Startseite Combined effects of Choquard and singular nonlinearities in fractional Kirchhoff problems
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Combined effects of Choquard and singular nonlinearities in fractional Kirchhoff problems

  • Fuliang Wang , Die Hu und Mingqi Xiang EMAIL logo
Veröffentlicht/Copyright: 3. Dezember 2020

Abstract

The aim of this paper is to study the existence and multiplicity of solutions for a class of fractional Kirchho problems involving Choquard type nonlinearity and singular nonlinearity. Under suitable assumptions, two nonnegative and nontrivial solutions are obtained by using the Nehari manifold approach combined with the Hardy-Littlehood-Sobolev inequality.

MSC 2010: 35A15; 35B38; 35D30

1 Introduction

In this paper, we study the following Choquard-Kirchhoff problem involving singular nonlinearity:

(1.1) L ( u ) = λ f ( x ) u β + R N g ( y ) | u ( y ) | q | x y | μ d y g ( x ) u q 1 i n R N , u < 0 i n R N ,

where N2 , 1<p<N/s with s(0,1) , 1<q<pμ,s* and

L ( u ) = a + b [ u ] s , p ( θ 1 ) p ( Δ ) p s u

with a>0,b0 , θ[1,2q) and

[ u ] s , p = R 2 N | u ( x ) u ( y ) | p | x y | N + p s d x d y 1 / p .

Here μ 0 , N , p μ , s = p 2 2 N μ N p s is the upper critical exponent in the sense of Hardy-Littlewood-Sobolev inequality, λ,μ>0 are two parameters,0<β<1 , and Δ p s is the fractional p Laplacian which, up to a normalization constant, is defined for any xN as

( Δ ) p s φ ( x ) = 2 lim ε 0 R N B ε ( x ) | φ ( x ) φ ( y ) | p 2 ( φ ( x ) φ ( y ) ) | x y | N + p s d y

for any φC0(N). Here Bε(x) denotes the ball in N centered at x with radius ε>0. For further details about the fractional Laplacian and its applications, we refer to [31]. Throughout our paper, the following assumptions will be satisfied:

  • f:N+ such that fLq(N), where q = p s p s 1 + β with p s = N p N p s is the fractional critical Sobolev exponent;

  • g L 2 μ p s p s q 2 μ ( R N ) , where 2μ*=2N2Nμ.

When s = 1 and b = 0, the equation (1.1) covers the Choquard-Pekar equation which has come forth in quantum physics of a polaron at rest [39] and which describes the modeling of an electron ensnared in its own hole [26], see also [32]. Equation (1.1) is also related with the fractional Kirchoff model which was first proposed by Fiscella and Valdinoci [11]. Indeed, the study of Kirchhoff–type problems, which arise in various models of physical and biological systems, have received more and more attention in recent years. Precisely, Kirchhoff in [24] extended the classical D’Alembert wave equation by considering the effects of the changes in the length of the strings during the vibrations and established a model given by the equation

ρ 2 u t 2 M 0 L u x 2 d x 2 u x 2 = 0 ,

where M(0L|ux|2dx)=(p0h+E2L0L|ux|2dx) and ρ, p0, h, E, L are constants. For recent results about fractional Kirchhoff problems, we refer to [10, 19, 21, 23, 22, 20, 28, 29, 30, 49, 50] and the references cited there.

In recent years, much attention has been focused on the existence and properties of nontrivial solutions for fractional Choquard equation involving fractional p-Laplacian, see for example [47, 37, 41, 51]. In [37], Mukherjee and Sreenadh studied the following subcritical Choquard system involving fractional p-Laplacian and perturbations

( Δ ) p s u + a 1 ( x ) u | u | p 2 = α ( | x | μ | u | q ) | u | q 2 u + β ( | x | μ | v | q ) | u | q 2 u + f 1 ( x ) in  R n , ( Δ ) p s v + a 2 ( x ) v | v | p 2 = γ ( | x | μ | v | q ) | v | q 2 v + β ( | x | μ | u | q ) | v | q 2 v + f 2 ( x ) in  R n .

The authors proved that the system admites at least two solutions by means of Nehari manifold and minimax methods. Pucci, Xiang and Zhang in [41] discussed the following Schrödinger-Choquard-Kirchhoff type fractional p-Laplacian equations with upper critical exponents

M ( u s p ) [ ( Δ ) p s u + V ( x ) | u | p 2 u ] = λ f ( x , u ) + R N | u | p μ , s | x y | μ d y | u | p μ , s 2 u i n R N ,

where p μ , s = p N p μ / 2 / N p s is the critical exponent in the sense of Hardy–Littlewood–Sobolev inequality and

u s = [ u ] s , p p + R N V ( x ) | u | p d x 1 / p .

The authors established the existence and asymptotic behavior of solutions for above problem in the cases of f satisfies superlinear and sublinear nonlinearities, respectively. The main techniques used in the paper are the mountain pass lemma and Ekeland’s variational principle. Recently, Yang et al. [51] considered the following problem

(1.2) ( Δ ) p s ( u ) = λ | u | r 2 u | x | a + γ Ω | u | q | x y | μ d y | u | q 2 u i n Ω u = 0 i n R N Ω ,

where Ω is a bounded domain in N with Lipschitz boundary, 1<p<,0<s<1,0<μ<N,N>sp,0asp,prpa* , and p 2 q p μ , s with p a = p N a N p s . The authors analyzed the minimizer of energy functional associated to the problem (1.2) on positive Nehari and sign-changing Nehari sets, and obtained the existence of positive and sign-changing solutions for the problem (1.2). Further discussions on Choquard equation can be found in the survey papers [33, 36] and the references cited there.

On the other hand, critical and subcritical fractional problems involving singular nonlinearity have received more and more attention. In [2], Barrios etal. considered the existence of solutions for the problem

( Δ ) s u = λ f ( x ) u γ + M u p i n Ω , u < 0 i n Ω , u = 0 i n R N Ω ,

where γ>0,p>1 and M{0,1} . The existence of solutions were obtained by using the approximation method by Boccardo and Orsina. Canino etal. [6] extended the above problem to the fractional p-Laplacian and considered the following problem

( Δ ) p s u = f ( x ) u γ i n Ω , u < 0 i n Ω , u = 0 i n R N Ω ,

where γ>0 . The authors considered two cases: γ(0,1) and γ>1 . The approximation method by Boccardo and Orsina was used to get the existence of solutions. Moreover, the uniqueness and symmetry of solutions were also investigated. In [12], Fiscella and Mishra studied the following Kirchhoff problem involving singular and critical nonlinearities

(1.3) L ( u ) = λ f ( x ) u q + g ( x ) u 2 s 1 i n Ω , u = 0 i n R N Ω , u < 0 i n Ω ,

where 0 > q > 1 , L u is defined as in (1.1) with p = 2 and θ [ 1 , 2 s / 2 ) , λ < 0 , and g is a sign-changing function. The authors analyzed the fibering map and gave the compactness property of the energy functional corresponding to the problem (1.3). The authors required b small enough in order to get some key estimates of the energy functional on the Nehari manifold. When λ is small enough, the authors obtained the existence of two positive solutions by using Nehari manifold method. For fractional Kirchhoff problems with singular nonlinearity, we also refer the interested readers to [44]. Very recently, Goel and Sreenadh [16] used the similar method as in [12] to discuss the following critical Choquard-Kirchhoff type problem

(1.4) K u = λ f ( x ) | u | q 2 u + Ω | u ( y ) | 2 μ | x y | μ d y | u | 2 μ 2 u i n Ω , u = 0 o n Ω ,

where K ( u ) = a + ϵ p ( Ω | u | 2 d x ) θ 1 Δ u with a>0,p>N2(N3) and θ [ 1 , 2 μ ) . Here 0<μ<N,1<q2 and λ is a positive parameter. The authors established the existence of two positive solutions for (1.4). They applied minimization argument on the Nehari sub-manifolds to obtain the first solution. To get the second solution, the authors divided the proof into two cases: μ<min{4,N} and μmin{4,N} . Furthermore, do Ó, Giacomoni and Mishra [9] studied fractional Kirchhoff system with critical and concave-convex nonlinearities.

