# Two Types of Solutions to a Class of(p,q)-Laplacian Systems with Critical Sobolev Exponents in RN

Authors: Jing Li; Caisheng Chen
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

We focus on the following elliptic system with critical Sobolev exponents:-div⁡∇up-2∇u+m(x)up-2u=λup⁎-2u+(1/η)Gu(u,v),x∈RN; -div⁡∇vq-2∇v+n(x)vq-2v=μvq⁎-2v+(1/η)Gv(u,v),x∈RN; u(x)>0,v(x)>0,x∈RN, where μ,λ>0,1<p≤q<N, either η∈(1,p) or η∈(q,p⁎), and critical Sobolev exponents p⁎=pN/(N-p) and q⁎=qN/(N-q). Conditions on potential functions m(x),n(x) lead to no compact embedding. Relying on concentration-compactness principle, mountain pass lemma, and genus theory, the existence of solutions to the elliptic system with η∈(q,p⁎) or η∈(1,p) will be established.


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## Body

## 1. Introduction and Main Result

We aim at the existence of nontrivial weak solutions for the following elliptic system:(1)-div⁡∇up-2∇u+mxup-2u=λup∗-2u+1ηGuu,v,x∈RN,-div⁡∇vq-2∇v+nxvq-2v=μvq∗-2v+1ηGvu,v,x∈RN,ux>0,vx>0,x∈RN, where μ,λ>0,1<p≤q<p∗<q∗, and either 1<η<p or q<η<p∗.Discussion regarding critical growth has been dominated research in various branches of mathematical physics. Here, we quote a few literatures on critical Sobolev exponent and their applications; see Brezis and Nirenberg [1], Wu [2–4], and Garcia-Azorero et al. [5] and the references therein.By applying variational arguments, Yin and Yang in [6] proved the existence of at least one or two positive solutions to the following system:(2)-Δpu+up-2u=f1λ1xuq∗-2u+2αα+βgμxuα-2uvβ,x∈Ω,-Δpv+vp-2v=f2λ2xvq∗-2v+2βα+βgμxuαvβ-2v,x∈Ω,u=v=0,on∂Ω, where Fiλi(x)=λifi+(x)+fi-(x)(i=1,2) are sign-changing functions.Subsequently, at least one nontrivial solution was established in the work [7]:(3)-Δpu=fxup∗-2u+λ∂F∂ux,u,v,inRN,-Δqv=gxvq∗-2v+λ∂F∂vx,u,v,inRN,ux,vx⟶0,asx⟶∞ via mountain pass lemma. More results can be found in [7–10].In papers cited above, the authors are fond of supposing that potential functionsm(x),n(x) satisfy such conditions:(F1)
Functionsm(x),n(x)∈C(RN) and satisfy m(x),n(x)≥c0>0 with some constant c0. Moreover, for every κ>0, meas⁡({x∈RN:m(x),n(x)≤κ})<∞, where “meas” denotes the Lebesgue measure in RN.(F2)
m(x)→+∞, n(x)→+∞, as n→∞.We point out that one of the above conditions has been used in many papers to guarantee the compact embeddingW1,p(RN)↪Lν(RN) for each ν∈[p,p∗). Nevertheless, throughout the paper, we are going to weaken the conditions and give assumptions as follows:(H1)
