Research ArticleMultiplicity Results for Variable-Order Nonlinear FractionalMagnetic Schrödinger Equation with Variable Growth

Jianwen Zhou ,1 Bianxiang Zhou,1 and Yanning Wang 2

1Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, China2School of Basic Medical Science, Kunming Medical University, Kunming, Yunnan 650500, China

Correspondence should be addressed to Yanning Wang; wangscanf2004@126.com

Received 16 February 2020; Accepted 16 April 2020; Published 13 July 2020

Guest Editor: Lishan Liu

Copyright © 2020 Jianwen Zhou et al. This is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

In this paper, we prove the multiplicity of nontrivial solutions for a class of fractional-order elliptic equation with magnetic field.Under appropriate assumptions, firstly, we prove that the system has at least two different solutions by applying the mountainpass theorem and Ekeland’s variational principle. Secondly, we prove that these two solutions converge to the two solutions ofthe limit problem. Finally, we prove the existence of infinitely many solutions for the system and its limit problems, respectively.

1. Introduction

In this paper, we consider the multiplicity of nontrivialsolutions of the following concave-convex elliptic equa-tion involving variable-order nonlinear fractional magneticSchrödinger equation:

−Δð Þs ·ð ÞA u +Vλ xð Þu = f xð Þ uj jp xð Þ−2u + g xð Þ uj jq xð Þ−2u inΩ,u = 0 inℝN \Ω,

(

ð1Þ

whereN ≥ 1; sð·Þ: ℝN ×ℝN ⟶ ð0, 1Þ is a continuous func-tion; Ω is a bounded subset in ℝN with N > 2sðx, yÞ for

all ðx, yÞ ∈Ω ×Ω; ð−ΔÞsð·ÞA is the variable-order fractionalmagnetic Laplace operator; the potential VλðxÞ = λV+ðxÞ −V−ðxÞ with V± = max f±V , 0g; λ > 0 is a parameter; mag-netic field A ∈ C0,αðℝN ,ℝNÞ with α ∈ ð0, 1�; f , g > 0 are twobounded nonnegative measurable function; p, q ∈ CðΩÞ; andu : ℝN ⟶ℂ. In [1], the fractional magnetic Laplacian hasbeen defined as

−Δð ÞsAu xð Þ = limr→0

ðBcr xð Þ

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þx − yj jN+2s dy, ð2Þ

for x ∈ℝN . In [2], the variable-order fractional magneticLaplace ð−ΔÞsð·Þ is defined as follows: for each x ∈ℝN ,

−Δð Þs ·ð Þφ xð Þ = 2P:VðℝN

φ xð Þ − φ yð Þx − yj jN+2s x,yð Þ dy, ð3Þ

along any φ ∈ C∞0 ðΩÞ. Inspired by them, we define the

variable-order fractional magnetic Laplacian ð−ΔÞsð·ÞA as fol-lows: for each x ∈ℝN ,

−Δð Þs ·ð ÞA u xð Þ = limr→0

ðBcr xð Þ

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þx − yj jN+2s x,yð Þ dy: ð4Þ

Since sð·Þ is a function, magnetic field A ∈ C0,αðℝN ,ℝNÞwith α ∈ ð0, 1�, we see that operator ð−ΔÞsð·ÞA is variable orderfractional magnetic Laplace operator. Especially, when sð·Þ ≡constant, ð−ΔÞsð·ÞA reduce to the usual fractional magneticLaplace operator. When sð·Þ ≡ constant, A = 0, ð−ΔÞsð·ÞA reduceto the usual fractional Laplace operator. Very recently, forsð·Þ = 1, pðxÞ, qðxÞ ≡ constant, and A = 0; in [3], underappropriate assumptions, the authors obtained the multiplic-ity and concentration of the positive solution of the followingindefinite semilinear elliptic equations involving concave-convex nonlinearities by the variational method:

HindawiJournal of Function SpacesVolume 2020, Article ID 7817843, 15 pageshttps://doi.org/10.1155/2020/7817843

−Δu + Vλ xð Þu = f xð Þ uj jq−2u + g xð Þ uj jp−2u inℝN ,u ≥ 0 inℝN :

(ð5Þ

For sð·Þ = α, pðxÞ, qðxÞ ≡ constant, and A = 0; in [4], theauthors obtained the existence, multiplicity, and concentrationof nontrivial solutions for the following indefinite fractionalelliptic equation by using Nehari manifold decomposition:

−Δð Þαu +Vλ xð Þu = a xð Þ uj jq−2u + b xð Þ uj jp−2u inℝN ,u ≥ 0 inℝN :

(

ð6Þ

When A = 0, V−ðxÞ = 0, and f ðxÞ, gðxÞ ≡ constant, theauthors in [2] give some sufficient conditions to ensure theexistence of two different weak solutions and use the varia-tional method and the mountain pass theorem to obtainthe two weak solutions of problem (12) which converge totwo solutions of its limit problems and the existence of infi-nitely many solutions to its limit problem:

−Δð Þs ·ð Þu + λV xð Þu = α uj jp xð Þ−2u + β uj jq xð Þ−2u inΩ,u = 0 inℝN \Ω:

(ð7Þ

For sð·Þ = s, pðxÞ, qðxÞ ≡ constant, in [1], the authorsstudy the existence of solutions for the following equationon the whole space by using the method of Nehari manifolddecomposition,and obtain some sufficient conditions for theexistence of nontrivial solutions of the following equation:

−Δð ÞsAu +Vλ xð Þu = f xð Þ uj jq−2u + g xð Þ uj jp−2u inℝN : ð8Þ

In recent years, with the continuous deepening of research,the fractional magnetic problem has attracted extensiveattention of researchers. More and more researchers havestudied the solvability of the fractional magnetic problem

(see [5–8]). We know that the fractional magnetic Laplacianoperator ð−ΔÞsA is introduced in literature [9]; ð−ΔÞsA comesfrom magnetic Laplacian ð∇−iAÞ2s. However, as far as weknow, up to now, few papers have studied the existence andmultiplicity of solutions for the variable-order fractionalmagnetic Schrödinger equation. Therefore, motivated bythe above literature, we are interested in the existence andmultiplicity of solutions to problem (1) with variable growth.As far as we know, this is the first time to study the existenceand multiplicity of nontrivial solutions for the variable-ordernonlinear fractional magnetic Schrödinger equation withvariable exponents.

It is worth noting that in this paper, we not only provethat there exist two different nontrivial solutions to problem(1), but we also show that the two nontrivial solutions ofproblem (1) converge to two solutions of the limit problemfor problem (1). The novelty of this paper is that, comparedwith [1], we write the original fractional magnetic Schrödin-ger equation without variable growth to a variable-order frac-tional magnetic Schrödinger equation with variable growth.In addition, compared with [2], we write the variable orderfractional Schrödinger equation to a variable order fractionalmagnetic Schrödinger equation.

Inspired by the above works, we assume that s : ℝN ×ℝN ⟶ ð0, 1Þ and V are continuous functions satisfying thefollowing:

(S1): 0 < s− ≔ minðx,yÞ∈ℝN×ℝN

sðx, yÞ ≤ s+ ≔ maxðx,yÞ∈ℝN×ℝN

sðx, yÞ< 1.

(S2): sð·Þ is symmetric, that is, sðx, yÞ = sðy, xÞ for all ðx,yÞ ∈ℝN ×ℝN .

(V1): X = int ððV+Þ−1ð0ÞÞ ⊂Ω is a nonempty boundeddomain and ~X = ðV+Þ−1ð0Þ.

(V2): there exists a nonempty open domain Ω0 ⊂ X suchthat V+ ≡ 0, V− ≡ 0 for all x ∈ �Ω0.

(V3): V+ is a continuous function onΩ and V− ∈ LN/2ðΩÞ.

(V4): there exists a constant ϑ0 > 1 such that

or all λ > 0, where Hsð·Þ0,AðΩ,ℂÞ is the Hilbert space related to

magnetic field A (see Section 2).For the variable exponents p, q, we assume that p, q ∈

Cð�ΩÞ and satisfy the following assumption:(H1): 2 < pðxÞ < 2N/ðN − 2sðx, xÞÞ for all x ∈ �Ω.(H2): 1 < qðxÞ < 2 for all x ∈ �Ω.In addition, we assume that f , g satisfy the following

assumption:(H3): f , g : ℝN ⟶ ½0,∞Þ are bounded nonnegative

measurable function such that f > 0, g > 0 on open intervalΩf ,Ωg ⊂Ω and

fk k∞ = fk kL∞ ℝNð Þ ≤p− 2 − q+ð Þ ϑ0 − 1ð Þ

max Cp+p , Cp−

p

n o2ϑ0 p+ − q+ð Þ

,

gk k∞ = gk kL∞ ℝNð Þ ≤q− p+ − 2ð Þ ϑ0 − 1ð Þ

max Cq+q , Cq−

q

n o2ϑ0 p+ − q+ð Þ

� 2 − q+ð Þ ϑ0 − 1ð Þ2A1ϑ0 p+ − q+ð Þ� � 2−q+ð Þ/ p+−2ð Þ

:

ð10Þ

Based on the hypothesis ðS2Þ, we can give the followingdefinition of weak solutions for problem (1).

infu∈Hs ·ð Þ

0,A Ω,ℂð Þ\ 0f g

Ðℝ2N u xð Þ − ei x−yð Þ·A x+y/2ð Þu yð Þ�� ��2� �

/ x − yj jN+2s x,yð Þ� �

dxdy + λÐΩV+u2dxÐ

ΩV−u2dx

≥ ϑ0, ð9Þ

2 Journal of Function Spaces

Definition 1.We say that u ∈ Eλ is a weak solution of problem(1), if

R

ðℝ2N

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ� �v xð Þei xyð Þ·A x+yð Þ/2ð Þv yð Þ� �

x − yj jN+2s x,yð Þ dxdy

+ λRðΩ

V+u�vdx −R

ðΩ

V−u�vdx

−R

ðΩ

f xð Þ uj jp xð Þ−2u + g xð Þ uj jq xð Þ−2u� �

�vdx = 0,

ð11Þ

for any v ∈ Eλ, where Eλ will be defined in Section 2.

