kirchhoff-Type Problem with Singularity and Critical Exponents

Authors: Wei Han; Yangyang Zhao
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We study in this paper the following singular Schrödinger-Kirchhoff-type problem with critical exponent-a+b∫Ω∇u2dxΔu+u=Q(x)u5+μxα-2u+f(x)(λ/uγ) in Ω,u=0 on ∂Ω, where a,b>0 are constants, Ω⊂R3 is a smooth bounded domain, 0<α<1, λ>0 is a real parameter, γ∈(0,1) is a constant, and 0<μ<aμ1 (μ1 is the first eigenvalue of -Δu=μxα-2u, under Dirichlet boundary condition). Under appropriate assumptions on Q and f, we obtain two positive solutions via the variational and perturbation methods.


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## Body

## 1. Introduction and Main Result

In this paper, we consider the following Schrödinger-Kirchhoff-type problem with Dirichlet boundary value conditions(1)-a+b∫Ω∇u2dxΔu+u=Qxu5+μxα-2u+fxλuγinΩ,u=0on∂Ω,where Ω is a smooth bounded domain in R3, 0<α<1, λ>0 is a real parameter, γ∈(0,1) is a constant, and 0<μ<aμ1 (μ1 is the first eigenvalue of -Δu=μ|x|α-2u, under Dirichlet boundary condition), and coefficient functions Q∈C(Ω¯), f∈C(Ω¯) are positive, which have at least one minimum. This kind of problem has been widely concerned. The Schrödinger-Kirchhoff-type problem in the effect of electromagnetic field on current motion in physics and the basic law of physical world material movement has been widely used. In recent years, the following elliptic problems have been extensively studied by many researchers:(2)-a+b∫RN∇u2dxΔu+Vxu=fx,u,u∈H1RN,where V:RN→R, f∈C(RN×R,R),N=1,2,3, and a,b>0 are constants. In [1], Li and Ye considered the existence of positive ground state solutions to the following Kirchhoff-type problem with pure power nonlinearities: (3)-a+b∫R3∇u2dxΔu+Vxu=up-1ux∈R3,u∈H1R3,u>0x∈R3.Furthermore, Chen and Tang [2] studied the following Kirchhoff-type equation: (4)-a+b∫R3∇u2dxΔu+Vxu=fx,ux∈R3,ux⟶0x⟶∞,and their result unifies both asymptotically cubic and supercubic cases, which generalizes and improves the existing ones. In [3], Huy and Quan investigated the equation -M(x,u)Δpu=λf(x,u,∇u)-g(x,u,∇u) in Ω, u=0 on ∂Ω. They proved the existence results for both nondegenerate and degenerate cases of the function M and combined the fixed point index theory with the cone theoretic argument which are key technical ingredients to obtain the main results. In [4], Figueiredo and Severo established the existence of a positive ground state solution for a Kirchhoff problem in R2 involving critical exponential growth. Moreover, Ricceri [5] obtained some results of a totally new type about a class of nonlocal problems. In [6], the authors studied the multiplicity of positive solutions for a class of Kirchhoff type of equations with the nonlinearity containing both singularity and critical exponents:(5)-a+b∫Ω∇u2dxΔu=u5+λuγinΩ,u=0on∂Ω.In the case that V(x)≡0 in (2), it is related to the stationary analogue of the equation (6)ρ∂2u∂t2-P0h+E2L∫0L∂u∂x2dx∂2u∂x2=0,presented by Kirchhoff in [7]. In [8], Lions introduced an abstract functional analysis framework to the following equation:(7)utt-a+b∫Ω∇u2dxΔu=fx,u,proposed by He and Zou in [9] as an existence of the classical D’Alembert’s wave equations for free vibration of elastic strings. After the vanguard work of Lions [8], where a functional analysis method was presented, the Kirchhoff-type equations started to arouse attention of researchers. D’Ancona and Spagnolo [10] studied the existence of a global classical periodic solution with real analytic data. Arosio and Panizzi [11] proved the well-posedness of the Cauchy problem concerned (7) in the Hadamard sense as a special case of an abstract second-order Cauchy problem in a Hilbert space. Perera and Zhang [12] obtained a nontrivial solution via Yang index and critical group. Caponi and Pucci [13] proved existence, multiplicity, and asymptotic behavior of entire solutions for a series of stationary Kirchhoff fractional p-Laplacian equations. In [14–16], the authors proved the existence of a nontrivial solution without the Ambrosetti-Rabinowitz condition. Moreover, we found in [17–19] that the authors considered the existence and multiplicity of solutions by Nehari manifold. In [20], Shen and Yao obtained two positive solutions about the following Kirchhoff problem:(8)-a+b∫R3∇u2dxΔu+u=Qxu4u+λfxuq-2uinR3,u∈H1R3,under some assumptions on the weight functions Q(x) and f(x). The related developments of the existence of solutions for the Kirchhoff-type problems also can be found in [21–24].However, up till now, no paper has appeared in the literature which discusses the singular Schrödinger-Kirchhoff-type problem with critical exponent. This paper attempts to fill in this gap in the literature; in this paper, we obtain the multiple positive solutions for the problem via the variational and perturbation methods. Throughout this paper, we make use of the following notations:(1) The norm inH01(Ω) is equipped with the norm u2=∫Ωa∇u2+u2dx,∀u∈H01(Ω), and the norm in Lp(Ω) is denoted by ·p.(2)un+x=max⁡unx,0, un-(x)=max⁡-unx,0.(3)C1, C2, and C3 denote various positive constants, which may vary from line to line.(4) We denote bySr (resp., Br) the sphere (resp., the closed ball) of center zero and radius r; that is, Sr={u∈H01(Ω):u=r}, Br={u∈H01(Ω):u≤r}. Let S be the Sobolev constant; namely,(9)S≔infu∈H01Ω∖0⁡∫Ω∇u2dx∫Ωu6dx1/3.We define the functional(10)Iu=12∫Ωa∇u2+u2dx+b4∫Ω∇u2dx2-μ2∫Ωxα-2u2dx-16∫ΩQxu+6dx-λ1-γ∫Ωfxu+1-γdx.In general, a function u is called a weak solution of (1) if u∈H01(Ω), and for all v∈H01(Ω), it holds that (11)∫Ωa∇u∇v+uvdx+b∫Ω∇u2dx∫Ω∇u∇vdx-μ∫Ωxα-2uvdx-∫ΩQxu+5vdx-λ∫Ωfxu+-γvdx=0,and for α>0, we consider the following perturbed problem:(12)-a+b∫Ω∇u2dxΔu+u=Qxu+5+μxα-2u+λfxu++αγinΩ,u=0on∂Ω.Solution (12) corresponds to critical point of C1-functional on H01(Ω) by (13)Iαu=12∫Ωa∇u2+u2dx+b4∫Ω∇u2dx2-μ2∫Ωxα-2u2dx-16∫ΩQxu+6dx-λ1-γ∫Ωfxu++α1-γ-α1-γdx.The main result can be described as follows.Theorem 1.
Assume thatγ∈(0,1), a>0, b≥0, and Q∈C(Ω¯), f∈C(Ω¯) are positive, which have at least one minimum. Then, there exists λ∗>0, such that, for any λ∈(0,λ∗), problem (1) has at least two different positive solutions.This paper is organized as follows. In Sections2 and 4, we give the proof of Theorem 1. In Section 3, we consider the existence of at least one mountain-pass solution of the perturbation problem.


