# Negative Energy Solutions for a New Fractionalpx-Kirchhoff Problem without the (AR) Condition

Authors: Weichun Bu; Tianqing An; Guoju Ye; Said Taarabti
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we investigate the following Kirchhoff type problem involving the fractionalpx-Laplacian operator. a−b∫Ω×Ωux−uypx,y/px,yx−yN+spx,ydxdyLu=λuqx−2u+fx,ux∈Ωu=0x∈∂Ω,, where Ω is a bounded domain in ℝN with Lipschitz boundary, a≥b>0 are constants, px,y is a function defined on Ω¯×Ω¯, s∈0,1, and qx>1, Lu is the fractional px-Laplacian operator, N>spx,y, for any x,y∈Ω¯×Ω¯, px∗=px,xN/N−spx,x, λ is a given positive parameter, and f is a continuous function. By using Ekeland’s variational principle and dual fountain theorem, we obtain some new existence and multiplicity of negative energy solutions for the above problem without the Ambrosetti-Rabinowitz ((AR) for short) condition.


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## Body

## 1. Introduction and the Main Results

In this article, we investigate the existence and multiplicity of solutions for the Kirchhoff type problem involving the fractionalpx-Laplacian operator. More precisely, we consider the following problem Pλ:
(1)Pλ:a−b∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdyLu=λuqx−2u+fx,ux∈Ω,u=0x∈∂Ω,where Ω is a bounded domain in ℝN with Lipschitz boundary and a≥b>0 are constants, s∈0,1 and N>spx,y for all x,y∈Ω¯×Ω¯. λ is a given positive parameter, and f is a continuous function. The operator Lu=−Δp·su is the fractional px-Laplacian operator, which is defined by
(2)Lu≔P.V.∫Ωux−uypx,y−2ux−uyx−yN+spx,ydy,x∈Ω,where P.V. stands for the Cauchy principal value. Especially, when p·≡constant, the operator Lu=−Δp·su reduced to the usual fractional p-Laplace operator. In the past few decades, many people studied the fractional p-Laplace operator problem, and we refer the readers to [1–9]. We point out that Behboudi et al. studied the existence of mountain pass solution for the nonlocal fractional p,q-Laplacian problem in [10] by using the variational method. The more general problem is the following px-Laplacian equation
(3)−Δpxsu=fx,uinΩ,u=0inℝN\Ω..The study for this problem and the related knowledge on fractional Sobolev spaces with variable exponent please refer to [11, 12] and the references therein.In the recent years, the fractional calculus and the related problems are studied widely, see for example, the literature [13, 14]. The elementary knowledge of fractional derivative and integral can be found in [15, 16]. Especially, the Caputo derivative is discussed in [17]. These references will be helpful for our research later.The following Kirchhoff equation was introduced by Kirchhoff in [18]:
(4)ρ∂2u∂t2−p0h+E2L∫0L∂u∂t2dx∂2u∂x2=0,where ρ,p0,h,E,L are constants that have physical meaning. We call the problem (Pλ) a new problem of Kirchhoff type because it contains the new Kirchhoff term:
(5)a−b∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy,which makes the problem Pλ more interesting, meaningful, and difficult. Up to now, the study on the Kirchhoff type problems involving px-Laplacian operator and the fractional px-Laplacian operator is very active, see [19–23]. Many results concerning the existence and multiplicity of solutions have been appeared. Especially, [24] proved the existence of two weak solutions by using the variational methods in Orlicz-Sobolev spaces, [25] studied a class of Kirchhoff nonlocal fractional equations, and obtained the existence of three solutions. We also mention that [26] discussed a class of p-Kirchhoff equations via the fountain theorem and dual fountain theorem, and [27] studied the existence of nonnegative solutions for a Kirchhoff type problem driven by a nonlocal integrodifferential operator.It is well known that the (AR) condition plays an important role in verifying the Palais-Smale condition. However, there are a lot of functions not satisfying this condition. Hence, many people pay attention to find the new reasonable conditions instead of the (AR) condition, see [28–31] and the references therein. Motivated by these work, we use Ekeland’s variational principle and dual fountain theorem to study the existence and multiplicity of negative energy solutions for a new fractional px-Kirchhoff problem without the (AR) condition. Our results generalize the related work in two ways. Firstly, we deal with the problem Pλ in the fractional framework and [32] consider only the integer framework. Secondly, compare to [33], we add a new Kirchhoff function and consider the case of variable exponents.Throughout this paper, the nonlinearityfx,t:Ω×ℝ⟶ℝ is a Carathe′odory function satisfying(f1): There exists a positive constantC such that
(6)fx,t≤C1+trx−1,forallx,t∈Ω×ℝ,where 1<rx<px∗=px,xN/N−spx,x;(f2):limt⟶0fx,tt/tp+=0, for x∈Ω uniformly;(f3):limt⟶∞Fx,t/t2p+=+∞, x∈Ω, where Fx,t=∫0tfx,sds≥0;(f4):): There exists positive constantsθ0 such that Fx,t≤1/p+fx,tt+θ0t2p−,forallx,t∈Ω×ℝ;(f5):fx,−t=−fx,t, forallx,t∈Ω×ℝ.Definition 1.
