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Existence and asymptotic behavior of bound states for a class of nonautonomous Schrödinger-Poisson system

  • This paper is concerned with the following Schrödinger-Poisson system

    (Pμ):Δu+u+K(x)ϕu=|u|p1u+μh(x)u, Δϕ=K(x)u2, xR3,

    where p(3,5), K(x) and h(x) are nonnegative functions, and μ is a positive parameter. Let μ1>0 be an isolated first eigenvalue of the eigenvalue problem Δu+u=μh(x)u, uH1(R3). As 0<μμ1, we prove that (Pμ) has at least one nonnegative bound state with positive energy. As μ>μ1, there is δ>0 such that for any μ(μ1,μ1+δ), (Pμ) has a nonnegative ground state u0,μ with negative energy, and u0,μ(n)0 in H1(R3) as μ(n)μ1. Besides, (Pμ) has another nonnegative bound state u2,μ with positive energy, and u2,μ(n)uμ1 in H1(R3) as μ(n)μ1, where uμ1 is a bound state of (Pμ1).

    Citation: Lirong Huang, Jianqing Chen. Existence and asymptotic behavior of bound states for a class of nonautonomous Schrödinger-Poisson system[J]. Electronic Research Archive, 2020, 28(1): 383-404. doi: 10.3934/era.2020022

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  • This paper is concerned with the following Schrödinger-Poisson system

    (Pμ):Δu+u+K(x)ϕu=|u|p1u+μh(x)u, Δϕ=K(x)u2, xR3,

    where p(3,5), K(x) and h(x) are nonnegative functions, and μ is a positive parameter. Let μ1>0 be an isolated first eigenvalue of the eigenvalue problem Δu+u=μh(x)u, uH1(R3). As 0<μμ1, we prove that (Pμ) has at least one nonnegative bound state with positive energy. As μ>μ1, there is δ>0 such that for any μ(μ1,μ1+δ), (Pμ) has a nonnegative ground state u0,μ with negative energy, and u0,μ(n)0 in H1(R3) as μ(n)μ1. Besides, (Pμ) has another nonnegative bound state u2,μ with positive energy, and u2,μ(n)uμ1 in H1(R3) as μ(n)μ1, where uμ1 is a bound state of (Pμ1).



    In this paper, we study a class of Schrödinger-Poisson system with the following version

    {Δu+u+K(x)ϕu=|u|p1u+μh(x)u in R3,Δϕ=K(x)u2 in R3, (1)

    where p(3,5), μ>0, K(x) and h(x) are nonnegative functions. System (1) can be looked on as a non-autonomous version of the system

    {Δu+u+ϕu=f(u) in R3,Δϕ=u2 in R3, (2)

    which has been derived from finding standing waves of the Schrödinger-Poisson system

    {iψtΔψ+ϕψ=f(ψ) in R3,Δϕ=|ψ|2 in R3.

    A starting point of studying system (1) is the following fact. For any uH1(R3) and KL(R3), there is a unique ϕuD1,2(R3) with

    ϕu(x)=14πR3K(y)|u(y)|2|xy|dy

    such that Δϕu=K(x)u2, see e.g. [11,20]. Inserting this ϕu into the first equation of the system (1), we get that

    Δu+u+K(x)ϕuu=|u|p1u+μh(x)u,uH1(R3). (3)

    Problem (3) can be also looked on as a usual semilinear elliptic equation with an additional nonlocal perturbation K(x)ϕuu. Our aim here is to prove some new phenomenon of (3) due to the presence of the term K(x)ϕuu. Before giving the main results, we state the following assumptions.

    (A1) h(x)0, h(x)0 in R3 and h(x)L32(R3)L(R3).

    (A2) There exist b>0 and H0>0 such that h(x)H0eb|x| for all xR3.

    (A3) K(x)0 and K(x)L2(R3)L(R3).

    (A4) There exist a>0 and K0>0 such that K(x)K0ea|x| for all xR3.

    From Lemma 2.1, we know that under the condition (A1), the following eigenvalue problem

    Δu+u=μh(x)u,u H1(R3)

    has a first eigenvalue μ1>0 and μ1 is simple. Denote

    F(u):=R3K(x)ϕu(x)|u(x)|2dx

    and introduce the energy functional Iμ:H1(R3)R associated with (3)

    Iμ(u)=12u2+14F(u)R3(1p+1|u|p+1+μ2h(x)u2)dx,

    where u2=R3(|u|2+u2)dx. From [11] and the Sobolev inequality, Iμ is well defined and IμC1(H1(R3),R). Moreover, for any vH1(R3),

    Iμ(u),v=R3(uv+uv+K(x)ϕuuv|u|p1uv+μh(x)uv)dx.

    It is known that there is a one to one correspondence between solutions of (3) and critical points of Iμ in H1(R3). Note that if uH1(R3) is a solution of (3), then (u,ϕu) is a solution of the system (1). If u0 and u is a solution of (3), then (u,ϕu) is a nonnegative solution of (1) since ϕu is always nonnegative. We call uH1(R3){0} a bound state of (3) if Iμ(u)=0. At this time (u,ϕu) is called a bound state of (1). A bound state u is called a ground state of (3) if Iμ(u)=0 and Iμ(u)Iμ(w) for any bound state w. In this case, we call (u,ϕu) a ground state of (1). The first result is about μ less than μ1.

    Theorem 1.1. Suppose that the assumptions of (A1) - (A4) hold and 0<b<a<2. If 0<μμ1, then problem (3) has at least one nonnegative bound state.

    The second result is about μ in a small right neighborhood of μ1.

    Theorem 1.2. Under the assumptions of (A1) - (A4), if 0<b<a<1, then there exists δ>0 such that, for any μ(μ1,μ1+δ),

    (1) problem (3) has at least one nonnegative ground state u0,μ with Iμ(u0,μ)<0. Moreover, u0,μ(n)0 strongly in H1(R3) for any sequence μ(n)>μ1 and μ(n)μ1;

    (2) problem (3) has another nonnegative bound state u2,μ with Iμ(u2,μ)>0. Moreover, u2,μ(n)uμ1 strongly in H1(R3) for any sequence μ(n)>μ1 and μ(n)μ1, where uμ1 satisfies Iμ1(uμ1)=0 and Iμ1(uμ1)>0.