In present paper, we are interested in the multiplicity of solutions for fractional Kirchhoff equations with Choquard and singular nonlinearities. Since the energy functional associated to (1.1) in general is not differentiable on Ds,p(N) , the usual critical point theory is not available. Inspired by [9, 45, 46], we shall use the Nehari manifold approach to get the existence of two solutions for (1.1). Clearly, equation (1.1) is different from the problems considered in the literature, since (1.1) deals with fractional p-Kirchhoff equations with Choquard type and singular nonlinearities. Thus, our equation and result are new. Definitely, we encounter some difficulties in analyzing the fiberling map and discussing the existence of local minimizes on Nehari manifold. Our discussions are more elaborate than the papers in the literature.

To introduce the main result of this paper more precisely, we first give the definition of weak solutions.

Definition 1.1

We say that uDs,p(N) is a (weak) solution of (1.1), if f(x)uβϕL1(N) and

( a + b [ u ] s , p p ( θ 1 ) ) u , ϕ s , p = λ R N f ( x ) u β ϕ d x + R 2 N g ( x ) g ( y ) ( u ( y ) ) q | x y | μ ( u ( x ) ) q 1 ϕ ( x ) d x d y

for all ϕDs,p(N) , where

u , ϕ s , p = R 2 N | u ( x ) u ( y ) | p 2 ( u ( x ) u ( y ) ) ( ϕ ( x ) ϕ ( y ) ) | x y | N + p s d x d y .

Let

(1.5) Λ 0 := p 1 + β 2 q p a ( 2 q p ) 2 q 1 + β 2 q 1 + β p 1 + β S 2 q 1 + β p 1 + β ( C g ( N , μ ) ) 1 p 2 q p ,
(1.6) Λ 00 := a ( p 1 + β ) S 2 q p C g ( N , μ ) ( 2 q 1 + β ) p 1 + β 2 q p S 1 β p a ( 2 q p ) ( 2 q 1 + β ) f p s p s 1 + β

and

(1.7) Λ := min { Λ 0 , Λ 00 } .

Here S>0 denotes the best constant of embedding from Ds,p(N) to L p s ( R N ) and Cg(N,μ)>0 will be given by (2.2).

Our result is the following theorem.

Theorem 1.1

Let N2 , 1<p<N/s with s(0,1) , 1<q<pμ,s* and θ[1,2q) . Assume that a>0,b0 , (f1) and (g1) hold. Then equation (1.1) has at least two nonnegative and nontrivial solutions for all λ(0,Λ*).

The rest of our paper is organized as follows. In Section 2, we recall some definitions and preliminaries which will be used in our discussion. In Section 3, the properties of fibering maps are analyzed. Furthermore, a compactness result is also given. In Section 4, two nontrivial and nonnegative solutions are obtained by applying the Nehari manifold approach.

2 Preliminaries

In this section, we recall some basic results on fractional Sobolev spaces and the Hardy-Littlehood-Sobolev inequalty. For the details, we refer to [8, 42, 43, 38].

Firstly, we denote by Ds,p(N) the usual fractional Sobolev space

D s , p ( R N ) := u L p s ( R N ) : u ( x ) u ( y ) | x y | N p + s L p ( R N × R N )

endowed with the norm

[ u ] s , p = R 2 N | u ( x ) u ( y ) | p | x y | N + p s d x d y 1 / p .

Theorem 2.1

(Hardy-Littlehood-Sobolev inequality, see [27]) Assume that 1<r,t< , 0<μ<N and

1r+1t+μN=2.

Then there exists C(N,μ,r,t)>0 such that

(2.1) R 2 N | u ( x ) | | v ( y ) | | x y | μ d x d y C ( N , μ , r , t ) u r v t

for all uLr(N) and vLt(N) . If r=t=2N/(2Nμ) , then

C ( N , μ , r , t ) = C ( N , μ ) = π μ 2 Γ ( N μ 2 ) Γ ( 2 N μ 2 ) Γ ( N 2 ) Γ ( N ) 1 + μ N .

In this case, the equality in (2.1) holds if and only if u=ch and

h(x)=A(γ2+|xx0|2)(2Nμ)2

for some A , 0 γ R and x0N .

Note that, by Theorem 2.1, we know

R 2 N g ( x ) g ( y ) | u ( x ) | q | u ( y ) | q | x y | μ d x d y

is finite if g|u|qL2μ*(N) for some 2μ*>1 defined as

2 μ = 2 N 2 N μ .

Hence, by the fractional Sobolev embedding theorem, if uDs,p(N) this occurs provided that 1 > 2 μ q > p s and g L p s 2 μ p s 2 μ q ( R N ) . Thus, q has to satisfy

q > ( N μ / 2 ) p N s p = p μ , s .

Hence, p μ , s is said to be the upper critical exponent in the sense of the Hardy–Littlewood–Sobolev inequality.

By Theorem 2.1 and g L p s 2 μ p s 2 μ q ( R N ) , we have

R 2 N g ( x ) g ( y ) | u ( x ) | q | u ( y ) | q | x y | μ d x d y C ( N , μ ) g ( x ) | u | q 2 μ 2 C ( N , μ ) g p s 2 μ p s 2 μ q 2 u p s 2 q f o r a l l u D s , p ( R N ) ,

where C(N,μ)>0 is a suitable constant. Here and from now on, we shortly denote by ν for the norm of Lebesgue space Lν(N) . Further, by the fractional Sobolev inequality, we get

(2.2) R 2 N g ( x ) g ( y ) | u ( x ) | q | u ( y ) | q | x y | μ d x d y C g ( N , μ ) S 2 q p [ u ] s , p 2 q f o r a l l u D s , p ( R N ) ,

where C g ( N , μ ) = C ( N , μ ) g p s 2 μ p s 2 μ q 2 and S>0 is given by

S = inf u D s , p ( R N ) { 0 } [ u ] s , p p u p s p .

3 Fibering map analysis

First, we define the energy functional Iλ:Ds,p(N) corresponding to equation (1.1) as

I λ ( u ) = a p [ u ] s , p p + b p θ [ u ] s , p p θ λ 1 β R N f ( x ) ( u + ) 1 β d x 1 2 q R N R N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y .

Here u+=max{u,0} . For each uDs,p(N) , we define the fibering map Φλ,u:+ as

Φλ,u(t)=Iλ(tu)

for all t>0 . A simple calculation gives that

Φ λ , u ( t ) = a t p 1 [ u ] s , p p + b t θ p 1 [ u ] s , p p θ λ t β R N f ( x ) ( u + ) 1 β d x t 2 q 1 R N R N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y

and

Φ λ , u ( t ) = a ( p 1 ) t p 2 [ u ] s , p p + b ( θ p 1 ) t θ p 2 [ u ] s , p p θ + λ β t β 1 R N f ( x ) ( u + ) 1 β d x ( 2 q 1 ) t 2 q 2 R N R N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y .

Define the Nehari manifold

(3.1) N λ = { u D s , p ( R N ) : a [ u ] s , p p + b [ u ] s , p p θ λ R N f ( x ) ( u + ) 1 β d x R N R N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y = 0 } .

Then u N λ if and only if Φ λ , u ( 1 ) = 0 . Thus, it is natural to divide N λ into three parts as

(3.2) N λ + = { u N λ : Φ λ , u ( 1 ) < 0 } ,
(3.3) N λ 0 = { u N λ : Φ λ , u ( 1 ) = 0 }

and

(3.4) N λ = { u N λ : Φ λ , u ( 1 ) > 0 } .