Let nonnegative functionG(u,v)∈C1(R2) be homogeneous with G(tu,tv)=tηG(u,v), where ∀(u,v)∈R2 and either η∈(1,p) or η∈(q,p∗). Furthermore, there exists β(x)>0 such that(4)0≤Gu,v,Guu,vu,Gvu,vv≤βxuη+vη,∀u,v∈R2,and for all(u,v),(ξ,ζ)∈R2(5)Gu,v-Gξ,ζ≤βxuη-1+vη-1+ξη-1+ζη-1u-ξ+v-ζ.(H2)
Functionsm(x),n(x)∈C(RN) and satisfy m(x),n(x)≥c0 with some constant c0. In addition, for every κ>0, meas⁡({x∈RN:m(x)/β(x),n(x)/β(x)≤κ})<∞, where “meas” denotes the Lebesgue measure in RN.(H3)
Functionβ(x) is positive and β(x)∈Llocσ(RN) with σ=p∗/(p∗-η), η∈(p,q∗). Moreover, limsupx→∞⁡β(x)<∞.(H4)
Functionβ(x) is positive and β(x)∈Lγ(RN) with γ=q∗/(q∗-η), η∈(1,p).Condition(H2) used in our paper is weaker than (F1) which leads to no compact embedding of W1,p(RN)↪Lν(RN). As far as the authors are aware, there is less literature dealing with such potential functions. Moreover, a more generalized function G(u,v) appeared in the paper.In order to state our main result, let us recall some Sobolev spaces and norms. LetX, Y be subspaces of W1,p(RN), W1,q(RN) with the norms(6)X=u∈W1,pRN:∫RNmxupdx<∞withuX=∫RN∇up+mxupdx1/p,Y=v∈W1,qRN:∫RNnxvqdx<∞withvY=∫RN∇vq+nxvqdx1/q,ut=∫RNutdx1/t,ut,m=∫RNmxuxtdx1/t,ut,n=∫RNnxuxtdx1/t,1<t<N.Obviously, the norm of reflexive Banach space E=X×Y is given by(7)u,vE=uX+vY,∀u,v∈E.Define(8)S1=infu∈X\0⁡uXp∫RNup∗p/p∗,S2=infv∈Y\0⁡vYq∫RNvq∗q/q∗,c∗=min⁡λNS1λN/P,μNS2μN/q.Corresponding functional of (1) defined in E is(9)Ju,v=1puXp+1qvYq-λp∗∫RNup∗dx-μq∗∫RNvq∗dx-1η∫RNGu,vdx.The main results are enunciated as follows.Theorem 1.
Suppose(H1), (H2), and (H3) with η∈(q,p∗) hold; then there exists Λ0>0 such that if 0<λ,μ<Λ0, system (1) has at least a nontrivial solution.Theorem 2.
Suppose(H1), (H2), and (H4) with η∈(1,p) hold; then there exists Λ1>0 such that if 0<λ,μ<Λ1, problem (1) has infinitely many solutions (un,vn)∈E with J(un,vn)<0, J(un,vn)→0, and (un,vn)→0 as n→∞.This work is structured as follows. Variational framework will be set up and some useful lemmas will be derived in Section2. The proofs of Theorems 1 and 2 are established in Sections 3 and 4, respectively.


## 2. Preliminaries

The main purpose of this section is to present that functionalJ satisfies PSc condition. Besides, some useful technical lemmas will be listed.Lemma 3.
Assume that(H1)–(H4) hold; if {(un,vn)}⊂E is a PSc sequence, then sequence {(un,vn)} is bounded in E.Proof.
LetJ(un,vn)→c, J′(un,vn)→0 in E∗ as n→∞.
Case  1. When η∈(q,p∗), we can choose q<θ<η such that(10)c+1+un,vnE≥Jun,vn-1θJ′un,vnun,vn=1p-1θunXp+1q-1θvnYq+1θ-1p∗∫RNλunp∗dx+1θ-1q∗∫RNμvnq∗dx+1θ-1η∫RNGun,vndx≥1p-1θunXp+1q-1θvnYq≥1q-1θmin⁡un,vnEp,un,vnEq.