We sayu∈X0 is a weak solution of problem Pλ, if
(7)a−b∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy×∫Ω×Ωux−uypx,y−2ux−uyφx−φyx−yN+spx,ydxd=λ∫Ωuqx−2uφdx+∫Ωfx,uφdx,for any φ∈X0, where X0 will be introduced in Section 2.
It is well known that a weak solution for problemPλ is a critical point of the following energy functional I defined on X0 by
(8)Iu≔a∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxuqxdx−∫ΩFx,udx,for all u∈X0. Moreover, we have
(9)I′u,φ≔a−b∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy×∫Ωux−uypx,y−2ux−uyφx−φyx−yN+spx,ydx−λ∫Ωuqx−2uφdx−∫Ωfx,uφdx,for any u,φ∈X0. Under our assumptions, I is well defined in X0 and I∈C1X0,ℝ. We say that a weak solution u for problem Pλ is a negative energy solution if the energy Iu<0.
In order to reduce our statements, we define the function space(10)C+Ω¯≔Hx:Hx∈CΩ¯,ℝ,1<Hx<∞forallx∈Ω¯,and q−,q+,p−,p+,r− will be introduced in Section 2. The main results of this paper are as follows.Theorem 2.
Assume that the functionf satisfies (f1)–(f4) and
(11)1<q−<q+<p−<p+<2p−<2p−<r−<2p+<p∗x.
Then, there exists aλ∗>0 such that for any 0<λ<λ∗, the problem Pλ has at least one solution u0 with negative energy.Theorem 3.
Assume that the functionf satisfies (f1)–(f5) and
(12)1<q−<q+<p−<p+<2p−<2p−<r−<2p+<p∗x.
Then, for anyλ>0, problem Pλ has infinitely many solutions un in X0 with negative energy converging to 0.
The rest of this paper is organized as follows. In Section2, some basic properties of the variable exponent fractional Sobolev spaces and Lebesgue spaces are given. In Section 3, it is proved that the functional satisfies the Cerami compactness condition in certain energy levels. In Section 4, by using Ekeland’s variational principle, we give the proof of Theorem 2. Finally, in Section 5, we prove Theorem 3 by using the dual fountain theorem.
Throughout this paper, for simplicity, we use lettersCii=1,2,⋯,N to denote positive constants in different cases, and we will specify them whenever it is necessary.


## 2. Preliminary Results

In this section, we recall some preliminary results of generalized Lebesgue spacesLqxΩ with variable exponent and generalized fractional Sobolev spaces Ws,hx,px,yΩ which will be used later. The readers can consult [34–36] and the references therein for more details. Let s∈0,1 and Ω be a bounded domain in ℝN with Lipschitz boundary, and qx,hx,px,y be continuous functions satisfying
(13)1<q−≔minx∈Ω¯qx≤qx≤q+≔maxx∈Ω¯qx<∞,(14)1<h−≔minx∈Ω¯hx≤hx≤h+≔maxx∈Ω¯hx<∞.We assume thatpx,y is symmetric; that is, px,y=py,x for all x,y∈Ω¯×Ω¯, such that
(15)1<p−≔minx,y∈Ω¯×Ω¯px,y≤px,y≤p+≔maxx,y∈Ω¯×Ω¯px,y<∞.For anyqx∈C+Ω¯, we introduce the variable exponent Lebesgue space as
(16)LqxΩ≔u:uismeasurable,∫Ωuqxdx<∞endowed with the so-called Luxemburg norm
(17)uqx≔infμ>0:∫Ωuμqxdx≤1.Define a mappingρ:LqxΩ⟶ℝ by
(18)ρqxu≔∫Ωuqxdx.Lemma 4 (See [37]).