Lemma 3.1

For uDs,p(N) and λ(0,Λ0) . Then there exist unique tmax=tmax(u)>0,t+=t+(u)>0,t=t(u)>0 with t+<tmax<t such that t + u N λ + , t u N λ and Iλ(t+u)=min0ttIλ(tu),Iλ(tu)=maxttmaxIλ(tu) .

Proof

For fixed uDs,p(N) , define ψλ,u:+ as

ψ λ , u ( t ) = a t p 2 q [ u ] s , p p + b t p θ 2 q [ u ] s , p p θ λ t β 2 q + 1 R N f ( x ) ( u + ) 1 β d x R 2 N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y .

A simple calculation gives that

(3.5) ψ λ , u ( t ) = a ( p 2 q ) t p 2 q 1 [ u ] s , p p + b ( p θ 2 q ) t p θ 2 q 1 [ u ] s , p p θ + ( β + 2 q 1 ) t β 2 q λ R N f ( x ) ( u + ) 1 β d x .

This means that ψ λ , u ( t ) = t p θ 2 q 1 h λ , u ( t ) , where

h λ , u ( t ) = a ( p 2 q ) t p p θ [ u ] s , p p + b ( p θ 2 q ) [ u ] s , p p θ + ( β + 2 q 1 ) t p θ + 1 β R N f ( x ) ( u + ) 1 γ d x .

Let h λ , u ( t ) = 0 . Then we have

t = ( β + 2 q 1 ) ( p θ + 1 β ) λ R N f ( x ) ( u + ) 1 β d x a ( 2 q p ) ( p θ p ) [ u ] s , p p 1 p 1 + β .

Observe that

lim t 0 + h λ , u ( t ) = a n d lim t + h λ , u ( t ) = b ( p θ 2 q ) [ u ] s , p p θ > 0 ,

Then there exists a unique t max < 0 such that hλ,u(tmax)=0 . Thus, we have ψ λ , u ( t max ) = 0 , ψλ,u(t) is increasing on (0,tmax) , decreasing on (tmax,+) .

In the following, we estimate ψλ,u(tmax) as follows

ψ λ , u ( t max ) = max t < 0 ψ λ , u ( t ) max t < 0 { a t p 2 q [ u ] s , p p t β 2 q + 1 λ R N f ( x ) ( u + ) 1 β d x R N R N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y } = p + β 1 2 q p 2 q p 2 q 1 + β 2 q + β 1 p + β 1 ( a [ u ] s , p p ) 2 q 1 + β p + β 1 ( λ R N f ( x ) ( u + ) 1 β d x ) 2 q p p 1 + β R 2 N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y .

It follows from the condition (f1) and Hölder’s inequality that

(3.6) λ R N f ( x ) ( u + ) 1 β d x S 1 β p λ f p s p s 1 + β [ u ] s , p 1 β .

By (3.6) and (2.2), we deduce that

ψ λ , u ( t max ) p + β 1 2 q p a ( 2 q p ) 2 q 1 + β 2 q 1 + β p 1 + β S 1 β p λ f p s p s 1 + β 2 q p p 1 + β C g ( N , μ ) S 2 q p [ u ] s , p 2 q := C ( λ ) [ u ] s , p 2 q ,

and C(λ)=0 if and only if

λ = Λ 0 := p 1 + β 2 q p a ( 2 q p ) 2 q 1 + β 2 q 1 + β p 1 + β S 2 q 1 + β p 1 + β ( C g ( N , μ ) ) 1 p 2 q p .

When λ(0,Λ0) , we have C(λ)>0 and ψλ,u(tmax)>0 . Thus we can find two zero points t+ and t of ψλ,u(t) with t+=t+(u)<tmax and t=t(u)>tmax . That is t+u,tuNλ . By (3.5), we have that ψλ,u'(t+)>0 and ψλ,u'(t)<0 . Therefore t+uNλ+ and tuNλ .

In view of 1β<pθ<2q , we derive that limt0+ψλ,u(t)= and limt+ψλ,u(t)<0 , which yield that ψλ,u(t)<0 for all t[0,t+) and t(t,+) , ψλ,u(t)>0 for all t(t+,t) . Consequently, Iλ(t+u)=min0ttIλ(tu) and Iλ(tu)=maxttmaxIλ(tu) .

Lemma 3.2

For all λ(0,Λ00) , we have Nλ0={0} .

Proof

Arguing by contradiction, we assume that there exists u N λ 0 { 0 } . Then we have

(3.7) a ( p 1 + β ) [ u ] s , p p + b ( p θ 1 + β ) [ u ] s , p p θ = ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y ,

and

(3.8) a ( 2 q p ) [ u ] s , p p + b ( 2 q p θ ) [ u ] s , p p θ = ( 2 q 1 + β ) λ R N f ( x ) ( u + ) 1 β d x .

Consider functional Tλ:Nλ defined as

T λ ( v ) = a ( p 1 + β ) [ v ] s , p p + b ( p θ 1 + β ) [ v ] s , p p θ 2 q 1 + β R 2 N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y

for all vNλ . By (3.7), we obtain that Tλ(u)=0 for all u N λ 0 { 0 } . It follows from the definition of T and (2.2) that

0 = T λ ( u ) a ( p 1 + β ) 2 q 1 + β [ u ] s , p p C g ( N , μ ) S 2 q p [ u ] s , p 2 q ,

which implies that

(3.9) [ u ] s , p a ( p 1 + β ) S 2 q p C g ( N , μ ) ( 2 q 1 + β ) 1 2 q p .

On the other hand, we deduce from (3.8) and (3.6) that

a ( 2 q p ) [ u ] s , p p > a ( 2 q p ) [ u ] s , p p + b ( 2 q p θ ) [ u ] s , p p θ λ ( 2 q 1 + β ) R N f ( x ) ( u + ) 1 β d x λ S 1 β p ( 2 q 1 + β ) f p s p s 1 + β [ u ] s , p 1 β ,

which yields that

(3.10) [ u ] s , p > S 1 β p λ ( 2 q 1 + β ) f p s p s 1 + β a ( 2 q p ) 1 p 1 + β .

Combining (3.9) with (3.10), we obtain

λ < a ( p 1 + β ) S 2 q p C g ( N , μ ) ( 2 q 1 + β ) p 1 + β 2 q p S 1 β p a ( 2 q p ) ( 2 q 1 + β ) f p s p s 1 + β := Λ 00 .

This contradicts the assumption λ<Λ00 . In conclusion, we prove that Nλ0={0} for all λ(0,Λ00) .

Lemma 3.3

Let λ(0,Λ00) . Then

[ U ] s , p < A 0 < A λ < [ u ] s , p f o r a l l u N λ + a n d U N λ ,

where

A 0 := a ( p 1 + β ) S 2 q p ( 2 q 1 + β ) C g ( N , μ ) 1 2 q p , A λ := ( 2 q 1 + β ) S 1 β p λ f p s p s 1 + β a ( 2 q p ) 1 p + β 1 .

Proof

For uNλ+ . By (f1) and (3.6), we get

a ( 2 q p ) [ u ] s , p p + b ( 2 q p θ ) [ u ] s , p p θ > ( 2 q 1 + β ) λ Ω f ( x ) ( u + ) 1 β d x ( 2 q 1 + β ) S 1 β p λ f p s p s 1 + β [ u ] s , p 1 β ,

which implies

[ u ] s , p > ( 2 q 1 + β ) S 1 β p λ f p s p s 1 + β a ( 2 q p ) 1 p + β 1 := A λ .