## 2. Existence of a First Solution of Problem (1)

In this section, our main work is to prove that problem (1) has a local minimum solution in H01(Ω).Lemma 2.
There exist constantsλ0>0 and R, ρ>0 such that, for all 0<λ<λ0, one has(14)Iuu∈SR≥ρ,infu∈BR⁡Iu<0.Proof.
By Hölder inequality and (9), we can obtain that(15)∫Ωfxu+1-γdx≤∫Ωfxu1-γdx≤maxx∈Ω¯⁡fx∫Ωu1-γ6/1-γdx1-γ/6∫Ω16/5+γdx5+γ/6≤maxx∈Ω¯⁡fxu61-γΩ5+γ/6≤Ω5+γ/6S-1-γ/2maxx∈Ω¯⁡fx∫Ω∇u2dx.Therefore, from (9) and (15), we have(16)Iu=12∫Ωa∇u2+u2dx+b4∫Ω∇u2dx2-μ2∫Ωxα-2u2dx-16∫ΩQxu+6dx-λ1-γ∫Ωfxu+1-γdx≥12a-μμ1∫Ω∇u2dx-16S-3∫ΩQx∇u6dx-λ1-γΩ5+γ/6S-1-γ/2maxx∈Ω¯⁡fx∫Ω∇u2dx≥∫Ω∇u1-γdx12a-μμ1∫Ω∇u1+γdx-16S-3∫ΩQx∇u5+γdx-λ1-γΩ5+γ/6S-1-γ/2∫Ωfxdx.We can set g(t)=(1/2)(a-μ/μ1)t1+γ-(1/6)S-3t5+γ∫ΩQ(x)dx for t>0; therefore, there exists a constant R=3(aμ1-μ)(1+γ)S3/μ1(5+γ)∫ΩQ(x)dx1/4>0, such that maxt>0⁡g(t)=g(R)>0. We can let λ0=((1-γ)S(1-γ)/2/|Ω|(5+γ)/6∫Ωf(x)dx)g(R); then, there exists a constant ρ>0 such that I(u)SR≥ρ for all 0<λ<λ0. For given R, we can deduce u∈BR with u+≠0; one has (17)limt→0⁡Itut1-γ=-λ1-γ∫Ωfxu+1-γdx<0.Hence, we have I(tu)<0 for all u+≠0 and t small enough. So, for ∫Ωdx sufficiently small, we have(18)m1=infu∈BR⁡Iu<0.This completes the proof of Lemma 2.Theorem 3.
Assumeλ∈(0,λ0), where λ0 is defined in Lemma 2. Hence, one obtains that problem (1) has a positive solution u0∈H01(Ω), with I(u0)<0.Proof.
First, there existsu0∈BR, such that I(u0)=m1<0. Actually, from (14), we can obtain that (19)12∫Ωa∇u2+u2dx+b4∫Ω∇u2dx2-μ2∫Ωxα-2u2dx-16∫ΩQxu+6dx≥ρ,foru∈SR,(20)12∫Ωa∇u2+u2dx+b4∫Ω∇u2dx2-μ2∫Ωxα-2u2dx-16∫ΩQxu+6dx≥0,foru∈BR.With the definition of m1, we have that there exists a minimizing sequence {un}⊂BR such that limn→∞⁡I(un)=m1<0. Therefore, {un} is bounded in BR. Hence, there exists a subsequence, still denoted by {un}. We assume that there exists u0∈H01(Ω) such that(21)un⇀u0,weakly  inH01Ω,un⟶u0,strongly  inLpΩ1≤p<2∗,unx⟶u0x,a.e.  inΩ.For γ∈(0,1), by Hölder inequality, one has(22)∫Ωfxun+1-γdx-∫Ωfxu0+1-γdx≤∫Ωfxun+1-γ-u0+1-γdx≤∫Ωfxun+-u0+1-γdx≤∫Ωfxdxun+-u0+21-γΩ1+γ/2.Therefore, by (21), we can obtain that(23)∫Ωfxun+1-γdx=∫Ωfxu0+1-γdx+o1.Let ωn=un-u0; by Brézis Lemma, we get(24)∫Ωa∇un2+un2dx=∫Ωa∇ωn2+ωn2dx+∫Ωa∇u02+u02dx+o1,(25)∫Ω∇un2dx2=∫Ω∇ωn2dx2+∫Ω∇u02dx2+2∫Ω∇ωn2dx∫Ω∇u02dx+o1,(26)∫ΩQxun+6dx=∫ΩQxωn+6dx+∫ΩQxu0+6dx+o1,(27)∫Ωxα-2un2dx=∫Ωxα-2ωn2dx+∫Ωxα-2u02dx+o1.If we set u0=0, it follows that ωn=un and ω0∈BR. If u0≠0, by (24), one has ωn∈BR for n large enough. Therefore, by (20), we have(28)12∫Ωa∇ωn2+ωn2dx+b4∫Ω∇ωn2dx2-μ2∫Ωxα-2ωn2dx-16∫ΩQxωn+6dx≥0.Hence, noticing (23)–(28), we get (29)m1=Iun+o1=Iu0+12∫Ωa∇ωn2+ωn2dx+b4∫Ω∇ωn2dx2+b2∫Ω∇ωn2dx∫Ω∇u02dx-μ2∫Ωxα-2ωn2dx-16∫ΩQxωn+6dx+o1≥Iu0+b2∫Ω∇ωn2dx∫Ω∇u02dx+o1≥Iu0+o1.Passing to the limit as n→∞, we have m1≥I(u0). Noticing that BR is closed and convex, so u0∈BR. Therefore, from (18), we get I(u0)=m1<0 and u0≠0. Thus, it can be seen that u0 is a local minimizer of I.