The spaceLqxΩ,uqx is separable, uniformly convex, and reflexive and its conjugate space is Lq′xΩ,uq′x, where q′x is the conjugate function of qx, i.e.,
(19)1qx+1q′x=1,forallx∈Ω,for all u∈LqxΩ,v∈Lq′xΩ, and the Hölder type inequality
(20)∫Ωuvdx≤1q−+1q′−uqxvq′x≤2uqxvq′xholds.
Note that for any functionq1x,q2x∈C+Ω¯ and q1x<q2x, there exists a continuous embedding Lq2xΩ↪Lq1xΩ for any x∈Ω¯.Lemma 5 (see [38]).
Suppose thatun,u∈LqxΩ. Then, the following properties hold
(21)uqx>1⇒uqxq−≤ρqxu≤uqxq+,uqx<1⇒uqxq+≤ρqxu≤uqxq−,uqx<1respectively,=1,>1⇔ρqxu<1respectively,=1,>1,unqx⟶0respectively,⟶∞⇔ρqxun⟶0respectively,⟶∞,limn⟶∞un−uqx=0⇔limn⟶∞ρqxun−u=0.
The fractional Sobolev spaces with variable exponentX=Ws,hx,px,yΩ are defined by
(22)Ws,hx,px,yΩ≔u:u∈LhxΩ,∫Ω×Ωux−uypx,yμpx,yx−yN+spx,ydxdy<∞,λ>0.
Foru∈X, let
(23)us,px,y≔infμ>0,∫Ω×Ωux−uypx,yμpx,yx−yN+spx,ydxdy≤1.
Define the corresponding variable exponent norm(24)uX≔uhx+us,px,yand set
(25)ρpx.yu≔∫Ω×Ωux−uypx,yx−yN+spx,ydxdy.
Then, the (X,·X) becomes an uniformly convex and reflexive Banach spaces (see [39]). Let X0=W0s,hx,px,yΩ denote the closure of C0∞Ω in X endowed with the norm
(26)uX0≔us,px,y,forallu∈X0,and then (X0,·X0) is also an uniformly convex and reflexive Banach spaces. X0∗ is the dual spaces of X0.Theorem 6 (see [39]).
LetΩ∈ℝN be a smooth bounded domain and s∈0,1, hx,px,y be continuous variable exponents with N>spx,y for x,y∈Ω×Ω and hx>px,x for x∈Ω¯, and (13) and (15) be satisfied. Assume that r′:Ω¯⟶1,∞ is a continuous function and r′−=minx∈Ω¯r′x such that
(27)px∗=px,xNN−spx,x>r′x≥r′−>1,for x∈Ω¯. Then, there exists a constant C0=C0N,s,p,h,r′,Ω such that for any u∈X,
(28)ur′x≤C0uX.
Thus, the spacesX is continuously embedded in Lr′xΩ for any r′x∈1,p∗. Furthermore, this embedding is compact. If u∈X0, then there exists a constant C0=C0N,s,p,h,r′,Ω such that
(29)ur′x≤C0us,px,y.Lemma 7 (See [40]).
For allu,φ∈X0, define the functional Lu:X0⟶X0∗ by
(30)Lu,φ≔∫Ω×Ωux−uypx,y−2ux−uyφx−φyx−yN+spx,ydxdy.
Then,(1)
Lu is a bounded and strictly monotone operator(2)
Lu is a mapping of type S+, i.e., if un⟶u in X0 and limsupn⟶∞Lun−Lu,un−u≤0, then un⟶u in X0(3)
Lu:X0⟶X0∗ is a homeomorphismLemma 8 (see [41]).
(1)
If1≤uX0<∞, then(31)uX0p−≤ρpx.yu≤uX0p+.(2)
If∥u∥X0<1, then(32)uX0p+≤ρpx.yu≤uX0p−.We state Ekeland’s variational principle and dual fountain theorem which will be used in the proofs of Theorems 2 and 3.Theorem 9 (Ekeland’s variational principle, see [42]).
LetX0 be a complete metric spaces with metric d, and let I:X0↦−∞,+∞ be a lower semicontinuous function, bounded from below and not identical to +∞. Let ε0 be given and U∈X0 be such that
(33)IU≤infX0I+ε0.
Then, there existsV∈X0 such that
(34)IU≤IV,dU,V≤1,and for each W∈X0, one has
(35)IU≤IW+ε0dV,W.
