# Ground State Solutions for a Nonlocal System in Fractional Orlicz-Sobolev Spaces

Authors: Hamza El-Houari; Hicham Moussa; Lalla Saâdia Chadli
Journal: International Journal of Differential Equations
 (2022)

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## Abstract

We consider an elliptic system driven by the fractionala.-Laplacian operator, with Dirichlet boundary conditions type. By using the Nehari manifold approach, we get a nontrivial ground state solution on fractional Orlicz–Sobolev spaces.


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## Body

## 1. Introduction

We consider the following fractional elliptic system(1)−Δa1.su=Fux,u,vinΩ,−Δa2.sv=Fvx,u,vinΩ,u=v=0onℝN/Ω,where Ω is a bounded open subset of ℝN with Lipschitz boundary ∂Ω, s∈0,1, and −Δai=1,2.s is the nonlocal fractional ai.-Laplacian operator of elliptic type introduced in [1] and is defined as(2)−Δai.sux=P.V∫ℝNaiux−uyx−ysux−uyx−ysdyx−yN,for all x∈ℝN, where P.V is the principal value and ai=1,2:ℝ+⟶ℝ+ are of class C2 and satisfies the following conditions: ϕ1limt⟶0tait=0,andlimt⟶∞tait=∞.ϕ2t⟶tait is strictly increasing.For example, when we choosea1t=tpx−2t and a2t=tqx−2t for t>0, then ai satisfies ϕ1 and ϕ2. In this case, the operator (2) is named fractional p.-Laplacian operator and reads as(3)−Δp.sux=limϵ⟶0∫ℝN\Bϵ|ux−uy|px−2ux−uy|x−y|N+s−1px,ydy,forx∈ℝN,and the system (1) reduces to the fractional p.,q.-Laplacian system studied in [2] and given as(4)−Δp.su=Fux,u,vinΩ,−Δq.sv=Fvx,u,vinΩ,u=v=0onℝN/Ω.The existence of solutions for systems like (3) has also received a wide range of interests. For this, we find that, in the literature, many researchers have studied this type of system using some important methods, such as variational method, Nehari manifold and fibering method, and three critical points theorem (see for instance [3–6]). This kind of operator can be used for many purposes, such as phase transition phenomena, population dynamics, and continuum mechanics (see for example [7–9]). It is natural to think about what results can be obtained when the a.-Laplace operator is changed to a fractional a.-Laplacian operator. From what we know is, there is a number of results obtained about the fractional Orlicz–Sobolev spaces, such as a nonlocal Kirchhoff problem (see [10]):(5)K∫Ω×ΩAux−uyx−ysdxdyx−yN−Δa.su=Fux,uinΩ,u=0onℝN/Ω,where K is the Kirchhoff function and A is an N function, i.e., A:ℝ+⟶ℝ+ is a convex function such that(6)At>0fort≠0,limt⟶0Att=0andlimt⟶0Att=∞.In a similar way, Fernández Bonder and Salort in [11] studied the following problem with a nonlinear nonhomogeneous reaction term:(7)−Δa.sux=fx,ux,gux+hxinΩ,u=0onℝN/Ω,where f is a Carathéodory function. Therefore, in [12], we have investigated the following type of fractional a.-Kirchhoff system(8)K1G1u−Δa1.su=Fux,u,vinΩ,K2G2v−Δa2.sv=Fux,u,vinΩ,u=v=0onℝN/Ω,where the functional Gi=1,2:W0s,AiΩ⟶ℝ is defined by(9)Giu=∫Ω×ΩAiux−uyx−ysdxdyx−yN.To the best of our knowledge, this is one of the first works devoted to the studies of the existence of a solution to the nonlocal problem (1) by using the suitable Nehari manifold method. The main difficulty in this work arises from the complicated method which we are applying to find the solution. Furthermore, the nonhomogeneities of the operator (2).To state the outcome, we assume the following condition on the functionai: ϕ3:(10)−1<li−2≔inft>0tait″ttait′≤supt>0tait″ttait′≔ni−2<N−2,where ni=1,2<minr1,r2,r1r2−1/r2+1,r2r1−1/r1+1,N. We can notice that by applying arguments as in [13], we arrive at ϕ3 which implies the condition ϕ3':(11)1<li≔inft>0t2aitAit≤supt>0t2aitAit≔ni<N.ϕ4: there exists 1<ηi≤maxr1,r2 such that(12)limt⟶+∞tηiAit=0,and F satisfies that F1: F:Ω×ℝ×ℝ⟶ℝ is a C1 function such that Fx,0,0=0 for all x∈Ω,(13)Fux,u,v≤c1ur1−1+c2vr2r1−1/r1,Fvx,u,v≤c1ur1r2−1/r2+c2vr2−1,F2: limu,v⟶+∞Fx,u,v/ur1+vr2=+∞,uniformlyforallxinΩ.F3: the functions(14)u↦Fux,u,vur1−2u+vr2−2vandv↦Fvx,u,vur1−2u+vr2−2vare increasing on Ω×ℝ/0×ℝ and Ω×\ℝ×\ℝ/0, respectively.This work is structured as follows. In Section2, we recall briefly some properties on Orlicz and fractional Orlicz–Sobolev spaces. Section 3 is related to specify the assumptions on the data and to show the existing results of the problem (1) and its proof based on Nehari manifolds.