Now, we will describe the first main result as follows.

Theorem 2. Assume that ðS1Þ, ðS2Þ, ðV1Þ − ðV4Þ, and ðH1Þ− ðH3Þ hold. LetN > 2s+. Then, the problem (1) allows at leasttwo different solutions for all λ > 0.

Theorem 3. Let u1λ and u2λ be two solutions obtained in

Theorem 2. Then, u1λ ⟶ u1 and u2λ ⟶ u2 in Hsð·Þ0,AðΩ,ℂÞ as

λ⟶∞, where u1 ≠ u2 are two nontrivial solutions of thefollowing problem:

−Δð Þs ·ð ÞA u − V− xð Þu = f xð Þ uj jp xð Þ−2u + g xð Þ uj jq xð Þ−2u inΩ0,u = 0 inℝN \Ω0:

(

ð12Þ

Remark 4. In general, if sð·Þ: ℝN ×ℝN ⟶ ð0, 1Þ is a con-tinuous function, magnetic field A ∈ C0,αðℝN ,ℝNÞ with α ∈ð0, 1�, then the variable-order fractional magnetic Laplaciancan be defined as follows: for each u ∈ C∞

0 ðΩ,ℂÞ,

along any v ∈ C∞0 ðΩ,ℂÞ.

2. Preliminaries and Notations

In this section, we first give the definition of the variable expo-nential Lebesgue space. Secondly, we define variable-orderfractional magnetic Sobolev spaces and prove the compactconditions between them. Finally, we give the variational set-ting for problem (1) and theorems that will be used later.

In this paper, we use jΩj to represent n-dimensionalLebesgue measure of a measurable set Ω ⊂ℝN . In addition,for each a ∈ℂ, we will use Ra to represent the real partof a and �a to represent the complex conjugate of a. LetN ≥ 1 andΩ ⊂ℝN be a nonempty set. A measurable functionr : �Ω⟶ ½1,∞Þ is called a variable exponent, and we definer+ = esssupx∈ΩrðxÞ, r− = essinf x∈ΩrðxÞ. If r+ is finite, then theexponent r is said to be bounded. The variable exponentLebesgue space is

Lr xð Þ Ω,ℂð Þ =u : Ω⟶ℂ is ameasurable function ;ðΩ

u xð Þj jr xð Þdx <∞

ð14Þ

with the Luxemburg norm

uk kLr xð Þ Ω,ℂð Þ = inf μ > 0 :

ðΩ

u xð Þj jμ

� �r xð Þdx ≤ 1

( ), ð15Þ

then LrðxÞðΩ,ℂÞ is a Banach space, and when r is bounded,we have the following relations

min uk kr−Lr xð Þ Ω,ℂð Þ, uk kr+Lr xð Þ Ω,ℂð Þn o

≤ðΩ

u xð Þj jr xð Þdx ≤max uk kr−Lr xð Þ Ω,ℂð Þ, uk kr+Lr xð Þ Ω,ℂð Þn o

:

ð16Þ

For bounded exponent, the dual space ðLrðxÞðΩ,ℂÞÞ′ canbe identified with Lr ′ðxÞðΩ,ℂÞ, where the conjugate exponentr′ is defined by r′ = r/ðr − 1Þ. If 1 < r− ≤ r+ <∞, then the var-iable exponent Lebesgue space LrðxÞðΩ,ℂÞ is a separable andreflexive. In particular,

L2 Ω,ℂð Þ =u : Ω⟶ℂ is ameasurable function ;ðΩ

u xð Þj j2dx <∞ ð17Þ

with the scalar product

u, vh iL2 Ω,ℂð Þ =R

ðΩ

u�vdx: ð18Þ

By Lemma 3.2.20 of [10] and k·kLrðxÞðΩ,ℂÞ = kj·jkLrðxÞðΩ,ℝÞ,we know that in the variable exponent Lebesgue space,

Hölder inequality is still valid. For all u ∈ LrðxÞðΩ,ℂÞ, v ∈Lr ′ðxÞðΩ,ℂÞ with rðxÞ ∈ ð1,∞Þ, the following inequalityholds:

−Δð Þs ·ð ÞA u, vD E

=R

ðℝ2N

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ� �v xð Þei xyð Þ·A x+yð Þ/2ð Þv yð Þ� �

x − yj jN+2s x,yð Þ dxdy, ð13Þ

3Journal of Function Spaces

ðΩ

uj j vj jdx ≤ 1r−

+ 1r′� �−

0@

1A uk kLr xð Þ Ω,ℂð Þ vk kLr ′ xð Þ Ω,ℂð Þ

≤ 2 uk kLr xð Þ Ω,ℂð Þ vk kLr ′ xð Þ Ω,ℂð Þ:

ð19Þ

Let Ω be a nonempty open subset of ℝN , and let sð·Þ:ℝN ×ℝN ⟶ ð0, 1Þ be a measurable function, and there existtwo constants 0 < s0 < s1 < 1 such that s0 < sðx, yÞ < s1 for allðx, yÞ ∈ℝN ×ℝN . Set

Hs ·ð Þ Ω,ℂð Þ =(u ∈ L2 Ω,ℂð Þ:

ðΩ

ðΩ

u xð Þ − u yð Þj j2x − yj jN+2s x,yð Þ dxdy

!1/2

<∞

):

ð20Þ

Equip Hsð·ÞðΩ,ℂÞ with the norm

uk k2Hs ·ð Þ Ω,ℂð Þ = uk k2L2 Ω,ℂð Þ +ðΩ

ðΩ

u xð Þ − u yð Þj j2x − yj jN+2s x,yð Þ dxdy: ð21Þ

Especially, if sð·Þ ≡ constant, then the space Hsð·ÞðΩ,ℂÞ isthe usual fractional Sobolev space HsðΩ,ℂÞ.

Lemma 8. Let Ω be a smooth bounded subset of ℝN and let sð·Þ: ℝN ×ℝN ⟶ ð0, 1Þ satisfying ðS1Þ and r : �Ω⟶ ð1,∞Þsatisfying 1 ≤ r ≤ 2N/ðN − 2sðx, xÞÞ. Then, there exists ~Cr =CðN , r+, s+, s−Þ > 0 such that kukLrðxÞðΩ,ℂÞ ≤ ~CrkukHsð·ÞðΩ,ℂÞ,for any u ∈Hsð·ÞðΩ,ℂÞ. That is, the embedding Hsð·ÞðΩ,ℂÞ°LrðxÞðΩ,ℂÞ is continuous. Moreover, Hsð·ÞðΩ,ℂÞ°LrðxÞðΩ,ℂÞis compact.

Proof. By Theorem 2.1 of [2], we know that Hsð·ÞðΩ,ℝÞ°LrðxÞðΩ,ℝÞ is continuous and compact, there exists ~Cr =CðN , r+, s+, s−Þ > 0 such that kukLrðxÞðΩ,ℝÞ ≤ ~CrkukHsð·ÞðΩ,ℝÞ.then, for any u ∈Hsð·ÞðΩ,ℂÞ, we have

uk kLr xð Þ Ω,ℂð Þ = uj jk kLr xð Þ Ω,ℝð Þ ≤ ~Cr uj jk kHs ·ð Þ Ω,ℝð Þ

= ~Cr uj jk k2L2 Ω,ℝð Þ +ðΩ

ðΩ

u xð Þj j − u yð Þj jj j2x − yj jN+2s x,yð Þ dxdy

!1/2

≤ ~Cr uk k2L2 Ω,ℂð Þ +ðΩ

ðΩ

u xð Þ − u yð Þj j2x − yj jN+2s x,yð Þ dxdy

!1/2

= ~Cr uk kHs ·ð Þ Ω,ℂð Þ:

ð22Þ

Hence, the embedding Hsð·ÞðΩ,ℂÞ°LrðxÞðΩ,ℂÞ is contin-uous and compact.

Let A : ℝN ⟶ℝN be a continuous function and A ∈L∞locðℝN ,ℝNÞ. For a function u : ℝN ⟶ℂ, define

u½ �2Hs ·ð Þ

A ℝN ,ℂð Þ ≔ðℝN

ðℝN

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ�� ��2x − yj jN+2s x,yð Þ dxdy

ð23Þ

and the corresponding norm denoted by kuk2Hsð·Þ

A ðℝN ,ℂÞ =kuk2L2ðℝN ,ℂÞ + ½u�2

Hsð·ÞA ðℝN ,ℂÞ. We consider the space H of mea-

surable functions u : ℝN ⟶ℂ such that kukHsð·Þ

A ðℝN ,ℂÞ <∞;

then, ðH, h·, · iHsð·ÞA ðℝN ,ℂÞÞ is a Hilbert space. Define Hsð·Þ

A ðℝN

,ℂÞ as the closure of C∞c ðℝN ,ℂÞ in H; then, Hsð·Þ

A ðℝN ,ℂÞ isa Hilbert space. Especially, if A = 0, then the space Hsð·Þ

A ðℝN

,ℂÞ is the variable-order fractional Sobolev space Hsð·ÞðℝN ,ℂÞ; if A = 0 and sð·Þ ≡ constant, then the space Hsð·Þ

A ðℝN ,ℂÞis the usual fractional Sobolev space HsðℝN ,ℂÞ.