The spaceX0 is a separable and reflexive real Banach space, and there exists ej∈X0 and ej∗∈X0∗ such that X0=spanej:j=1,2,⋯¯,X0∗=spanej∗:j=1,2,⋯¯ and
ej∗,ei=1,i≠j;0,i=j.
SetXi=spanei, we define Ak=⊕i=1kXi,Bk=⊕i=k∞Xi¯.Theorem 10 (Dual Fountain Theorem, see [43]).
Assume thatI∈C1X0,ℝ satisfies I−u=Iu, and for every k≥k0, there exist ρk>γk>0 such that
(B1)ak=infIu:u∈Bk;u=ρk≥0;
(B2)bk=maxIu:u∈Ak;u=γk<0;
(B3)dk=infIu:u∈Bk;u≤γk⟶0,ask⟶∞;
(B4)I satisfies the Cec∗ condition for every c∈dk0,0.
Then,I has a sequence of negative critical values converging to 0.


## 3. Cerami Compactness Condition

We discuss the compactness properties of our energy functionalI related to the Cec condition and Cec∗ condition.Definition 11.
LetI∈C1X,ℝ, and we say that I satisfies the Cerami condition at the level c∈ℝ(Cec for short), if any sequence un⊂X0 with
(36)Iun⟶c,1+unX0I′un⟶0,asn⟶∞possesses a convergent subsequence in X0.Definition 12.
LetI∈C1X,ℝ, and we say that I satisfies Cec∗ condition at the level c∈ℝ(Cec∗ for short), if any sequence unj−nj⊂X0, namely, unj−∈Anj, with
(37)Iunj−⟶c,1+unj−X0I′unj−⟶0,asnj⟶∞possesses a convergent subsequence in X0.
Assume thatun is a bounded sequence in X0. By Theorem 6, there exists u∈X0 such that
(38)un⇀uinX0,un⟶ua.e.inΩ,un⟶uinLqxΩ.Lemma 13.
Letun,u,v∈X0 such that (38) holds. Then, passing to a subsequence, the following properties hold
(39)limn⟶∞∫Ωunqx−2unun−udx=0,limn⟶∞∫Ωfx,unun−udx=0.Proof.
(i)
By Hölder’s inequality (Lemma4),Theorem 6, and (38), we have(40)∫Ωunqx−2unun−udx≤∫Ωunqx−1un−udx≤unqx−1qx/qx−1un−uqx⟶0,asn⟶∞,and thus
(41)limn⟶∞∫Ωunqx−2unun−udx=0.(ii)
By virtue of conditions (f1) and (f2), for anyε∈0,1, there exists Cε>0 such that(42)fx,un≤εunp+−1+Cεunrx−1,combining with Hölder’s inequality, Theorem 6 and (38),and it follows that
(43)∫Ωfx,unun−udx≤∫Ωεunp+−1un−u+Cεunrx−1un−udx≤εunp+−1p+/p+−1un−up++Cεunrx−1rx/rx−1un−urx⟶0,asn⟶∞,which implies that
(44)limn⟶∞∫Ωfx,unun−udx=0.
The following lemma about theCec condition will play an important role in the proof of our main results.Lemma 14.
Let the functionf satisfy (f1), (f3), and (f4), then the energy functional I satisfies the Cec condition, where precisely c<a2/2b.Proof.
Step 1. We prove that un is bounded in X0. Let un be a Cec sequence, Iun⟶c,1+unX0I′un⟶0, which implies that
(45)c=Iun+o1,I′un,un=o1,where o1⟶0 as n⟶∞.
We claim thatun is a bounded sequence. Suppose to the contrary that
(46)unX0⟶∞,asn⟶∞.
Denoteωn≔un/unX0, then ωn∈X0 with ωnX0=1. Up to subsequences, for some ω∈X0, we get ωn⇀ω in X0,ωn⟶ωa.e. in Ω,ωn⟶ωinLqxΩ. There are only two cases need to be discussed ω=0 and ω≠0.