## 2. Some Preliminary Results

The reader is referred to [1, 14–17] for more details on Orlicz and fractional Orlicz-Sobolev space.We point out that anN function A∈Δ2. If for a certain constant k>0,(15)A2t≤kAt,foreveryt>0.The relationship-relatedA and A¯ is given by A¯t≔supr≥0tr−Ar, where A¯ is the complementary to A. We assume that(16)H0∫01Ai−1tt1+s/Ndt<∞andH∞∫1+∞Ai−1tt1+s/Ndt=+∞,fors∈0,1.Then, we introduced the conjugateN function A∗, given by the following expression of its inverse in ℝ+:(17)Ai∗−1t=∫0tAi−1rrN+s/Ndrfort≥0.LetΦ and A be two N functions. The notation Φ≪A means that, for each ε>0,(18)ΦtAεt⟶0ast⟶∞.The Orlicz spaceLAΩ is defined as follows:(19)LAΩ=u:Ω⟶ℝmesurablesuchthat∫ΩAλuxdx+∞forsomeλ0.The usual norm onLAΩ is uA=infλ>0/∫ΩAux/λdx≤1.We recall that the Hölder inequality holds(20)∫Ωuxvxdx≤uAvA¯forallu∈LAΩandv∈LA¯Ω.And the Young inequality reads as follows:(21)rt≤Ar+A¯tforallt,r≥0andx∈Ω.After this, we list a few inequalities that will be used for our proofs. The proof is provided in [18].Lemma 1.
Letξ0t=mintli,tni and ξ1t=maxtli,tni, and Ai is an N function; then, these assertions are equivalent as follows:(22)1<li:=inft>0taitAit≤supt>0taitAit:=ni<+∞,(23)ξ0tAiρ≤Aiρt≤ξ1tAiρ,∀t,ρ≥0.Lemma 2.
IfAi is an N function satisfies (22), then we have(24)ξ0uAi≤∫ΩAiudx≤ξ1uAi,∀u∈LAiΩ.Lemma 3.
LetA¯i be the complement of Ai and ξ2t=mintl¯i,tn¯i, ξ3t=maxtl¯i,tn¯i, t≥0 where l¯i=li/li−1 and n¯i=ni/ni−1. If Ai is an N function and (22) holds, then A¯i satisfies(25)ξ2tA¯iρ≤A¯iρt≤ξ3tA¯iρ,∀t,ρ≥0.(26)ξ2uA¯i≤∫ΩA¯iudx≤ξ3uA¯i,∀u∈LA¯iΩ.Lemma 4.
We haveAi≪Ai∗, i.e., limt⟶∞Aikt/Ai∗t=0,foreveryconstantk>0.Remark 1.
By Lemma1, Lemma 3, and (22), we show that Ai,A¯i∈Δ2.
We now look at the definition of the fractional Orlicz–Sobolev spacesWs,AiΩ (see[1]), and we set some properties on these spaces. The fractional Orlicz–Sobolev spaces Ws,AiΩ defined as follows:(27)Ws,AiΩ=u∈LAiΩ:∫Ω×ΩAiλux−uyx−ysx−y−Ndxdy<∞,where Ω in ℝN and s∈0,1. This space is equipped with the norm(28)us,Ai=uAi+us,Ai,where .s,Ai is the Gagliardo seminorm, defined by(29)us,Ai=infλ>0:∫Ω×ΩAiux−uyλx−ysx−y−Ndxdy≤1.
To deal with the problem under consideration, we choose(30)W0s,AiΩ≔u∈Ws,AiℝN:u=0a.eℝN/Ω.
In these spaces, the generalized Poincaré inequality reads as follows (see [19]):(31)uAi≤Cus,Ai,∀u∈W0s,AiΩ,where C is a positive constant. Then, W0s,AiΩ,us,Ai is a Banach space whose norm is equivalent to us,Ai. Also it is a separable (resp. reflexive) space if and only if Ai∈Δ2 (resp. Ai∈Δ2 and A¯i∈Δ2). Furthermore, if Ai∈Δ2 and Ait is convex, then the space Ws,AiΩ is uniformly convex Theorem 1 (see [1]).Theorem 1 (see [19]).
LetAi be an N function s∈0,1, Ω⊂ℝN, and C0,1− regularity with bounded boundary. If H0, H∞, and 2.5 hold, then the embedding Ws,AiΩLAi∗Ω is continuous, and the embedding Ws,AiΩLBΩ is compact for any N function B≪Ai∗.Remark 2.
The assumptionϕ4 implies that there is ηi∈1,maxr1,r2 such that tηi≪Ai; then, by Lemma 4 and Theorem 1, the following embedding Ws,AiΩLηiΩ are compact. i.e., there exists a constant Cηi>0 such that(32)uηi≤Cηius,Aiforallu∈Ws,AiΩ.
Now, we defined our working spaceW≔W0s,A1Ω×W0s,A2Ω under the norm(33)u,v≔us,A1+vs,A2≃us,A1+vs,A2.
So, by the above arguments,W,u,v is a separable and reflexive Banach space.