In order to define weak solutions of problem (1), weintroduce the functional space

Hs ·ð Þ0,A Ω,ℂð Þ = u ∈Hs ·ð Þ

A ℝN ,ℂ� �

: u = 0 a:e:inℝN \Ωn o

, ð24Þ

equipping Hsð·Þ0,AðΩ,ℂÞ with the scalar product

which induces the following norm kukHsð·Þ0,AðΩ,ℂÞ ≔

hu, ui1/2Hsð·Þ

0,AðΩ,ℂÞ: Hence, Hsð·Þ0,AðΩ,ℂÞ generalizes to the

variable-order fractional Sobolev space (see [2]) and themagnetic framework the space introduced in [9]. Next, we

state and prove some properties of space Hsð·Þ0,AðΩ,ℂÞ, which

will be useful in the sequel.

Lemma 6. There exists a constant C2 > 0, depending only onN , s1 and Ω, such that

uk kHs ·ð Þ0,A Ω,ℂð Þ ≤ uk kHs ·ð Þ

A ℝN ,ℂð Þ ≤ C2 uk kHs ·ð Þ0,A Ω,ℂð Þ, ð26Þ

for any u ∈Hsð·Þ0,AðΩ,ℂÞ. Thus, kuk

Hsð·Þ0,AðΩ,ℂÞ is an equivalent

norm of Hsð·ÞA ðℝN ,ℂÞ.

u, vh iHs ·ð Þ0,A Ω,ℂð Þ ≔R

ðℝ2N

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ� �v xð Þei xyð Þ·A x+yð Þ/2ð Þv yð Þ� �

x − yj jN+2s x,yð Þ dxdy, ð25Þ

4 Journal of Function Spaces

Proof. For any u ∈Hsð·Þ0,AðΩ,ℂÞ, by Lemma 3.1 in [9], we have

the pointwise diamagnetic inequality

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ��� ���

≥ u xð Þj j − u yð Þj jj j, for a:e:x, y ∈ℝN ,ð27Þ

from which we immediately have

ðℝN

ðℝN

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ�� ��2x − yj jN+2s x,yð Þ dxdy

≥ðℝN

ðℝN

u xð Þj j − u yð Þj jj j2x − yj jN+2s x,yð Þ dxdy

≥ðℝN\Ω

ðΩ

u xð Þj j − u yð Þj jj j2x − yj jN+2s x,yð Þ dy

!dx

=ðℝNΩ

ðΩ

u yð Þj j2x − yj jN+2s x,yð Þ dy

!dx:

ð28Þ

Since Ω is bounded, there exists r > 1/2 such that Ω ⊂ Brand jBr \Ωj > 0; then, we have

ðℝN \Ω

ðΩ

u yð Þj j2x − yj jN+2s x,yð Þ dy

!dx

≥ðBrΩ

ðΩ

u yð Þj j22rj jN+2s x,yð Þ dy

!dx

≥ðBr\Ω

ðΩ

u yð Þj j22rj jN+2s1

dy

!dx

= 12rð ÞN+2s1

ðBr\Ω

ðΩ

u yð Þj j2dy� �

dx

= Br \Ωj j2rð ÞN+2s1

uk k2L2 Ω,ℂð Þ:

ð29Þ

Thus, we obtain

uk k2L2 Ω,ℂð Þ ≤2rð ÞN+2s1

Br \Ωj jðℝ2N

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ�� ��2x − yj jN+2s x,yð Þ dxdy

= C1 uk k2Hs ·ð Þ

0,A Ω,ℂð Þ,

uk k2Hs ·ð Þ

A ℝN ,ℂð Þ = uk k2L2 Ω,ℂð Þ + u½ �2Hs ·ð Þ

A ℝN ,ℂð Þ≤ C1 uk k2

Hs ·ð Þ0,A Ω,ℂð Þ + uk k2

Hs ·ð Þ0,A Ω,ℂð Þ

= C22 uk k2

Hs ·ð Þ0,A Ω,ℂð Þ,

ð30Þ

where C1 = ð2rÞN+2s1 /jBr \Ωj and C22 = C1 + 1.

In addition,

uk k2Hs ·ð Þ

0,A Ω,ℂð Þ ≤ðΩ

u xð Þj j2dx +ðℝ2N

� u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ�� ��2x − yj jN+2s x,yð Þ dxdy

= uk k2Hs ·ð Þ

A ℝN ,ℂð Þ:

ð31Þ

Combining the above two aspects, we have

uk kHs ·ð Þ0,A Ω,ℂð Þ ≤ uk kHs ·ð Þ

A ℝN ,ℂð Þ ≤ C2 uk kHs ·ð Þ0,A Ω,ℂð Þ, ð32Þ

which implies that kukHsð·Þ

0,AðΩ,ℂÞ is the equivalent norm of a

norm kukHsð·Þ

A ðℝN ,ℂÞ.

Lemma 2.3. Let Ω be a bounded subset of ℝN . Assume thatsð·Þ: ℝN ×ℝN ⟶ ð0, 1Þ is a continuous function satisfyingðS1Þ and r : �Ω⟶ ½1,∞Þ is a continuous function satisfying

1 ≤ r ≤ 2N/ðN − 2sðx, xÞÞ. If u ∈Hsð·Þ0,AðΩ,ℂÞ, then

Hs ·ð Þ0,A Ω,ℂð Þ°Hs ·ð Þ Ω,ℂð Þ ð33Þ

is continuous and

Hs ·ð Þ0,A Ω,ℂð Þ°Lr xð Þ Ω,ℂð Þ ð34Þ

is compact, that is, there exists Cr = CðN , r+, s+, s−Þ > 0 suchthat

uk kLr xð Þ Ω,ℂð Þ ≤ Cr uk kHs ·ð Þ0,A Ω,ℂð Þ: ð35Þ

Proof. For any u ∈Hsð·Þ0,AðΩ,ℂÞ, we have

uk k2Hs ·ð Þ Ω,ℂð Þ =ðΩ

u xð Þj j2dx +ðΩ

ðΩ

u xð Þ − u yð Þj j2x − yj jN+2s x,yð Þ dxdy

≤ðΩ

u xð Þj j2dx

+ 2ðΩ

ðΩ

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ�� ��2x − yj jN+2s x,yð Þ dxdy

+ 2ðΩ

ðΩ

u yð Þj j2 ei x−yð Þ·A x+yð Þ/2ð Þ − 1�� ��2x − yj jN+2s x,yð Þ dxdy

= 2 uk k2Hs ·ð Þ

A Ω,ℂð Þ + 2J ,

ð36Þ

5Journal of Function Spaces

where

J =ðΩ

ðΩ

u yð Þj j2 ei x−yð Þ·A x+yð Þ/2ð Þ − 1�� ��2x − yj jN+2s x,yð Þ dxdy

=ðΩ

u yð Þj j2ðΩ∩ x−yj j>1f g

ei x−yð Þ·A x+yð Þ/2ð Þ − 1�� ��2

x − yj jN+2s x,yð Þ dx

!dy

+ðΩ

u yð Þj j2ðΩ∩ x−yj j≤1f g

ei x−yð Þ·A x+yð Þ/2ð Þ − 1�� ��2

x − yj jN+2s x,yð Þ dx

!dy

= J1 + J2:

ð37Þ

Since jeit − 1j ≤ 2, we have

J1 ≤ 4ðΩ

u yð Þj j2ðΩ∩ x−yj j>1f g

1x − yj jN+2s x,yð Þ dx

!dy

≤ 4ðΩ

u yð Þj j2ðΩ∩ x−yj j>1f g

1x − yj jN+2s0

dx

!dy

= 4ðΩ

u yð Þj j2ðΩ∩ zj j>1f g

1zj jN+2s0

dz

!dy

≤ C3

ðΩ

u yð Þj j2dy = C3 uk k2L2 Ω,ℂð Þ:

ð38Þ

In view of Ω which is bounded, there exists a compactset K ⊂ℝN such that Ω ⊂ K . By Lemma 2.2 of [11], weknow that A is locally bounded and K ⊂ℝN is compact,

jeiðx−yÞ·Aððx+yÞ/2Þ − 1j2 ≤ C4jx − yj2, for jx − yj ≤ 1, x, y ∈ K:

Thus, we obtain

J2 ≤ðKu yð Þj j2

ðK∩ x−yj j≤1f g

ei x−yð Þ·A x+yð Þ/2ð Þ − 1�� ��2

x − yj jN+2s x,yð Þ dx

!dy

≤ðKu yð Þj j2

ðK∩ x−yj j≤1f g

C4x − yj jN+2s x,yð Þ−2 dx

!dy

≤ðKu yð Þj j2

ðK∩ x−yj j≤1f g

C4x − yj jN+2s1−2

dx

!dy

≤ C4

ðKu yð Þj j2

ðK∩ zj j≤1f g

1zj jN+2s1−2

dz

!dy

≤ C5

ðKu yð Þj j2dy = C5

ðK\Ω

u yð Þj j2dy + C5

ðΩ

u yð Þj j2dy

= C5

ðΩ

u yð Þj j2dy = C5 uk k2L2 Ω,ℂð Þ:

