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The Brezis-Nirenberg type double critical problem for a class of Schrödinger-Poisson equations

  • Received: 01 August 2020 Revised: 01 October 2020 Published: 14 December 2020
  • Primary: 35J15, 35J20; Secondary: 35J50

  • In this paper, we study the following Schrödinger-Poisson equations with double critical exponents:

    {Δu=|u|4u+ϕ|u|3u+λu,inΩ,Δϕ=|u|5,inΩ,u=ϕ=0,onΩ,

    where Ω is a bounded domain in R3 with Lipschitz boundary, λ is a real parameter satisfying suitable conditions. Using variational methods, we show the existence and nonexistence of nontrivial solutions for the Schrödinger-Poisson equations.

    Citation: Li Cai, Fubao Zhang. The Brezis-Nirenberg type double critical problem for a class of Schrödinger-Poisson equations[J]. Electronic Research Archive, 2021, 29(3): 2475-2488. doi: 10.3934/era.2020125

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  • In this paper, we study the following Schrödinger-Poisson equations with double critical exponents:

    {Δu=|u|4u+ϕ|u|3u+λu,inΩ,Δϕ=|u|5,inΩ,u=ϕ=0,onΩ,

    where Ω is a bounded domain in R3 with Lipschitz boundary, λ is a real parameter satisfying suitable conditions. Using variational methods, we show the existence and nonexistence of nontrivial solutions for the Schrödinger-Poisson equations.



    In this paper, we are concerned with the existence of solutions for the following Schrödinger-Poisson equations

    {Δu=|u|4u+ϕ|u|3u+λu,inΩ,Δϕ=|u|5,inΩ,u=ϕ=0,onΩ, (1)

    where Ω is a bounded domain in R3 with Lipschitz boundary, λ is real parameter. In recent years, many people have studied the Brezis-Nirenberg type problem. For example, the following Brezis-Nirenberg type equation

    Δu=|u|22u+λu,inΩ, (2)

    where 2=2NN2 is the critical exponent for the embedding of H10(Ω) to Lp(Ω). In [6], H. Brezis and L. Nirenberg showed that equation (2) has a nontrivial solution if N4,λ(0,λ1), where λ1 is the first eigenvalue of Δ in the bounded domain Ω. While, if N=3, then there exists a constant λ(0,λ1) such that for any λ(λ,λ1), equation (2) with Dirichlet boundary data has a positive solution. Moreover, if Ω is a ball, equation (2) has a positive solution if and only if λ(λ14,λ1). Other conclusions about equation (2) can be founded in [8,9]. In [5], the following form of Schrödinger-Poisson equations with a critical exponent on a bounded domain were studied :

    {Δu=qϕ|u|3u+λu,inBR,Δϕ=q|u|5,inBR,u=ϕ=0,onBR, (3)

    where BR(0) is a ball centered in 0 in R3 with radius R. They obtained a positive ground state solution for λ(310λ1,λ1) and q>0. In addition, Qi Zhang [29] studied the following equations

    {Δu+ηϕu=μuγ,inΩ,Δϕ=u2,inΩ,u>0,inΩ,u=ϕ=0onΩ, (4)

    where ΩR3. By using the variational method and the Nehari manifold, the author proved the existence and uniqueness of positive solution for μ>0, η=1 and the existence of two positive solutions for μ>0, η=1 respectively.

    Further, Fashun Gao and Minbo Yang [12] considered the following general Choquard equation in a bounded domain ΩRN with Lipschitz boundary,

    Δu=(Iμ|u|2μ)|u|2μ2u+λu,inΩ. (5)

    They proved that if N4, then for each λ>0 equation (5) has a nontrivial solution in H10(Ω), and if N=3, then there exists λ such that equation (5) has a nontrivial solution for λ>λ, where λ is not an eigenvalue of Δ with homogeneous Dirichlet boundary data. In general, equation (5) is related to the following nonlinear Choquard equation

    Δu+V(x)u=(1|x|μ|u|p)|u|p2u,inR3. (6)

    A large number of results have been obtained about the existence and qualitative properties of solutions to equation (6) in recent decades. For example, the existence and uniqueness of the ground states about equation (6) have been obtained in [18]. Later, in [19], Lions proved the existence of a sequence of radially symmetric solutions. The regularity, positivity and radial symmetry of the ground states and decay property at infinity about equation (6) have been studied in [10,20,21]. Other results about equation (6) are available in [1,2,3].