## 3. Nehari Manifolds in Fractional Orlicz–Sobolev Spaces

We observe that the energy functionalℐ on W corresponding to system (1) is(34)ℐu,v≔G1u+G2v−ℱu,v,forallu,v∈W,where ℱu,v=∫ΩFx,u,vdx, and Gi=1;2.:W0s,AiΩ⟶ℝ is the function defined in (9). Now, let us prove that the energy functional ℐ is well defined. To do this, we need to see the following:Remark 3.
By (13) and the fact that(35)Fx,u,v=∫0uFpx,p,vdp+∫0vFtx,0,tdt∀x,u,v∈Ω×ℝ×ℝ,we show that(36)Fx,u,v≤∫0uFpx,p,vdp+∫0vFtx,0,tdt≤c1∫0upr1−1dp+c2∫0uvr2r1−1/r1dp+c2∫0vtr2−1dt=c1r1ur1+c2|uv|r2r1−1/r1+c2r2vr2≤c1r1ur1+c2r1ur1+c2r1−1r1vr2+c2r2vr2=c3ur1+c4vr2,where c3=c1+c2/r1andc4=c2r2r1−1+r1/r1r2.
The fractionalai.-Laplacian operator specified in (2) is indeed defined between W0s,AiΩ and its dual space W0s,AiΩ∗. Indeed, in [1], Theorem 6.12, the following expression is found:(37)Gi′u,v=∫Ω×Ωaiux−uyx−ysux−uyvx−vyx−y2sdxdyx−yN,=−Δai.su,v,for all u,v∈W0s,AiΩ. Throughout the document, we note by hu=ux−uy/x−ys and dμ=dxdy/x−yN (i.e., regular Borel measure on the set Ω×Ω).Lemma 5.
The functionℐ is well defined, and it is the C1W,ℝ, and we have(38)ℐ′u,v,u¯,v¯=−Δa1.su,u¯+−Δa2.su,v¯−ℱ′u,v,u¯,v¯,for all u¯,v¯∈W.Proof.
First, we can see that(39)Gi′u,v=−Δai.su,v=∫Ω×Ωaihuhuhvdμ,for all v∈Ws,AiΩ. It follows from (39) for each u∈Ws,AiΩ that Gi′u∈Ws,AiΩ∗. Next, we prove that Gi∈C1Ws,AiΩ,ℝ. On the one hand, let uk⊂Ws,AiΩ with uk⟶u strongly in Ws,AiΩ, for v∈Ws,AiΩ; we have hv∈LAiΩ×Ω,dμ and by Hölder inequality,(40)∫Ω×Ωaihukhuk−aihuhuhvdμ≤2aihukhuk−aihuhuLA¯ihvLAi
On the other hand, ifun⟶u in Ws,AiΩ, then hun⟶hu in LAiΩ×Ω,dμ, so by Dominated convergence theorem, there exists a subsequence hunk and a function h in LAiΩ×Ω,dμ such that(41)aihunkhunk≤aihh∈LΓ¯iΩ×Ωa.einΩ×Ω.
And(42)aihunkhunk⟶aihuhua.einΩ×Ω.
Again by dominated convengeance theorem, we obtain that(43)supvs,A¯i≤1∫Ω×Ωaihunhun−aihuhuhvdμ⟶0.
Combining (43) in (40), we get Gi′ is continuous. Now, we turn to prove that(44)ℱ′u,v,u¯,v¯=∫ΩFux,u,vu¯+Fvx,u,vv¯dxforallu¯,v¯∈W.
By (32) and (36), we have(45)ℱu,v≤∫ΩFx,u,vdx≤c3∫Ωur1dx+c4∫Ωvr2dx=c3ur1r1+c4vr2r2≤c3Cr1r1us,A1r1+c4Cr2r2vs,A2r2.
Then,ℱ is well defined in W. Now, by (13) and (32) and the similar argument in [[20] Lemma 3.1,] we see that (44) is holds. This completes the proof.
To find the critical points ofℐ, we will minimize the energy functional ℐ on the constraint of Nehari manifold(46)N=u,v∈W/0,0:ℐ′u,v,u,v=0=u,v∈W/0,0:G1′u,u+G2′v,v=ℱ′u,v,u,v.
The main result of this section is as follows:Theorem 2.
We assume thatϕ1-ϕ4 and F1-F3 hold true. Then, the system (1) has a nontrivial ground state solution u,v in the way that there is u,v∈N such that(47)∫Ω×Ωa1huhuhu¯dμ+∫Ω×Ωa2hvhvhv¯dμ=∫ΩFux,u,vu¯dx+∫ΩFvx,u,vv¯dx,ℐu,v=infu^,v^∈Nℐu^,v^,where u¯,v¯∈W.Lemma 6 (see [16] Lemma 4.1).
The following properties are true:(48)Giuus,Ai≤1,forallu∈W0s,Ai/0Ω,ξ0us,Ai≤Giu≤ξ1us,Ai,forallu∈W0s,AiΩ.Lemma 7 (see [21]).
We suppose thatA1t is convex, uk⇀u in W0s,A1Ω and limsupG1′uk,uk−u≤0. Then, uk⟶u∈W0s,A1Ω. Similary, we can obtain that, vk⟶v in W0s,A2Ω. Therefore, uk,vk⟶u,v in W.Lemma 8.
We suppose thatϕ1-ϕ4, (22), and F1 hold true. Then, we have(49)i∫ΩFux,uk,vkuk−udx⟶0andii∫ΩFvx,uk,vkvk−vdx⟶0.Proof.
Letuk,vk be sequence in W such that uk,vk⇀u,v. By (13) and Hölder inequality, we have(50)∫ΩFux,uk,vkuk−udx≤∫ΩFux,uk,vkuk−udx≤c1∫Ωukr1−1uk−udx+c2∫Ωvkr2r1−1/r1uk−udx≤c1∫Ωukr1dxr1−1/r1∫Ωuk−ur1dx1/r1+c2∫Ωuk−ur1dx1/r1∫Ωvkr2dxr1−1/r1.
Sinceuk,vk⇀u,v in W, it is clear that uk⇀u in W0s,A1Ω and vk⇀v in W0s,A2Ω, then uks,A1 and vks,A2 are bounded. By Remark 2, there exists G>0 such that(51)uks,A1,vks,A2,ukr1,vkr2≤G,forn∈ℕ.
Combining (50) and (51), then we have(52)∫ΩFux,uk,vkuk−udx≤c1∫Ωuk−ur1dx1/r1ukr1r1−1+c2∫Ωuk−ur1dx1/r1×vkr2r2r1−1/r1≤c1∫Ωuk−ur1dx1/r1Gr1−1+c2∫Ωuk−ur1dx1/r1×Gr2r1−1/r1.
Hence, the compact embeddingWs,AiΩLriΩ (see Remark 2) make sure that uk⟶u in LriΩ; then, for any positive ϵ, we can choose n0∈ℕ such that(53)∫Ωuk−ur1dx1/r1<ϵfor allk>n0.