ð39Þ

Equations (36)–(39) together with Lemma 6, we have

uk k2Hs ·ð Þ Ω,ℂð Þ ≤ 2 uk k2Hs ·ð Þ

A Ω,ℂð Þ + 2C3 uk k2L2 Ω,ℂð Þ

+ 2C5 uk k2L2 Ω,ℂð Þ ≤ C6 uk k2Hs ·ð Þ

A Ω,ℂð Þ≤ C6 uk k2

Hs ·ð ÞA ℝN ,ℂð Þ ≤ C7 uk k2

Hs ·ð Þ0,A Ω,ℂð Þ,

ð40Þ

which implies that the embedding Hsð·Þ0,AðΩ,ℂÞ↪Hsð·ÞðΩ,ℂÞ

is continuous. In addition, by Lemma 5, we know that Hsð·Þ

ðΩ,ℂÞ↪LrðxÞðΩ,ℂÞ is compact. Therefore, the embedding

Hsð·Þ0,AðΩ,ℂÞ↪LrðxÞðΩ,ℂÞ is compact.Next, we give the variational setting for problem (1). For

λ > 0, we need the following scalar product and norm:

Let E = fu ∈Hsð·Þ0,AðΩ,ℂÞ: Ð

ΩV+u2dx <∞g be equipped

with the norm kukE = kuk1 (that is, λ = 1 in kukλ). Obvi-ously, kukE ≤ kukλ for λ ≥ 1. Set Eλ = ðE, k·kλÞ. Moreover,for rðxÞ ∈ ð1, 2N/ðN − 2sðx, xÞÞÞ, we can getð

Ω

u xð Þj jr xð Þdx ≤max uk kr+Lr ·ð Þ Ω,ℂð Þ, uk kr−Lr ·ð Þ Ω,ℂð Þn o

≤max Cr+r uk kr+

Hs ·ð Þ0,A Ω,ℂð Þ, C

r−r uk kr−

Hs ·ð Þ0,A Ω,ℂð Þ

≤max Cr+r uk kr+λ , Cr−

r uk kr−λn o

:

ð42Þ

For simplicity, we let kuk2λ,V ≔ kuk2Hsð·Þ

0,AðΩ,ℂÞ +ÐΩVλu

2dx.

Therefore, by condition ðV4Þ,

uk k2λ ≥ uk k2λ,V ≥ϑ0 − 1ϑ0

uk k2λ, for all λ ≥ 0: ð43Þ

Associated with problem (1), we consider the energyfunctional Ψλ : Eλ ⟶ℝ,

Ψλ uð Þ = 12 uk k2λ −

12

ðΩ

V−u2dx

−ðΩ

f xð Þp xð Þ uj jp xð Þ + g xð Þ

q xð Þ uj jq xð Þ� �

dx

= 12 uk k2λ,V −

ðΩ

f xð Þp xð Þ uj jp xð Þ + g xð Þ

q xð Þ uj jq xð Þ� �

dx:

ð44Þ

In fact, one can verify that Ψλ is well-defined of class C1

in Eλ and

u, vh iλ ≔R

ðℝ2N

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ� �v xð Þei xyð Þ·A x+yð Þ/2ð Þv yð Þ� �

x − yj jN+2s x,yð Þ dxdy + λ ·RðΩ

V+u�vdx, uk kλ ≔ u, uh i1/2λ : ð41Þ

6 Journal of Function Spaces

for all u, v ∈ Eλ. Therefore, if u ∈ Eλ is a critical point of Ψλ,then u is a solution of problem (1).

Now we give the theorems that we need later.

Theorem 8 (see [2, 12]). Let X be a real infinite dimensionalBanach space and I ∈ C′ðXÞ a functional satisfying the ðPSÞccondition as well as the following three properties:

(1) Ið0Þ = 0, and there exists two constants ρ, δ > 0 suchthat IðuÞ ≥ δ for all u ∈ X with kuk = ρ

(2) I is even

(3) For all finite dimensional subspaces Y ⊂ X, there existsR = RðYÞ > 0 such that IðuÞ ≤ 0 for all u ∈ X \ BRðYÞ,where BRðYÞ = fu ∈ Y : kuk ≤ Rg. Then, I poses anunbounded sequence of critical values characterizedby a minimax argument

Theorem 9 (see [13]). Let X be a real Banach space and I ∈C1ðX,ℝÞ. Suppose I satisfies the (PS) condition, which is evenand bounded from below, and Ið0Þ = 0. If for any k ∈N , thereexists a k-dimensional subspace Xk of X and ρk > 0 such thatsupXk∩Sρk

I < 0, where Sρk = fu ∈ X : kukX = ρkg, then at least

one of the following conclusions holds:

(i) There exists a sequence of critical points fukg satisfy-ing IðukÞ < 0 for all k and kukkX ⟶ 0 as k⟶∞

(ii) There exists r > 0 such that for any 0 < a < r, thereexists a critical point u such that kukX = a and IðuÞ = 0

It is easy to verify that Eλ is a separable Hilbert space. LetXi = spanfeig, then Eλ = ⊕ i≥1Xi. we define

Yk = ⊕ ki=1Xi, Zk = ⊕ ∞

i=k+1Xi: ð46Þ

Theorem 10 (see [14], fountain theorem). Suppose that I ∈C1ðE,ℝÞ satisfying Ið−uÞ = IðuÞ. Assume that for every k ∈N , there exist rk > γk > 0 such that

(D1): ak =max fIðuÞ: u ∈ Yk, kuk = rkg ≤ 0.(D2): bk = inf fIðuÞ: u ∈ Zk, kuk = γkg⟶∞ as k⟶∞.(D3): I satisfies ðPSÞc condition for every c > 0.

Then, I has an unbounded sequence of critical valueswhich have the form

ck = infr∈Γk

maxu∈Bk

I η uð Þð Þ, ð47Þ

where Γk = fη ∈ CðBk, EÞ: η is equivariant and ηj∂Bk = idg.

Theorem 11 (see [15], dual fountain theorem). Suppose thatI ∈ C1ðE,ℝÞ satisfying Ið−uÞ = IðuÞ. Assume that for every k≥ k0, there exist rk > γk > 0 such that

(D4): ak = inf fIðuÞ: u ∈ Zk, kuk = rkg ≥ 0.(D5): bk =max fIðuÞ: u ∈ Yk, kuk = γkg < 0.(D6): dk =max fIðuÞ: u ∈ Zk, kuk ≤ γkg⟶ 0 as k⟶

∞.(D7): I satisfies ðPSÞ∗c condition for every c ∈ ½dk0 , 0�.Then, I has a sequence of negative critical values converg-

ing to 0.

3. Proof of Theorem 1

In this part, we first recall that definitions of functional Ψλ

satisfies the ðPSÞc condition and ðPSÞ∗c condition in Eλ atthe level c ∈ℝ and use the usual mountain pass theorem(see [2]) to find a ðPSÞc sequence in Eλ. Second, we show thatfunctional Ψλ satisfies the ðPSÞ∗c condition in Eλ at the levelc < c0. Finally, we give the proof of problem (1).

Definition 12 (see [2]). Let I ∈ C1ðE,ℝÞ and c ∈ℝ. The func-tional I satisfies the ðPSÞc condition if any sequence fung ⊂ Esuch that IðunÞ⟶ c and I ′ðunÞ⟶ 0 as n⟶∞ admits astrongly convergent subsequence in E.

Definition 13 (see [16]). Let I ∈ C1ðE,ℝÞ and c ∈ℝ. Thefunctional I satisfies the ðPSÞ∗c condition (with respect to Yn)if any sequence fung ⊂ E such that fung ∈ Yn, IðunÞ⟶ c

and I ′jYnðunÞ⟶ 0 as n⟶∞ admits a strongly convergent

subsequence in E.

Remark 14. From Remark 2.1 in [16], we get that the ðPSÞ∗ccondition means the ðPSÞc condition.

Theorem 15 (Theorem 3.1, [2]). Let E be a real Banach spaceand I ∈ C1ðE,ℝÞ with Ið0Þ = 0: Suppose that

(i) there exist δ > 0 and ρ > 0 such that IðuÞ ≥ δ for eachu ∈ E subject to kukE = ρ

R

ðℝ2N

u xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu yð Þ� �v xð Þei xyð Þ·A x+yð Þ/2ð Þv yð Þ� �

x − yj jN+2s x,yð Þ dxdy + λRðΩ

V+u�vdx

−RðΩ

V−u�vdx −R

ðΩ

f xð Þ uj jp xð Þ−2u + g xð Þ uj jq xð Þ−2u� �

�vdx = 0,ð45Þ

7Journal of Function Spaces

(ii) there exists e ∈ E with kekE > ρ such that IðeÞ < 0

Define Γ = fγ ∈ C1ð½0, 1�, EÞ: γð0Þ = 1, γð1Þ = eg. Then,

c = infγ∈Γ

max0≤t≤1

I γ tð Þð Þ ≥ δ, ð48Þ

and there exists a ðPSÞc sequence fungn ⊂ E.

In order to obtain our main results by using the mountainpass theorem, we first prove that Ψλ satisfies the mountainpass geometry (i) and (ii).