Case  2. When η∈(1,p), we can choose q<θ<p∗ such that(11)c+1+un,vnE≥Jun,vn-1θJ′un,vnun,vn=1p-1θunXp+1q-1θvnYq+1θ-1p∗∫RNλunp∗dx+1θ-1q∗∫RNμvnq∗dx-1η-1θ∫RNGun,vndx≥1p-1θunXp+1q-1θvnYq-1η-1θ∫RNβxunη+vnηdx≥1q-1θmin⁡un,vnEp,un,vnEq-1η-1θβxϑunp∗η-1η-1θβxγvnq∗η≥1q-1θmin⁡un,vnEp,un,vnEq-1η-1θS1-η/pβxϑunXη-1η-1θS2-η/qβxγvnYη≥1q-1θmin⁡un,vnEp,un,vnEq-max⁡1η-1θS1-η/pβxϑ,1η-1θS2-η/qβxγun,vnEη,where ϑ=p∗/(p∗-η)≥γ=q∗/(q∗-η); the above inequalities imply the sequence {(un,vn)} is bounded in E and the proof is finished.As the consequence of Lemma3, there exist a constant M>0, (u,v)∈E, and sequence {(un,vn)}∈E such that(12)un,vnE≤M,∀n∈N,u,vE≤Mun,vn⇀u,vweaklyinE,unx,vnx⟶ux,vxa.e.inRNun,vn⟶u,vstronglyinLlocsRN×LlocrRN,∀s∈1,p∗,r∈1,q∗.Lemma 4.
Let(H1), (H2), and (H3) be satisfied with p≤η<p∗. Suppose sequence {un} is bounded in X; then there exists u∈X∩Lη(RN,β) such that un→u strongly in Lη(RN,β).Proof.
Since{un} is bounded in X, there exist u∈X and M1>0 such that(13)un+u≤M1;un⟶ustronglyinLlocsRN,∀s∈1,p∗.Thus, we have that un→u strongly in Lp(Bi,β), where Bi={x∈RN:x<i} and Bic={x∈RN:x≥i}.
Step  1. We want to show that, for ∀ϵ>0, there exists R>1 such that(14)∫BRcβxunpdx<ϵ. We can fix ϵ>0 and choose t∈(1,N/(N-p)) and a0, κ, where(15)a0=supX\0⁡unptpunXp,κ>2M1pϵ≥2ϵsupn≥1⁡unXp.
DenoteAi={x∈Bi:m(x)/β(x)≤κ}; then Ai⊂Ai+1 for i=1,2,…, and A∞={x∈RN:m(x)/β(x)≤κ}=⋃i=1∞Ai. What is more, there exist i0≥1 and β0>0 such that βx≤β0 for x≥i0 by condition (H3) and meas⁡(A∞) is bounded by condition (H2); then, for every ϵ>0, there exists i1≥i0 such that for i≥i1(16)0≤meas⁡A∞-meas⁡Ai=meas⁡Ωic≤ϵ2a0β0M1pt/t-1,where(17)Ωi=x∈Bic:mxβx>κ,Ωic=x∈Bic:mxβx≤κ.By (15) and (H2), we observe that(18)∫Ωiβxunpdx≤1κ∫Ωimxunpdx≤1κ∫RNmxunpdx≤1κunXp≤ϵ2,∫Ωicβxunpdx≤∫Ωicunptdx1/t∫Ωicβxt/t-1dxt-1/t≤a0unXpβxLt/t-1Ωic≤a0β0unXpmeasΩict-1/t≤ϵ2.Therefore, we get(19)∫Bicβxunpdx=∫Ωiβxunpdx+∫Ωicβxunpdx≤ϵ,and by Fatou’s lemma(20)∫Bicβxupdx≤liminfn→∞⁡∫Bicβxunpdx≤ϵ.Thus, from (13) and (20), we can get that un→u strongly in Lp(RN,β).
Step  2. For p<η<p∗, let η=rp+(1-r)p∗, r∈(0,1). By (19), we have that(21)∫Bicβxunηdx≤∫Bicβxrunrpβx1-run1-rp∗dx≤β01-runLpBic,βrpunLp∗Bic1-rp∗≤β01-rS1-1-rp∗/punX1-rp∗unLpBic,βrp≤β01-rS1-1-rp∗/pM11-rp∗ϵ=Cϵ.
By Fatou’s lemma, we can see that(22)∫Bicβxuηdx≤lim⁡infn→∞⁡∫Bicβxunηdx≤Cϵ.