Furthermore,(54)∫ΩFux,uk,vkuk−udx≤c1∫Ωuk−ur1dx1/r1Gr1−1+c2∫Ωuk−ur1dx1/r1×Gr2r1−1/r1≤ϵc1Gr1−1+c2Gr2r1−1/r1,for all k>n0 and for small n0. Since ϵ is an arbitrary one, by joining (52) with (54), we concluded that item i helds. With a similar discussion to the one above, we can show that ii is proven to be true.Lemma 9.
We assume thatϕ1-ϕ3, F1-F3. Then, there exists a constant C>0 such that u,v≥C for each u,v∈N.Proof.
The proof is augmented by contradiction. We suppose that there exists a subsequence denoted byuk,vk∈N such that uk,vk≤1/k, for each integer k≥1. Using ϕ3', the construction of N (see (46)), (13), the Hölder inequality, and (32), we have(55)∫Ω×ΩA1hukdμ+∫Ω×ΩA2hvkdμ≤∫Ω×Ωa1hukhuk2dμ+∫Ω×Ωa2hukhvk2dμ=∫ΩFux,uk,vkuk+Fvx,uk,vkvkdx≤c1∫Ωukr1dx+c2∫Ωukr1dx1/r1∫Ωvkr2dxr1−1/r1+c1∫Ωvkr2dx+c2∫Ωvkr2dx1/r2∫Ωvkr1dxr2−1/r2=c1ukr1r1+c2ukr1vkr2r2r1−1/r1+c1vkr2r2+c2vkr2ukr1r1r1−1/r2≤c1Cr1r1uks,A1r1+c2Cr1Cr2r2r1−1/r1uks,A1vks,A2r1r1−1/r2+c1Cr2r2vks,A2r2+c2Cr2Cr1r1r2−1/r2vks,A2uks,A1r1r1−1/r2≤c5uks,A1r1+uks,A1vks,A2r2r1−1/r1+vks,A2r2+vks,A2uks,A1r1r2−1/r2,where c5=maxc1Cr1r1,c2Cr1Cr2r2r1−1/r1,c1Cr2r2,c2Cr2Cr1r1r2−1/r2.
According to Lemma1, Remark 2, and for k large enough, last inequality rewrites(56)uks,A1n1+vks,A2n2≤c5uks,A1r1+uks,A1vks,A2r2r1−1/r1+vks,A2r2+vks,A2uks,A1r1r2−1/r2.
Dividing the last expression byuks,A1n1+vks,A2n2, we get to(57)1≤c5uks,A1r1+vks,A2r2uks,A1n1+vks,A2n2+c5vks,A2uks,A1r1r2−1/r2+uks,A1vks,A2r2r1−1/r1uks,A1n1+vks,A2n2.
The fact thatni<r1,r2,r1r2−1/r2+1,r2r1−1/r1+1, we get that for a large k,(58)1≤c5ok1,which is a contradiction.Lemma 10.
We assume thatϕ1-ϕ3 and F1-F3 hold true. Then, the function Ji:W0s,AiΩ⟶ℝ given by(59)Jiw=∫Ω×Ωaihwhw2dμ,w∈W0s,AiΩ,is C1 and Ji′w,v=∫Ω×Ω2aihw+ai′hwhwhwhvdμ,w,v∈W0s,Ai.Proof.
We set(60)ht=aihu+thvhu+thv2,0≤t≤1.
So, by definition ofai, we have h∈C1 and(61)h′t=2aihu+thv+ai′hu+thvhu+thvhu+thvhv.
Furthermore, there isθ∈ℝ with 0<θ<t≤1 such that(62)ht−h0t=h′t.
Thus,(63)Ji′u,v=limt⟶0∫Ω×Ωaihu+thvhu+thv2−aihuhu2tdμ=limt⟶0∫Ω×Ωh′θdμ.
We recall that byϕ3':(64)ai′tt≤maxli−2,ni−2ait,0≤t<∞,and as a consequence,(65)limt⟶0ai′tt2=limt⟶0maxli−2,ni−2ait=0.
Usingϕ1-ϕ3 and 0≤θ≤1, we have(66)h′θ≤2aihu+θhv+ai′hu+θhvhu+θhvhu+θhvhv≤2+maxli−2,ni−2aihu+hvhu+hvhv∈L1Ω×Ω,dμ.
Indeed, using Young’s inequality, (15) and the fact A¯itait≤Ai2t, we infer that(67)aihu+hvhu+hvhv≤Aihu+A¯iaihu+hvhu+hv≤Aihu+Ai2hu+hv≤Aihu+2niAihu+hv≤1+2niAihu+hv∈L1Ω×Ω,dμ.
Hence,(68)h′θ≤2+maxli−2,ni−21+2niAihu+hv:=g1∈L1Ω×Ω,dμ.
Now, let us see thatJi′ is continuous. Let ukk∈ℕ⊂W0s,AiΩ be a sequence such that uk⟶u.
Then,(69)∫Ω×ΩAihuk−Aihudμ⟶0,huk⟶huinΩ×Ω,huk≤g2a.e.inΩforsomeg2∈L1Ω×Ω,dμ.
By arguments as above,(70)2aihuk+ai′hukhukhukhv≤2aihuk+ai′hukhukhukhv≤2aihukhuk+ai′hukhuk2hv≤Caihukhukhv∈L1Ω×Ω,dμ,where C≔2+maxli−2,ni−2. By (64) and (65), we get that(71)2aihuk+ai′hukhukhukhv⟶2aihu+ai′huhuhuhva.einΩ.
Applying the Lebesgue dominated convergence theorem,(72)limk⟶∞J′uk−J′u,v=0.
Hence, we haveJ∈C1W0s,AiΩ,ℝ which completes the proof of Lemma 10.
At this step, in view of determining the behavior ofℐ on N, we introduce the fibering map Υu,v:ℝ+⟶ℝ associated to the Nehari manifold given by(73)Υu,vt=ℐtu,tv,forallu,v∈W.
Using (38), then the first derivative of the map Υu,v is given by(74)Υu,v′t=ℐ′tu,tv,u,v=∫Ω×Ωta1thuhu2dμ+∫Ω×Ωta2thvhv2dμ−ℱ′tu,tv,u,v.Lemma 11.
We suppose thatϕ1-ϕ3 and F1-F3 hold. Then,(75)limt⟶0Υu,vttn1+tn2>0,limt⟶∞Υu,vttn1+tn2=−∞,limt⟶0Υu,v′ttn1−1+tn2−1>0,limt⟶∞Υu,v′ttn1−1+tn2−1=−∞.Proof.
By equation (36), there exists c3,c4>0 such that(76)Fx,u,v≤c3ur1+c4vr2,for a given ϵ,t∈0,1, and by Poincaré’s inequality and Lemma 6, we have(77)Υu,vt=G1tu+G2tv−ℱtu,tv>ϵG1tu+ϵG2tv−∫ΩFx,tu,tvdx.
Using Lemma 2.3 in [10] and (23), we infer that(78)Υu,vt>ϵλ1∫ΩA1tudx+ϵλ2∫ΩA2tvdx−c3∫Ωtur1dx−c4∫Ωtvr2dx≥tn1ϵλ1∫ΩA1udx+tn2ϵλ2∫ΩA2vdx−tr1+tr2×2maxc3∫Ωur1dx,c4∫Ωvr2dx.
With the above arguments, we have for allu,v∈W/0,0,(79)Υu,vttn1+tn2>minϵλ1∫ΩA1udx,ϵλ2∫ΩA2vdx−tr1+tr2tn1+tn2×2maxc3∫Ωur1dx,c4∫Ωvr2dx.
We note thatni=1,2<minr1,r2, so the last inequality (79) rewrites as follows:(80)Υu,vttn1+tn2>minελ1∫ΩA1udx,ελ2∫ΩA2vdx+o1.