Lemma 16. Assume that (S1), (V1)–(V4), and (H1)–(H3) hold.Then, for each λ > 0, there exists ρ > 0 and τ > 0 such that

Ψλ uð Þ > τ for all u ∈ Eλ with uk kλ = ρ: ð49Þ

Proof. In view of (42) and the fractional Sobolev inequality,for each u ∈ Eλ, one has

ðΩ

f xð Þp xð Þ uj jp xð Þ + g xð Þ

q xð Þ uj jq xð Þ� �

dx

≤fk k∞p−

ðΩ

uj jp xð Þdx + gk k∞q−

ðΩ

uj jq xð Þdx

≤fk k∞p−

max Cp+p uk kp+λ , Cp−

p uk kp−λn o

+ gk k∞q−

max Cq+q uk kq+λ , Cq−

q uk kq−λn o

,

ð50Þ

where Cp, Cq are two constants of embedding from variable-

order fractional Sobolev space Hsð·Þ0,AðΩ,ℂÞ to LpðxÞðΩ,ℂÞ and

LqðxÞðΩ,ℂÞ, respectively. Making use of (43) and (50), weobtain that

Ψλ uð Þ = 12 uk k2λ −

12

ðΩ

V−u2dx

−ðΩ

f xð Þp xð Þ uj jp xð Þ + g xð Þ

q xð Þ uj jq xð Þ� �

dx

= 12 uk k2λ,V −

ðΩ

f xð Þp xð Þ uj jp xð Þ + g xð Þ

q xð Þ uj jq xð Þ� �

dx

≥12ϑ0 − 1ϑ0

uk k2λ − fk k∞max Cp+

p , Cp−p

n op−

uk kp+λ

− gk k∞max Cq+

q , Cq−q

n oq−

uk kq+λ

= ϑ0 − 12ϑ0

uk k2λ − A1 uk kp+λ − A2 uk kq+λ

= uk kq+λϑ0 − 12ϑ0

uk k2−q+λ − A1 uk kp+−q+λ − A2

� �,

ð51Þ

for each u ∈ Eλ with kukλ ≥ 1.

Set

A1 =fk k∞ max Cp+

p , Cp−p

n op−

,

A2 =gk k∞ max Cq+

q , Cq−q

n oq−

:

ð52Þ

Define Φ1ðtÞ: ½0,∞Þ⟶ℝ as follows

Φ1 tð Þ =Φ2 tð Þtq+ for all t ≥ 0, ð53Þ

where

Φ2 tð Þ = ϑ0 − 12ϑ0

t2−q+− A1t

p+−q+ − A2: ð54Þ

As long as

A2 <2 − q+ð Þ ϑ0 − 1ð Þ2A1ϑ0 p+ − q+ð Þ

� � 2−q+ð Þ/ p+−2ð Þ ϑ0 − 1ð Þ p+ − 2ð Þ2ϑ0 p+ − q+ð Þ , ð55Þ

that is,

A2−q+1 Ap+−2

2 ≤2 − q+ð Þ ϑ0 − 1ð Þ2ϑ0 p+ − q+ð Þ

� �2−q+ ϑ0 − 1ð Þ p+ − 2ð Þ2ϑ0 p+ − q+ð Þ

� �p+−2,

ð56Þ

that is,

gk k∞ ≤q− p+ − 2ð Þ ϑ0 − 1ð Þ

max Cq+q , Cq−

q

n o2ϑ0 p+ − q+ð Þ

� 2 − q+ð Þ ϑ0 − 1ð Þ2A1ϑ0 p+ − q+ð Þ� � 2−q+ð Þ/ p+−2ð Þ

,ð57Þ

we can easily show that for t =~t = ½ð2 − q+Þðϑ0 − 1Þ/2A1ϑ0ðp+ − q+Þ�1/ðp+−2Þ, we have

Φ2 ~t� �

=maxt≥0

Φ2 tð Þ > 0: ð58Þ

By

fk k∞ ≤p− 2 − q+ð Þ ϑ0 − 1ð Þ

max Cp+p , Cp−

p

n o2ϑ0 p+ − q+ð Þ

, ð59Þ

it is easy to derive that

~t = 2 − q+ð Þ ϑ0 − 1ð Þ2A1ϑ0 p+ − q+ð Þ� �1/ p+−2ð Þ

≥ 1: ð60Þ

By letting ρ =~t > 0 and τ =Φð~tÞ > 0, we can easily getΨλðuÞ > τ for all u ∈ Eλwith kukλ = ρ:

8 Journal of Function Spaces

Lemma 17. Suppose that ðS1Þ, ðV1Þ − ðV4Þ, and ðH1Þ − ðH3Þhold. Then, there exists e ∈ Eλ with kekλ > ρ such that ΨλðeÞ< 0 for all λ > 0, where ρ > 0 is given by Lemma 16.

Proof. Notice that f : ℝN ⟶ ½0,∞Þ is a bounded nonnega-tive measurable function such that f > 0 on open intervalΩf ⊂Ω; then, we can select w0 ∈ Eλ such that

w0k kλ = 1 andðΩ

f xð Þ w0 xð Þj jp xð Þdx > 0: ð61Þ

For all t ≥ 1, combining (43) with (44), we have

Ψλ tw0ð Þ = 12 tw0k k2λ −

12

ðΩ

V− tw0ð Þ2dx

−ðΩ

f xð Þp xð Þ tw0j jp xð Þ + g xð Þ

q xð Þ tw0j jq xð Þ� �

dx

= 12 tw0k k2λ,V −

ðΩ

f xð Þp xð Þ tw0j jp xð Þ + g xð Þ

q xð Þ tw0j jq xð Þ� �

dx

≤12 tw0k k2λ −

ðΩ

f xð Þp xð Þ tw0j jp xð Þ + g xð Þ

q xð Þ tw0j jq xð Þ� �

dx

≤t2

2 w0k k2λ −tp

−

p+

ðΩ

f xð Þ w0j jp xð Þdx:

ð62Þ

Since p− > 2, then there exists t∗ ≥ 1 large enough suchthat kt∗w0kλ > ρ and Ψλðt∗w0Þ < 0. By letting e = t∗w0, wecan easily reach the conclusion.

Define

cλ = infγ∈Γ1

max0≤t≤1

Ψλ γ tð Þð Þ,

c Ω0ð Þ = infγ∈Γ2

max0≤t≤1

ΨλjHs ·ð Þ0,A Ω0,ℂð Þ γ tð Þð Þ,

ð63Þ

where ΨλjHsð·Þ0,AðΩ0,ℂÞ is a restriction of Ψλ on Hsð·Þ

0,AðΩ0,ℂÞ and

Γ1 = γ ∈ C 0, 1½ �, Eλð Þ: γ 0ð Þ = 0, γ 1ð Þ = ef g,Γ2 = γ ∈ C 0, 1½ �,Hs ·ð Þ

0,A Ω0,ℂð Þ� �

: γ 0ð Þ = 0, γ 1ð Þ = en o

:

ð64Þ

Observe that

ΨλjHs ·ð Þ0,A Ω0,ℂð Þ uð Þ = 1

2 uk k2Hs ·ð Þ

0,A Ω0,ℂð Þ −12

ðΩ0

V−u2dx

−ðΩ0

f xð Þp xð Þ uj jp xð Þ + g xð Þ

q xð Þ uj jq xð Þ� �

dx,

ð65Þ

for all u ∈Hsð·Þ0,AðΩ0,ℂÞ:Obviously, cðΩ0Þ is independent of λ.

From the proofs of Lemma 16 and Lemma 17, we can easilyderive thatΨλjHsð·Þ

0,AðΩ0,ℂÞ satisfies the mountain pass geometry.

Since Hsð·Þ0,AðΩ0,ℂÞ ⊂ Eλ for all λ > 0, we have 0 < τ ≤ cλ ≤

cðΩ0Þ for all λ > 0. Evidently, for any t ∈ ½0, 1�, te ∈ Γ2.Consequently, there exists c0 > 0 such that

c Ω0ð Þ ≤max0≤t≤1

Ψλ teð Þ ≤ c0 <∞, ð66Þ

being p− > 2. Then,

0 < τ ≤ cλ ≤ c Ω0ð Þ < c0, ð67Þ

for all λ > 0. In view of Lemma 16, Lemma 17, and Theorem15, it is easy to get that for all λ > 0, there exists fung ⊂ Eλsuch that

Ψλ unð Þ⟶ cλ > 0,Ψλ′ unð Þ⟶ 0, as n⟶∞:

ð68Þ

Lemma 18. Assume that (S1),(V1)–(V4) and (H1)–(H3) hold.Then, Ψλ satisfies the ðPSÞ∗c condition in Eλ for all c < c0 andλ > 0.

Proof. Assume that fung be a ðPSÞ∗c sequence in Eλ withc < c0; then, fung ∈ Yn, ΨλðunÞ⟶ cλ andΨλ

′jYnðunÞ⟶ 0

as n⟶∞: It follows from (42) and (43) and the Hölderinequality that

cλ + o 1ð Þ unk kλ≥Ψλ unð Þ − 1

p−Ψλ′ unð Þ, un

D E= 12 unk k2λ,V

−ðΩ

�f xð Þp xð Þ unj jp xð Þ + g xð Þ

q xð Þ unj jq xð Þ�dx −

1p−

unk k2λ,V

+ 1p−

ðΩ

f xð Þ unj jp xð Þ + g xð Þ unj jq xð Þ� �

dx

= 12 −

1p−

� �unk k2λ,V −

ðΩ

f xð Þ 1p xð Þ −

1p−

� �

� unj jp xð Þdx −ðΩ

g xð Þ 1q xð Þ −

1p−

� �unj jq xð Þdx

≥p− − 22p−

ϑ0 − 1ϑ0

unk k2λ −1q−

−1p−

� �ðΩ

g xð Þ unj jq xð Þdx

≥p− − 22p−

ϑ0 − 1ϑ0

unk k2λ −1q−

−1p−

� �

� gk k∞ max Cq+q unk kq+λ , Cq−

q unk kq−λn o

:

ð69Þ

On the contrary, we suppose that fung is not boundedin Eλ. Then, there exists a subsequence still denoted by