    The proofs of Theorem 1.1 and Theorem 1.2 are based on critical point theory. There are several difficulties in the road of getting critical points of Iμ in H1(R3) since we are dealing with the problem in the whole space R3, the embedding from H1(R3) into Lq(R3) (2<q<6) is not compact, the appearance of a nonlocal term K(x)ϕuu and the non coercive linear part. To explain our strategy, we review some related known results. For the system (2), under various conditions of f, there are a lot of papers dealing with the existence and nonexistence of positive solutions (u,ϕu)H1(R3)×D1,2(R3), see for example [2,23] and the references therein. The lack of compactness from H1(R3)Lq(R3) (2<q<6) was overcome by restricting the problem in H1r(R3) which is a subspace of H1(R3) containing only radial functions. The existence of multiple radial solutions and non-radial solutions have been obtained in [2,13]. See also [6,15,16,17,18,19,24,29,30] for some other results related to the system (2).

    While for nonautonomous version of Schrödinger-Poisson system, only a few results are known in the literature. Jiang et.al.[21] have studied the following Schrödinger-Poisson system with non constant coefficient

    {Δu+(1+λg(x))u+θϕ(x)u=|u|p2u in R3,Δϕ=u2in R3,lim|x|ϕ(x)=0,

    in which the authors prove the existence of ground state solution and its asymptotic behavior depending on θ and λ. The lack of compactness was overcome by suitable assumptions on g(x) and λ large enough. The Schrödinger-Poisson system with critical nonlinearity of the form

    {Δu+u+ϕu=V(x)|u|4u+μP(x)|u|q2uin R3,Δϕ=u2in R3,2<q<6,μ>0

    has been studied by Zhao et al. [31]. Besides some other conditions, Zhao et. al. [31] assume that V(x)C(R3,R), lim|x|V(x)=V(0,) and V(x)V for xR3 and prove the existence of one positive solution for 4<q<6 and each μ>0. It is also proven the existence of one positive solution for q=4 and μ large enough. Cerami et. al. [11] study the following type of Schrödinger-Poisson system

    {Δu+u+L(x)ϕu=g(x,u) in R3,Δϕ=L(x)u2 in R3. (4)

    Besides some other conditions and the assumption L(x)L2(R3), they prove the existence and nonexistence of ground state solutions. We emphasize that L(x)L2(R3) will imply suitable compactness property of the coupled term L(x)ϕu. Huang et. al. [20] have used this property to prove the existence of multiple solutions of (4) when g(x,u)=a(x)|u|p2u+μh(x)u and μ stays in a right neighborhood of μ1. The lack of compactness was overcome by suitable assumptions on the sign changing function a(x). While for (3), none of the aboved mentioned properties can be used. We have to analyze the energy level of the functional Iμ such that the Palais-Smale ((PS) for short) condition may hold at suitable interval. Also for (3), another difficulty is to find mountain pass geometry for the functional Iμ in the case of μμ1. We point out that for the semilinear elliptic equation

    Δu=a(x)|u|p2u+˜μk(x)u, in RN, (5)

    Costa et.al.[14] have proven the mountain pass geometry for the related functional of (5) when ˜μ˜μ1, where ˜μ1 is the first eigenvalue of Δu=˜μk(x)u in D1,2(RN). Costa et. al. have managed to do these with the help of the condition RNa(x)˜ep1dx<0, where ˜e1 is a positive eigenfunction corresponding to ˜μ1. In the present paper, it is not possible to use such kind of condition. We will develop further the techniques in [20] to prove the mountain pass geometry. A third difficulty is to look for a ground state of (3). A usual method of getting a ground state is by minimizing the functional Iμ over the Nehari set {uH1(R3){0} : Iμ(u),u=0}. But in the case of μ>μ1, one can not do like this because we do not know if 0 belongs to the boundary of this Nehari set. To overcome this trouble, we will minimize the functional over the set {uH1(R3){0} : Iμ(u)=0}.

    This paper is organized as follows. In Section 2, we give some preliminaries. Special attentions are focused on several lemmas analyzing the Palais-Smale conditions of the functional Iμ, which will play an important role in the proofs of Theorem 1.1 and Theorem 1.2. In Section 3, we prove Theorem 1.1. And Section 4 is devoted to the proof of Theorem 1.2.

    Notations. Throughout this paper, o(1) is a generic infinitesimal. The H1(R3) denotes dual space of H1(R3). Lq(R3) (1q+) is a Lebesgue space with the norm denoted by uLq. The Sp+1 is defined by

    Sp+1=infuH1(R3){0}R3(|u|2+|u|2)dx(R3|u|p+1dx)2p+1.

    For any ρ>0 and xR3, Bρ(x) denotes the ball of radius ρ centered at x. C or Cj (j=1, 2, ) denotes various positive constants, whose exact value is not important.

    In this section, we give some preliminary lemmas, which will be helpful to analyze the (PS) conditions for the functional Iμ. Firstly, for any uH1(R3) and KL(R3), defining the linear functional

    Lu(v)=R3K(x)u2vdx,vD1,2(R3),

    one may deduce from the Hölder and the Sobolev inequalities that

    |Lu(v)|Cu2L125vL6Cu2L125vD1,2. (6)

    Hence, for any uH1(R3), the Lax-Milgram theorem implies that there exists a unique ϕuD1,2(R3) such that Δϕ=K(x)u2 in D1,2(R3). Moreover it holds that

    ϕu(x)=14πR3K(y)u2(y)|xy|dy.