First, let us consider the caseω=0. In view of (f4), (8), (9), (45), (46), and Theorem 6, we obtain
(47)0=1unX02p−Iun−1p+I′un,un=1unX02p−a∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxunqxdx−∫ΩFx,undx−ap+∫Ω×Ωunx−unypx,yx−yN+spx,ydxdy+bp+∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy×∫Ω×Ωunx−unypx,yx−yN+spx,ydxdy+λp+∫Ωunqxdx+1p+∫Ωfx,unundx≥1unX02p−ap+∫Ω×Ωunx−unypx,yx−yN+spx,ydxdy−b2p−2∫Ω×Ωunx−unypx,yx−yN+spx,ydxdy2−λq−∫Ωunqxdx−∫ΩFx,undx−ap+∫Ω×Ωunx−unypx,yx−yN+spx,ydxdy+bp+2∫Ω×Ωunx−unypx,yx−yN+spx,ydxdy×∫Ω×Ωunx−unypx,yx−yN+spx,ydxdy+λp+∫Ωunqxdx+1p+∫Ωfx,unundx≥1unX02p−bp+2−b2p−2unX02p−−λq−−λp+Cq−unX0q−+∫Ω1p+fx,unun−Fx,undx≥1unX02p−bp+2−b2p−2unX02p−−λq−−λp+Cq−unX0q−−θ0unX02p−∫Ωωn2p−dx≥bp+2−b2p−2−λq−−λp+Cq−unX0q−−2p−−θ0∫Ωωn2p−dx,which implies 0>b/p+2−b/2p−2, and this is a contradiction according to the conditions p+<2p− and q−<2p− (see (11)).
In the case ofω≠0, setting Ω1≔x∈Ω:ω≠0, then limn⟶∞ωn=limn⟶∞un/∥un∥X0=ω≠0, from (46) implies that un=ωnunX0⟶+∞, for x∈Ω1, we have un⟶∞ as n⟶∞. In view of (f3), we see that
(48)Fx,ununX02p+=Fx,unun2p+un2p+unX02p+=Fx,unun2p+ωn2p+⟶+∞,a.e.inΩ1;so, using Fatou’s Lemma, we have
(49)limn⟶∞∫Ω1Fx,ununX02p+dx⟶+∞,a.e.inΩ1.
From (f1) and (f3), there existM′,M′′,M>0, such that ∣Fx,un∣≤M′unrx and Fx,un≥−M′′un2p+ as un⟶∞, and then we have Fx,un≥−Munr−, hence
(50)∫Ω\Ω1Fx,ununX02p+dx≥−∫Ω\Ω1Munr−dxunX02p+≥−Munr−unX02p+,by (46), which implies that
(51)liminfn⟶∞∫Ω\Ω1Fx,ununX02p+dx≥0.
Using (8), (45), (46), (49), (51), and Theorem 6, we obtain
(52)0=limn⟶∞−c+o1unX02p+=limn⟶∞−IununX02p+≥limn⟶∞1unX02p+−ap−unX0p++b2p+2unX02p−+λCqq+unX0q−+∫ΩFx,undx≥−ap−limn⟶∞unX0p+−2p++b2p+2limn⟶∞unX02p−−2p++λCqq+limn⟶∞unX0q−−2p++limn⟶∞∫Ω1Fx,ununX02p+dx+liminfn⟶∞∫Ω\Ω1Fx,ununX02p+dx⟶+∞,asn⟶∞,and this is a contradiction according to the condition q−<p−<p+ (see (11)). Then, the sequence un is bounded in X0.
Step 2. We prove that un has a convergent subsequence in X0. Since 1+∥un∥I′un⟶0,asn⟶∞, un is bounded in X0, we have
(53)I′un,un−u≤I′un,un+I′un,u≤I′unX0unX0+I′unX0uX0,therefore, we obtain
(54)limn⟶∞I′un,un−u=0,or
(55)I′un,un−u=a−b∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy×∫Ω×Ωunx−unypx,y−2unx−uny−ux+uyx−yN+spx,ydxdy−λ∫Ωunqx−2unun−udx−∫Ωfx,unun−udx,and we deduce from Lemma 13 that
(56)a−b∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy×∫Ω×Ωunx−unypx,y−2unx−uny−ux+uyx−yN+spx,ydxdy⟶0.
Sinceun is bounded in X0, passing to a subsequence, we may assume
(57)∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy⟶t0≥0asn⟶∞.(i)
Ift0=0, then un strongly converges to u=0 in X0(ii)
Ift0>0,t0≠a/b, then(58)a−b∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy⟶0is not true. Because un is bounded in X0, so,
(59)a−b∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdyis bounded.