9Journal of Function Spaces

fung such that kunkλ ⟶∞ as n⟶∞. Then, it followsfrom (69) that

cλunk k2λ

+ o 1ð Þ 1unk kλ

≥p− − 22p−

ϑ0 − 1ϑ0

−1q−

−1p−

� �

� gk k∞ max Cq+q unk kq+−2λ , Cq−

q unk kq−−2λ

n o,

ð70Þ

which leads to a contradiction. Hence, fung is boundedin Eλ for all λ > 0. Therefore, there exist a subsequence offung still denoted by fung and u0 in Eλ such that

un ⇀ u0 in Eλ, un ⟶ u0 a:e:inΩ,

unj jr xð Þ−2un ⇀ u0j jr xð Þ−2u0 in Lr′ xð Þ Ω,ℂð Þ,

ð71Þ

where r′ðxÞ = rðxÞ/rðxÞ − 1. The next step is to show thatun ⟶ u0 in Eλ. By Lemma 7, we can get un ⟶ u0 inLrðxÞðΩ,ℂÞ. Thus,

limn→∞

ðΩ

un − u0j jp xð Þdx = 0, ð72Þ

limn→∞

ðΩ

un − u0j jq xð Þdx = 0: ð73Þ

Making use of Hölder inequality, we can obtain

ðΩ

f xð Þ unj jp xð Þ−2un − u0j jp xð Þ−2u0� �

unu0ð Þ��� ���dx≤ fk k∞

ðΩ

unj jp xð Þ−2un − u0j jp xð Þ−2u0� �

unu0ð Þ��� ���dx

= fk k∞ðΩ

unj jp xð Þ−2un − u0j jp xð Þ−2u0��� ��� unu0ð Þ

��� ���dx= fk k∞

ðΩ

unj jp xð Þ−2un − u0j jp xð Þ−2u0��� ��� un − u0j jdx

≤ fk k∞ðΩ

unj jp xð Þ−2un − u0j jp xð Þ−2u0��� ���p xð Þ/ p xð Þ−1ð Þ

dx� � p xð Þ−1ð Þ/p xð Þ

�ðΩ

un − u0j jp xð Þdx� �1/p xð Þ

:

ð74Þ

Combining (H3), (71), and (72), we have

limn→∞

ðΩ

f xð Þ unj jp xð Þ−2un − u0j jp xð Þ−2u0� �

unu0ð Þdx = 0: ð75Þ

Similarly, we have

limn→∞

ðΩ

g xð Þ unj jq xð Þ−2un − u0j jq xð Þ−2u0� �

unu0ð Þdx = 0: ð76Þ

We notice that un ⇀ u0 in Eλ and Ψλ′ðunÞ⟶ 0; then,

we obtain

limn→∞

Ψλ′ unð Þ −Ψλ

′ u0ð Þ, un − u0D E

= 0: ð77Þ

Therefore,

o 1ð Þ = Ψλ′ unð Þ −Ψλ

′ u0ð Þ, un − u0D E

= un, un − u0h iλ,V −R

ðΩ

�f xð Þ unj jp xð Þ−2un

+ g xð Þ unj jq xð Þ−2un�unu0ð Þdx − u0, un − u0h iλ,V

+R

ðΩ

f xð Þ u0j jp xð Þ−2u0 + g xð Þ u0j jq xð Þ−2u0� �

unu0ð Þdx

= un − u0, un − u0h iλ,V −R

ðΩ

f xð Þ�unj jp xð Þ−2un

− u0j jp xð Þ−2u0�unu0ð Þdx −R

ðΩ

g xð Þ

� unj jq xð Þ−2un − u0j jq xð Þ−2u0� �

unu0ð Þdx,ð78Þ

which means that

limn→∞

un − u0k kλ,V = 0: ð79Þ

It follows from (43) that limn→∞

kun − u0kλ = 0:

Proof of Theorem 2. In view of Lemma 16, Lemma 17, andTheorem 15, we can easily infer that for all λ > 0, there existsa ðPSÞcλ sequence fung for Ψλ on Eλ. It derives from Lemma18 and 0 < cλ < cðΩ0Þ < c0 for all λ > 0 that there exists a sub-sequence of fung still denoted by fung and u1λ ∈ Eλ such thatun ⟶ u1λ in Eλ. Furthermore, ΨλðunÞ⟶ cλ ≥ τ and u1λ is asolution of problem (1).

The next step is to prove that system (1) has another solu-tion. Set

�cλ = inf Ψλ uð Þ: u ∈ Bρ

�, ð80Þ

where Bρ = fu ∈ Eλ : kukλ < ρg and ρ > 0 is given by Lemma16. Then, �cλ < 0 for all λ > 0. For this purpose, we first provethere exists v0 ∈ Eλ such that Ψλðσv0Þ < 0 for all σ > 0

10 Journal of Function Spaces

sufficiently small. Let v0 ∈Hsð·Þ0,AðΩ,ℂÞ such that

ÐΩgðxÞ

jv0jqðxÞdx > 0. Making use of the hypothesis ðH2Þ and (43),we obtain that for σ ∈ ð0, 1Þ small enough,

Ψλ σv0ð Þ = σ2

2 v0k k2λ −12

ðℝN

V− σv0ð Þ2dx

−ðΩ

f xð Þp xð Þ σv0j jp xð Þ + g xð Þ

q xð Þ σv0j jq xð Þ� �

dx

≤σ2

2 v0k k2λ,V − σp+ðΩ

f xð Þp xð Þ v0j jp xð Þdx

− σq+ðΩ

g xð Þq xð Þ v0j jq xð Þdx

≤σ2

2 v0k k2λ −σq

+

q+

ðΩ

g xð Þ v0j jq xð Þdx < 0:

ð81Þ

Consequently, there exists v0 ∈ Eλ such that Ψλðσv0Þ < 0for all σ > 0 sufficiently small.

Applying Lemma 16 and the Ekeland variational princi-ple to �Bρ, there exists a sequence fung such that

�cλ ≤Ψλ unð Þ ≤�cλ +1n,

Ψλ vð Þ ≥Ψλ unð Þ − un − vk kλn

,ð82Þ

for all v ∈ Bρ. Now we prove that kunkλ < ρ for n enoughlarge. On the contrary, we suppose that kunkλ = ρ for infi-nitely many n. Without loss of generality, we can supposethat kunkλ = ρ for n ∈N . It follows from Lemma 16 that

Ψλ unð Þ ≥ τ > 0: ð83Þ

Combining with (82), we obtain �cλ ≥ τ > 0, which is con-tradictive with �cλ < 0. Next, we prove Ψλ

′ðunÞ⟶ 0 in E∗λ as

n⟶∞. Let

yn = un + σv, for all v ∈ B1 ≔ u ∈ Eλ : uk kλ = 1 �

, ð84Þ

where σ > 0 small enough such that 2σρ + σ2 ≤ ρ2 − kunk2λfor fixed n large. Then,

ynk k2λ = unk k2λ + 2σρ un, vh iλ + σ2

≤ unk k2λ + 2σρ + σ2 = ρ2,ð85Þ

which implies that yn ∈ �Bρ. Hence, by using (50), we get

Ψλ ynð Þ ≥Ψλ unð Þ − un − ynk kλn

, ð86Þ

that is,

Ψλ un + σvð Þ −Ψλ unð Þσ

≥ −1n: ð87Þ

Set σ⟶ 0+; we obtain hΨλ′ðunÞ, vi ≥ −1/n for each fixed

n large. Similarly, choosing σ < 0 and jσj small enough andrepeating the process above, we can easily get that

Ψλ′ unð Þ, v

D E≤1n, ð88Þ

for each fixed n large.In short, we have

limn→∞

supv∈B1

Ψλ′ unð Þ, v�� �� = 0, ð89Þ

which immediately concludes that Ψλ′ðunÞ⟶ 0 in E∗

λ asn⟶∞. Therefore, fung is a ðPSÞ�cλ sequence for the func-tional Ψλ. Making use of a similar proof as Lemma 18, thereexists u2λ such that un ⟶ u2λ in Eλ. Therefore, we obtain anontrivial solution u2λ of problem (1) satisfying

Ψλ u2λ� �

≤ ξ < 0,

u2� ��

λ< ρ:

ð90Þ

Hence, it is easy to conclude that

Ψλ u2λ� �

=�cλ ≤ ξ < 0 < τ < cλ =Ψλ u1λ� �

, for all λ > 0, ð91Þ

which completes the proof.

4. Proof of Theorem 3

In this section, we mainly give the proof of Theorem 3. Inaddition, inspired by [2, 17], we obtain the method to proveTheorem 3.