Next, we prove thatu0 is a solution of (1) and u0>0. By the above expression, we can obtain that u0 is a local minimizer of I. So, for any ψ∈H01(Ω), ψ≥0, set t>0 sufficiently small, such that u0+tψ∈BR; we have(30)0≤Iu0+tψ-Iu0=12∫Ωa∇u0+tψ2+u0+tψ2dx+b4∫Ω∇u0+tψ2dx2-μ2∫Ωxα-2u0+tψ2dx-16∫ΩQxu0+tψ+6dx-λ1-γ∫Ωfxu0+tψ+1-γdx-12∫Ωa∇u02+u02dx-b4∫Ω∇u02dx2+μ2∫Ωxα-2u02dx+16∫ΩQxu0+6dx+λ1-γ∫Ωfxu0+1-γdx≤12∫Ωa∇u0+tψ2+u0+tψ2dx+b4∫Ω∇u0+tψ2dx2-12∫Ωa∇u02+u02dx-b4∫Ω∇u02dx2=∫Ω∇u0+tψ2dx-∫Ω∇u02dxa2+b4∫Ω∇u0+tψ2dx+∫Ω∇u02dx+12∫Ωu0+tψ2-u02dx.Currently, we shall prove that u0 is a solution of (1). Indeed, by (30), one has (31)λ1-γ∫Ωfxu0+tψ+1-γ-u0+1-γdx≤a2∫Ω∇u0+tψ2dx-∫Ω∇u02dx+12∫Ωu0+tψ2dx-∫Ωu02dx+b4∫Ω∇u0+tψ2dx2-∫Ω∇u02dx2-μ2∫Ωxα-2u0+tψ2dx-∫Ωxα-2u02dx-16∫ΩQxu0+tψ+6-u0+6dx=∫Ω∇u0+tψ2dx-∫Ω∇u02dxa2+b4∫Ω∇u0+tψ2dx+∫Ω∇u02dx+12∫Ωu0+tψ2-u02dx-μ2∫Ωxα-2u0+tψ2-xα-2u02dx-16∫ΩQxu0+tψ+6-u0+6dx.Dividing by t>0 and taking the limit t→0+, one gets(32)λ1-γliminft→0+⁡∫Ωfxu0+tψ+1-γ-u0+1-γtdx≤∫Ωa∇u0∇ψ+u0ψdx+b∫Ω∇u02dx∫Ω∇u0∇ψdx-μ∫Ωxα-2u0ψdx-∫ΩQxu0+5ψdx.Note that(33)λ1-γ∫Ωfxu0+tψ+1-γ-u0+1-γtdx=λ∫Ωfxu0++ξtψ+-γψdx,where ξ→0+ and ((u0+ξtψ)+)-γψ→(u0+)-γψ a.e. x∈Ω as t→0+, for ((u0+ξtψ)+)-γψ≥0. Therefore, using Fatou Lemma, one has (34)λ∫Ωfxu0+-γψdx≤λ1-γliminft→0+⁡∫Ωfxu0+tψ+1-γ-u0+1-γtdx.Hence, by (32) and the above estimate, we can obtain that(35)∫Ωa∇u0∇ψ+u0ψdx+b∫Ω∇u02dx∫Ω∇u0∇ψdx-μ∫Ωxα-2u0ψdx-∫ΩQxu0+5ψdx-λ∫Ωfxu0-γψdx≥0,ψ≥0.For I(u0)<0 and (14), we can get u0∉SR; then, ∫Ω∇u0dx<R. For u0, there is δ1∈(0,1) such that (1+t)u0∈BR for t≤δ1. Define k:[-δ1,δ1]→R by k(t)=I((1+t)u0). Actually, when t=0, k(t) achieves its minimum; namely,(36)k′tt=0=∫Ωa∇u02+u02dx+b∫Ω∇u02dx2-μ∫Ωxα-2u02dx-∫ΩQxu0+6dx-λ∫Ωfxu0+1-γdx=0.Assume ϕ∈H01(Ω) and ε>0. Define Ψ∈H01(Ω) by Ψ=(u0++εϕ)+. So, from (35) and (36), we can get(37)
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.By the measure of the domain of integration u0++εϕ≤0→0 as ε→0, it can be seen that (38)limε→0⁡∫u0++εϕ≤0a+b∇u02∇u0∇ϕ+u0ϕdx=0.Then, dividing by ε and letting ε→0 in (37), one has(39)a+b∫Ω∇u02dx∫Ω∇u0∇ϕdx+∫Ωu0ϕdx-μ∫Ωxα-2u0ϕdx-∫ΩQxu0+5ϕdx-λ∫Ωfxu0+-γϕdx≥0.With the arbitrariness of ϕ, this inequality also holds for -ϕ; that is,(40)a+b∫Ω∇u02dx∫Ω∇u0∇ϕdx+∫Ωu0ϕdx-μ∫Ωxα-2u0ϕdx-∫ΩQxu0+5ϕdx-λ∫Ωfxu0+-γϕdx=0.
On the one hand, we can take the test functionϕ=u0- in (40); we have ∫Ω∇u0-dx=0, and then we get u0≥0. Therefore, u0 is a nonzero solution of (1).
On the other hand, letuα be a family of positive solutions of (12); with an easy computation, we see that (41)-a+b∫Ω∇uα2dxΔuα+uα=Qxuα5+μxα-2uα+λfxuα+αγ≥min⁡1,Qmmin⁡1,λfm2γ,where Qm=min⁡Q(x), fm=min⁡f(x), which means that u0∈H01(Ω) and satisfies (42)-Δu0≥min⁡1,Qmmin⁡1,λfm/2-u0a+b∫Ω∇u02dx,inΩ.Noticing u0≥0, u0≠0, thus, we have u0>0, in Ω by the strong maximum principle.
From the above discussion, we can deduce thatu0 is a positive solution of (1) with I(u0)=m1<0. This completes the proof of Theorem 3.