    Clearly ϕu(x)0 for any xR3. We also have that

    ϕu2D1,2=R3|ϕu|2dx=R3K(x)ϕuu2dx. (7)

    Using (6) and (7), we obtain that

    ϕuL6CϕuD1,2Cu2L125Cu2. (8)

    Then we deduce that

    R3K(x)ϕu(x)u2(x)dxCu4. (9)

    Hence on H1(R3), both the functional

    F(u)=R3K(x)ϕu(x)u2(x)dx (10)

    and

    Iμ(u)=12u2+14F(u)R3(1p+1|u|p+1+μ2h(x)u2)dx (11)

    are well defined and C1. Moreover, for any vH1(R3),

    Iμ(u),v=R3(uv+uv+K(x)ϕuuv|u|p1uvμh(x)uv)dx.

    The following Lemma 2.1 is a direct consequence of [28,Lemma 2.13].

    Lemma 2.1. Assume that the hypothesis (A1) holds. Then the functional uH1(R3)R3h(x)u2dx is weakly continuous and for each vH1(R3), the functional uH1(R3)R3h(x)uvdx is weakly continuous.

    Using the spectral theory of compact symmetric operators on Hilbert space, the above lemma implies the existence of a sequence of eigenvalues (μn)nN of

    Δu+u=μh(x)u,inH1(R3)

    with μ1<μ2 and each eigenvalue being of finite multiplicity. The associated normalized eigenfunctions are denoted by e1,e2, with ei=1, i=1,2,. Moreover, one has μ1>0 with an eigenfunction e1>0 in R3. In addition, we have the following variational characterization of μn:

    μ1=infuH1(R3){0}u2R3h(x)u2dx,μn=infuSn1{0}u2R3h(x)u2dx,

    where Sn1={span{e1,e2,,en1}}.

    Next we analyze the (PS) condition of the functional Iμ in H1(R3). The following definition is standard.

    Definition 2.2. For dR, the functional Iμ is said to satisfy (PS)d condition if for any (un)nNH1(R3) with Iμ(un)d and Iμ(un)0, the (un)nN contains a convergent subsequence in H1(R3). The functional Iμ is said to satisfy (PS) conditions if Iμ satisfies (PS)d condition for any dR.

    Lemma 2.3. Let (un)nNH1(R3) be such that Iμ(un)dR and Iμ(un)0, then (un)nN is bounded in H1(R3).

    Proof. For n large enough, we have that

    d+1+o(1)un=Iμ(un)14Iμ(un),un=14un2μ4R3h(x)u2ndx+p34(p+1)R3|un|p+1dx. (12)

    Note that p+1p1>32 for p(3,5). Then for any ϑ>0, we obtain from hL32(R3)L(R3) that

    R3h(x)u2ndx(R3|un|p+1dx)2p+1(R3|h(x)|p+1p1dx)p1p+12ϑp+1R3|un|p+1dx+p1p+1ϑ2p1R3|h(x)|p+1p1dx.

    Choosing ϑ=p32μ, we get

    d+1+o(1)un14un2D(p,h)μp+1p1, (13)

    where D(p,h)=p14(p+1)(p32)2p1R3|h(x)|p+1p1dx. Hence (un)nN is bounded in H1(R3).

    The following lemma is a variant of Brezis-Lieb lemma. One may find the proof in [20].

    Lemma 2.4. [20] If a sequence (un)nNH1(R3) and unu0 weakly in H1(R3), then

    limnF(un)=F(u0)+limnF(unu0).

    Lemma 2.5. There is a δ1>0 such that for any μ[μ1,μ1+δ1), any solution u of (3) satisfies

    Iμ(u)>p12(p+1)Sp+1p1p+1.

    Proof. Since u is a solution of (3), we get that

    Iμ(u)=12(u2μR3h(x)u2dx)+14F(u)1p+1R3|u|p+1dx=p12(p+1)(u2μR3h(x)u2dx)+p34(p+1)F(u).

    Noticing that u2μ1R3h(x)u2dx for any uH1(R3), we deduce that for any u0,

    Iμ1(u)p34(p+1)F(u)>0.

    Next, we claim: there is a δ1>0 such that for any μ[μ1,μ1+δ1), any solution u of (3) satisfies

    Iμ(u)>p12(p+1)Sp+1p1p+1.

    Suppose this claim is not true, then there is a sequence μ(n)>μ1 with μ(n)μ1 and solutions uμ(n) of (3) such that

    Iμ(n)(uμ(n))p12(p+1)Sp+1p1p+1.

    Note that Iμ(n)(uμ(n))=0. Then we deduce that for n large enough,

    Iμ(n)(uμ(n))+o(1)uμ(n)Iμ(n)(uμ(n))14Iμ(n)(uμ(n)),uμ(n)14uμ(n)2D(p,h)(μ(n))p+1p1.

    This implies that (uμ(n))nN is bounded in H1(R3). Since for any nN, uμ(n)2μ1R3h(x)(uμ(n))2dx, we obtain that as μ(n)μ1

    uμ(n)2μ(n)R3h(x)(uμ(n))2dx(1μ(n)μ1)uμ(n)20

    because (uμ(n))nN is bounded in H1(R3). Noting that

    Iμ(n)(uμ(n))=p12(p+1)(uμ(n)2μ(n)R3h(x)(uμ(n))2dx)+p34(p+1)F(uμ(n)),

    we deduce that

    lim infnIμ(n)(uμ(n))p34(p+1)lim infnF(uμ(n))0,

    which contradicts to the

    Iμ(n)(uμ(n))p12(p+1)Sp+1p1p+1.

    This proves the claim and the proof of Lemma 2.5 is complete.

    Lemma 2.6. If μ[μ1,μ1+δ1), then Iμ satisfies (PS)d condition for any d<0.

    Proof. Let (un)nNH1(R3) be a (PS)d sequence of Iμ with d<0. Then for n large enough,

    d+o(1)=12un2μ2R3h(x)u2ndx+14F(un)1p+1R3|un|p+1dx

    and

    Iμ(un),un=un2μR3h(x)u2ndx+F(un)R3|un|p+1dx.