Proof of Theorem 3. For each sequence fλng such that 1 ≤λn ⟶∞ as n⟶∞, set uðiÞn to be the critical points of Ψλobtained in Theorem 2, where i = 1, 2. Therefore, one has

Ψλnu 2ð Þn

� �≤ ξ < 0 < τ < cλn =Ψλn

u 1ð Þn

� �< c0,

Ψλn′ u 1ð Þ

n

� �=Ψλn

′ u 2ð Þn

� �= 0

ð92Þ

11Journal of Function Spaces

Ψλnu ið Þn

� �= 12 u ið Þ

n

��� ���2λn−12

ðΩ

V− u ið Þn

� �2dx

−ðΩ

f xð Þp xð Þ u ið Þ

n

��� ���p xð Þdx −

ðΩ

g xð Þq xð Þ u ið Þ

n

��� ���q xð Þdx

= 12 u ið Þ

n

��� ���2λn ,V

−ðΩ

f xð Þp xð Þ u ið Þ

n

��� ���p xð Þdx

−ðΩ

g xð Þq xð Þ u ið Þ

n

��� ���q xð Þdx ≥

12ϑ0 − 1ϑ0

u ið Þn

��� ���2λn

−ðΩ

g xð Þp xð Þ u ið Þ

n

��� ���p xð Þdx −

ðΩ

g xð Þq xð Þ u ið Þ

n

��� ���q xð Þdx

≥p− − 22p−

ϑ0 − 1ϑ0

u ið Þn

��� ���2λn

+ 1p−

−1q−

� �ðΩ

g xð Þ u ið Þn

��� ���q xð Þdx

≥p− − 22p−

ϑ0 − 1ϑ0

u ið Þn

��� ���2λn−

1q−

−1p−

� �

� gk k∞ max Cq+q u ið Þ

n

��� ���q+λn, Cq−

q u ið Þn

��� ���q−λn

:

ð93ÞIn view of (92) and (93), it gains

u ið Þn

��� ���λn≤ c2, ð94Þ

where c2 > 0 is independent of λn. So we can suppose that

uðiÞn ⇀ uðiÞ in Hsð·Þ0,AðΩ,ℂÞ and uðiÞn ⟶ ui in LrðxÞðΩ,ℂÞ. Mak-

ing use of Fatou’s lemma, we can easily obtain

ðΩ

V+ u ið Þ��� ���2dx ≤ liminf

n→∞

ðΩ

V+ u ið Þn

��� ���2dx ≤ liminfn→∞

u ið Þn

��� ���2λn

λn= 0:

ð95Þ

Consequently, uðiÞ = 0 a.e. in ℝN \ ðV+Þ−1ð0Þ and uðiÞ ∈Hsð·Þ

0,AðΩ0,ℂÞ.Next, we will prove that uðiÞn ⟶ uðiÞ in Hsð·Þ

0,AðΩ,ℂÞ.Indeed, combining uðiÞn ⟶ uðiÞ in LrðxÞðΩ,ℂÞ and ðH3Þ,one has

ðΩ

f xð Þ u ið Þn − u ið Þ

��� ���p xð Þdx ≤ fk k∞

ðΩ

u ið Þn − u ið Þ

��� ���p xð Þdx,

ðΩ

g xð Þ u ið Þn − u ið Þ

��� ���q xð Þdx ≤ gk k∞

ðΩ

u ið Þn − u ið Þ

��� ���q xð Þdx:

ð96Þ

Then,

limn→∞

ðΩ

f xð Þ u ið Þn − u ið Þ

��� ���p xð Þdx = 0, ð97Þ

limn→∞

ðΩ

g xð Þ u ið Þn − u ið Þ

��� ���q xð Þdx = 0: ð98Þ

We notice that

limn→∞

Ψλn′ u ið Þ

n

� �, u ið Þ

n

D E= lim

n→∞Ψλn

′ u ið Þn

� �, u ið Þ

D E= 0: ð99Þ

Therefore, we have

u ið Þn

��� ���2λn=ðΩ

V− u ið Þn

� �2dx

+ðΩ

f xð Þ u ið Þn

��� ���p xð Þ+ g xð Þ u ið Þ

n

��� ���q xð Þ� �dx + o 1ð Þ,

ð100Þ

u ið Þn , u ið Þ

D Eλn=R

ðΩ

V−u ið Þn u ið Þdx

+R

ðΩ

f xð Þ u ið Þn

��� ���p xð Þ−2u ið Þn u ið Þdx

+R

ðΩ

g xð Þ u ið Þn

��� ���q xð Þ−2u ið Þn u ið Þdx + o 1ð Þ:

ð101ÞBy (97)–(101), we have

limn→∞

u ið Þn

��� ���2λn= lim

n→∞u ið Þn , u ið Þ

D Eλn= u ið Þ��� ���2

Hs ·ð Þ0,A Ω,ℂð Þ

: ð102Þ

On the other hand, the weak lower semicontinuity ofnorm yields that

u ið Þ��� ���2

Hs ·ð Þ0,A Ω,ℂð Þ

≤ liminfn→∞

u ið Þn

��� ���2Hs ·ð Þ

0,A Ω,ℂð Þ

≤ limsupn→∞

u ið Þn

��� ���2Hs ·ð Þ

0,A Ω,ℂð Þ

≤ limn→∞

u ið Þn

��� ���2λn:

ð103Þ

To sum up, we can see that

limsupn→∞

u ið Þn

��� ���Hs ·ð Þ

0,A Ω,ℂð Þ≤ u ið Þ��� ���

Hs ·ð Þ0,A Ω,ℂð Þ

: ð104Þ

By Proposition 3.32 of [18], we can obtain that uðiÞn ⟶

uðiÞ in Hsð·Þ0,AðΩ,ℂÞ. We notice that lim

n→∞hΨλn

′ðuðiÞn Þ, vi = 0, forany v ∈ C∞

0 ðΩ0,ℂÞ. Hence,

R

ðℝ2N

u ið Þ xð Þ − ei x−yð Þ·A x+yð Þ/2ð Þu ið Þ yð Þ� �v xð Þei xyð Þ·A x+yð Þ/2ð Þv yð Þ� �

x − yj jN+2s x,yð Þ dxdy

−R

ðΩ0

V−u ið Þ�vdx −R

ðΩ0

�f xð Þ u ið Þ

��� ���p xð Þ−2u ið Þ

+ g xð Þ u ið Þ��� ���q xð Þ−2

u ið Þ��vdx = 0:

ð105Þ

12 Journal of Function Spaces

Since the density of C∞0 ðΩ0,ℂÞ in Hsð·Þ

0,AðΩ0,ℂÞ, we canobtain that uðiÞ is a weak solution of problem (12).

Together with (92), uðiÞ = 0 a.e. inℝN \ ðV+Þ−1ð0Þ and theconstants ξ, τ are independent of λ; we have

12 u 1ð Þ��� ���2

Hs ·ð Þ0,A Ω0,ℂð Þ

−12

ðΩ0

V− u 1ð Þ� �2

dx

−ðΩ0

f xð Þp xð Þ u 1ð Þ

��� ���p xð Þ+ g xð Þq xð Þ u 1ð Þ

��� ���q xð Þ� �dx ≥ τ > 0,

12 u 2ð Þ��� ���2

Hs ·ð Þ0,A Ω0,ℂð Þ

−12

ðΩ0

V− u 2ð Þ� �2

dx

−ðΩ0

f xð Þp xð Þ u 2ð Þ

��� ���p xð Þ+ g xð Þq xð Þ u 2ð Þ

��� ���q xð Þ� �dx ≤ ξ < 0,

ð106Þ

which implies that uðiÞ ≠ 0 and uð1Þ ≠ uð2Þ.Now we consider the case where f ðxÞ, gðxÞ ≡ constants;

that is

−Δð Þs ·ð ÞA u + Vλ xð Þu = a uj jp xð Þ−2u + b uj jq xð Þ−2u inΩ,u = 0 inℝN \Ω,

(

ð107Þ

where a, b are two nonnegative constants. Correspondingly,the energy functional Ψλ: Eλ ⟶ℝ is

Ψλ uð Þ = 12 uk k2λ −

12

ðΩ

V−u2dx

−ðΩ

ap xð Þ uj jp xð Þ + b

q xð Þ uj jq xð Þ� �

dx

= 12 uk k2λ,V −

ðΩ

ap xð Þ uj jp xð Þ + b

q xð Þ uj jq xð Þ� �

dx:

ð108Þ

Next, we mainly prove the existence of infinitely manysolutions to problem (107) by using four different methods.

Theorem 19. Assume that (S1), (S2), (V1)–(V4), and(H1)–(H3) hold. Let N > 2s+. Then, problem (107) has infi-nitely many solutions.

Proof.Method 1: It is easy to verify that functionalΨλ is evenand satisfies Ψλð0Þ = 0. Furthermore, Lemma 18 shows thatfunctional Ψλ is bounded from below in Eλ and satisfies theðPSÞ condition. For any k ∈N and ρk > 0, let Sρk = fu ∈ Eλ,kukλ = ρkg; then, for any u ∈ Sρk , one has

Ψλ uð Þ = 12 uk k2λ,V −

ðΩ

ap xð Þ uj jp xð Þdx −

ðΩ

bq xð Þ uj jq xð Þdx

≤12 uk k2λ −

bq+

ðΩ

uj jq xð Þdx

≤12 uk k2λ −

bq+

min uk kq+Lq xð Þ Ω,ℂð Þ, uk kq−Lq xð Þ Ω,ℂð Þ

n o:

ð109Þ

We find that there exists CSρk> 0 such that kukLqðxÞðΩÞ ≥

CSρkkukλ, since all norms are equivalent on finite dimen-

sional Banach space. Then, by 1 < qðxÞ < 2, it gains

supu∈Xk∩Sρk

Ψλ uð Þ ≤ 12 uk k2λ −

bp+

min Cq+

Sρk, Cq−

Sρk

n ouk kq+λ :

ð110Þ

Letting kukλ = ρk small enough, we can obtain supu∈Xk∩Sρk

ΨλðuÞ < 0 =Ψλð0Þ. Furthermore, we assert that (ii) of Theo-rem 9 does not work. In fact, (109) means ΨλðuÞ ≠ 0 sinceΨλðtuÞ < 0 as t small enough with the given u ∈ Eλ. Thus,by Theorem 9, we get that problem (1) has a sequence ofsolutions fukg with kukkλ ⟶ 0 as k⟶∞. In short, prob-lem (1) has infinitely many solutions for all λ > 0.