## 3. Existence of a Mountain-Pass Solution of Problem (12)

In this section, we consider the existence of at least one mountain-pass solution of the perturbation problem.Lemma 4.
Suppose thatλ<λ0. For R,ρ>0, where λ0, R, and ρ are as in Lemma 2, the functional Iα satisfies the following conditions:(i)
Iα(u)≥ρ>0, if u∈SR.(ii)
There existsu1∈H01(Ω) such that ∫Ω|∇u1|dx>R, and Iα(u1)<ρ.Proof.
(i) Notice the subadditivity oft1-γ; namely,(43)u++α1-γ-α1-γ≤u+1-γ,∀u∈H01Ω.From (43), we can get Iα(u)≥I(u),∀u∈H01(Ω). Hence, when λ<λ0, from (14), we can get the conclusion.
(ii) For everyu+∈H01(Ω), u+≠0, and t>0, one has (44)Iαtu=t22∫Ωa∇u2+u2dx+b4t4∫Ω∇u2dx2-μ2∫Ωxα-2u2dx-t66∫ΩQxu+6dx-λ1-γ∫Ωfxtu++α1-γ-α1-γdx≤t22∫Ωa∇u2+u2dx+b4t4∫Ω∇u2dx2-t66∫ΩQxu+6dx⟶-∞,as t→+∞. Hence, we can consider u1 such that ∫Ω∇u1dx>R and Iα(u1)<ρ. This completes the proof of Lemma 4.We define(45)Λ=abS36QM+b3S624QM2+aSb2S4+4aSQM6QM+b2S4b2S4+4aSQM24QM2.Lemma 5.
Letα∈(0,1); then Iα satisfies the (PS)c condition for every c<Λ-Dλ2/(1+γ), where D=(1+γ)31-γ/2|Ω|5+γ/6(μ1(1-γ))1-γ/2fm/2(2-γ)/2((aμ1-μ)S)(1-γ)/22/(1+γ), and S is as in (9).Proof.
Let{vn}⊂H01(Ω), so(46)Iαvn⟶c,Iα′vn⟶0,asn⟶∞.We shall claim that vn is bounded in H01(Ω). Otherwise, we can suppose that ∫Ω∇vndx→∞, as n→∞. From (46), (43), and (15), one has (47)1+c+o1∫Ω∇vndx≥Iαvn-16Iα′vn,vn=13∫Ωa∇vn2+vn2dx+b12∫Ω∇vn2dx2-μ3∫Ωxα-2vn2dx-λ1-γ∫Ωfxvn++α1-γ-α1-γdx+λ6∫Ωfxvnvn++αγdx≥a3∫Ω∇vn2dx+b12∫Ω∇vn2dx2-μ3∫Ωxα-2vn2dx-λ1-γ∫Ωfxvn+1-γdx≥13a-μμ1∫Ω∇vn2dx+b12∫Ω∇vn2dx2-λ1-γΩ5+γ/6S-1-γ/2∫Ωfx∇vn+1-γdx.By 1-γ<2<4, it can be seen that the last inequality is unreasonable. Clearly, vn is bounded in H01(Ω). Up to a subsequence, there exists v∗∈H01(Ω) such that(48)vn⇀v∗,weakly  inH01Ω,vn⟶v∗,strongly  inLpΩ1≤p<6,vnx⟶v∗x,a.e.  inΩ,there  existsk∈L1Ωsuch  that  for  alln,vnx≤kxa.e.  inΩ.Since α>0 and v∗/(v∗+α)γ≤v∗x/αγ, using the dominated convergence theorem and (48), we get(49)limn→∞⁡∫Ωfxvn++α-γv∗dx=∫Ωfxv∗++α-γv∗dx.Furthermore, from α>0 and (48), we get(50)vnvn++αγ≤vnvn++αγ≤vnαγ≤1αγkx.Hence, it follows from the dominated convergence theorem that(51)limn→∞⁡∫Ωfxvn++α-γvndx=∫Ωfxv∗++α-γv∗dx.