    Then we can prove that (un)nN is bounded in H1(R3). Without loss of generality, we may assume that unu0 weakly in H1(R3) and unu0 a. e. in R3. Denoting wn:=unu0, we obtain from Brezis-Lieb lemma and Lemma 2.4 that for n large enough,

    un2=u02+wn2+o(1),
    F(un)=F(u0)+F(wn)+o(1)

    and

    unp+1Lp+1=u0p+1Lp+1+wnp+1Lp+1+o(1).

    Using Lemma 2.1, we also have that R3h(x)u2ndxR3h(x)u20dx as n. Therefore

    d+o(1)=Iμ(un)=Iμ(u0)+12wn2+14F(wn)1p+1R3|wn|p+1dx. (14)

    Noticing Iμ(un),ψ0 for any ψH1(R3), we obtain that Iμ(u0)=0. From which we deduce that

    u02μR3h(x)u20dx+F(u0)=R3|u0|p+1dx. (15)

    Since (un)nN is bounded in H1(R3), we obtain from Iμ(un)0 that

    o(1)=un2μR3h(x)u2ndx+F(un)R3|un|p+1dx.

    Combining this with (15) as well as Lemma 2.1, we obtain that

    o(1)=wn2+F(wn)R3|wn|p+1dx. (16)

    Recalling the definition of Sp+1, we have that u2Sp+1u2Lp+1 for any u H1(R3). Now we distinguish two cases:

    (i) R3|wn|p+1dx0 as n;

    (ii) R3|wn|p+1dx0 as n.

    Suppose that the case (ⅰ) occurs. We may obtain from (16) that

    wn2Sp+1(wn2+F(wn)o(1))2p+1.

    Hence we get that for n large enough,

    wn2Sp+1p1p+1+o(1). (17)

    Therefore using (14), (16) and (17), we deduce that for n large enough,

    d+o(1)=Iμ(un)=Iμ(u0)+12wn2+14F(wn)1p+1R3|wn|p+1dx.=Iμ(u0)+p12(p+1)wn2+p34(p+1)F(wn)>p12(p+1)Sp+1p1p+1+p12(p+1)wn2+p34(p+1)F(wn)>0, (18)

    which contradicts to the condition d<0. This means that the case (ⅰ) does not occur. Therefore the case (ⅱ) occurs. Using (16), we deduce that wn20 as n. Hence we have proven that unu0 strongly in H1(R3).

    Next we give a mountain pass geometry for the functional Iμ.

    Lemma 2.7. There exist δ2>0 with δ2δ1, ρ>0 and α>0, such that for any μ[μ1,μ1+δ2), Iμ|Bρα>0.

    Proof. For any uH1(R3), there exist tR and vS1 such that

    u=te1+v,whereR3(ve1+ve1)dx=0. (19)

    Hence we deduce that

    u2=(te1+v)2L2+te1+v2L2=t2+v2, (20)
    μ2R3h(x)v2dxv2,μ1R3h(x)e21dx=e12=1 (21)

    and

    μ1R3h(x)e1vdx=R3(ve1+ve1)dx=0. (22)

    We first consider the case of μ=μ1. Denoting θ1:=(μ2μ1)/2μ2>0, then by the relations from (19) to (22), we obtain that

    Iμ1(u)=12u2+14F(u)μ12R3h(x)u2dx1p+1R3|u|p+1dx=12te1+v2+14F(te1+v)μ12R3h(x)(te1+v)2dx1p+1R3|te1+v|p+1dx12(1μ1μ2)v2+14F(te1+v)1p+1R3|te1+v|p+1dxθ1v2+14F(te1+v)C1|t|p+1C2vp+1.

    Next we estimate the term F(te1+v). Using the expression of F(u), we have that

    F(te1+v)=14πR3×R3K(x)K(y)(te1(y)+v(y))2(te1(x)+v(x))2|xy|dydx.

    Since

    (te1(y)+v(y))2(te1(x)+v(x))2=t4(e1(y))2(e1(x))2+(v(y))2(v(x))2+2t3(e1(y)(e1(x))2v(y)+e1(x)(e1(y))2v(x))+2t(e1(x)v(x)(v(y))2+e1(y)v(y)(v(x))2)+t2((e1(x))2(v(y))2+4e1(y)e1(x)v(y)v(x)+(e1(y))2(v(x))2),

    we know that

    |R3×R3K(x)K(y)(e1(y)(e1(x))2v(y)+e1(x)(e1(y))2v(x))|xy|dydx|Cv; (23)
    |R3×R3K(x)K(y)(2(e1(x))2(v(y))2+4e1(y)e1(x)v(y)v(x))|xy|dydx|Cv2 (24)

    and

    |R3×R3K(x)K(y)(e1(x)v(x)(v(y))2+e1(y)v(y)(v(x))2)|xy|dydx|Cv3. (25)

    Hence

    Iμ1(u)θ1v2+θ2|t|4C1|t|p+1C2vp+1C3|t|3vC4|t|2v2C5|t|v3+14F(v),

    where θ2=14R3K(x)ϕe1e21dx. Note that

    t2v22p+1|t|p+1+p1p+1v2(p+1)p1,
    |t|v31p+1|t|p+1+pp+1v3(p+1)p

    and for some q0 with 2<q0<4, we also have that

    |t|3v1q0vq0+q01q0|t|3q0q01.