    Also, a great deal of work on elliptic equations can been founded in [4,11,13,14,15,16,17,23,25,28,30].

    Recently, in [7], we studied the Brezis-Nirenberg type double critical problem for the Choquard equation

    Δu=α|u|22u+β(Iμ|u|2μ)|u|2μ2u+λu,inΩ, (7)

    where α and β satisfy suitable conditions, we obtained the existence of nontrivial solutions and nonexistence. As far as we know, there are only a few results on the Brezis-Nirenberg type for the Schrödinger-Poisson equations with double critical exponents, the main difficulty is the lack of compactness of (PS) sequence.

    In the following, we will give some notations which will be used throughout this paper. Let Lp(Ω)(1p<+) be the usual Lebesgue space with the norm |u|p=(Ω|u|pdx)1p, H10(Ω) be the closure of C0(Ω) in H1(R3), where

    H1(R3):={uL2(R3):uL2(R3)}.

    u:=(Ω|u|2dx)12 is the norm of H10(Ω). D1,20(Ω) is the closure of C0(Ω) in D1,2(R3), where

    D1,2(R3)={uL6(R3):uL2(R3)}

    with the corresponding norm (R3|u|2dx)12.

    Define

    S:=infD1,2(R3){0}R3|u|2dx(R3|u|6dx)13.

    By [26], we have

    S(Ω):=infD1,20(Ω){0}Ω|u|2dx(Ω|u|6dx)13=S.

    Firstly, we observe that by the Lax-Milgram theorem, for given uH10(Ω), there exists a unique solution ϕ=ϕuD1,20(Ω) satisfying Δϕu=|u|5 in a weak sense. By [5,14], it is easy to obtain the following properties.

    Lemma 1.1. The following properties hold:

    (i) ϕu0 for all uH10(Ω);

    (ii) for each uH10(Ω),

    |ϕu|D1,2S12|u|56

    and

    Ωϕu|u|5S1|u|106;

    (iii) if unu in H10(Ω), then, up to a subsequence, ϕunϕu in D1,20(Ω).

    Next we will denote the sequence of eigenvalues of the operator Δ on Ω with homogeneous Dirichlet boundary data by

    0<λ1λiλi+1

    and λi+ as i+. {ei}iN+L(Ω) will be the sequence of eigenfunctions corresponding to {λi}. We recall that this sequence is an orthonormal basis of L2(Ω) and an orthogonal basis of H10(Ω). Denote

    Ei+1:={uH10(Ω):(u,ej)H10=0,j=1,2,,i}, (8)

    while Yi:=span{e1,e2,,ei} denote the linear subspace generated by the first i eigenfunctions of Δ for any iN+. It is easy to see that Yi is finite dimensional and YiEi+1=H10(Ω).

    Now we introduce the energy functional associated to (1) by

    I(u)=12Ω|u|2dx16Ω|u|6dx110Ωϕu|u|5dxλ2Ω|u|2dx.

    It is easy to know that I belongs to C1(H10(Ω),R). Moreover, u is a weak solution of (1) if and only if u is a critical point of functional I.

    Definition 1.2. Let I be a C1 functional defined on Banach space X, if I(un)c, and I(un)0, as n+, then {un} is called a Palais-Smale sequence of I at c. If every Palais-Smale sequence at c has a convergent subsequence, then we call that I satisfies the Palais-Smale condition at the level c.

    Now we state the main results of this paper.

    Theorem 1.3. (i) Assume that ΩR3 is a bounded domain with Lipschitz boundary, if Ω contains a ball with radius R such that 13540π2R2<λ1, then for λ(13540π2R2,λ1), (1) has a nontrivial solution.

    (ii) There exists a constant λ such that (1) has a nontrivial solution for λ>λλ1, provided λ is not an eigenvalue of Δ with homogeneous Dirichlet boundary data.

    Remark 1. If Ω is the ball in R3 centered in 0 with radius R, then π2R2=λ1, the first eigenvalue of Δ in BR(0), thus we obtain that if λ(13540λ1,λ1), (1) has a nontrivial solution. Furthermore, if Ω has s a good approximation of a ball, then the condition of (ii) can be satisfied, due to the continuous dependence of λ1 on Ω.

    Theorem 1.4. If λ<0, Ω is a smooth and strictly star-shaped domain with respect to the origin in R3, then (1) admits no nontrivial solution.

    This paper is organized as follows. In Section 2, we give some preliminaries which can be applied to Theorem 1.3. In Sections 3,4, we will prove Theorem 1.3. Section 5 is devoted to the proof of Theorem 1.4.