Method 2. To start with, we assert that for any finitedimensional subspace X of Eλ, there exists r1 = r1ðXÞ suchthat ΨλðuÞ < 0 for all u ∈ Eλ \ Br1

ðXÞ, where Br1ðXÞ = fu ∈

Eλ : kukλ < r1g: Indeed, for each t ≥ 1, we can easily get that

Ψλ tuð Þ = t2

2 uk k2λ −ðΩ

ap xð Þ tuj jp xð Þdx −

ðΩ

bq xð Þ tuj jq xð Þdx

≤t2

2 uk k2λ −atp

−

p+

ðΩ

uj jp xð Þdx

≤t2

2 uk k2λ −atp

−

p+min uk kp+Lp xð Þ Ω,ℂð Þ, uk kp−Lp xð Þ Ω,ℂð Þ

n o:

ð111Þ

We observe that there exists CX > 0 such that kukLpðxÞðΩÞ≥ CXkukλ, since all norms are equivalent on finite dimen-sional Banach space X. Then, by p− > 2, it gains

Ψλ tuð Þ ≤ t2

2 uk k2λ −atp

−

p+min Cp+

X , Cp−

X

n ouk kp+λ

⟶ −∞as t⟶∞:

ð112Þ

Thus, there exists r1 > 0 large enough such thatΨλðuÞ < 0for all u ∈ Eλ, with kukλ = r2 and r2 ≥ r1. Consequently, theassertion is valid.

From Lemma18, we know thatΨλ satisfies the ðPSÞc con-dition for any c ∈ℝ. Obliviously,Ψλð0Þ = 0 andΨλ is an evenfunctional. In short, it follows from Theorem 8 that there

13Journal of Function Spaces

exists an unbounded sequence of solutions of problem (1) forall λ > 0.

Lemma 19 (see Lemma 4.1, [10]). Let 1 < rðxÞ < 2N/ðN − 2sðx, xÞÞ for all x ∈ �Ω. For any k ∈N , define

βk ≔ uk kLr xð Þ Ω,ℂð Þ : u ∈ Zk, uk kλ = 1n o

: ð113Þ

Then, βk ⟶ 0, as k⟶∞.

Method 3. By Remark 14 and Lemma 18, we know that Ψλsatisfies the ðPSÞc condition for any c ∈ℝ. To start with, wewill prove (D1) is satisfied. It follows from (36) and (43) that

Ψλ uð Þ = 12 uk k2λ,V −

ðΩ

ap xð Þ uj jp xð Þdx −

ðΩ

bq xð Þ uj jq xð Þdx

≤12 uk k2λ −

ap+

ðΩ

uj jp xð Þdx

≤12 uk k2λ −

ap+

min uk kp+Lp xð Þ Ω,ℂð Þ, uk kp−Lp xð Þ Ω,ℂð Þ

n o:

ð114Þ

We find that there exists CYk> 0 such that kukLpðxÞðΩÞ ≥

CYkkukλ, since all norms are equivalent on finite dimensional

Banach space Yk. Then,

Ψλ uð Þ ≤ 12 uk k2λ −

ap+

min Cp+

Yk, Cp−

Yk

n ouk kp−λ : ð115Þ

Then, by p− > 2, we can easily obtain that (D1) is satisfiedfor kukλ = rk > 1 big enough. Next, we will show (D2) is ful-filled. In view of (43) and Lemma 20, for u ∈ Eλ, one has

Ψλ uð Þ = 12 uk k2λ −

12

ðℝN

V−u2dx

−ðΩ

ap xð Þ uj jp xð Þ + b

q xð Þ uj jq xð Þ� �

dx

= 12 uk k2λ,V −

ðΩ

ap xð Þ uj jp xð Þ + b

q xð Þ uj jq xð Þ� �

dx

≥ϑ0 − 12ϑ0

uk k2λ −ap−

ðΩ

uj jp xð Þdx −bq−

ðΩ

uj jq xð Þdx

≥ϑ0 − 12ϑ0

uk k2λ −ap−

βp−

k uk kp+λ −bq−

βq−

k uk kq+λ

= uk kq+λϑ0 − 12ϑ0

uk k2−q+λ −ap−

βp−

k uk kp+−q+λ −bq−

βq−

k

� �

= uk kq+λ uk k2−q+λ

ϑ0 − 12ϑ0

−ap−

βp−

k uk kp+−2λ

� �−

bq−

βq−

k

� �

= γq+

k γ2−q+

k

ϑ0 − 12ϑ0

−ap−

βp−

k γp+−2k

� �−

bq−

βq−

k

� �:

ð116Þ

Choosing γk = ððp−ðϑ0 − 1ÞÞ/8aϑ0βp−

k Þ1/ðp+−2Þ

. Combingwith Lemma 20, we know that βk ⟶ 0 as k⟶∞. Thus,one has γk ⟶ +∞ as k⟶∞ and

Ψλ uð Þ ≥ γq+

k

38ϑ0 − 1ϑ0

γ2−q+

k −bq−

βq−

k

� �⟶ +∞, ð117Þ

as k⟶∞. In conclusion, (D2) is fulfilled. It is easy to checkthat satisfying Ψλð−uÞ =ΨλðuÞ. Thus, by Theorem 10, wecan obtain that problem (1) has infinitely many solutionsfor all λ > 0.

Method 4. First, we will show that (D4) is fulfilled. By (42)and (43), one has

Ψλ uð Þ ≥ 12ϑ0 − 1ϑ0

uk k2λ −ap−

max Cp+p uk kp+λ , Cp−

p uk kp−λn o

−bq−

max uk kq+Lq xð Þ Ω,ℂð Þ, uk kq−Lq xð Þ Ω,ℂð Þ

n o:

ð118Þ

Since 1 < q− ≤ q+ < 2 < p− ≤ p+ < 2N/ðN − 2sðx, xÞÞ, wecan choose r3 ∈ ð0, 1Þ small enough such that for all u ∈ Eλwith kukλ ≤ r3,

18ϑ0 − 1ϑ0

uk k2λ ≥ap−

max Cp+p uk kp+λ , Cp−

p uk kp−λn o

ð119Þ

hold. Then,

Ψλ uð Þ ≥ 38ϑ0 − 1ϑ0

uk k2λ −bq−

βq−

k uk kq−λ : ð120Þ

Choosing

rk =8bϑ0β

q−

k

3q− ϑ0 − 1ð Þ

!1/ 2−q−ð Þ: ð121Þ

Combing with Lemma 20, we know that βk ⟶ 0 ask⟶∞. Thus, one has rk ⟶ 0 as k⟶∞. Hence, thereexists �k such that rk ≤ r3 as k ≥ �k. Consequently, for k ≥ �kand u ∈ Zk with kukλ = rk, we can obtain that ΨλðuÞ ≥ 0.Next, we will show (D5) is fulfilled. For any u ∈ Yk,kukλ = γk with rk > γk > 0, we get

Ψλ uð Þ = 12 uk k2λ,V −

ðΩ

ap xð Þ uj jp xð Þdx −

ðΩ

bq xð Þ uj jq xð Þdx

≤12 uk k2λ −

ap+

ðΩ

uj jp xð Þdx −bq+

ðΩ

uj jq xð Þdx

≤12 uk k2λ −

ap+

min uk kp+Lp xð Þ Ω,ℂð Þ, uk kp−Lp xð Þ Ω,ℂð Þ

n o

−bq+

min uk kq+Lq xð Þ Ω,ℂð Þ, uk kq−Lq xð Þ Ω,ℂð Þ

n o:

ð122Þ

14 Journal of Function Spaces

We find that there exists CYk> 0 such that

kukLqðxÞðΩÞ ≥ CYkkukλ and kukLpðxÞðΩÞ ≥ CYk

kukλ, since allnorms are equivalent on finite dimensional Banach spaceYk. Then,

Ψλ uð Þ ≤ 12 uk k2λ −

ap+

min Cp+

Ykuk kp+λ , Cp−

Ykuk kp−λ

n o

−bq+

min Cq+

Ykuk kq+λ , Cq−

Ykuk kq−λ

n o≤12 uk k2λ −

ap+

min Cp+

Yk, Cp−

Yk

n ouk kp+λ

−bq+

min Cq+

Yk, Cq−

Yk

n ouk kq+λ < 0,

ð123Þ

as γk > 0 small enough. Now we check (D6) is fulfilled. Itfollows from (D4) that for k ≥ �k and u ∈ Zk with kukλ ≤ rk,

Ψλ uð Þ ≥ 38ϑ0 − 1ϑ0

uk k2λ −bq−

βq−

k uk kq−λ

≥ −bq−

βq−

k uk kq−λ ≥ −bq−

βq−

k rq−

k ,ð124Þ

thanks to βk ⟶ 0 as k⟶∞. Thus, one has rk ⟶ 0 as k⟶∞. Thus, (D6) is also satisfied. In conclusion, by Theo-rem 11, we can obtain that problem (1) has infinitely manysolutions for all λ > 0.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there is no conflict of interestsregarding the publication of this paper.

Authors’ Contributions

All authors contributed equally to the manuscript and typed,read, and approved the final manuscript.

Acknowledgments

This work is supported by the Natural Sciences Founda-tion of Yunnan Province under Grant 2018FE001(-136),2017zzx199, the National Natural Sciences Foundation ofPeople's Republic of China under Grants 11961078 and11561072, the Yunnan Province, Young Academic and Tech-nical Leaders Program (2015HB010), the Natural SciencesFoundation of Yunnan Province under Grant 2016FB011

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