Letωn=vn-v∗; then, ∫Ω∇ωndx→0 as n→∞. Otherwise, there exists a subsequence (still denoted by ωn) such that limn→∞⁡∫Ω∇ωndx=l (where l is a positive constant). Therefore, by limn→∞⁡Iα′vn,v∗=0 and (49), one has(52)∫Ωa∇v∗2+v∗2dx+b∫Ω∇v∗2dx2+bl2∫Ω∇v∗2dx-μ∫Ωxα-2v∗2dx-∫ΩQxv∗+6dx-λ∫Ωfxv∗++α-γv∗dx=0.By Brézis-Lieb Lemma and (46), we get(53)∫Ωa∇v∗2+v∗2dx+∫Ωa∇ωn2+ωn2dx+b∫Ω∇v∗2dx2+b∫Ω∇ωn2dx2+2b∫Ω∇ωn2dx∫Ω∇v∗2dx-μ∫Ωxα-2v∗2dx-μ∫Ωxα-2ωn2dx-∫ΩQxv∗+6dx-∫ΩQxωn+6dx-λ∫Ωfxvn++α-γvndx=o1.Thus, from the above equality and (51) and (52), we get(54)∫Ωa∇ωn2+ωn2dx+b∫Ω∇ωn2dx2+b∫Ω∇ωn2dx∫Ω∇v∗2dx-μ∫Ωxα-2ωn2dx-∫ΩQxωn+6dx=o1.For ∫Ω(ωn+)6dx≤∫Ωωn6dx, by (54) and (9), one has(55)∫Ωa-μμ1∇ωn2+ωn2dx+b∫Ω∇ωn2dx2+b∫Ω∇ωn2dx∫Ω∇v∗2dx≤S-3∫ΩQx∇ωn6dx+o1.Taking the limit n→∞, we have(56)a-μμ1l2+bl4+bl2∫Ω∇v∗2dx≤S-3l6∫ΩQxdx,and then we get(57)l2≥bS3+b2S6+4S3∫ΩQxdxa-μ/μ1+b∫Ω∇v∗2dx2∫ΩQxdx.Furthermore, from Brézis-Lieb Lemma and (46) and (54) and letting n→∞, we obtain that(58)Iαv∗=c-13a-μμ1l2-b12l4-b3l2∫Ω∇v∗2dx-13∫Ωωn2dx.
On the one hand, using (54), (43), and (15) and Young’s inequality, one has(59)Iαv∗=12∫Ωa∇v∗2+v∗2dx+b4∫Ω∇v∗2dx2-μ2∫Ωxα-2v∗2dx-16∫ΩQxv∗+6dx-λ1-γ∫Ωfxv∗++α1-γ-α1-γdx≥13∫Ωa∇v∗2+v∗2dx+b12∫Ω∇v∗2dx2-b6l2∫Ω∇v∗2dx-μ3∫Ωxα-2v∗2dx-λ1-γ∫Ωfxv∗++α1-γ-α1-γdx≥a3∫Ω∇v∗2dx-μ3∫Ωxα-2v∗2dx-λ1-γ∫Ωfxv∗1-γdx-b4l2∫Ω∇v∗2dx≥13a-μμ1∫Ω∇v∗2dx-λ1-γΩ5+γ/6S-1-γ/2∫Ωfx∇v∗1-γdx-b4l2∫Ω∇v∗2dx≥μa-13μ1∫Ω∇v∗2dx-1-γ22aμ1-μ3μ11-γ1-γ/2∫Ω∇v∗1-γdx2/1-γ-1+γ2λΩ5+γ/63μ11-γ1-γ/2fm2aμ1-μS1-γ/22/1+γ-b4l2∫Ω∇v∗2dx=-Dλ2/1+γ-b4l2∫Ω∇v∗2dx.
On the other hand, by (58), (57), and c<Λ-Dλ2/(1+γ), one has(60)Iαv∗≤c-13a-μμ1l2-112bl4-13bl2∫Ω∇v∗2dx≤c-13a-μμ1bS3+b2S6+4S3QMa-μ/μ1+b∫Ω∇v∗2dx2QM-b12b2S6+2S3QMa-μ/μ1+b∫Ω∇v∗2dx+bS4b2S4+4SQMa-μ/μ1+b∫Ω∇v∗2dx2QM2-b3∫Ω∇v∗2dxbS3+b2S6+4S3QMa-μ/μ1+b∫Ω∇v∗2dx2QM≤c-abS36QM-b3S624QM2-aSb2S4+4S3QMa-μ/μ1+b∫Ω∇v∗2dx6QM-b2S4b2S4+4S3QMa-μ/μ1+b∫Ω∇v∗2dx24QM2≤c-abS36QM-b3S624QM2-aSb2S4+4aSQM6QM-b2S4b2S4+4aSQM24QM2-b4l2∫Ω∇v∗2dx=c-Λ-b4l2∫Ω∇v∗2dx<-Dλ2/1+γ-b4l2∫Ω∇v∗2dx,which contradicts inequality (59). Then, it follows that vn→v∗ in H01(Ω) as n→∞. From the above discussion, Iα satisfies the (PS)c condition. This completes the proof of Lemma 5.For allε>0, we consider uε(x)=(3ε)1/4/(ε+|x|2)1/2,x∈RN, which is a solution of the critical problem (61)-Δuε=uε5,inRN.Let(62)Cε=3ε1/4,yεx=uεxCε.Pick a cut-off function φ∈C0∞(Ω) such that φ(x)=1 for x<R0, and φ(x)=0 for x>2R0, 0≤φ(x)≤1. Set ωε(x)=φ(x)yε(x), vε(x)=ωε(x)/|ωε|6; then, |vε|66=1. Thus, we have the following results (see the details in [25]):(63)∫Ω∇vε2dx≤S+Cε1/2,∫Ω∇vε2dx3≤S3+Cε1/2,∫Ω∇vε2dx4≤S4+Cε1/2,∫Ω∇vε2dx6≤S6+Cε1/2,Oε1-γ/4≤∫Ωvε1-γdx≤Oε1-γ/4,∫Ωxα-2vε1-γdx=Oεα/2.Lemma 6.