For UNλ . It follows from (2.2) that

a ( p 1 + β ) [ U ] s , p p + b ( p θ 1 + β ) [ U ] s , p p θ > ( 2 q 1 + β ) Ω Ω g ( x ) g ( y ) ( U + ( x ) ) q ( U + ( y ) ) q | x y | μ d x d y ( 2 q 1 + β ) C g ( N , μ ) S 2 q p [ U ] s , p 2 q ,

which leads to

[ U ] s , p < a ( p 1 + β ) S 2 q p ( 2 q 1 + β ) C g ( N , μ ) 1 2 q p := A 0 .

Clearly, A0>Aλ since λ(0,Λ00) . Thus, the proof is complete.□

Lemma 3.4

Let λ(0,Λ00) . Then Nλ is a closed subset of Nλ .□

Proof

The desired result follows by combining Lemma 3.2 and Lemma 3.3.

Lemma 3.5

For λ>0 and uNλ± . There exist R>0 and a continuous function ζ:BR(0)+ such that

ζ ( v ) < 0 , ζ ( 0 ) = 1 , ζ ( v ) ( u + v ) N λ ± f o r a l l v B R ( 0 ) ,

where B R ( 0 ) = { v D s , p ( R N ) : [ v ] s , p > R } .

Proof

We only show the proof for the case uNλ+ while the proof of the case uNλ is similar. Define F : D s , p ( R N ) × R + R as

F ( v , t ) = a t p 1 + β [ u + v ] s , p p + b t p θ 1 + β [ u + v ] p θ λ R N f ( x ) ( ( u + v ) + ) 1 β d x t 2 q 1 + β R 2 N g ( x ) g ( y ) ( ( u + v ) + ( x ) ) q ( ( u + v ) + ( y ) ) q | x y | μ d x d y .

Since uNλ+Nλ , it follows that

(3.11) F ( 0 , 1 ) = a [ u ] s , p p + b [ u ] s , p p θ λ R N f ( x ) ( u + ) 1 β d x R 2 N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y = 0.

and

(3.12) F t | ( 0 , 1 ) = a ( p 1 + β ) [ u ] s , p p + b ( p θ 1 + β ) [ u ] s , p p θ ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y < 0.

Applying the implicit function theorem at the point (0,1) , there exists R ^ < 0 such that for every v D s , p ( R N ) with [ v ] s , p > R ^ , the equation F(v,t)=0 has a unique solution t=ζ(v)>0 . By (3.11), we have ζ(0)=1 . This together with F(v,ζ(v))=0 for every vDs,p(N) with [ v ] s , p > R ^ yields that

0 = a ζ p 1 + β ( v ) [ u + v ] s , p p + b ζ p θ 1 + β ( v ) [ u + v ] s , p p θ λ R N f ( x ) ( ( u + v ) + ) 1 β d x ζ 2 q 1 + β R 2 N g ( x ) g ( y ) ( ( u + v ) + ( x ) ) q ( ( u + ψ ) + ( y ) ) q | x y | μ d x d y = 1 ζ 1 β ( v ) ( a [ ζ ( v ) ( u + v ) ] s , p p + b [ ζ ( v ) ( u + v ) ] s , p p θ λ R N f ( x ) ( ζ ( v ) ( u + v ) + ) 1 β d x R 2 N g ( x ) g ( y ) ( ζ ( v ) ( u + v ) + ( x ) ) q ( ζ ( v ) ( u + v ) + ( y ) ) q | x y | μ d x d y ) ,

which means that

ζ ( v ) ( u + v ) N λ f o r a l l v D s , p ( R N ) w i t h [ v ] s , p > R ^ .

Observe by (3.12) that

F t | ( v , ζ ( v ) ) = a ( p 1 + β ) ζ p 2 + β ( v ) [ u + v ] s , p p + b ( p θ 1 + β ) ζ p θ 2 + β ( v ) [ u + v ] s , p p θ ( 2 q 1 + β ) ζ p θ 2 + β ( v ) R 2 N g ( x ) g ( y ) ( ( u + v ) + ( x ) ) q ( ( u + v ) + ( y ) ) q | x y | μ d x d y = 1 ζ 2 β ( v ) [ a ( p 1 + β ) [ ζ ( v ) ( u + v ) ] s , p p + b ( p θ 1 + β ) [ ζ ( v ) ( u + v ) ] s , p p θ ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( ζ ( v ) ( u + v ) + ( x ) ) q ( ζ ( v ) ( u + v ) + ( y ) ) q | x y | μ d x d y ] .

Then we can choose sufficiently small R ( 0 , R ^ ) such that

a ( p 1 + β ) [ ζ ( v ) ( u + v ) ] s , p p + b ( p θ 1 + β ) [ ζ ( v ) ( u + v ) ] s , p p θ

( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( ζ ( v ) ( u + v ) + ( x ) ) q ( ζ ( v ) ( u + v ) + ( y ) ) q | x y | μ d x d y < 0 ,

for all vDs,p(N) with [v]s,p<R . Thus,

ζ ( v ) ( u + v ) N λ + f o r a l l v B R ( 0 ) .

The proof is completed.□

Lemma 3.6

The functional Iλ is coercive and bounded from below on Nλ .

Proof

For uNλ , it follows from θp<2q and (3.6) that

I λ ( u ) = a ( 2 q p ) 2 p q [ u ] s , p p + b ( 2 q p θ ) 2 p θ q [ u ] s , p p θ 1 1 β 1 2 q λ R N f ( x ) ( u + ) 1 β d x a ( 2 q p ) 2 p q [ u ] s , p p 1 1 β 1 2 q S β 1 p λ f p s p s 1 + β [ u ] s , p 1 β .

Since p>1β , we know that Iλ is coercive on Nλ . Define

G ( t ) = a ( 2 q p ) 2 p q t p 1 1 β 1 2 q S β 1 p λ f p s p s 1 + β t 1 β ,

then G(t) attains its minimum at

t min = 2 q 1 + β a ( 2 q p ) S β 1 p λ f p s p s 1 + β 1 p 1 + β .

Thus

I λ ( u ) a ( 2 q p ) 2 p q β 1 p + β 1 1 1 β 2 q S β 1 p λ f p s p s 1 + β p p p + β 1 1 β p 1 β .

Therefore, we get that Iλ is bounded from below on Nλ .□

Set

c λ + = inf u N λ + { 0 } I λ ( u ) a n d c λ = inf u N λ I λ ( u ) .

By Lemma 3.2 and 3.4, we know that Nλ+{0} and Nλ are two closed sets in E. Using Ekeland’s variational principle 1, we can extract a minimizing sequence (un)nNλ+{0}(Nλ) with

(3.13) c λ + ( c λ ) > I λ ( u n ) > c λ + ( c λ ) + 1 n , I λ ( v ) I λ ( u n ) 1 n [ v u n ] s , p

for all vNλ+{0}(Nλ) . In view of Lemma 3.6, we derive that the sequence (un)n is bounded in Nλ with [un]s,pC1 for some C1>0 . Thus, up to a subsequence still denoted by (un)n we may assume that there exists u0Ds,p(N) such that

(3.14) u n u 0 w e a k l y i n D s , p ( R N ) u n u 0 a . e . i n R N .

Lemma 3.7

For λ(0,Λ0) . Then there exists a constant C2>0 such that the following conclusions hold:

  1. if (un)nNλ+ , for each n , we have

    a ( p 1 + β ) [ u n ] s , p p + b ( p θ 1 + β ) [ u n ] s , p p θ ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y C 2 ;
  2. if (un)nNλ , for each n , we have

    a ( p 1 + β ) [ u n ] s , p p + b ( p θ 1 + β ) [ u n ] s , p p θ ( 2 q 1 + β R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y C 2 .

Proof

We only prove the case (i), while the proof of (ii) follows similarly. Since (un)nNλ+ , it suffices to prove that

(3.15) lim inf n a ( 2 q p ) [ u n ] s , p p + b ( 2 q p θ ) [ u n ] s , p p θ > ( 2 q 1 + β ) λ R N f ( x ) ( u 0 + ) 1 β d x .