(iii)
Ift0>0,t0=a/b, then a−b∫Ω×Ωunx−unypx,y/px,yx−yN+spx,ydxdy⟶0. We define a function by(60)ψu=λ∫Ω1qxuqxdx+∫ΩFx,udx,forallu∈X0,and then
(61)ψ′u,φ=λ∫Ωuqx−2uφdx+∫Ωfx,uφdx,forallu,φ∈X0,and it follows that
(62)ψ′un−ψ′u,φ=λ∫Ωunqx−2un−uqx−2uφdx+∫Ωfx,un−fx,uφdx.
Sinceun⟶u in LqxΩ, which implies that
(63)unqx−2un⟶uqx−2uinLqx/qx−1Ω,according to the Hölder’s inequality, we get
(64)∫Ωunqx−2un−uqx−2uφdx≤∫Ωunqx−2un−uqx−2uφdx≤unqx−2un−uqx−2uqx/qx−1φqx⟶0.Hence, we obtain
(65)∫Ωunqx−2un−uqx−2uφdx=0.
With a slight modification of the above proof, we can also prove the following conclusion; so, we have omitted the details. By Hölder’s inequality, (38) and (42), we deduce that
(66)limn⟶∞∫Ωfx,un−fx,uφdx=0.
Combining with (65) and (66), we get
(67)ψ′un−ψ′uX0∗⟶0andψ′un⟶ψ′u.
While(68)I′un,φ=a−b∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy×∫Ω×Ωunx−unypx,y−2unx−unyφx−φyx−yN+spx,ydxdy−ψ′un,φ,since I′un,φ⟶0, and a−b∫Ω×Ωunx−unypx,y/px,yx−yN+spx,ydxdy⟶0, it follows that ψ′un⟶0 as n⟶∞, i.e., by virtue of (61), we obtain
(69)−λuqx−2u−fx,u=0,a.e.x∈Ωby the fundamental lemma of the variational method (see [43]). It follows that u=0. So,
(70)ψun=λ∫Ω1qxunqxdx+∫ΩFx,undx⟶λ∫Ω1qxuqxdx+∫ΩFx,udx=0.
Hence, fort0=a/b, we have
(71)Iun=a∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxunqxdx−∫ΩFx,undx⟶a22b.
This is a contradiction withIun⟶c<a2/2b; then, a−b∫Ω×Ωunx−unypx,y/px,yx−yN+spx,ydxdy⟶0 is not true and similarly to (ii), and we obtain
(72)a−b∫Ω×Ωunx−unypx,ypx,yx−yN+spx,ydxdythat is bounded. So, it can be concluded from the above discussion (ii) and (iii) that
(73)∫Ω×Ωunx−unypx,y−2unx−uny−ux+uyx−yN+spx,ydxdy⟶0.
Thus, invoking theS+ condition (see (2) of Lemma 7), we deduce that the un converges to u in X0.
Therefore, the energy functionalI satisfies the Cec condition.Lemma 15.
Let the functionf satisfy (f1), (f3), and (f4), and then the energy functional I satisfies the Cec∗ condition, where precisely c<a2/2b.Proof.
Consider a sequenceunj−nj⊂X0 such that unj−∈Anj, Iunj−⟶c,I′unj−Anj′⟶0, in X0∗, as nj⟶∞. By the same method used in the proof of Lemma 14, we can prove that unj−nj has a strongly convergent subsequence in X0. The details are omitted.Remark 16.
Since theCec and Cec∗ conditions hold for the energy c<a2/2b, we discuss the negative energy solutions for the problem Pλ.


## 4. Proof of Theorem2

We firstly prove two lemmas.Lemma 17.
Let assumptions (f1), (f2), and (f3) hold, and then there are constantsρ,λ∗ and α>0 such that Iu≥α, where λ∈0,λ∗ and uX0=ρ.Proof.
For anyε∈0,1 and λ>0, by (f1) and (f2), there exists Cε>0 such that
(74)Fx,u≤εp+up++Cεrxurx.
Letu∈X0 be such that u=ρ∈0,1 and combining with (74), we deduce that
(75)Iu=a∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxuqxdx−∫ΩFx,udx≥a∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxuqxdx−∫Ωεp+up+dx−∫ΩCεrxurxdx≥ap+ρpx.yu−b2p−2ρpx.y2u−λq−ρqxu−εp+ρp+u−Cεr−ρrxu.
By Theorem6, Lemma 5, and Lemma 8, there exist positive constants Cq−,Cp,Cr such that
(76)uqxq−≤Cq−uX0q−,up+p+≤CpuX0p+,urxr−≤CruX0r−.