For each measure subsetΩ⊂Bi, we have that(23)∫Ωβxun-uηdx≤βLσΩun-uLp∗Ωη≤βLσΩun-uLp∗RNη≤S1-η/p2M1ηβLσΩ.Thus, one can get that {β(x)un-uη} is uniformly integral and ∫Biβ(x)un-uηdx is bounded. Then, we have that(24)limn→∞⁡∫Biβxun-uηdx=0.Therefore, from (22) and (24), we can obtain(25)limn→∞⁡∫RNβxun-uηdx=0.Similar argument can be applied to get that vn→v strongly in Lη(RN,β). Thus, one can easily get that limn→∞⁡∫RNG(un,vn)dx=∫RNG(u,v)dx.Lemma 5.
Assume that(H1) and (H4) hold with η∈(1,p); if {(un,vn)} satisfies (12), then(26)limn→∞⁡∫RNGun,vndx=∫RNGu,vdx.Proof.
By(H1) and (H4), we get that(27)∫BRβxunηdx⟶∫BRβxuηdx,n⟶∞,where BR={x∈RN:|x|<R} and BRc=RN\BR¯ for any R>0. Moreover, we obtain from Hölder inequality that(28)∫BRcβxunηdx≤∫BRcβxp∗/p∗-ηp∗-η/p∗∫BRcunp∗η/p∗≤S1-η/pβxLϑBRcunEη≤S1-η/pMηβxLϑBRc⟶0,asR⟶∞,where ϑ=p∗/(p∗-η)≥γ=q∗/(q∗-η). By Fatou’s lemma, we have(29)∫BRcβxuηdx≤lim⁡infn→∞⁡∫BRcβxunηdx≤MηβxLγBRc⟶0,asR⟶∞.Then, from (28) and (29), we can deduce ∫RNβ(x)|un|ηdx→∫RNβ(x)|u|ηdx. Similarly, we can get ∫RNβ(x)|vn|ηdx→∫RNβ(x)|v|ηdx. Thus, the proof is completed.Denote(30)Pn=J′un,vnun-u,vn-v=∫RN∇unp-2∇un∇un-u+mxunp-2unun-udx+∫RN∇vnq-2∇vn∇vn-v+nxvnq-2vnvn-udx-λ∫RNunp∗-2unun-udx-μ∫RNvnq∗-2vnvn-vdx-1η∫RNGuun,vnun-u+Gvun,vnvn-vdx. Then the fact J′(un,vn)→0 in E∗ implies Pn→0 as n→∞. Similarly, the fact that (un,vn)⇀(u,v) in E implies that Qn→0, where(31)Qn=∫RN∇up-2∇u∇un-u+mxup-2uun-udx+∫RN∇vq-2∇v∇vn-v+nxvq-2vvn-udx.Note that(32)Pn-Qn=Tn-λ∫RNunp∗-2unun-udx-μ∫RNvnq∗-2vnvn-vdx-1η∫RNGuun,vnun-u+Gvun,vnvn-vdx,where(33)Tn=∫RN∇unp-2∇un-∇up-2∇u∇un-udx+∫RNmxunp-2un-up-2uun-udx+∫RN∇vnq-2∇vn-∇vq-2∇v∇vn-vdx+∫RNnxvnq-2vn-vq-2vvn-vdx.If Tn→0, this implies un-uX→0 and vn-vY→0 according to the following lemma.Lemma 6 (see [11]).