Hence,(81)limt⟶0Υu,vttn1+tn2>minϵλ1∫ΩA1udx,ϵλ2∫ΩA2vdx≥0.
Now, for item (ii), by using (23), we have(82)Υu,vttn1+tn2≤max∫ΩA1udx,∫ΩA2vdx−∫ΩFx,tu,tvtn1+tn2dx.
Moreover, we can see that(83)∫ΩFx,tu,tvtn1+tn2dx≥∫ΩFx,tu,tvtun1+tvn2×tun1+tvn2tn1+tn2dx≥∫ΩFx,tu,tvtun1+tvn2×minun1+vn2dx≥∫ΩFx,tu,tvtur1+tvr2×minun1+vn2dx.
In fact from the conditionF2 and (83), we infer that(84)∫ΩFx,tu,tvtn1+tn2dx⟶+∞.
Combining (84) in (82), we proved item (ii). Now, for (iii), we first show that, by equation (13), there exists c1,c2>0 such that(85)∫ΩFux,tu,tvudx+∫ΩFvx,tu,tvvdx≤c1∫Ωtur1dx+c2∫Ωtvr2r1−1/r1udx+c1∫Ωtur1r2−1/r2vdx+c2∫Ωtvr2dx≤c1tr1∫Ωur1dx+c2tr2r1−1/r1∫Ω|v|r2r1−1/r1udx+c1tr1r2−1/r2∫Ωur1r2−1/r2vdx+c2tr2∫Ωvr2dx≤c1tr1+tr2max∫Ωur1dx,∫Ωvr2dx+c2tr2r1−1/r1+tr1r2−1/r2×max∫Ωvur1r2−1/r2dx,∫Ω|u‖v|r2r1−1/r1dx.
Now, usingϕ3′ and (85) in (78), we have(86)Υu,v′t≥l1t∫Ω×ΩA1|thu|dμ+l2t∫Ω×ΩA2|thv|dμ−c1tr1+tr2max∫Ω|u|r1dx,∫Ω|v|r2dx−c2tr2r1−1r1+tr1r2−1r2max∫Ω|v‖u|r1r2−1r2dx,∫Ω|u‖v|r2r1−1r1dx.
In fact from Lemma 2.3 in [10] and (23), we infer that(87)Υu,v′t≥l1tn1−1λ1∫ΩA1udx+l2tn2−1λ2∫ΩA2vdx−c1tr1+tr2max∫Ωur1dx,∫Ωvr2dx−c2tr2r1−1/r1+tr1r2−1/r2max∫Ωvur1r2−1/r2dx,∫Ωuvr2r1−1/r1dx.
Then,(88)Υu,v′ttn1−1+tn2−1≥minl1λ1∫ΩA1udx,l2λ2∫ΩA2vdx−c1tr1+tr2tn1−1+tn2−1×max∫Ωur1dx,∫Ωvr2dx−c2tr2r1−1/r1+tr1r2−1/r2tn1−1+tn2−1×max∫Ωvur1r2−1/r2dx,∫Ωuvr2r1−1/r1dx.
Sinceni=1,2<minr1,r2,1+r1r2−1/r2,1+r2r1−1/r1, we conclude the result for item (iii). For the last item, we have(89)Υu,v′ttn1−1+tn2−1≤maxn1∫Ω×ΩA1hudμ,n2∫Ω×ΩA2hvdμ−∫ΩFux,tu,tvutn1−1+tn2−1dx−∫ΩFvx,tu,tvvtn1−1+tn2−1dx≤maxn1∫Ω×ΩA1hudμ,n2∫Ω×ΩA2|hv|dμ−∫ΩFux,tu,tvutr1−1+tr2−1dx−∫ΩFvx,tu,tvvtr1−1+tr2−1dx.
Using (35), we have(90)Fx,u,vur1+vr2=∫0uFpx,p,vur1+vr2dp+∫0vFzx,p,vur1+vr2dz≤∫0uFpx,p,vpr1+vr2dp+∫0vFzx,p,vur1+zr2dz≤∫0uFpx,p,vur1−1+vr2×pr1−1+vr2−1pr1+vr2dp+∫0vFzx,p,vur1−1+vr2×ur1−1+zr2−1ur1+zr2dz.
We can easily see that(91)ur1−1+zr2−1ur1+zr2≤1andpr1−1+vr2−1pr1+vr2≤1.
As a consequence ofF3, we infer that(92)Fzx,0,zzr2−1≤Fvx,u,vur1−1+vr2−1andFpx,p,vpr1−1+vr2−1≤Fux,u,vur1−1+vr2−1.
It follows using (91) and (92) and inequalities (90) that(93)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dp+∫0vFvx,u,vur1−1+vr2−1dz=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
UsingF2, we conclude that(94)Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1⟶+∞whenu,v⟶+∞.
At this point, last limit just above and (89), we infer the item (iv).Lemma 12.
We assume thatϕ1-ϕ3 and F1-F3 hold true; then, for each u,v∈W/0,0, there is an only t∗=tu,v>0 such that t∗u,t∗v∈N. Moreover, ℐu,v>0 for each u,v∈N.Proof.
On the one hand, letu,v∈W/0,0. By Lemma 11, we have Υu,v′t>0 for t small enough and Υu,v′t<0 for t large enough. On the other hand, the map t↦Υu,v′t is continuous, and there is at least one number t∈0,∞ such that Υu,v′t=0, which means that tu,tv∈N. Now, let us see that there is an only t∗=tu,v>0 such that Υu,v′t=0. Using (78), we have(95)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dp+∫0vFvx,u,vur1−1+vr2−1dz=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
Indeed,(96)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dq+∫0vFvx,u,vur1−1+vr2−1dx=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
We remark thatϕ3 implies that(97)li−2≤tai′tait≤ni−2,where ni<minr1,r2; then, we have(98)li−2≤tai′tait<minr1−2,r2−2.
Using inequality (98) in (97), we infer that(99)ddtta1thuhu2+ta2thvhv2tr1−1+tr2−1≤0.
Hence,(100)ddtΥu,v′ttr1−1+tr2−1≤−∫ΩddtFux,tu,tvu+Fvx,tu,tvvtr1−1+tr2−1dx.
It can be seen that(101)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dv+∫0vFvx,u,vur1−1+vr2−1dy=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
According toF3 and (101), we get(102)ddtΥu,v′ttr1−1+tr2−1<0.