    Therefore we deduce that

    Iμ1(u)θ1v2+θ2|t|4C3q0vq0C3(q01)q0|t|3q0q012C4p+1|t|p+1(p1)C4p+1v2(p+1)p1C5p+1|t|p+1pC5p+1v3(p+1)pC|t|p+1Cvp+1. (26)

    From q0>2 and 3q0q01>4 (since q0<4), we know that there are positive constants θ3, θ4 and ˜θ3, ˜θ4 such that

    Iμ1(u)θ3v2+θ4|t|4

    provided that v˜θ3 and |t|˜θ4. Hence there are positive constants θ5 and ˜θ5 such that

    Iμ1(u)θ5u4foru2˜θ25. (27)

    Set ˉδ:=min{μ12θ5˜θ25,  μ2μ1}>0 and δ2:=min{ˉδ,δ1}. Then for any μ[μ1,μ1+δ2), we deduce from (27) that

    Iμ(u)=Iμ1(u)+12(μ1μ)Rh(x)u2dxθ5u4μμ12μ1u2=u2(θ5u2μμ12μ1)u2(12θ5˜θ2514θ5˜θ25)=14θ5˜θ25u2

    for 12˜θ25u2˜θ25. Choosing ρ2=12˜θ25 and α=14θ5˜θ25ρ2, we finish the proof of Lemma 2.7.

    In this section, our aim is to prove Theorem 1.1. For 0<μ<μ1, it is standard to prove that the functional Iμ contains mountain pass geometry. For μ=μ1, as we have seen in Lemma 2.7, with the help of the competing between the Poisson term K(x)ϕuu and the nonlinear term, the 0 is a local minimizer of the functional Iμ1 and Iμ1 contains mountain pass geometry. To get a mountain pass type critical point of the functional Iμ, it suffices to prove the (PS)d condition by the mountain pass theorem of [3]. In the following we will focus our attention to the case of μ=μ1, since the case of 0<μ<μ1 is similar.

    Proposition 3.1. Let the assumptions (A1)(A4) hold and 0<b<a<2. Define

    dμ1=infγΓ1supt[0,1]Iμ1(γ(t))

    with

    Γ1={γC([0,1],H1(R3)) : γ(0)=0, Iμ1(γ(1))<0}.

    Then dμ1 is a critical value of Iμ1.

    Before proving Proposition 3.1, we analyze the (PS)dμ1 condition of Iμ1. Let U(x) be the unique positive solution of Δu+u=|u|p1u in H1(R3). We know that for any ε(0,1), there is a CC(ε)>0 such that U(x)Ce(1ε)|x|.

    Lemma 3.2. If the assumptions (A1)(A4) hold and 0<b<a<2, then the dμ1 defined in Proposition 3.1 satisfies dμ1<p12(p+1)Sp+1p1p+1.

    Proof. It suffices to find a path γ(t) starting from 0 such that

    supt[0,1]Iμ1(γ(t))<p12(p+1)Sp+1p1p+1.

    Define UR(x)=U(xRθ) with θ=(0,0,1). Note that for the UR defined as above, the Iμ1(tUR) as t+ and Iμ1(tUR)0 as t0. We know that there is a unique TR>0 such that tIμ1(tUR)|t=TR=0, which is

    UR2μ1h(x)U2Rdx+T2RF(UR)Tp1RUp+1Rdx=0.

    If TR0 as R, then UR2μ1h(x)U2Rdx0 as R, which is impossible. If TR as R, then as R,

    1T2R(UR2μ1h(x)U2Rdx)+F(UR)=Tp3RUp+1Rdx,

    which is impossible either. Hence we only need to estimate Iμ1(tUR) for t in a finite interval and we may write

    Iμ1(tUR)g(t)+CF(UR),

    where

    g(t)=t22(UR2μ1R3h(x)U2Rdx)|t|p+1p+1R3Up+1Rdx.

    Noting that under the assumptions (A1)(A4), we obtain that for R large enough,

    F(UR)(R3K(x)65U125Rdx)56(R3ϕ6URdx)16C(R3e65a|x+Rθ|(U(x))125dx)56C(R3e65aRe(65a125(1ε))|x|dx)56CeaR (28)

    since 0<a<2. We can also prove that

    R3h(x)U2Rdx=R3h(x+Rθ)U2(x)dxCR3eb|x+Rθ|U2(x)dxCR3eb|x|bRU2(x)dxCebRR3eb|x|U2(x)dxCebR. (29)

    It is now deduced from (28) and (29) that

    supt>0Iμ1(tUR)supt>0g(t)+CeaRp12(p+1)(UR2μ1R3h(x)U2Rdx)p+1p1(UR2Lp+1)p+1p1+CeaRp12(p+1)Sp+1p1p+1CebR+o(ebR)+CeaR<p12(p+1)Sp+1p1p+1

    for R large enough since 0<b<a. The proof is complete.

    Lemma 3.3. Under the assumptions (A1)(A4), Iμ1 satisfies (PS)d condition for any d<p12(p+1)Sp+1p1p+1.

    Proof. Let (un)nNH1(R3) be a (PS)d sequence of Iμ1 with d<p12(p+1)Sp+1p1p+1. Then we have that for n large enough,

    d+o(1)=12un2μ12R3h(x)u2ndx+14F(un)1p+1R3|un|p+1dx

    and

    Iμ1(un),un=un2μ1R3h(x)u2ndx+F(un)R3|un|p+1dx.

    Hence we can deduce that (un)nN is bounded in H1(R3). Going if necessary to a subsequence, we may assume that unu0 weakly in H1(R3) and unu0 a. e. in R3. Denote wn:=unu0. We then obtain from Brezis-Lieb lemma and Lemma 2.4 that for n large enough,

    un2=u02+wn2+o(1),F(un)=F(u0)+F(wn)+o(1)

    and

    unp+1Lp+1=u0p+1Lp+1+wnp+1Lp+1+o(1).

    Since R3h(x)u2ndxR3h(x)u20dx as n, we deduce that

    d+o(1)=Iμ1(un)=Iμ1(u0)+12wn2+14F(wn)1p+1R3|wn|p+1dx. (30)

    From Iμ1(un),ψ0 for any ψH1(R3), one may deduce that Iμ1(u0)=0. Therefore

    u02μ1R3h(x)u20dx+F(u0)=R3|u0|p+1dx

    and then

    Iμ1(u0)p12(p+1)(u02μ1R3h(x)u20dx)+p34(p+1)F(u0)0.