There exists constantC>0 such that, for all x,y∈RN,(34)xp-2x-yp-2y,x-y≥Cx+yp-2x-y2,if1<p<2,xp-2x-yp-2y,x-y≥Cx-yp,ifp≥2,where (·,·) denotes the usual scalar product in RN.In the following, we will claim(35)∫RNλunp∗-2unun-udx⟶0,∫RNμvnq∗-2vnvn-vdx⟶0,asn⟶∞.Based on concentration-compactness principle [12] and (12), it follows that(36)∇unp⇀μ≥∇up+∑j∈Jμjδxj,unp∗⇀ν=unp∗+∑j∈Jνjδxj,μj≥S1νjp/p∗,where δxj is the Dirac measure at xj and μj, νj are nonnegative bounded measures. Analogously, we get ∇vnq⇀μ¯ and vnq∗⇀ν¯. Concentration at ∞ of {un} is defined by(37)μ∞=limR→∞⁡limsupn→∞⁡∫x>R∇unpdx,ν∞=limR→∞⁡limsupn→∞⁡∫x>Runp∗dx.Analogously, the concentration at ∞ of {vn} is given by μ¯∞ and ν¯∞.Step 1.
Forj∈J, we define ϕj∈C0∞(RN) as follows:(38)ϕjx=1,ifx∈Bxj,ϵ,0,ifx∈RN\Bxj,2ϵ,∇ϕj≤C,theother.One can see that {unϕj} is bounded in X; we obtain limn→∞⁡〈J′(un,vn),(unϕj,0)〉=0. Namely,(39)-limn→∞⁡∫RN∇unp-2un∇un∇ϕjdx=limn→∞⁡∫RN∇unpϕj+mxunpϕjdx-λ∫RNunp∗ϕjdx-1η∫RN∂G∂uun,vnunϕjdx.The fact that un⇀u produces ∫RN(∂G/∂u)(un,vn)unϕjdx→∫RN(∂G/∂u)(u,v)uϕjdx. In the light of Lions’ principle, we acquire(40)∫RN∇unpϕjdx⟶∫RNϕjdμ,∫RNunp∗ϕjdx⟶∫RNϕjdν.Moreover(41)limϵ→0⁡limn→∞⁡∫RN∇unp-2un∇un∇ϕjdx≤limϵ→0⁡limn→∞⁡∫RN∇unpdxp-1/p∫RNun∇ϕjpdx1/p≤Mp-1limϵ→0⁡limn→∞⁡∫RNun∇ϕjpdx1/p≤Mp-1limϵ→0⁡limn→∞⁡∫B2ϵxjunp∗dx1/p∗∫B2ϵxj∇ϕjNdx1/N≤Climϵ→0⁡∫B2ϵxjup∗dx1/p∗=0.At the same time, we have(42)limϵ→0⁡limn→∞⁡∫RN∂G∂uun,vnunϕjdx≤limϵ→0⁡limn→∞⁡∫B2ϵxjβxunη+vnηϕjdx⟶0,limϵ→0⁡limn→∞⁡∫RNunpϕjdx⟶0.From the aforementioned equations, it follows that(43)0=limϵ→0⁡limn→∞⁡I′un,vn,unϕj,0=μj-λνj;combining with (36), we obtain(44)νj=0orνj≥S1λN/P.Step 2.
By choosing a suitable cut-off functionφ(x)∈C∞(RN,[0,1]) such that(45)φx=0,ifx≤1,1,ifx≥2,we set φR(x)=φ(x/R); then {φRun} is bounded in X and limR→∞⁡〈J′(un,vn),(unφR,0)〉=0.
In a similar manner, we gain(46)-limn→∞⁡∫RN∇unp-2un∇un∇φRdx=limn→∞⁡∫RN∇unpφR+mxunpφRdx-λ∫RNunp∗φRdx-1η∫RN∂G∂uun,vnunφRdx.Note that(47)limR→∞⁡limsupn→∞⁡∫RN∇unp-2un∇un∇φRdx≤limR→∞⁡limsupn→∞⁡∫RN∇unpdxp-1/p∫RNun∇φRpdx1/p≤Mp-1limR→∞⁡limsupn→∞⁡∫RNun∇φRpdx1/p≤Mp-1limR→∞⁡∫R≤x≤2Rup∗dx1/p∗∫R≤x≤2R∇φRNdx1/N≤C′limR→∞⁡∫R≤x≤2Rup∗dx1/p∗=0,which deduces(48)v∞=0orv∞≥S1λN/P.Next we will prove νj≥(Sp/λ)N/P and v∞≥(Sp/λ)N/P are impossible.Case 1.