    Lemma 2.1. [27] Let N3, q(1,+) and {un} be a bounded sequence in Lq(RN). If unu almost everywhere in RN as n, then unu weakly in Lq(RN).

    Lemma 2.2. Let λ>0. If {un} is a (PS)c sequence of I, then {un} is bounded in H10(Ω). Let u0H10(Ω) be the weak limit of {un}, then u0 is a weak solution of (1).

    Proof. Since {un} is a (PS)c sequence of I, we know that there exists C>0 such that

    |I(un)|C,|(I(un),unun)|C.

    In order to prove the boundedness of {un}, we will consider the two cases: 0<λ<λ1 and λ[λi,λi+1) for some iN+.

    Case 1. 0<λ<λ1.

    For n large enough, we get

    C(1+un)I(un)16(I(un),un)13(un2λ|un|22)δ13un2

    for δ1=λ1λλ1>0. Thus {un} is bounded in H10(Ω).

    Case 2. λ[λi,λi+1) for some iN+.

    Let θ(16,12). For n large enough, we have

    C(1+un)I(un)θ(I(un),un)(12θ)(un2λ|un|22)+(θ16)|un|66
    +(θ110)Ωϕu|u|5dx(12θ)(xn2+yn2λ|xn|22λ|yn|22)+(θ16)|un|66(12θ)(δ2xn2+(λ1λ)|yn|22)+(θ16)|un|66(12θ)(δ2xn2+(λ1λ)|yn|22)+(θ16)|un|66

    for δ2=λi+1λλi+1>0, un=xn+yn, xnEi+1, ynYi. Then we have

    (12θ)(λλ1)|yn|22+C(1+un)(12θ)δ2xn2+(θ16)|un|66(12θ)δ2xn2+C(θ16)|un|62(12θ)δ2xn2+C(θ16)|yn|62.

    So, from the fact that Yi is finite dimensional, we can deduce that {un} is bounded in H10(Ω). Since H10(Ω) is reflexive, up to a sequence, still denoted by {un}, there exists u0H10(Ω) such that unu0 in H10(Ω). Since the proof that u0 is a weak solution is almost the same as that of Lemma 2.4 of [12], so we omit it here. Especially, we take h=u0H10(Ω) as a test function in (1), then

    Ω|u0|2dx=λΩu20dx+Ωϕu|u|5dx+Ωu60dx.

    And we can deduce that

    I(u0)25Ωϕu|u|5dx+13Ωu60dx0.

    This completes the proof.

    Lemma 2.3. Let λ>0. If {un} is a (PS) sequence at c with

    0<c<13530(512)12S32,

    then there exists u0H10(Ω) such that u00 is a weak limit of {un}.

    Proof. From Lemma 2.2, we know that {un} is bounded in H10(Ω). Up to a subsequence, we can assume that it is weakly convergent. Suppose by contradiction that un0 in H10(Ω). Then we have

    cI(un)=12Ω|un|2dx16Ωu6ndx110Ωϕun|un|5dx+on(1) (9)

    and

    on(1)=(I(un),un)=Ω|un|2dxΩu6ndxΩϕun|un|5dx+on(1). (10)

    From (9), up to subsequences, we know that there exists a non-negative constant a such that

    Ω|un|2dxa

    and

    Ωu6ndx+Ωϕun|un|5dxa,

    as n+. If a=0, then by c>0, we know that this is a contradiction. If a0, from (9) and (10), for convenience, let bn=Ωu6ndx, ln=Ωϕun|un|5dx. Without loss of generality, we may assume bnb and lnl, as n. Recall that, t>0,

    Ω|un|6dx=Ωϕunundxt22Ω|un|2dx+12t2Ω|ϕun|2dx=12t2Ωϕun|un|5dx+t22Ω|un|2dx,

    thus, as n, we have

    b12t2l+t22a.

    Taking t2=512, since b+l=a, we get

    l352a.

    Then from (9), we can deduce

    c=25l+13b=13a+115l13530a.

    On the other hand, it follows from Lemma 1.1 that

    aS6a5+S3a3.

    Hence, we have either a=0 or a21+52S3. In each case it will come to a contradiction. So the proof is completed.

    Lemma 3.1. If λ(0,λ1), then the functional I satisfies the following properties:

    (i) There exist γ1,ρ1>0 such that I(u)γ1 for u=ρ1.

    (ii) There exists u0H10(Ω) with u0ρ1 such that I(u0)<0.