Therefore by (102), we have that t↦Υu,v′t/tr1−1+tr2−1 is a decreasing function that vanishes once in 0,∞. So, there exists a unique t=tu,v>0 such that Υu,v′t/tr1−1+tr2−1=0. Therefore, the function Υu,v has one critical point, namely t∗=tu,v>0 and t∗u,t∗v∈N. Furthermore, by Lemma 11, it follows that t∗ is a maximal point of Υu,v on 0,∞ and, in fact Υu,vt∗>0, implying that ℐt∗u,t∗v>0. The arguments above also show that Υu,v′′t<0, for every u,v∈W/0,0. And finally, because u,v∈N if and only if t∗=1, we conclude that ℐu,v>0 for every u,v∈N. This completes the proof.Proposition 1.
We assume thatϕ1-ϕ3andF1-F3 hold true. Then N is a C1-submanifold of W, i.e., any critical point of ℐ/N is a critical point of ℐ.Proof.
We consider the functionalJt:W⟶ℝ defined by(103)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dx+∫0vFvx,u,vur1−1+vr2−1dp=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1,where(104)ℐtu,v=∫Ω×Ωta1thuhu2dμ+∫Ω×Ωta2thvhv2dμ.
According to Lemma10, we can see that ℐt∈C1, and by using equation (59), we show that(105)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dz+∫0vFvx,u,vur1−1+vr2−1dq=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
We also show that(106)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dp+∫0vFvx,u,vur1−1+vr2−1dz=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
Now, we set(107)Qu,v=ℐ′u,v,u,vu,v∈W.
By using Lemma10, it follows that Q∈C1. Furthermore, from Lemma 11, we infer that t=1 is the global maximum of Υu,v, and also in the proof of Lemma 12, we see that(108)Q′u,v,u,v=Υ′u,v′1<0,u,v∈N.
Using the fact thatN=Q−10 and 0 is a regular value for Q, the set N is a C1-submanifold of W. Now, we assume that u,v∈N is a critical point of ℐ/N. According to the theorem of Lagrange multiplier, there exists a real constant λ such that(109)ℐ′u,v=λQ′u,v,Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dp+∫0vFvx,u,vur1−1+vr2−1dz=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
Using (108), we infer that λ=0. Therefore, ℐu,v≡0 so that u,v is a free critical point of ℐ. This completes the proof.Lemma 13.
We assume thatϕ1-ϕ3, F1-F3 hold. Let uk,vk∈N be a minimizing sequence of ℐ over the Nehari manifold N. Then, uk,vk is bounded in W.Proof.
Letuk,vk∈N be a minimizing sequence that is uk,vk∈N and ℐuk,vk⟶CN. To prove the boundedness of uk,vk, we argue by contradiction and consider that there exists a subsequence of uk,vk, always denoted by uk,vk such that uk,vk⟶+∞. We approach the problem in two cases.Case 1.
We suppose thatuks,A1⟶+∞ and also vks,A2⟶+∞. Let u¯k=uk/uks,A1 and v¯k=vk/vks,A2. Then, u¯k,v¯k is bounded in separable and reflexive Banach space W. By Theorem 1 and Remark 2, there exists a point u¯,v¯∈W such that(i)
u¯k⇀u¯ in W0s,A1Ω; and u¯k⟶u¯ in a.e in Ω(ii)
v¯k⇀v¯ in W0s,A2Ω and v¯n⟶v¯ in a.e in Ω
We claim thatu¯,v¯≠0,0 has nonzero Lebesgue measure. Firstly, we assume by the way of contradiction that both u¯≠0 and v¯≠0 have zero Lebesgue measure, that is u¯=0 in W0s,A1Ω and v¯=0 in W0s,A2Ω. Since uk,vk∈N, then(110)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dq+∫0vFvx,u,vur1−1+vr2−1dx=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
LetG>0 be a constant. Now, we observe that(111)CN+ok1=ℐuk,vk≥ℐGu¯k,Gv¯k=G1Gu¯k+G2Gv¯k−ℱu¯k,Gv¯k,
Now usinga, b, and the continuous of the function F, we infer that(112)ℱGu¯k,Gv¯k⟶0x∈Ω.
Combining (111), (112), and using Lemma 6, we get(113)Fx,u,vur1+vr2≤∫0vFvx,u,vur1−1+vr2−1dz=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
Passing to the limit in (113), we get(114)CN≥2ξ0G>0,G>0,which is impossible. Therefore, u¯≠0 and v¯≠0 has nonzero Lebesgue measure. It is remembered that we are assuming that uk,vk⟶+∞ and ℐuk,vk⟶CN. Hence, ℐuk,vk/uks,A1n1+vks,A2n2=ok1.
On the one hand, employing Lemma1, we have(115)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dp=Fux,u,vuur1−1+vr2−1.
Passing to the limit above, we have(116)limk⟶∞sup∫ΩFx,uk,vkuks,A1n1+vks,A2n2dx≤1.
On the other hand, applying Fatou’s lemma and the fact thatu¯≠0 and v¯≠0 has nonzero Lebesgue measure, we have(117)Fx,u,vur1+vr2∫0uFux,u,vzr1−1+vr2−1dp≤∫0vFvx,u,vur1−1+vr2−1dz=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1.