If we fixη∈(q,p∗), there exists j0 such that νj0≥(Sp/λ)N/P; then one can get(49)Jun,vn-J′un,vnunp,vnq=1p-1p∗∫RNλunP∗dx+1q-1q∗∫RNμvnq∗dx+1η∫RN1pGuun+1qGvvn+Gun,vndx≥λN∫RNunP∗dx+μN∫RNvnq∗dx.Let n→∞; we gain(50)c≥λN∫RNuP∗dx+λN∑j∈Jνj+μN∫RNvq∗dx+μN∑j∈Jν¯j≥λN∑j∈Jνj+μN∑j∈Jν¯j≥λNS1λN/P+μNS2μN/q,which gives rise to a contradiction, so νj=0 for all j∈J. With the similar argument, we can get μ∞=ν∞=0.Case 2.
If we fixη∈(1,p), one can see(51)0>c=limn→∞⁡Jun,vn-J′un,vnunp∗,vnq∗=1p-1p∗unXp+1q-1q∗vnYq-1η-1q∗∫RNGun,vndx≥1Nmin⁡un,vnEp,un,vnEq-1η-1q∗∫RNβxunη+vnηdx≥1Nmin⁡un,vnEp,un,vnEq-1η-1q∗βxϑS1-η/punXη+βxγS2-η/qvnYη≥1Nmin⁡un,vnEp,un,vnEq-1η-1q∗max⁡βxϑS1-η/p,βxγS2-η/qun,vnEη;if min⁡{un,vnEp,un,vnEq}=(un,vn)Ep we can obtain(52)un,vnE≤N1η-1q∗max⁡βxϑS1-η/p,βxγS2-η/q1/p-η;otherwise,(53)un,vnE≤N1η-1q∗max⁡βxϑS1-η/p,βxγS2-η/q1/q-η;moreover,(54)0>c=limR→∞⁡limn→∞⁡Jun,vn-J′un,vnunp∗φR,vnq∗φR≥1p-1p∗μ∞+1q-1q∗μ¯∞-1η-1q∗max⁡βxϑS1-η/p,βxγS2-η/qun,vnEη=1Nμ∞+μ∞¯-1η-1q∗max⁡βxϑS1-η/p,βxγS2-η/qp/p-ηNη/p-η≥1NS1λN/p+S2μN/q-1η-1q∗max⁡βxϑS1-η/p,βxγS2-η/qp/p-ηNη/p-η.We can easily get that there exists Λ0>0 being small such that when 0<λ, μ<Λ0, the right side is positive, which gives rise to a contradiction.
Up to now, we have shown that(55)limn→∞⁡∫RNunp∗=∫RNup∗limn→∞⁡∫RNvnq∗=∫RNvq∗.
In view of (12) and the Brezis-Lieb Lemma [13], we acquire(56)limn→∞⁡∫RNun-up∗dx=0,limn→∞⁡∫RNvn-vp∗dx=0.
As a result of the above, it yieldsun-u,vn-vE→0, which implies the functional of (1) fulfils PSc condition.


## 3. Proof of Theorem1

The purpose of this section is to give proof of Theorem1 which mainly relies on mountain pass lemma as follows.Lemma 7 (see [14]).
LetX be a real Banach space, J∈C1(E,R), J(0)=0. If (P1)
there exist constantsr,b>0 such that J(u)≥b for all u∈X with uX=r,(P2)
there ise, eX>r, such that J(e)≤0,(P3)
the functionalJ(u) satisfies PS condition: every sequence {un}⊂X, such that J(un)→c for some c, and J′un→0 in X∗ as n→∞, is relatively compact in X,
thenJ possesses a critical value c≥b which can be characterized as (57)c=infg∈Γ⁡maxt∈0,1⁡Jgt,where (58)Γ=g∈C0,1,X∣g0=0,g1=e.In the following, we want to show that(59)c<c∗=min⁡λNS1λN/P,μNS2μN/q.We suppose c∗=(λ/N)S1/λN/P and we can get the other case with similar arguments. Let ω=u/up∗∈X\{0}; then we get the following lemma.Lemma 8.