    Proof. (i) Since λ(0,λ1), by the Sobolev embedding, for all uH10(Ω){0} we get

    I(u)12Ω|u|2dxλ2λ1Ω|u|2dx16|u|66110S1|u|10612(1λλ1)u216S3u6110S6u10.

    So we can choose some γ1,ρ1>0 such that I(u)γ1 for u=ρ1.

    (ii) For each u1H10(Ω){0}, we have

    I(tu1)t22Ω|u1|2dxλt22Ωu21dxt66Ωu61dxt1010Ωϕu1|u1|5dx<0

    for t>0 large enough. Hence, we can take u0=tu1 for t>0 large to end the proof.

    By Lemma 3.1 and the mountain pass theorem without (PS) condition, there exists a (PS) sequence {un} such that I(un)c and I(un)0 in H1(Ω) at the minimax level

    c:=infγΓmaxt[0,1]I(γ(t))>0,

    where

    Γ:={γC([0,1],H10(Ω)):γ(0)=0,I(γ(1))<0}.

    Without loss of generality, we may assume that 0Ω and BRΩBkR for some positive R,k. By [6] and [5], we consider a fixed function ψ such that ψ(r)=cos(πr2R), r[0,R), and ψ(r)=0, r[R,kR]. Set r=|x| and

    uϵ(r)=ψ(r)(ϵ+r2)12.

    Then we have

    Ω|uϵ|2dx=BkR|uϵ|2dx=BR|uϵ|2dx=SKϵ12+wR0|ψ(r)|2dr+O(ϵ12), (11)
    (Ωu6ϵdx)13=(BkRu6ϵdx)13=(Ru6ϵdx)13=Kϵ12+O(ϵ12), (12)

    and

    Ωu2ϵdx=BkRu2ϵdx=BkRu2ϵdx=wR0ψ2(r)dr+O(ϵ12), (13)

    where K is a positive constant and w is the area of the unitary sphere in R3. Note that

    Ωu6dx151Ωϕ|u|5dx+514Ω|u|2dx,

    hence we can introduce the new functional J:H1(Ω)R defined by

    J(u):=13520Ω|u|2dxλ2Ωu2dx2+3530Ωu6dx.

    Obviously we have I(u)J(u) for any uH1(Ω).

    Proof of Theorem 1.3(i). Firstly, for vH10(Ω){0}, we have

    0<maxt0I(tv)maxt0J(tv)=maxt0{13520t2Ω|v|2dxλt22Ωv2dx2+3530t6Ωv6dx},

    Set

    y(t):=13520t2b1t22c12+3530t6d1,

    where

    b1:=Ω|v|2dx,
    c1=λΩv2dx
    d1:=Ωv6dx.

    Let

    y=13510b1tc1t2+355d1t5=0,

    which yields that there exists unique t>0 such that t4=13510b1c12+355d1 and y(t) is the maximum of y. If we can verify that ymax<13530(512)12S32, by the definition of c, we know that c<13530(512)12S32. Then let {un} be the (PS) sequence, which follows from Lemma 2.3 that unu00. Thus (1) has a nontrivial solution. So it is enough to verify that

    ymax<13530(512)12S32.

    Set v=uϵ, compute maxt0J(tuϵ)=J(tϵuϵ), where tϵ is the unique positive solution of the equation ddtJ(tuϵ)=0, we obtain that

    t4ϵ=13510BR|uϵ|2dxλBRu2ϵdx2+355BRu6ϵdx,

    which yields that

    J(tϵuϵ)13(13510BR|uϵ|2dxλBRu2ϵdx)32(2+355BRu6ϵdx)12. (14)

    In addition, when r[0,R], we get

    R0|ψ(r)|2dr=π24R2R0ψ2(r)dr. (15)

    Then we take (11), (12), (13) and (15) into (14), by the simple computation, we have if λ(13540π2R2,λ1), we can deduce that c<13530(512)12S32 for ϵ small enough. Then by using Lemma 2.3, the conclusion is true in case λ(13540π2R2,λ1).

    In this section, we will consider the cross eigenvalue case, i.e., suppose λ[λi,λi+1) for some iN+, where λi is the i-th eigenvalue of Δ on Ω with boundary condition u=0, and ei is the i-th eigenfunctions corresponding to the eigenvalue λi. For the cross eigenvalue case, we need to use the following linking theorem [22].