Sinceα∈(0,1) and λ>0 sufficiently small, there exists u1∈H01(Ω) such that(64)supt≥0⁡Iαtu1<Λ-Dλ2/1+γ,where Λ and D are as in Lemma 5.Proof.
Usingλ<Λ/D(1+γ)/2, we have Λ-Dλ2/(1+γ)>0 for any λ∈(0,(Λ/D)(1+γ)/2). For Iα(tvε)→∞, as t→∞, then there exists tε>0 such that Iα(tεvε)=maxt>0⁡Iα(tvε). By Lemma 4, we obtain that maxt>0⁡Iα(tvε)≥ρ>0, and then Iα(tεvε)≥ρ>0. So, by the continuity of Iα, we suppose that there exist positive constants t0,T0 such that 0<t0≤tε≤T0. Set Iα(tεvε)=A(tεvε)+B(tεvε), where (65)Atεvε=tε22∫Ωa∇vε2+vε2dx+b4tε4∫Ω∇vε2dx2-16tε6∫ΩQxdx,Btεvε=λ1-γ∫Ωfxα1-γ-tεvε+α1-γdx-μ2tε2∫Ωxα-2vε2dx.Then,(66)Btεvε≤-μ2tε2∫Ωxα-2vε2dx.Note that(67)μ2tε2∫Ωxα-2vε2dx=c2εα2,c2>0,where c2 is independent of ε, λ. From the above information, there holds(68)supt≥0⁡Iαtvε≤Λ+c1ε1/2+Btεvε≤Λ+c1ε1/2-c2εα/2.Set ε=λ4/(1+γ), Λ0=c2/(c1+D)(1+γ)/2(1-α); when λ∈(0,Λ0), it follows that (69)supt≥0⁡Iαtvε≤Λ-Dλ2/1+γ.This completes the proof of Lemma 6.Theorem 7.
Sinceα∈(0,1), there exists λ∗>0 such that λ∈(0,λ∗), and problem (12) has at least a positive solution vα∈H01(Ω) such that Iα(vα)>ρ, where ρ is as in Lemma 2.Proof.
Setλ∗=min⁡λ0,λ0¯; then, Lemmas 4–6 hold for 0<λ<λ∗. From Lemma 4, Iα satisfies the geometry of the mountain-pass lemma. Using the mountain-pass lemma, there exists a sequence vn⊂H01(Ω), such that(70)Iαvn⟶cα>ρ,Iα′vn⟶0,where(71)cα=infγ∈Γ⁡maxt∈0,1⁡Iαγt,Γ=γ∈C0,1,H01Ω:γ0=0,γ1=u1.Furthermore, by Lemmas 4 and 6, one has(72)0<ρ<cα≤maxt∈0,1⁡Iαtu1≤supt≥0⁡Iαtu1<Λ-Dλ2/1+γ.
Using Lemma5, we obtain that {vn}⊂H01(Ω) has a convergent subsequence, still denoted by {vn}, and we may suppose that vn→vα in H01(Ω) as n→∞. Then, from (74) and (72), we get(73)Iαvα=limn→∞⁡Iαvn=cα>ρ>0.Therefore, by (73), we deduce that vα≠0. Hence, by the continuity of Iα′, we know that vα is a solution of (12); namely,(74)∫Ωa∇vα∇φ+vαφdx+b∫Ω∇vα2dx∫Ω∇vα∇φdx-μ∫Ωxα-2vαφdx-∫ΩQxvα+5φdx-λ∫Ωfxφvα+γdx=0,for all φ∈H01(Ω). We can take the test function φ=vα- in (74); one gets ∫Ω∇vα-dx=0, so vα>0 and vα≠0. From the strong maximum principle, we have that vα is a positive solution of (12). The proof is completed.