Under hypotheses(H1), (H2), and (H3) with η∈(q,p∗), there exists Λ0>0 such that if 0<λ<Λ0, we have 0<supt≥0⁡J(tω,0)<c∗.Proof.
For(60)Jtω,0=tppωXp-tp∗p∗∫RNλωp∗dx-1η∫RNGtω,0dx=tppωXp-tp∗p∗∫RNλωp∗dx-tηη∫RNGω,0dxwith p<η<p∗, we can easily get J(tω,0)→-∞ as t→∞, J(t0ω,0)<0 with t0 large enough, and J(tω,0)≥r>0 with t sufficiently small. So 0<supt≥0J(tω,0)=J(tλω,0) for some tλ>0; namely,(61)Jtλω,0=tλppωXp-tλp∗p∗∫RNλωp∗dx-tληη∫RNGω,0dx.We can have J′(tλω,0)(tλω,0)=0; that is, (62)tλpωXp-tλp∗∫RNλωp∗dx-1η∫RNGutλω,0tλω,0dx=0.By direct computation tλpωXp≥tλp∗∫RNλωp∗dx, we can get tλ≤S1/λ(N-p)/p2.
Meanwhile(63)c≤maxt≥0⁡Jtω,0=maxt≥0⁡tλppωXp-tλp∗p∗∫RNλωp∗dx-tληη∫RNGω,0dx≤maxt≥0⁡tλppωXp-tλp∗p∗∫RNλωp∗dx.Denote F(t)=(tp/p)ωXp-(tp∗/p∗)∫RNλωp∗dx, t>0; it can achieve its maximum at (64)t=ωXp∫RNλωp∗dxN-p/p2=S1λN-p/p2;(65)supt>0⁡Ft=FS1λ=ωXppS1λN-p/p2p-∫RNλωp∗dxp∗S1λN-p/p2p∗=ωXppS1λN-p/p-∫RNλωp∗dxp∗S1λN/p=λNS1λN/P=c∗thus we get 0<supt≥0J(tω)<c∗.Proof of Theorem1.
Obviously, functionalJ(u,v) satisfies conditions (P1)–(P3) of the Lemma 7 we can define c=infg∈Γ⁡maxt∈[0,1]⁡J(g(t)), where Γ denotes the class of all continuous paths g(t) joining the origin to t0ω. Thus, mountain pass lemma can be applied to conclude that J(u,v) has a critical point with corresponding critical value c. So, system (1) has at least one nontrivial solution.