Arguing by contradiction, we assume that

lim inf n a ( 2 q p ) [ u n ] s , p p + b ( 2 q p θ ) [ u n ] s , p p θ = ( 2 q 1 + β ) λ R N f ( x ) ( u 0 + ) 1 β d x .

Since (un)nNλ+ , we have

a ( 2 q p ) [ u n ] s , p p + b ( 2 q p θ ) [ u n ] s , p p θ > ( 2 q 1 + β ) λ R N f ( x ) ( u n + ) 1 β d x .

By f L p s p s 1 + β ( R N ) and Vitali’s convergence theorem, one can prove that

lim n R N f ( x ) ( u n + ) 1 β d x = R N f ( x ) ( u 0 + ) 1 β d x

Then

lim inf n a ( 2 q p ) [ u n ] s , p p + b ( 2 q p θ ) [ u n ] s , p p θ > lim sup n a ( 2 q p ) [ u n ] s , p p + b ( 2 q p θ ) [ u n ] s , p p θ ( 2 q 1 + β ) λ R N f ( x ) ( u 0 + ) 1 β d x ,

which yields that

lim n a ( 2 q p ) [ u n ] s , p p + b ( 2 q p θ ) [ u n ] s , p p θ = ( 2 q 1 + β ) λ R N f ( x ) ( u 0 + ) 1 β d x .

Hence, there exists A>0 such that [un]s,ppA as n . Consequently, we get

(3.16) a ( 2 q p ) A + b ( 2 q p θ ) A θ = ( 2 q 1 + β ) λ R N f ( x ) ( u 0 + ) 1 β d x .

By Lemma 3.1, for λ(0,Λ0) , we have

0 > p 1 + β 2 q p 2 q p 2 q 1 + β 2 q 1 + β p 1 + β ( a [ u n ] s , p p ) 2 q 1 + β p 1 + β ( λ R N f ( x ) ( u n + ) 1 β d x ) 2 q p p 1 + β R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y .

Hence

(3.17) 0 p 1 + β 2 q p 2 q p 2 q 1 + β 2 q 1 + β p 1 + β ( a A ) 2 q p p 1 + β ( λ R N f ( x ) ( u 0 + ) 1 β d x ) 2 q p p 1 + β lim n R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y .

Since (un)nNλ+ , it follows from (3.16) that

(3.18) lim n R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y = a A ( p 1 + β ) 2 q 1 + β + b A θ ( p θ 1 + β ) 2 q 1 + β .

Substituting (3.16) and (3.18) into (3.17), we arrive at

p 1 + β 2 q p 2 q p 2 q 1 + β 2 q 1 + β p 1 + β ( a A ) 2 q 1 + β p 1 + β [ a A ( 2 q p ) 2 q 1 + β + b A θ ( 2 q p θ ) 2 q 1 + β ] p 2 q p 1 + β a A ( p 1 + β ) 2 q 1 + β b A θ ( p θ 1 + β ) 2 q 1 + β 0 ,

this together with 2q>p implies that

p 1 + β 2 q p 2 q p 2 q 1 + β 2 q 1 + β p 1 + β ( a A ) 2 q 1 + β p 1 + β [ a A ( 2 q p ) 2 q 1 + β ] p 2 q p 1 + β a A ( p 1 + β ) 2 q 1 + β b A θ ( p θ 1 + β ) 2 q 1 + β 0.

It follows that

b A θ ( p θ 1 + β ) 2 q 1 + β a A ( p 1 + β ) 2 q 1 + β p 1 + β 2 q p 2 q p 2 q 1 + β 2 q 1 + β p 1 + β ( a A ) 2 q 1 + β p 1 + β a A ( 2 q p ) 2 q 1 + β p 2 q p 1 + β = 0 ,

which is impossible. Therefore, the proof is complete.□

For φDs,p(N) with φ0 . Recalling the constants C1>0 with [un]s,pC1 and C2>0 given in Lemma 3.7. Thus, for n sufficiently large so that

(3.19) 1 β n C 1 C 2 .

Lemma 3.5 guarantees to extract a sequence of functions (ζn)n satisfying ζn(0)=1 and ζn(tφ)(un+tφ)Nλ± for t>0 sufficiently small. By (un)nNλ and ζn(tφ)(un+tφ)Nλ , we get

(3.20) a [ u n ] s , p p + b [ u n ] s , p p θ λ R N f ( x ) ( u n + ) 1 β d x R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y = 0

and

(3.21) a ζ n p ( t φ ) [ u n + t φ ] s , p p + b ζ n p θ ( t φ ) [ u n + t φ ] s , p p θ λ ζ n 1 β ( t φ ) R N f ( x ) ( ( u n + t φ ) + ) 1 β d x ζ n 2 q ( t φ ) R 2 N g ( x ) g ( y ) ( ( u n + t φ ) + ( x ) ) q ( ( u n + t ψ ) + ( y ) ) q | x y | μ d x d y = 0 ,

where ζn'(0) denotes the derivative of ζn' at the point 0 with (ζn'(0),φ) for every φDs,p(N) . If it does not exist, ζn'(0) should be replaced by limkζn(tnφ)1tn for some sequence ( t n ) n = 1 with limntn=0 and tn>0 .

Lemma 3.8

Let λ(0,Λ0) and (un)nNλ± . Then (ζn'(0),φ) is uniformly bounded for every φDs,p(N) with φ0 .

Proof. We only prove the case that (un)nNλ+ . In view of (3.20) and (3.21), we have

0 = a [ ζ n p ( t φ ) 1 ] [ u n + t φ ] s , p p + a ( [ u n + t φ ] s , p p [ u n ] s , p p ) + b [ ζ n p θ ( t φ ) 1 ] [ u n + t φ ] s , p p θ + b ( [ u n + t φ ] s , p p θ [ u n ] s , p p θ ) [ ζ n 1 β ( t φ ) 1 ] λ R N f ( x ) ( ( u n + t φ ) + ) 1 β d x λ R N f ( x ) [ ( ( u n + t φ ) + ) 1 β ( u n + ) 1 β ] d x
[ ζ n 2 q ( t φ ) 1 ] R 2 N g ( x ) g ( y ) ( ( u n + t φ ) + ( x ) ) q ( ( u n + t φ ) + ( y ) ) q | x y | μ d x d y R 2 N g ( x ) g ( y ) ( ( u n + t φ ) + ( x ) ) q ( ( u n + t φ ) + ( y ) ) q | x y | μ ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y .

Dividing the above estimate by t>0 and letting t0+ , we obtain

0 p a ( ζ n ( 0 ) , φ ) [ u n ] s , p p + p a u n , φ s , p + p b θ ( ζ n ( 0 ) , φ ) [ u n ] s , p p θ + p b θ [ u n ] s , p p ( θ 1 ) u n , φ s , p ( 1 β ) ( ζ n ( 0 ) , φ ) λ R N f ( x ) ( u n + ) 1 β d x 2 q ( ζ n ( 0 ) , φ ) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y q R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 φ ( x ) ( u n + ( y ) ) q | x y | μ d x d y q R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q φ ( y ) ( u n + ( y ) ) q 1 | x y | μ d x d y = ( ζ n ( 0 ) , φ ) ( p a [ u n ] s , p p + p b θ [ u n ] s , p p θ ( 1 β ) λ R N f ( x ) ( u n + ) 1 β d x 2 q R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y ) + ( p a + p b θ [ u n ] s , p p θ 1 ) u n , φ s , p 2 q R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 φ ( x ) ( u n + ( y ) ) q | x y | μ d x d y .