## 4. Existence of a Second Solution of Problem (1)

In this section, we obtained a second solution to problem (1) as a limit point of a family of solutions {vα} of (12).Theorem 8.
Assumeλ∈(0,λ∗), where λ∗ is as in Theorem 7; problem (1) admits a positive solution v∗ such that I0(v0)>0.Proof.
Set{vα} as a family of positive solutions of (12); we can deduce that (75)-a+b∫Ω∇vα2dxΔvα+vα=Qxvα5+μxα-2vα+λfxvα+αγ≥min⁡1,Qmmin⁡1,λfm2γ,where Qm=min⁡Q(x), fm=min⁡f(x), and then we have(76)-Δvα≥min⁡1,Qmmin⁡1,λfm/2γ-vαa+b∫Ω∇vα2dx.
Noticing thate is a weak solution of the problem (77)-Δu=1inΩ,u=0on∂Ω,hence it follows from the strong maximum principle that e>0 in Ω, and for every Ω0⊂⊂Ω, there exists e0>0 such that eΩ0≥e0. Then, from the comparison principle, we obtain (78)vα≥min⁡1,Qmmin⁡1,λfm/2γea+b∫Ω∇vα2dx+e.In particular, from eΩ0≥e0>0, we get(79)vαΩ0≥min⁡1,Qmmin⁡1,λfm/2γe0a+b∫Ω∇vα2dx+e0>0.Furthermore, from (72), (43), and (15), we have(80)Λ-Dλ2/1+γ≥Iαvα-16Iα′vα,vα=13∫Ωa∇vα2+vα2dx+b12∫Ω∇vα2dx2-μ3∫Ωxα-2vα2dx-λ1-γ∫Ωfxvα++α1-γ-α1-γdx+λ6∫Ωfxvαvα++αγdx≥a3∫Ω∇vα2dx+b12∫Ω∇vα2dx2-μ3∫Ωxα-2vα2dx-λ1-γ∫Ωfxvα+1-γdx≥13a-μμ1∫Ω∇vα2dx+b12∫Ω∇vα2dx2-λ1-γΩ5+γ/6S-1-γ/2∫Ωfx∇uα+1-γdx.Therefore, {vα} is bounded in H01(Ω), and thus there exists a subsequence, still denoted by {vα}. We suppose that there exists v0∈H01(Ω) such that (81)vα⇀v0,weakly  inH01Ω,vα⟶v0,strongly  inLpΩ1≤p<6,vαx⟶v0x,a.e.  inΩ.Currently, we shall prove that vα→v0 in H01(Ω) as α→0.
In fact, we letωα=vα-v0; then, ∫Ω∇ωdx→0 as α→0. Otherwise, there exists a subsequence (still denoted by ωα) such that limα→0⁡∫Ω∇ωαdx=l>0. Notice that 0≤vα/(vα+α)γ≤vα1-γ; then, by the Vitali convergence theorem, we have (82)limα→0⁡∫Ωfxvαvα+αγdx=∫Ωfxv01-γdx.Therefore, by 〈Iα′(vα),vα〉=0 and Brézis-Lieb Lemma, one has(83)∫Ωa∇ωα2+ωα2dx+∫Ωa∇v02+v02dx+b∫Ω∇ωα2dx2+b∫Ω∇v02dx2+2b∫Ω∇ωα2dx∫Ω∇v02dx-μ∫Ωxα-2ωα2dx-μ∫Ωxα-2v02dx-∫ΩQxωα6dx-∫ΩQxv06dx-λ∫Ωfxv01-γdx=o1.For vα⇀v0 in H01(Ω) as α→0, we can take the test function φ=ϕ∈H01(Ω)∩C0(Ω) (here C0(Ω) is the subset of C(Ω) consisting of functions with compact support in Ω); taking the limit α→0, one has v0(x)>0 (by (79)); namely,(84)∫Ωa∇v0∇ϕ+v0ϕdx+b∫Ω∇v02dx∫Ω∇v0∇ϕdx+bl2∫Ω∇v0∇ϕdx=μ∫Ωxα-2v0ϕdx+∫ΩQxv05ϕdx+λ∫Ωfxv0-γϕdx.We can assume that (84) holds for any ϕ∈H01(Ω). Indeed, since H01(Ω)∩C0(Ω) is dense in H01(Ω), for any ϕ∈H01(Ω), there exists a sequence {ϕn}∈H01(Ω)∩C0(Ω) such that ϕn→ϕ, as n→∞. Since n,m∈N+ sufficiently large, by (84) written for ϕ=ϕn-ϕm, we have(85)∫Ωa∇v0∇ϕn-ϕm+v0ϕn-ϕmdx+b∫Ω∇v02dx∫Ω∇v0∇ϕn-ϕmdx+bl2∫Ω∇v0∇ϕn-ϕmdx=μ∫Ωxα-2v0ϕn-ϕmdx+∫ΩQxv05ϕn-ϕmdx+λ∫Ωfxv0-γϕn-ϕmdx.
On the one hand, forϕn→ϕ, by (85), we can deduce that {ϕn/v0γ} is a Cauchy sequence in L1(Ω), and thus there exists η∈L1(Ω) such that ϕn/v0γ→η(x), and then we get ϕn/v0γ→η(x) in measure. From Riesz Theorem, {ϕn/v0γ} has a sequence, namely, {ϕn/v0γ}, such that(86)ϕnv0γ⟶ηx,a.e.x∈Ω.