    Now using an argument similar to the proof of (16), we obtain that

    o(1)=wn2+F(wn)R3|wn|P+1dx. (31)

    By the relation u2Sp+1u2Lp+1 for any uH1(R3), we proceed our discussion according to the following two cases:

    (I) R3|wn|p+1dx0 as n;

    (II) R3|wn|p+1dx0 as n.

    Suppose that the case (I) occurs. Then up to a sbusequence, we may obtain from (31) that

    wn2Sp+1(wn2+F(wn)o(1))2p+1,

    which implies that for n large enough,

    wn2Sp+1p1p+1+o(1).

    It is deduced from this and (30) that dp12(p+1)Sp+1p1p+1, which is a contradiction. Therefore the case (II) must occur. This and (31) imply that wn0. Hence we have proven that Iμ1 satisfies (PS)d condition for any d<p12(p+1)Sp+1p1p+1.

    Proof of Proposition 3.1. Since 0 is a local minimizer of Iμ1 and for v0, Iμ1(sv) as s+, Lemma 3.2, Lemma 3.3 and the mountain pass theorem [3] imply that dμ1 is a critical value of Iμ1.

    Proof of Theorem 1.1. By Proposition 3.1, the dμ1 is a critical value of Iμ1 and dμ1>0. The proof of nonnegativity for at least one of the corresponding critical point is inspired by the idea of [1]. In fact, since Iμ1(u)=Iμ1(|u|) for any uH1(R3), for every nN, there exists γnΓ1 with γn(t)0 (a.e. in R3) for all t[0,1] such that

    dμ1maxt[0,1]Iμ1(γn(t))<dμ1+1n. (32)

    By Ekeland's variational principle [5], there exists γnΓ1 satisfying

    {dμ1maxt[0,1]Iμ1(γn(t))maxt[0,1]Iμ1(γn(t))<dμ1+1n;maxt[0,1]γn(t)γn(t)<1n; there existstn[0,1]such thatzn=γn(tn) satisfies:Iμ1(zn)=maxt[0,1]Iμ1(γn(t)),andIμ1(zn)1n. (33)

    By Lemma 3.2 and Lemma 3.3 we get a convergent subsequence (still denoted by (zn)nN). We may assume that znz in H1(R3) as n. On the other hand, by (33), we also arrive at γn(tn)z in H1(R3) as n. Since γn(t)0, we conclude that z0, z0 in R3 with Iμ1(z)>0 and it is a nonnegative bound state of (3) in the case of μ=μ1.

    In this section, we always assume the conditions (A1)(A4). We will prove the existence of ground state and bound states of (3) as well as their asymptotical behavior with respect to μ. We emphasize that if 0<μ<μ1, then one may consider a minimization problem like

    inf{Iμ(u) : uM},M={uH1(R3) : Iμ(u),u=0}

    to get a ground state solution. But for μμ1, we can not do like this because for μ>μ1, we do not know if 0M. To overcome this difficulty, we define the set of all nontrivial critical points of Iμ in H1(R3):

    N={uH1(R3){0}:Iμ(u)=0}.

    And then we consider the following minimization problem

    c0,μ=inf{Iμ(u):uN}. (34)

    Lemma 4.1. Let δ2 and ρ be as in Lemma 2.7 and μ(μ1,μ1+δ2). Define the following minimization problem

    d0,μ=infu<ρIμ(u).

    Then the d0,μ is achieved by a nonnegative function w0,μH1(R3). Moreover this w0,μ is a nonnegative solution of (3).

    Proof. Firstly, we prove that <d0,μ<0 for μ(μ1,μ1+δ2). Keeping the expression of Iμ(u) in mind, we obtain from the Sobolev inequality that

    Iμ(u)=12u2μ2R3h(x)u2dx+14F(u)1p+1R3|u|p+1dx12u2μ2μ1u2Cup+1>

    as u<ρ. Next, for any t>0, we have that

    Iμ(te1)=t22e12μt22R3h(x)e21dx+t44F(e1)tp+1p+1R3|e1|p+1dx.

    It is now deduced from μ1R3h(x)e21dx=e12 that

    Iμ(te1)=t22(1μμ1)e12+t44F(e1)tp+1p+1R3|e1|p+1dx.

    Since μ>μ1, we obtain that for t small enough, the Iμ(te1)<0. Thus we have proven that <d0,μ<0 for μ(μ1,μ1+δ2).

    Secondly, let (vn)nN be a minimizing sequence, that is, vn<ρ and Iμ(vn)d0,μ as n. By the Ekeland's variational principle, we can obtain that there is a sequence (un)nNH1(R3) with un<ρ such that as n,

    Iμ(un)d0,μandIμ(un)0.

    Then we can prove that (un)nN is bounded in H1(R3). Using Lemma 2.6, we obtain that (un)nN contains a convergent subsequence, still denoted by (un)nN, such that unu0 strongly in H1(R3). Noticing the fact that if (vn)nN is a minimizing sequence, then (|vn|)nN is also a minimizing sequence, we may assume that for each nN, the un0 in R3. Therefore we may assume that u00 in R3. The Iμ(un)0 and unu0 strongly in H1(R3) imply that Iμ(u0)=0. Hence choosing w0,μu0, we know that w0,μ is a nonnegative solution of the (3).

    We emphasize that the above lemma does NOT mean that w0,μ is a ground state of (3). But it does imply that N for any μ(μ1,μ1+δ2). Now we are in a position to prove that the c0,μ defined in (34) can be achieved.

    Lemma 4.2. For μ(μ1,μ1+δ2), the c0,μ is achieved by a nontrivial v0,μH1(R3), which is a nontrivial critical point of Iμ and hence a solution of the (3).

    Proof. By Lemma 4.1, we know that N for μ(μ1,μ1+δ2). Hence we have that c0,μ<0. Next we prove that the c0,μ>.