By (3.20), we have

0 ( ζ n ( 0 ) , φ ) ( a ( p 1 + β ) [ u n ] s , p p + b ( p θ 1 + β ) [ u n ] s , p p θ ( 2 q 1 + β ) R 2 N ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y ) + ( p a + p b θ [ u n ] s , p p ( θ 1 ) ) u n , φ s , p 2 q R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 φ ( x ) ( u n + ( y ) ) q | x y | μ d x d y .

Then by Lemma 3.7-(i) and the boundedness of the sequence (un)n , we obtain that (ζn'(0),φ) is bounded from below for every φDs,p(N) with φ0 .

Next, we show that (ζn'(0),φ) is bounded from above. Arguing by contradiction, we assume that (ζn'(0),φ)= . Since

(3.22) | ζ n ( t φ ) 1 | [ u n ] s , p + ζ n ( t φ ) [ t φ ] s , p [ ( ζ n ( t φ ) 1 ) u n + ζ n ( t φ ) t φ ] s , p = [ ζ n ( t φ ) ( u n + t φ ) u n ] s , p

and ζn(tφ)>ζn(0)=1 for sufficiently large n. By the definition of ζn'(0) and (3.13), we get

[ ζ n ( t φ ) 1 ] [ u n ] s , p n + ζ n ( t φ ) [ t φ ] s , p n 1 n [ ζ n ( t φ ) ( u n + t φ ) u n ] s , p I λ ( u n ) I λ [ ζ n ( t φ ) ( u n + t φ ) ] = a 1 1 β 1 p ( [ u n + t φ ] s , p p [ u n ] s , p p ) + b 1 1 β 1 p θ ( [ u n + t φ ] s , p p θ [ u n ] s , p p θ ) + a 1 1 β 1 p ( ζ n p ( t φ ) 1 ) [ u n + t φ ] s , p p + b 1 1 β 1 p θ [ ζ n p θ ( t φ ) 1 ] [ u n + t φ ] s , p p θ 1 1 β 1 2 q ζ n 2 q ( t φ ) R 2 N g ( x ) g ( y ) [ ( ( u n + t φ ) + ( x ) ) q ( ( u n + t φ ) + ( y ) ) q | x y | μ ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ ] d x d y
1 1 β 1 2 q ( ζ n 2 q ( t φ ) 1 ) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y .

Dividing above inequality by t>0 and letting t0+ , we get

( ζ n ( 0 ) , φ ) [ u n ] s , p n + [ φ ] s , p n a p 1 + β 1 β + b p θ 1 + β 1 β [ u n ] s , p p ( θ 1 ) u n , φ s , p + a p 1 + β 1 β ( ζ n ( 0 ) , φ ) [ u n ] s , p p + b p θ 1 + β 1 β ( ζ n ( 0 ) , φ ) [ u n ] s , p p θ 2 q 1 + β 1 β R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 φ ( x ) ( u n + ( y ) ) q | x y | μ d x d y 2 q 1 + β 1 β ( ζ n ( 0 ) , φ ) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y = ( ζ n ( 0 ) , φ ) 1 β [ a ( p 1 + β ) [ u n ] s , p p + b ( p θ 1 + β ) [ u n ] s , p p θ ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y ] + ( a p 1 + β 1 β + b p θ 1 + β 1 β [ u n ] s , p p ( θ 1 ) ) u n , φ s , p 2 q 1 + β 1 β R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 φ ( x ) ( u n + ( y ) ) q | x y | μ d x d y .

That is

[ φ ] s , p n ( ζ n ( 0 ) , φ ) 1 β [ a ( p 1 + β ) [ u n ] s , p p + b ( p θ 1 + β ) [ u n ] s , p p θ ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y 1 β n [ u n ] s , p ] + a p 1 + β 1 β + b p θ 1 + β 1 β [ u n ] s , p p ( θ 1 ) u n , φ s , p 2 q 1 + β 1 β R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 φ ( x ) ( u n + ( y ) ) q | x y | μ d x d y ,

which contradicts with our assumption that (ζn'(0),φ)= . From Lemma 3.7 and (3.19), we have

a ( p 1 + β ) [ u n ] s , p p + b ( p θ 1 + β ) [ u n ] s , p p θ ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y 1 β n [ u n ] s , p C 2 ( 1 β ) n C 1 < 0.

This ends the proof.□

Lemma 3.9

For λ(0,Λ0) and (un)nNλ± . Then, for every φDs,p(N) and n , we have

f ( x ) ( u n + ) β φ L 1 ( R N ) ,

and

( a + b [ u n ] s , p p ( θ 1 ) ) u n , φ s , p λ R N f ( x ) ( u n + ) β φ d x R 2 N g ( x ) g ( y ) ( u n + ( y ) ) q ( u n + ( x ) ) q 1 φ ( x ) | x y | μ d x d y
(3.23):= H ( u n , φ ) = o n ( 1 ) ,

as n .

Proof

Let φDs,p(N) with φ0 . By (3.13) and (3.22), we have

( ζ n ( t φ ) 1 ) [ u n ] s , p n + ζ n ( t φ ) [ t φ ] s , p n I λ ( u n ) I λ ( ζ n ( t φ ) ( u n + t φ ) ) = ζ n p ( t φ ) 1 p a [ u n ] s , p p θ ζ n p ( t φ ) 1 p b [ u n ] s , p p θ ζ n p ( t φ ) p a ( [ u n + t φ ] s , p p [ u n ] s , p p ) ζ n p θ ( t φ ) p θ b ( [ u n + t φ ] s , p p θ [ u n ] s , p p θ ) + ζ n 1 β ( t φ ) 1 1 β λ R N f ( x ) ( ( u n + t φ ) + ) 1 β d x + λ 1 β R N f ( x ) [ ( ( u n + t φ ) + ) 1 β ( u n + ) 1 β ] d x + ζ n 2 q ( t φ ) 1 2 q R 2 N g ( x ) g ( y ) ( ( u n + t φ ) + ( x ) ) q ( ( u n + t φ ) + ( y ) ) q | x y | μ d x d y + 1 2 q R 2 N g ( x ) g ( y ) ( ( u n + t φ ) + ( x ) ) q ( ( u n + t φ ) + ( y ) ) q ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y .

Dividing by t < 0 and letting t0+ , we obtain

| ( ζ n ( 0 ) , φ ) | [ u n ] s , p n + [ φ ] s , p n a ( ζ n ( 0 ) , φ ) [ u n ] s , p p b ( ζ n ( 0 ) , φ ) [ u n ] s , p p θ ( a + b [ u n ] s , p p ( θ 1 ) ) u n , φ s , p + ( ζ n ( 0 ) , φ ) λ R N f ( x ) ( u n + ) 1 β d x + lim inf t 0 + λ 1 β R N f ( x ) ( ( u n + t φ ) + ) 1 β ( u n + ) 1 β t d x + ( ζ n ( 0 ) , φ ) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y + R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + ( y ) ) q φ ( x ) | x y | μ d x d y .

Since unNλ , it follows that

| ( ζ n ( 0 ) , φ ) | [ u n ] s , p n + [ φ ] s , p n ( a + b [ u n ] s , p p ( θ 1 ) ) u n , φ s , p + lim inf t 0 + λ 1 β R N f ( x ) ( ( u n + t φ ) + ) 1 β ( u n + ) 1 β t d x + R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + ( y ) ) q φ ( x ) | x y | μ d x d y

Note that

f ( x ) ( ( u n + t φ ) + ) 1 β ( u n + ) 1 β t 0

in N and u n n is bounded in Ds,p(N) . Using Fatou’s lemma and Lemma 3.8, we know that

lim inf t 0 + R N f ( x ) ( ( u n + t φ ) + ) 1 β ( u n + ) 1 β t d x

is finite, and

λ R N f ( x ) ( u n + ) β φ d x lim inf t 0 + λ 1 β R N f ( x ) ( ( u n + t φ ) + ) 1 β ( u n + ) 1 β t d x 1 n | ( ζ n ( 0 ) , φ ) | [ u n ] s , p + [ φ ] s , p + ( a + b [ u n ] s , p p ( θ 1 ) ) u n , φ s , p R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + ( y ) ) q φ ( x ) | x y | μ d x d y 1 n ( C 1 C 3 + [ φ ] s , p ) + ( a + b [ u n ] s , p p ( θ 1 ) ) u n , φ s , p R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + ( y ) ) q φ ( x ) | x y | μ d x d y

with C3>0 given by the boundedness of ζ n ' 0 , φ , which implies that

(3.24) ( a + b [ u n ] s , p p ( θ 1 ) ) u n , φ s , p λ R N f ( x ) ( u n + ) β φ d x R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + ( y ) ) q φ ( x ) | x y | μ d x d y o n ( 1 ) .