It is clear thatu¯kn1v¯kn2/maxu¯kn1,v¯kn2⟶minu¯n1,v¯n2≠0 when k⟶+∞. Then, by using F2 and last inequalities just above, we conclude(118)Fx,u,vur1+vr2≤∫0uFux,u,vzr1−1+vr2−1dp+∫0vFvx,u,vur1−1+vr2−1dz=Fux,u,vuur1−1+vr2−1+Fvx,u,vvur1−1+vr2−1,which is impossible.Case 1.
We suppose thatuks,A1≤C or vks,A2≤C for some C>0 and all k∈ℕ. Without loss of generality, we assume that uks,A1⟶+∞ and vks,A2≤C, for some C>0 and for all k∈ℕ. Let u¯k=uk/uks,A1 and v¯k=vk/uks,A1, then v¯ks,A2⟶0 and u¯ks,A1⟶1. By Theorem 1 and Remark 2, there exists a point u¯,v¯∈W such that(i)
u¯k⇀u¯ in W0s,A1Ω and u¯k⟶u¯ in a.e in Ω(ii)
v¯k⇀v¯ in W0s,A2Ω; and v¯k⟶v¯ in a.e in Ω
By similar argument in Case 1, we prove thatu¯≠0 has nonzero Lebesgue measure. It is recalled that we are assuming that uk,vk⟶+∞ and ℐuk,vk⟶CN. Hence,(119)ℐuk,vkuks,A1n1+Cn2=ok1.
Applying Lemma1 and the last equation just above, we get(120)∫ΩFx,uk,vkuks,A1n1+Cn2dx=1uks,A1n1+Cn2∫Ω×ΩA1hukdμ+1uks,A1n1+Cn2∫Ω×ΩA2hukdμ+ok1≤uks,A1n1uks,A1n1+Cn2+vks,A2n2uks,A1n1+Cn2+ok1≤uks,A1n1uks,A1n1+Cn2+Cn2uks,A1n1+Cn2+ok1=1+ok1.
Passing to the limit to the inequalities just above, we have(121)limk⟶∞sup∫ΩFx,uk,vkuks,A1n1+Cn2dx≤1.
In other way, using Fatou’s lemma and the fact thatu¯≠0, we have(122)limk⟶∞inf∫ΩFx,uk,vkuks,A1n1+Cn2dx≥∫Ωlimk⟶∞infFx,uk,vkuks,A1n1+Cn2dx=∫Ωlimk⟶∞infFx,uk,vkukn1/u¯kn1+Cn2vkn2/v¯kn2uks,A1n2dx≥∫Ωlimk⟶∞infFx,uk,vkukn1+|vk|n2×uks,A1n2ukn1|vk|n2maxuks,A1n2ukn2,Cn2u¯kn1dx.
It is clear to see that(123)uks,A1n2u¯kn1vkn2maxuks,A1n2vkn2,Cn2u¯kn1≠0.
According toF2, (123), and last inequalities just above, we conclude(124)limk⟶∞inf∫ΩFx,uk,vkuks,A1n1+Cn2dx=+∞,which is impossible. Thus, uk,vk is bounded in W. The proof is complete.Lemma 14.
We assume thatϕ1-ϕ3, F1-F3 hold. Then, there exists u,v∈N such that(125)CN=ℐu,v>0.Proof.
Letuk,vk∈N be a minimizing sequence of ℐ over the Nehari manifold N. By Lemma 13, there exists u,vW such that(i)
uk⇀u in W0s,A1Ω and vk⇀v in W0s,A2Ω(ii)
We claim thatu,v≠0,0. We assume on the contrary that u,v=0,0. The fact that uk,vk∈N and by using ϕ3, we have(126)0≤∫Ω×ΩA1hukdμ+∫Ω×ΩA1hvkdμ≤∫Ω×Ωa1hukhuk2dμ+∫Ω×Ωa2hvkhvk2dμ≤∫ΩFux,uk,vkuk+Fvx,uk,vkvkdx.
Applying Lemma8, we get(127)∫ΩFux,uk,vkuk+Fvx,uk,vkvkdx=ok1.
As a consequence of Lemma6, uk,vk⟶0 which is contradicting Lemma 9. That proves the claim.
Accordingly, from Lemma10, we have that u,v∈W↦ℐ′u,v,u,v is weakly lower continuous. Hence,(128)ℐ′u,v,u,v≤liminfk⟶+∞ℐ′uk,vk,uk,vk=0.
We recall thatΥu,v′1=ℐ′u,v,u,v≤0. By Lemma 12, there is t∈0,1 such that Υu,v′tu,tv=0. Hence, tu,tv∈N.(i)
We claim thatt=1 so that u,v is in N.
Furthermore, by contrary, we assume thatt∈0,1. Then,(129)CN≤ℐtu,tv=ℐtu,tv−1Kℐ′tu,tv,tu,tv=∫Ω×ΩA1thudμ+∫Ω×ΩA2thvdμ−∫ΩFx,tu,tvdx−1K∫Ω×Ωa1thuthu2dμ−1K∫Ω×Ωa2thvthv2dμ+1K∫ΩFux,tu,tvtu+Fvx,tu,tvtvdx=∫Ω×ΩA1thu−1Ka1thuthu2dμ+∫Ω×ΩA2thv−1Ka2thvthv2dμ+∫Ω1KFux,tu,tvtu−F¯ux,tu,tvdx+∫Ω1KFvx,tu,tvtv−F¯vx,0,tvdx,where K=mini=1,2ri, F¯ux,tu,tv=∫0tuFpx,p,tvdp and F¯vx,0,tv=∫0tvFzx,0,zdz Now, we can see by using F3 that(130)ddp1KFpx,p,zp−F¯px,p,z=tKKddpFpx,p,zpK−1≥tKKddpFpx,p,zpK−1+zK−1>0,p>0,x∈Ω.
Again by usingF3, we have(131)Fzx,p,zpK−1+zK−1≥Fzx,0,zzK−1forallp,z>0,which implies(132)Fzx,0,z≤Fzx,p,zzK−1pK−1+zK−1≤Fzx,p,z.
Then, we infer that(133)ddz1KFzx,p,zz−F¯zx,0,z≥ddz1KFzx,p,zz−F¯zx,p,z=tKKddzFzx,p,zzK−1>0,z>0,x∈Ω,and then the functions w↦1/KFwx,w,zw−F¯wx,w,z and z↦1/KFzx,w,zz−F¯zx,0,z are increasing in Ω×0,∞×0,∞ and Ω×0,∞×0,∞, respectively.
Moreover, the functions(134)t↦Ait−1Kaitt2areincreasingin0,∞see22.Then, we conclude that(135)CN<∫Ω×ΩA1hu−1Ka1huhu2dμ+∫Ω×ΩA2hv−1Ka2hvhv2dμ+∫Ω1KFux,u,vu−F¯ux,u,vdx+∫Ω1KFvx,u,vv−F¯vx,0,vdx.
Now, using the weak lower continuity of the functions(136)u↦∫Ω×ΩAihu−1Kaihuhu2dμ,and the continuity of the function F, we infer that CN<liminfk⟶+∞∫Ω×ΩA1huk−1/Ka1hukhuk2dμ+∫Ω×ΩA2hvk−1/Ka2hvkhvk2dμ+∫Ω1/KFux,uk,vkuk−F¯ux,uk,vkdx+∫Ω1/KFvx,uk,vkvk−F¯vx,0,vkdx]=limk⟶∞ℐuk,vk−1/Kℐ′uk,vk,uk,vk=CN.which is impossible. Thus, t=1, and then u,v∈N. This finishes the proof. Now, we have all tools to prove our result.