    For any uN, since Iμ(u)=0, then Iμ(u),u=0. Then we can deduce that

    Iμ(u)=Iμ(u)14Iμ(u),u14u2D(p,h)μp+1p1.

    Therefore the c0,μ>.

    Now let (un)nNN be a sequence such that

    Iμ(un)c0,μandIμ(un)=0.

    Since <c0,μ<0, we know from Lemma 2.6 that (un)nN contains a convergent subsequence in H1(R3) and then we may assume without loss of generality that unv0 strongly in H1(R3). Therefore we have that Iμ(v0)=c0,μ and Iμ(v0)=0. Choosing v0,μv0 and we finish the proof of the Lemma 4.2.

    Next, to analyze further the (PS)d condition of the functional Iμ, we have to prove a relation between the minimizer w0,μ obtained in Lemma 4.1 and the minimizer v0,μ obtained in Lemma 4.2.

    Lemma 4.3. There exists δ3(0,δ2] such that for any μ(μ1,μ1+δ3), the v0,μ obtained in Lemma 4.2 can be chosen to coincide the w0,μ obtained in Lemma 4.1.

    Proof. The proof is divided into two steps. In the first place, for u0 and Iμ1(u)=0, we have that

    u2μ1R3h(x)u2dx+F(u)=R3|u|p+1dx

    and hence

    Iμ1(u)=p12(p+1)(u2μ1R3h(x)u2dx)+p34(p+1)F(u).

    Since u2μ1R3h(x)u2dx for any uH1(R3), we obtain that

    Iμ1(u)p34(p+1)F(u)>0.

    In the second place, denoted by u0,μ a ground state obtained in Lemma 4.2. For any sequence μ(n)>μ1 and μ(n)μ1 as n, we have that u0,μ(n) satisfies

    Iμ(n)(u0,μ(n))=0

    and we also have that

    c0,μ(n)=Iμ(n)(u0,μ(n))<0.

    Hence we deduce that (u0,μ(n))nN is bounded in H1(R3). Since Iμ(n)(u0,μ(n))=0, one also has that

    Iμ(n)(u0,μ(n))=p12(p+1)(u0,μ(n)2μ(n)R3h(x)(u0,μ(n))2dx)+p34(p+1)F(u0,μ(n)).

    Using the definition of μ1, we obtain that, as n,

    u0,μ(n)2μ(n)R3h(x)(u0,μ(n))2dx(1μ(n)μ1)u0,μ(n)20

    because (u0,μ(n))nN is bounded in H1(R3). Next since (u0,μ(n))nN is bounded in H1(R3), we may assume without loss of generality that u0,μ(n)˜u0 weakly in H1(R3).

    Claim. As n, the u0,μ(n)˜u0 strongly in H1(R3) and ˜u0=0.

    Proof of the Claim. From u0,μ(n)˜u0 weakly in H1(R3), we may assume that u0,μ(n)˜u0 a. e. in R3. Using these and the fact of Iμ(n)(u0,μ(n))=0, we deduce that Iμ1(˜u0)=0. Then similar to the proof in Lemma 2.6, we obtain that

    o(1)+Iμ(n)(u0,μ(n))=Iμ(n)(˜u0)+12˜wn2+14F(˜wn)1p+1R3|˜wn|p+1dx, (35)

    where ˜wn:=u0,μ(n)˜u0.

    Now we distinguish two cases:

    (i) R3|˜wn|p+1dx0 as n;

    (ii) R3|˜wn|p+1dx0 as n.

    Suppose that the case (ⅰ) occurs. We may deduce from a proof similar to Lemma 2.6 that

    Iμ(n)(u0,μ(n))+o(1)Iμ1(˜u0)+p12(p+1)Sp+1p1p+1,

    which is a contradiction because Iμ1(˜u0)>p12(p+1)Sp+1p1p+1 by Lemma 2.5 and the fact of Iμ(n)(u0,μ(n)) <0. Therefore the case (ii) occurs, which implies that u0,μ(n)˜u0 strongly in H1(R3) (the proof is similar to those in Lemma 2.6). From this we also deduce that F(˜wn)F(˜u0).

    Next we prove that ˜u0=0. Arguing by a contradiction, if ˜u00, then we know from Iμ(n)(u0,μ(n))=0 that

    lim infnIμ(n)(u0,μ(n))p34(p+1)F(˜u0)>0,

    which is also a contradiction since Iμ(n)(u0,μ(n))<0. Therefore ˜u0=0.

    Hence there is δ3(0,δ2] such that for any μ(μ1,μ1+δ3), u0,μ<ρ, which implies that c0,μ=d0,μ. Using Lemma 4.1, we can get a nonnegative ground state of (3), called w0,μ and c0,μ=d0,μ=Iμ(w0,μ). The proof is complete.

    Remark 4.4. The proof of Lemma 4.3 implies that (1) of Theorem 1.2 holds.

    In the following, we are going to prove the existence of another nonnegative bound state solution of (3). To obtain this goal, we have to analyze further the (PS)d condition of the functional Iμ.

    Lemma 4.5. Under the assumptions of (A1)(A4), if μ(μ1,μ1+δ3), then Iμ satisfies (PS)d condition for any d<c0,μ+p12(p+1)Sp+1p1p+1.

    Proof. Let (un)nNH1(R3) be a (PS)d sequence of Iμ with d<c0,μ+p12(p+1)Sp+1p1p+1. Then we have that for n large enough,

    d+o(1)=12un2μ2R3h(x)u2ndx+14F(un)1p+1R3|un|p+1dx

    and

    Iμ(un),un=un2μR3h(x)u2ndx+F(un)R3|un|p+1dx.

    Similar to the proof in Lemma 2.3, we can deduce that (un)nN is bounded in H1(R3). Going if necessary to a subsequence, we may assume that unu0 weakly in H1(R3) and unu0 a. e. in R3. Denote wn:=unu0. We then obtain from Brezis-Lieb lemma and Lemma 2.4 that for n large enough,

    un2=u02+wn2+o(1),
    F(un)=F(u0)+F(wn)+o(1)

    and

    unp+1Lp+1=u0p+1Lp+1+wnp+1Lp+1+o(1).