In the following, we prove that (3.24) holds for any φDs,p(N) . In (3.24), we choose φ=ωϵ+ with ωϵ=un++ϵφ and φDs,p(N) as test function. Then

(3.25) o n ( 1 ) ( a + b [ u n ] s , p p ( θ 1 ) ) u n , ω ϵ + ω ϵ s , p λ R N f ( x ) ( u n + ) 1 β ( ω ϵ + ω ϵ ) d x R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + ( y ) ) q ( ω ϵ + ω ϵ ) ( x ) | x y | μ d x d y .

Observe that

u n , ω ϵ + ω ϵ s , p [ u n ] s , p p + ϵ u n , φ s , p + u n , ω ϵ s , p .

Hence, denote Ωϵ={xN:ωϵ(x)0} . In view of (3.35) and (3.3), we deduce n

o n ( 1 ) ( a + b [ u n ] s , p p ( θ 1 ) ) [ u n ] s , p p λ R N f ( x ) ( u n + ) 1 β d x R N R N g ( x , y ) ( u n + ( x ) ) p μ , s ( u n + ( y ) ) p μ , s | x y | μ d x d y ] + ϵ [ ( a + b [ u n ] s , p p ( θ 1 ) ) u n , φ s , p λ R N f ( x ) ( u n + ) β φ d x R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + ( y ) ) q φ ( x ) | x y | μ d x d y ] + ( a + b [ u n ] s , p p ( θ 1 ) ) u n , ω ϵ s , p + R N Ω ϵ g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + + ϵ φ ) ( x ) ( u n + ( y ) ) q | x y | μ d x d y = ϵ [ ( a + b [ u n ] s , p p ( θ 1 ) ) u n , φ s , p λ R N f ( x ) ( u n + ) β φ d x
(3.26) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + ( y ) ) q φ ( x ) | x y | μ d x d y ] + ( a + b [ u n ] s , p p ( θ 1 ) ) u n , ω ϵ s , p + R N Ω ϵ g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + + ϵ φ ) ( x ) ( u n + ( y ) ) q | x y | μ d x d y .

Furthermore,

u n , ω ϵ s , p = Ω ϵ × Ω ϵ | u n ( x ) u n ( y ) | p 2 ( u n ( x ) u n ( y ) ) ( ω ϵ ( x ) ω ϵ ( y ) ) | x y | N + p s d x d y + 2 Ω ϵ × ( R N Ω ϵ ) | u n ( x ) u n ( y ) | p 2 ( u n ( x ) u n ( y ) ) ( ω ϵ ( x ) ω ϵ ( y ) ) | x y | N + p s d x d y 2 ϵ Ω ϵ × R N | u n ( x ) u n ( y ) | p 2 | ( u n ( x ) u n ( y ) ) ( φ ( x ) φ ( y ) ) | | x y | N + p s d x d y 2 C ϵ Ω ϵ × R N φ ( x ) φ ( y ) | x y | N + p s p p d x d y 1 P .

For any δ < 0 , there exists Rδ large enough such that

( s u p p φ ) × ( R N B R δ ) φ ( x ) φ ( y ) | x y | N + p s p p d x d y > δ 2 .

In view of the definition of Ω1,ϵ , we have Ω1,ϵsuppφ and | Ω 1 , ϵ × B R δ | 0 as ϵ0+ . Hence, there exist κδ>0 and ϵδ>0 such that for every ϵ(0,ϵδ] ,

| Ω 1 , ϵ × B R δ | > κ δ , a n d Ω ϵ × B R δ φ ( x ) φ ( y ) | x y | N + p s p p d x d y > δ 2 .

Consequently, for every ϵ(0,ϵδ] ,

Ω ϵ × R N φ ( x ) φ ( y ) | x y | N + p s p p d x d y > δ ,

it follows that

(3.27) lim ϵ 0 + Ω ϵ × R N φ ( x ) φ ( y ) | x y | N + p s p p d x d y = 0.

Now, we show that

(3.28) lim ϵ 0 + 1 ϵ R N Ω ϵ g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + + ϵ φ ) ( x ) ( u n + ( y ) ) q | x y | μ d x d y = 0.

In fact,

R N Ω ϵ g ( x ) g ( y ) ( u n + ( x ) ) q 1 ( u n + + ϵ φ ) ( x ) ( u n + ( y ) ) q | x y | μ d x d y R N Ω ϵ g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y + ϵ R N Ω ϵ g ( x ) g ( y ) ( u n + ( x ) ) q 1 φ ( x ) ( v n + ( y ) ) q | x y | μ d x d y
+ ϵ Ω ϵ Ω ϵ g ( x ) g ( y ) ( u n + ( x ) ) q 1 φ ( x ) ( u n + ( y ) ) q 1 φ ( y ) | x y | μ d x d y 1 2 R N R N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y 1 2 C C g ( N , μ ) Ω ϵ ( u n + ( x ) ) p s d x q p s + C ϵ C g ( N , μ ) Ω ϵ ( ( u n + ( x ) ) q 1 φ ( x ) ) p s q d x q p s C C g ( N , μ ) Ω ϵ ( u n + ( x ) ) p s d x q p s + C ϵ C g ( N , μ ) Ω ϵ ( u n + ( x ) ) p s ) d x q 1 p s Ω ϵ | φ ( x ) | p s d x 1 p s C C g ( N , μ ) ϵ q Ω 1 , ϵ | φ ( x ) | p s d x q p s + C ~ ϵ C g ( N , μ ) Ω ϵ | φ ( x ) | p s d x 1 p s

Dividing by ϵ the above estimate and using that |Ωϵ|0 as ϵ0+ , we can get (3.28).

Dividing (3.26) by ϵ , together with (3.27)-(3.28), and letting ϵ0+ , we deduce that (3.24) holds. According to the arbitrariness of φ , we show that (3.23) holds.□

Lemma 3.10

For λ(0,Λ0) , u n n N λ ± , and Iλ(un)c as n . Then, the sequence u n n contains a subsequence strongly convergent to u0 in Ds,p(N) .

Proof

By (3.14), u n n and u n n are both bounded in Ds,p(N) . Using (3.23) with φ=un as n , we have

lim n ( a + b [ u n ] s , p p ( θ 1 ) ) u n , u n s , p = 0 ,

which together with a < 0 yields that u n s , p 0 as n . Consequently, we can assume that u n n is a sequence of nonnegative functions. Furthermore, we can extract a subsequence, still denoted by u n n such that

(3.29) u n u 0 0 i n L p s ( R N ) , [ u n ] s , p χ , u n u 0 i n L l o c ν ( R N ) , f o r e a c h ν [ 1 , p s ) , u n u 0 a . e . i n R N ,

as n . If χ=0 , then un0 in Ds,p(N) as n . Hence, we assume that χ < 0 . By [4, Lemma 3.2], [40, Lemma 2.4], and [34, Lemma 2.3] we obtain

(3.30) [ u n ] s , p p = [ u n u 0 ] s , p p + [ u 0 ] s , p p + o ( 1 ) ,

and

(3.31) R 2 N g ( x ) g ( y ) ( u n + ( x ) ) q ( u n + ( y ) ) q | x y | μ d x d y = R 2 N g ( x ) g ( y ) ( ( u n ( x ) u 0 ( x ) ) + ) q ( ( u n ( y ) u 0 ( y ) ) + ) q | x y | μ d x d y + R 2 N g ( x ) g ( y ) ( u 0 + ( x ) ) q ( u 0 + ( y ) ) q | x y | μ d x d y + o ( 1 ) .