Hence, we obtain(77)Iu≥ap+uX0p+−b2p−2uX02p−−λCq−q−uX0q−−εCpp+uX0p+−CεCrr−uX0r−=uX0p+ap+−b2p−2uX02p−−p+−λCq−q−uX0q−−p+−εCpp+−CεCrr−uX0r−−p+.
Chooseε=a/2Cp>0 and take
(78)It=a2p+−b2p−2t2p−−p+−CεCrr−tr−−p+,forallt≥0.
Note thatp+<2p−<r− (see (11)), and we can conclude that there exists a constant ρ>0 such that
(79)Iρ=maxt≥0It>0.
Therefore, takingλ∗≔q−/2Cq−ρq−−p+Iρ>0, we have
(80)Iu≥12ρp+Iρ≕α>0,whenever ∥u∥X0=ρ and λ∈0,λ∗. This completes the proof.Lemma 18.
Let assumptions (f1), (f2,) and (f3) hold, and then(81)−∞<infIu:u∈B¯ρ<0,where ρ is given by Lemma 17 and B¯ρ≔u∈X0:uX0≤ρ.Proof.
By (f2) and (f3), for anyM1>0, there exists CM1>0 such that
(82)Fx,u≥M1u2p+−CM1up+,forallx,u∈Ω×ℝ.
Letϕ∈X0,ϕ>0andt>0, combining with (82), we have
(83)Itϕ=a∫Ω×Ωtϕx−tϕypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωtϕx−tϕypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxtϕqxdx−∫ΩFx,tϕdx≤a∫Ω×Ωtϕx−tϕypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωtϕx−tϕypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxtϕqxdx−∫ΩM1tϕ2p+dx+∫ΩCM1tϕp−dx≤atp−p−∫Ω×Ωϕx−ϕypx,yx−yN+spx,ydxdy−bt2p+2p+2∫Ω×Ωϕx−ϕypx,yx−yN+spx,ydxdy2−λtq+q+∫Ωϕqxdx−M1t2p+∫Ωϕ2p+dx+CM1tp+∫Ωϕp+dx.
Sinceq+<p−<p+ (see (11)), for t>0, it is small enough. Hence, we obtain that
(84)−∞<infIu:u∈B¯ρ<0,where ρ is given by Lemma 17 and B¯ρ≔u∈X0:uX0≤ρ.
Now, we give the proof of Theorem2.Proof.
By Lemma17 and Lemma 18, we know that
(85)−∞<c=infB¯ρIu<0<α<inf∂BρIu,whenever λ∈0,λ∗. Set
(86)1n∈0,inf∂BρIu−infB¯ρIu,n∈ℕ.
So, by Theorem9 (Ekeland’s variational principle), there exists un∈B¯ρ such that
(87)c≤Iun≤c+1n,(88)Iun≤Iu+1nun−uX0,for all u∈B¯ρ,un≠u. In that way, by (86)–(88), we have
(89)Iun≤c+1n≤infu∈B¯ρIu+1n<infu∈∂BρIu,so that un∈Bρ.
Define the functionalMn:B¯ρ⟶ℝ by
(90)Mnu=Iu+1nun−uX0,forallu∈B¯ρ.
Like that, (88) implies that Mnun<Mnu for all u∈B¯ρ, with un≠u, and thus un is a strict local minimum of Mn. Furthermore, for all v∈B1, where B1=v∈X0:vX0=1, taking t>0 small enough, we have
(91)Mnun+tv−Mnunt≥0,which implies that
(92)Iun+tv−Iunt+1nvX0≥0.
Thus,(93)I′un,v+1nvX0≥0forallv∈B1.
In the above inequality, replacingv with −v, we have
(94)−I′un,v+1nvX0≥0forallv∈B1.
Hence,(95)I′unX0∗≤1n.
Passing to the limit in (87) and (95), we conclude that there is a sequence un⊂B¯ρ such that
(96)Iun⟶c<0andI′unX0∗⟶0,asn⟶∞.
To sum up, by Lemma14, there exists a strongly convergent sequence un, still denoted by un, such that un⟶u0 in X0. Consequently, u0 is a solution of Pλ, with Iu0<0. The proof of Theorem 2 is completed.


## 5. Proof of Theorem3

In order to prove Theorem3, we need the following result.Lemma 19 (see [44], Lemma 18).