On the other hand, forϕn/v0γ→ϕ/v0γ a.e. in Ω, then, from (86), one has ϕ/v0γ=η(x). Therefore, ∫Ω(ϕn/v0γ)dx→∫Ω(ϕ/v0γ)dx as n→∞. Thus, we can take the test function ϕ=ϕn in (84); taking the limit n→∞, hence we infer that (84) holds for ϕ∈H01(Ω).
In particular, we can takeϕ=v0 as the test function in (84); one has(87)∫Ωa∇v02+v02dx+b∫Ω∇v02dx2+bl2∫Ω∇v02dx-μ∫Ωxα-2v02dx-∫ΩQxv06dx-λ∫Ωfxv01-γdx=0.By (83) and (87), we deduce that(88)∫Ωa∇ωα2+ωα2dx+b∫Ω∇ωα2dx2+b∫Ω∇ωα2dx∫Ω∇v02dx-μ∫Ωxα-2ωα2dx-∫ΩQxωα6dx=o1.It follows from (9) and (88) that(89)a-μμ1l2+bl4+bl2∫Ω∇v02dx≤S-3l6∫ΩQxdx,and then we get(90)l2≥bS3+b2S6+4S3∫ΩQxdxa-μ/μ1+b∫Ω∇v02dx2∫ΩQxdx.Furthermore, from Brézis-Lieb Lemma and (88), we obtain that(91)Iv0=Iαvα-13∫Ωa-μμ1∇ωα2+ωα2dx-b12∫Ω∇ωα2dx2-b3l2∫Ω∇ωα2dx+o1.
On the one hand, by (87), (15), and Young’s inequality, one has(92)Iv0≥a3∫Ω∇v02dx-μ3∫Ωxα-2v02dx-λ1-γ∫Ωfxv01-γdx-b4l2∫Ω∇v02dx≥13a-μμ1∫Ω∇v02dx-λ1-γΩ5+γ/6S-1-γ/2∫Ωfx∇v01-γdx-b4l2∫Ω∇v02dx≥μa-13μ1∫Ω∇v02dx-1-γ22aμ1-μ3μ11-γ1-γ/2∫Ω∇v01-γdx2/1-γ-1+γ2λΩ5+γ/63μ11-γ1-γ/2fm2aμ1-μS1-γ/22/1+γ-b4l2∫Ω∇v02dx=-Dλ2/1+γ-b4l2∫Ω∇v02dx.
On the other hand, sinceIα(vα)<Λ-Dλ2/(1+γ), then from (91) and (90) and taking α→0, we obtained that(93)Iαv∗≤Λ-Dλ2/1+γ-13a-μμ1l2-112bl4-13bl2∫Ω∇v∗2dx≤Λ-Dλ2/1+γ-13a-μμ1bS3+b2S6+4S3QMa-μ/μ1+b∫Ω∇v∗2dx2QM-b12b2S6+2S3QMa-μ/μ1+b∫Ω∇v∗2dx+bS4b2S4+4SQMa-μ/μ1+b∫Ω∇v∗2dx2QM2-b3∫Ω∇v∗2dxbS3+b2S6+4S3QMa-μ/μ1+b∫Ω∇v∗2dx2QM≤Λ-Dλ2/1+γ-abS36QM-b3S624QM2-aSb2S4+4S3QMa-μ/μ1+b∫Ω∇v∗2dx6QM-b2S4b2S4+4S3QMa-μ/μ1+b∫Ω∇v∗2dx24QM2≤Λ-Dλ2/1+γ-abS36QM-b3S624QM2-aSb2S4+4aSQM6QM-b2S4b2S4+4aSQM24QM2-b4l2∫Ω∇v∗2dx=Λ-Dλ2/1+γ-Λ-b4l2∫Ω∇v∗2dx<-Dλ2/1+γ-b4l2∫Ω∇v∗2dx,which contradicts inequality (92). Hence, vα→v0 in H01(Ω) as α→0; that is, l=0; by (84), we deduce that v0 is a positive solution of (1).
Finally, we only indicate thatI(v0)>0. For vα→v0 as α→0, by Theorem 7, it follows that I(v0)=limα→0⁡Iα(vα)>ρ>0. This completes the proof of Theorem 8.


## 5. Conclusions

In this paper, we studied the existence and multiplicity of positive solutions for the Schrödinger-Kirchhoff-type problem with singularity and critical exponents using the variational method, perturbation method, and the mountain-pass lemma.


## 6. Discussion

In the future, a natural question is whether multiplicity of positive solutions for (1) can be established with the form of Laplacian Δpu=div⁡(∇up-2∇u) (the p-Laplacian operator of u, 1<p<3). Another generalization is whether multiplicity of positive solutions for (1) can be established with multiple space dimensions in the form of Ω⊂RN(N>3).



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