### 3.1. Proof of Theorem2

Letuk,vk∈N be a minimizing sequence for ℐ over N. By the proof of Lemma 13, there is u,v∈N⊂W such that(137)uk⇀uinW0s,A1Ω,vk⇀vinW0s,A2Ω.Applying Lemma7, we get(138)uk⟶uinW0s,A1Ω,vk⟶vinW0s,A2Ω.Since Lemma5 infer that ℐ∈C1, it follows that ℐ′uk,vk⟶ℐ′u,v. By Lemma 14, u,v∈N and(139)CN=ℐu,v=minNI>0.By Proposition1 the set N is a C1 submanifold of W so that u,v is a critical point of ℐ/N. Again Proposition 1 shows that u,v is a critical point of ℐ.


## 3.1. Proof of Theorem2

Letuk,vk∈N be a minimizing sequence for ℐ over N. By the proof of Lemma 13, there is u,v∈N⊂W such that(137)uk⇀uinW0s,A1Ω,vk⇀vinW0s,A2Ω.Applying Lemma7, we get(138)uk⟶uinW0s,A1Ω,vk⟶vinW0s,A2Ω.Since Lemma5 infer that ℐ∈C1, it follows that ℐ′uk,vk⟶ℐ′u,v. By Lemma 14, u,v∈N and(139)CN=ℐu,v=minNI>0.By Proposition1 the set N is a C1 submanifold of W so that u,v is a critical point of ℐ/N. Again Proposition 1 shows that u,v is a critical point of ℐ.



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*Source: 3849217-2022-05-20.xml*