Assume thatζx∈C+Ω¯,ζx<p∗x, for any x∈Ω¯ and denote
(97)βk=supu∈Bk,uX0=1uζx,then limk⟶∞βk=0.
We give the proof of Theorem3.Proof.
The functionf satisfies hypotheses (f1)–(f5),and it is obvious that I−u=Iu for all u∈X0. Together with Lemma 15, we only need to verify that the energy functional I:X0⟶ℝ satisfies all conditions (B1)–(B3) of Theorem 10.
Step 1. Verification of condition (B1). According to (f1) and (f2), there exist C2>0,C3>0, such that
(98)Fx,u≤C2p+up++C3rxurx.
For anyu∈Bk with u≤min1,1/C0, combining with (98) Theorem 6 and Lemma 8, there exist positive constants Cq−,Cp,Cr such that we obtain
(99)Iu=a∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxuqxdx−∫ΩFx,udx≥a∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxuqxdx−∫ΩC2p+up+dx−∫ΩC3rxurxdx≥ap+ρpx.yu−b2p−2ρpx.y2u−λq−ρqxu−C2p+ρp+u−C3r−ρrxu≥ap+uX0p+−b2p−2uX02p−−λCq−q−uX0q−−C2Cpp+uX0p+−C3Crr−uX0r−.
According to Lemma19 and the conditions q−<p+<2p− (see (12)), it follows that
(100)Iu≥ap+uX0p+−b2p−2uX02p−−λCq−βkq−q−uX0q−−C2Cpβkp+p+uX0p+−C3Crβkr−r−uX0r−≥ap+−b2p−2uX02p−−λCq−+C2Cpβkq−q−uX0q−−C3Crβkr−r−uX0r−,
sincea≥b and p+<2p−<2p−<r−(see (12)) imply that a/p+−b/2p−2=2p−2a−bp+/2p−2p+>0.
Hence, we deduce(101)Iu≥2p−2a−bp+2p−2p+−C3Crβkr−r−uX0r−−λCq−+C2Cpβkq−q−uX0q−.
Choosing2p−2a−bp+/4p−2p+>C3Crβkr−/r−, we have
(102)Iu≥2p−2a−bp+4p−2p+uX0r−−λCq−+C2Cpβkq−q−uX0q−.
Obviously, there exists a large enoughk such that
(103)Iu≥uX0q−2p−2a−bp+4p−2p+uX0r−−q−−λCq−+C2Cpβkq−q−.
We choose(104)ρk≔4λCq−+C2Cpp+p−2q−2ap−2−bq−p+βkq−1/r−−q−,by Lemma 19, and we know that ρk⟶0 as k⟶∞. So, there exists k0>0, such that k>k0 and u∈Bk with ∥u∥X0=ρk, and we have Iu≥0. Therefore, the condition (B1) of Theorem 10 holds.
Step 2. Verification of condition (B2). By (f2) and (f3), there exist C4,C5>0 such that
(105)Fx,u≥C4u2p+−C5up+,forallx,u∈Ω×ℝ.
Foru∈Ak, when u=γk with ρk>γk>0, we have
(106)Iu=a∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxuqxdx−∫ΩFx,udx≤a∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy−b2∫Ω×Ωux−uypx,ypx,yx−yN+spx,ydxdy2−λ∫Ω1qxuqxdx−∫ΩC4u2p+dx+∫ΩC5up+dx≤ap−ρpx.yu−b2p+2ρpx.y2u−λq+ρqxu−C4ρ2p+u+C5ρp+u.
Since all norms on the finite-dimensional spaceAk are equivalent, we obtain
(107)Iu≤ap−uX0p−−b2p+2+C4C2p+uX02p+−λCq+q+uX0q++C5Cp+uX0p+.
According to the conditionq+<p−<p+ (see (12)), for some γk=∥u∥X0 that is small enough, we deduce that
(108)bk=maxu∈Ak,uX0=γkIu≤0.
Therefore, the condition (B2) of Theorem10 holds.
Step 3. Verification of condition (B3). By verification of condition (B1), one has that for k≥k0 and u∈Bk with uxX0≤γk,
(109)Iu≥−λ+C2Cpβkq−q−uX0q−≥−λ+C2Cpβkq−q−γkq−,
sinceγk⟶0 as k⟶∞. Therefore, the condition (B3) of Theorem 10 also holds.
By above three steps and Lemma15, I satisfies all conditions of Theorem 10. The proof of Theorem 3 is completed.



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*Source: 8888078-2021-03-23.xml*