## 4. Proof of Theorem2

In this section, we want to apply genus theory to get the main conclusion of Theorem2. Let X be a Banach space and Σ be the class of closed and symmetric subsets with respect to the origin of X-{0}. For Z∈Σ, genus γ(Z) is given by(66)γZ=min⁡k∈N:∃φ∈CZ,Rk\0,φx=-φ-x. Since β(x)∈Lγ(RN) with γ=q∗/(q∗-η), η∈(1,p), thus we can have (67)∫RNGu,vdx≤∫RNβxuη+vηdx≤βxϑup∗η+βxγvq∗η≤βxϑS1-η/puXη+βxγS2-η/qvYη≤max⁡βxϑS1-η/p,βxγS2-η/qu,vEη,where ϑ=p∗/(p∗-η)≥γ.Moreover,(68)∫RNλp∗up∗+μq∗vq∗dx≤λp∗S1-p∗/puXp∗+μq∗S2-q∗/qvYq∗≤max⁡λp∗S1-p∗/p,μq∗S2-q∗/qmax⁡u,vEp∗,u,vEq∗. The corresponding functional J(u,v) of (1) turns into(69)Ju,v=1puXp+1qvYq-λp∗∫RNup∗dx-μq∗∫RNvq∗dx-1η∫RNGu,vdx≥1qmin⁡u,vEp,u,vEq-max⁡λp∗S1-p∗/p,μq∗S2-q∗/qmax⁡u,vEp∗,u,vEq∗-1ηmax⁡βxϑS1-η/p,βxγS2-η/qu,vEη=1qmin⁡u,vEp,u,vEq-b1max⁡u,vEp∗,u,vEq∗-b2u,vEη,where b1=max⁡{(λ/p∗)S1-p∗/p,(μ/q∗)S2-q∗/q} and b2=(1/η)max⁡{βxϑS1-η/p,β(x)γS2-η/q}.Define(70)hx=1qmin⁡xp,xq-b1max⁡xp∗,xq∗-b2xη;we can observe that there exists Λ1>0 being small such that if 0<λ,μ<Λ1, there exists 0<x0<x1 such that h(x) becomes -,0,+,0,- if x∈(0,x0),x0,(x0,x1),x1,(x1,+∞), respectively.We set up the truncated functional [15](71)Fu,v=1puXp+1qvYq-ψu,v∫RNλp∗up∗+μq∗vq∗dx-1η∫RNGu,vdx,where ψ(u,v)=ψ(z)=ϖ(zE) and ϖ:R+→[0,1] is a nonincreasing C∞ function such that ϖ(x)=1 for x≤x0 and ϖ(x)=0 for x≥x1. Obviously, F(u,v) is even and F(u,v) satisfies PSc with c<0.Lemma 9.
Assume(H1), (H2), and (H4) with η∈(1,p) hold. Given n∈N, there exists ϵ=ϵ(n)<0 such that(72)γu,v∈E:Fu,v≤ϵ≥n.Proof.
Fixn, and suppose En is n-dimensional subspace of E. We can take (un,vn)∈En, with (un,vn)E=1. Thus, for 0<ρ<x0, we have(73)Fρun,ρvn=Jρun,ρvn≤1pρp-min⁡λp∗,μq∗ρq∗∫RNup∗+vq∗dx-1ηρη∫RNGu,vdx;we can choose ϵ<0 and ς<x0 small enough such that F(ςu,ςv)≤ϵ. Let Sς={(u,v)∈E,(u,v)E=ς}; then, Sς∩En⊂γ({u∈E:F(u,v)≤ϵ}). Hence, γ({(u,v)∈E:F(u,v)≤ϵ})≥γ(Sς∩En)=n.Lemma 10.
DenoteΣk={Z∈Σ:γ(Z)≥k} and set (74)ck=infZ∈Σk⁡supu,v∈Z⁡Fu,v,Bc=u,v∈E:F′u,v=0,Fu,v=c;then, if c=ck=ck+1=⋯=ck+r, γ(Bc)≥r+1Proof.
For convenience, we denoteFc={{(u,v)∈E:F(u,v)≤c}}. Suppose c=ck=ck+1=⋯=ck+r; we observe that c<0 and Bc is a compact set. If γ(Bc)≤r, there exists a symmetric and closed set W1,Bc⊂W1 such that γ(W1)≤r. According to deformation lemma, one can get an odd homeomorphism ϱ:E→E such that(75)ϱFc+δ\W1⊂Fc-δ,forδ>0.Based on the definition c=ck+r, there exists W2∈Σk+r such that sup(u,v)∈W2⁡F(u,v)<c+δ and(76)ϱW2\W1⊂ρFc+δ\W1⊂Fc-δ.Hence we gain(77)γW2\W1¯≥γW2-γW1≥k,γϱW2\W1¯≥γW2\W1¯≥k.Therefore, ϱ(W2\W1¯)∈Σk and it is in contradiction to (76). Lemma 10 and Kajikiya’s [16] symmetric mountain pass lemma give the proof of Theorem 2.



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