    Theorem 4.1. Let E be a real Banach space with E=VX, where V is finite dimensional. Suppose IC1(E,R), satisfies (PS), and

    (i) there are constants ρ,α>0 such that I|BρXα;

    (ii) there is an eB1X and R>ρ such that if Q(ˉBRV){re|0<r<R}, then I|Q0.

    Then I possesses a critical value cα which can be characterized as

    c=infhΓmaxuQI(h(u)),

    where

    Γ={hC(ˉQ,E)|h=idonQ}.

    Lemma 4.2. If λ[λi,λi+1) for some iN, then the functional I satisfies the following properties:

    (i) there exist γ2,ρ2>0 such that Iγ2, for any uEi+1 with u=ρ2;

    (ii) I(u)<0 for any uYi{0};

    (iii) let F be a finite dimensional subspace of H10(Ω), then there exists R>ρ2 such that I(u)0, for any uF with uR.

    Proof. (i) Since λ[λi,λi+1), by Lemma 1.1 and the Sobolev embedding, for all uEi+1{0} we have

    I(u)12Ω|u|2dxλ2λi+1Ω|u|2dx16S3u6110S6u1012(1λλi+1)u216S3u6110S6u10.

    We can choose some γ2,ρ2>0 such that Iγ2 for uEi+1 with u=ρ2.

    (ii) Let uYi, i.e., u=ij=1ljej, where ljR, j=1,,i. Since {ej}jN is an orthonormal basis of L2(Ω) and H10(Ω), we get

    Ω|u|2dx=ij=1l2j|ej|22,Ωu2dx=ij=1l2j.

    Then we can get

    I(u)=12ij=1l2j(|ej|22λ)16Ωu6dx110Ωϕu|u|5dx<12ij=1l2j(λjλ)0.

    (iii) Choose uF{0}. Since all norms on finite dimensional space are equivalent, then

    I(u)=12u2λ2|u|2216|u|66110Ωϕu|u|5dx12u2Cu6,

    where C is a positive constant. So, I as u+. Hence, there exists R>ρ2 such that I(u)0, for any uF with uR and the proof of (iii) ends.

    It is well known that U(x)=[N(N2)]N24(1+|x|2)N22 is a minimizer for S.

    Without loss of generality, we may assume that 0Ω and BδΩBkδ for some positive k. Let φC0 such that if xBδ, φ=1; and if xR3Bkδ, φ=0. xR3, 0φ1 and |Dφ(x)|C=const.

    For ϵ>0, define

    Uϵ(x):=ϵ12U(xϵ),uϵ(x):=φ(x)Uϵ(x),

    then in the view of [12], we have

    |Uϵ|22=|Uϵ|66=S32,

    and as ϵ0+, we can have

    Ω|uϵ|2dx=S32+O(ϵ), (16)
    Ω|uϵ|6dx=S32+O(ϵ3), (17)

    and

    Ω|uϵ|2dxC0ϵ, (18)

    where C0 is a positive constant.

    Define the linear space for any iN+,

    Gi,ϵ:=span{e1,,ei,uϵ}. (19)

    Now we give the proof of Theorem 1.3 for the general case.

    Proof of Theorem 1.3(ii). Suppose that λ[λi,λi+1) for iN+, then from the definition of Gi,ϵ, we know

    u=v+tuϵ,uGi,ϵ,

    where vYi. Then

    Gi,ϵ=YiRuϵ.

    From Lemma 4.2, we know that I satisfies the geometric structure of the linking theorem, i.e.,

    infuEi+1,u=ρ2I(u)γ2>0,supuYi{0}I(u)<0,supuGi,ϵ,uRI(u)0,

    where γ2 and R are defined in Lemma 4.2. Define the linking critical level of I, i.e.,

    ˉc=infγΓmaxuVI(γ(u))>0,

    where

    Γ:={γC(ˉV,H10(Ω)):γ=idonV}

    and

    V:=(ˉBRYi){ruϵ:r(0,R)}.

    For any γΓ, we get

    ˉcmaxuVI(γ(u))

    and in particular, if we take γ=id on ˉV, then

    ˉcmaxuVI(u).