It follows from (3.29), (3.30) and (3.31) that

o ( 1 ) = ( a + b [ u n ] s , p p ( θ 1 ) ) u n , u n u 0 s , p λ R N f ( x ) ( u n + ) β ( u n u 0 ) d x R N R N g ( x ) g ( y ) ( ( u n ( x ) ) + ) q 1 ( ( u n ( y ) ) + ) q | x y | μ ( u n ( x ) u 0 ( x ) ) d x d y = ( a + b χ p ( θ 1 ) ) ( χ p [ u 0 ] s , p p ) λ Ω f ( x ) ( u n + ) β ( u n u 0 ) d x R 2 N g ( x ) g ( y ) ( ( u n ( x ) ) + ) q ( ( u n ( y ) ) + ) q | x y | μ d x d y R 2 N g ( x ) g ( y ) ( ( u 0 ( x ) ) + ) q ( ( u 0 ( y ) ) + ) q | x y | μ d x d y + o ( 1 ) = ( a + b χ p ( θ 1 ) ) [ u n u 0 ] s , p p λ R N f ( x ) ( u n + ) β ( u n u 0 ) d x R 2 N g ( x ) g ( y ) ( ( u n ( x ) u 0 ( x ) ) + ) q ( ( u n ( y ) u 0 ( y ) ) + ) q | x y | μ d x d y + o ( 1 ) .

Hence

(3.32) ( a + b χ p ( θ 1 ) ) lim n [ u n u 0 ] s , p p = λ lim n R N f ( x ) ( u n + ) β ( u n u 0 ) d x + lim n R 2 N g ( x ) g ( y ) ( ( u n ( x ) u 0 ( x ) ) + ) q ( ( u n ( y ) u 0 ( y ) ) + ) q | x y | μ d x d y .

Since β(0,1) and f L p s p s 1 + β R N , we deduce from the Vitali’s convergence theorem that

(3.33) lim n R N f ( x ) ( u n + ) 1 β d x = R N f ( x ) ( u 0 + ) 1 β d x .

By virtue of Lemma 3.9, we obtain

f ( x ) u n β u 0 L 1 ( R N )

for each n . It follows from Fatou’s lemma that

(3.34) lim n R N λ f ( x ) ( u n + ) β u 0 d x λ R N f ( x ) ( u 0 + ) 1 β d x .

Similarly, by g L p s 2 μ p s 2 μ q R N and Vitali’s convergence theorem, one can deduce that

g ( x ) | u n u 0 | q 0 s t r o n g l y i n L 2 μ ( R N ) .

Furthermore, it follows from (2.2) that

(3.35) lim n R 2 N g ( x ) g ( y ) ( ( u n ( x ) u 0 ( x ) ) + ) q ( ( u n ( y ) u 0 ( y ) ) + ) q | x y | μ d y d x = 0.

From (3.32)-(3.35), we get

(3.36) ( a + b χ p ( θ 1 ) ) lim n [ u n u 0 ] s , p p = 0 ,

which together with a>0 yields that unu0 strongly in Ds,p(N) .

4 Proof of the main result

Theorem 4.1

Let 0<λ<Λ*=min{Λ0,Λ00} . Assume f satisfies (f1) and g satisfies (g1) . Then equation (1.1) has a nontrivial and nonnegative solution in Nλ+

Proof

Firstly, we show that cλ+=infuNλ+Iλ(u)<0 . In fact, for λ>0 and uNλ+Nλ , we obtain

I λ ( u ) = a 1 p 1 1 β [ u n ] s , p p + b 1 p θ 1 1 β [ u n ] s , p p θ 1 2 q 1 1 β R 2 N g ( x ) g ( y ) ( u + ( x ) ) q ( u + ( y ) ) q | x y | μ d x d y .

Since 2q>θp and Iλ(0)=0 , we obtain

I λ ( u ) 1 p q ( 1 β ) [ a ( p 1 + β ) [ u n ] s , p p + b ( p θ 1 + β ) [ u n ] s , p p θ p ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( u + ( x ) ) q ( v + ( y ) ) q | x y | μ d x d y ] > 0.

Now, fix λ > Λ = m i n Λ 0 , Λ 00 . According to Ekeland’s variational principle and Lemma 3.2, there exists a minimizing sequence u n n N λ + 0 satisfying (3.11) and (3.12). Hence I λ u n c λ + > 0 as n , which yields that u n n N λ + . As a consequence, using Lemma 3.10 with c=cλ+ , we know that unu0 in Ds,p(N) , up to a subsequence. Furthermore, according to Lemma 3.7 and (3.12), we get

a ( p 1 + β ) [ u 0 ] s , p p + b ( p θ 1 + β ) [ u 0 ] s , p p θ p ( 2 q 1 + β ) R 2 N g ( x ) g ( y ) ( u 0 + ( x ) ) q ( u 0 + ( y ) ) q | x y | μ d x d y < 0 ,

which means that u0Nλ+ , and cλ+ is attained at u0 by Iλ is continuous on Ds,p(N) .

Letting n in (3.23), together with Fatou’s lemma, we obtain the inequality H(u0,φ)0 for every test function φDs,p(N) with φ0 . Taking a test function Φ ϵ + in above inequality with ωϵ=u0++ϵφ,ϵ>0 and φDs,p(N) . Repeating the same discussion as (3.23)-(3.28) with u0 instead of un , we know that H(u0,φ)0 for any φDs,p(N) , which yields that λ f ( x ) ( u 0 + ) β φ L 1 ( R N ) for any φDs,p(N) and u0 satisfies (3.2). It follows from Lemma 3.2 that u 0 0 . Moreover, Using (3.2) with test function φ 1 = u 0 , we deduce that ( a + b [ u 0 ] s , p p ( θ 1 ) ) [ u 0 ] s , p p = 0 , which means that u0 is a nontrivial and nonnegative solution of equation (1.1).□

Theorem 4.2

Let 0 > λ > Λ = m i n Λ 0 , Λ 00 . Assume f satisfies (f1) and g satisfies (g1) . Then equation (1.1) has a nontrivial and nonnegative solution in N λ

Proof

Since Nλ is a closed set in Ds,p(N) , we can extract the minimizing sequence U n n N λ satisfying the Ekeland variational principle for infuNλIλ(u) . Since U n n is bounded in Ds,p(N) , after taking a subsequence, we suppose that U n n U 0 in Ds,p(N) . By Lemma 3.10, we know that UnU0 in Ds,p(N) , up to a subsequence, if necessary, so that UnNλ and Iλ(U0)=cλ . Repeating the same argument as in the proof of Theorem 4.1, U0 satisfies H(U0,φ)0 , so that λ f ( x ) ( U 0 + ) β φ L 1 ( R N ) for any φDs,p(N) and U0 satisfies (3.2). Combining this with Lemma 3.2, we deduce that U0 is a nontrivial solution of equation (1.1). The proof is complete.□

Proof of Theorem 1.1

Combining Theorem 4.1 and Theorem 4.2, we know that equation (1.1) admits at least two nontrivial and nonnegative solutions u0 and U0 . Since N λ + N λ = , we know that u0 and U0 are distinct.□

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Received: 2019-06-29
Accepted: 2020-09-02
Published Online: 2020-12-03

© 2021 Fuliang Wang et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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