    Using Lemma 2.1, we also have that R3h(x)u2ndxR3h(x)u20dx as n. Therefore we deduce that

    d+o(1)=Iμ(un)=Iμ(u0)+12wn2+14F(wn)1p+1R3|wn|p+1dx. (36)

    Since Iμ(un),ψ0 for any ψH1(R3), we know that Iμ(u0)=0. Moreover we have that

    Iμ(u0)c0,μ

    and

    u02μR3h(x)u20dx+R3ϕu0u20=R3|u0|p+1dx.

    Note that (un)nN is bounded in H1(R3). The Brezis-Lieb lemma, Lemma 2.4 and

    o(1)=un2μR3h(x)u2ndx+F(un)R3|un|p+1dx

    imply that

    o(1)=wn2+F(wn)R3|wn|p+1dx. (37)

    Using u2Sp+1u2Lp+1 for any uH1(R3), we distinguish two cases:

    (I) R3|wn|p+1dx0 as n;

    (II) R3|wn|p+1dx0 as n.

    Suppose (I) occurs. Up to a subsequence, we may obtain from (37) that

    wn2Sp+1(wn2+F(wn)o(1))2p+1.

    Hence we get that for n large enough,

    wn2Sp+1p1p+1+o(1). (38)

    Therefore using (36) and (38), we deduce that for n large enough,

    d+o(1)=Iμ(un)=Iμ(u0)+12wn2+14F(wn)1p+1R3|wn|p+1dx=Iμ(u0)+p12(p+1)wn2+p34(p+1)F(wn)c0,μ+p12(p+1)wn2+p34(p+1)F(wn)>c0,μ+p12(p+1)Sp+1p1p+1, (39)

    which contradicts to the assumption d<c0,μ+p12(p+1)Sp+1p1p+1. Therefore the case (Ⅱ) must occur, i.e., R3|wn|p+1dx0 as n. This and (37) imply that wn0. Hence we have proven that Iμ satisfies (PS)d condition for any d<c0,μ+p12(p+1)Sp+1p1p+1.

    Next, for the w0,μ obtained in Lemma 4.3, we define

    d2,μ=infγΓ2supt[0,1]Iμ(γ(t))

    with

    Γ2={γC([0,1],H1(R3)) : γ(0)=w0,μ, Iμ(γ(1))<c0,μ}.

    Lemma 4.6. Suppose that the conditions (A1)(A4) hold and 0<b<a<1. If μ(μ1,μ1+δ3), then

    d2,μ<c0,μ+p12(p+1)Sp+1p1p+1.

    Proof. It suffices to find a path starting from w0,μ and the maximum of the energy functional over this path is strictly less than c0,μ+p12(p+1)Sp+1p1p+1. To simplify the notation, we denote w0:=w0,μ, which corresponds to the critical value c0,μ. We will prove that there is a T0 such that the path γ(t)=w0+tT0UR is what we need, here UR(x)U(xRθ) is defined as before. Similar to the discussion in the proof of Lemma 3.2, we only need to estimate Iμ(w0+tUR) for positive t in a finite interval. By direct calculation, we have that

    Iμ(w0+tUR)=12(w0+tUR2μR3h(x)|w0+tUR|2dx)+14F(w0+tUR)1p+1R3|w0+tUR|p+1dx=Iμ(w0)+A1+A2+A3+t22UR2μ2R3h(x)U2Rdx,

    where

    A1=w0,tURμtR3h(x)w0URdx,
    A2=14(F(w0+tUR)F(w0))

    and

    A3=1p+1R3(|w0|p+1|w0+tUR|p+1)dx.

    Since is a solution of (3), we have that

    From an elementary inequality:

    we deduce that

    For the estimate of , using the expression of and the symmetry property of the integral with respect to and , we can obtain that

    Since is a nonnegative solution of (3) and , we obtain from the assumption on that

    Similarly we can deduce that for large enough,

    Since for large enough, we obtain that

    for large enough since . The proof is complete.

    Proposition 4.7. Under the conditions (A1)-(A4), if and be the minimizer obtained in Lemma 4.3, then the is a critical value of .

    Proof. Since for , we know from Lemma 4.1 and Lemma 4.3 that the is a local minimizer of . Moreover, one has that as . Therefore Lemma 4.5, Lemma 4.7 and the mountain pass theorem of [3] imply that is a critical value of .

    Proof of Theorem 1.2. The conclusion (1) of Theorem 1.2 follows from Lemma 4.3 and Remark 4.4. It remains to prove (2) of Theorem 1.2. By Proposition 4.7, the is a critical value of and . The proof of nonnegativity for at least one of the corresponding critical point is inspired by the idea of [1]. In fact, since for any , for every , there exists with (a.e. in ) for all such that

    (40)

    By Ekeland's variational principle, there exists satisfying

    (41)

    By Lemma 4.6 we get a convergent subsequence (still denoted by ). We may assume that strongly in as . On the other hand, by (41), we also arrive at strongly in as . Since , we conclude that , in with and it is a nonnegative solution of problem (3).

    Next, let be the nonnegative solution given by the above proof, that is, and . We claim that for any sequence and , there exist a sequence of solution of (3) with and a with such that strongly in . In fact, denoted by the minimizer corresponding to , according to the definition of and the proof of Lemma 4.6, we deduce that for large enough,

    and

    (42)

    Next, similar to the proof in Lemma 2.3, we can deduce that is bounded in . Going if necessary to a subsequence, we may assume that weakly in and a. e. in . Then we have that . Moreover . If does not converge strongly to in , then using an argument similar to the proof of Lemma 4.5, we may deduce that

    which contradicts to (42). Hence strongly in and hence . The proof is complete by choosing .

    The author thanks the unknown referee for helpful comments.



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