    Then we have

    I(u)=12Ω|u|2dxλ2Ωu2dx16Ωu6dx110Ωϕu|u|5dx=12Ω|(v+tuϵ)|2dxλ2Ω|v+tuϵ|2dx16Ω|v+tuϵ|6dx110Ωϕv+tuϵ|v+tuϵ|5dx(λiλ)2|v|22+t22(Ω|uϵ|2dxλΩ|uϵ|2dx)+C1t|uϵ|1|v|2t66Ω|uϵ|6dxΩ|tuϵ|5v(x)dxt1010Ωϕuϵ|uϵ|5dxΩϕtuϵ|tuϵ|4vdx(λiλ)2|v|22+C1t|v|2O(ϵ12)+t22(Ω|uϵ|2dxλΩu2ϵdx)t66Ωu6ϵdxt1010Ωϕuϵ|uϵ|5dx+|v|2t5O(ϵ12)+|v|2t9O(ϵ12)C22(λλi)O(ϵ)+t22(Ω|uϵ|2dxλΩu2ϵdx)t66Ωu6ϵdxt1010Ωϕuϵ|uϵ|5dx, (20)

    where in the view of boundedness of t, by calculating the maximum value of the quadratic function, we get

    (λiλ)2|v|22+C1t|v|2O(ϵ12)+|v|2t5O(ϵ12)+|v|2t9O(ϵ12)(λiλ)2|v|22+C|v|2O(ϵ12)C22(λλi)O(ϵ).

    Then take (16), (17), and (18) into (20), we have

    Ω|uϵ|2dxλΩu2ϵdxS32+O(ϵ)λC0ϵ.

    Since Δϕuϵ=|uϵ|5, we deduce that

    Ωu6ϵdx=Ωϕuϵuϵdx12Ω|ϕuϵ|2dx+12Ω|uϵ|2dx=12Ωϕuϵ|uϵ|5dx+12Ω|uϵ|2dx.

    For ϵ>0 sufficiently small, we deduce

    Ωϕuϵ|uϵ|5dx2Ωu6ϵdxΩ|uϵ|2dx=S32+O(ϵ).

    We have

    C22(λλi)O(ϵ)+t22(Ω|uϵ|2dxλΩu2ϵdx)t66Ωu6ϵdxt1010Ωϕuϵ|uϵ|5dxC3ϵ+t22S32t66S32t1010S32+t22(O(ϵ)λC0ϵ)t66O(ϵ3)+t1010O(ϵ)<t22S32t66S32t1010S3213510(512)12S32

    for ϵ small enough. Hence there exists λ>0 such that λ>λ, we can get

    maxuGi,ϵI(u)<13530(512)12S32.

    Taking λλ1, it follows from the linking theorem and Lemma 2.3 that (1) has a nontrivial solution uH10(Ω) in case λ>λ and λλi, where λi is an eigenvalue of Δ with homogeneous Dirichlet boundary data.

    In this section, we will prove the nonexistence of nontrivial solutions for (1) in case λ<0.

    Proof of Theorem 1.4. By [26,24], similar to the proof in [5], let ΩR3 be a star shaped domain and u,ϕC2(Ω)C1(Ω) be a nontrivial solution of (1). We multiply the first equation of (1) by xu and the second one by xϕ, we obtain that

    0=(Δu+λu+ϕ|u|3u+|u|4u)(xu)=div[(u)(xu)x|u|22+λ2xu2+15xϕ|u|5+x|u|6]+12|u|232λu235ϕ|u|515(xϕ)|u|512|u|6

    and

    0=(Δϕ+|u|5)(xϕ)=div[(ϕ)(xϕ)x|ϕ|22]+12|ϕ|2+(xϕ)|u|5.

    Assume that ν is the unit exterior normal to Ω, integrating on Ω, by boundary conditions u=uνν and ϕ=ϕνν on Ω, we have

    12u2212Ω|uν|2xν=3λ2|u|2235Ωϕ|u|5dx15Ω(xϕ)|u|512|u|6 (21)

    and

    12ϕ2212Ω|ϕν|2xν=Ω(xϕ)|u|5. (22)

    Taking (22) into (21), we get

    12u2212Ω|uν|2xν=3λ2|u|2235Ωϕ|u|5dx+110ϕ22+110Ω|ϕν|2xν12Ω|u|6.

    In addition, since (u,ϕ) is a nontrivial solution for (1), we see that

    u22=λ|u|22+Ωϕ|u|5+|u|66

    and ϕ22=Ωϕ|u|5. Combing the above argument, we deduce that

    λ|u|22+12Ω|uν|2xν+110Ω|ϕν|2xν=0.

    In the view of λ<0, we know that this is a contradiction. So we have u=0, that is, (1) admits no nontrivial solution.

    The authors would like to express their sincere gratitude to the referees for careful reading the manuscript and valuable suggestions. The work was supported by the National Natural Science Foundation of China (Nos. 11671077).



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