Research article

Normalized ground states for fractional Kirchhoff equations with critical or supercritical nonlinearity

  • Received: 31 December 2021 Revised: 28 February 2022 Accepted: 15 March 2022 Published: 31 March 2022
  • MSC : 35J65, 47J05, 47J30

  • The aim of this paper is to study the existence of ground states for a class of fractional Kirchhoff type equations with critical or supercritical nonlinearity

    (a+bR3|()s2u|2dx)()su=λu+|u|q2u+μ|u|p2u, xR3,

    with prescribed L2-norm mass

    R3u2dx=c2

    where s(34, 1), a,b,c>0, 6+8s3<q<2s, p2s (2s=632s), μ>0 and λR as a Langrange multiplier. By combining an appropriate truncation argument with Moser iteration method, we prove that the existence of normalized solutions for the above equation when the parameter μ is sufficiently small.

    Citation: Huanhuan Wang, Kexin Ouyang, Huiqin Lu. Normalized ground states for fractional Kirchhoff equations with critical or supercritical nonlinearity[J]. AIMS Mathematics, 2022, 7(6): 10790-10806. doi: 10.3934/math.2022603

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  • The aim of this paper is to study the existence of ground states for a class of fractional Kirchhoff type equations with critical or supercritical nonlinearity

    (a+bR3|()s2u|2dx)()su=λu+|u|q2u+μ|u|p2u, xR3,

    with prescribed L2-norm mass

    R3u2dx=c2

    where s(34, 1), a,b,c>0, 6+8s3<q<2s, p2s (2s=632s), μ>0 and λR as a Langrange multiplier. By combining an appropriate truncation argument with Moser iteration method, we prove that the existence of normalized solutions for the above equation when the parameter μ is sufficiently small.



    In this paper, we study mainly the existence of ground states to the Kirchhoff type problem with critical or supercritical nonlinearity

    {(a+bR3|()s2u|2dx)()su=λu+|u|q2u+μ|u|p2u, xR3,R3u2=c2, (1.1)

    where  s(34, 1), a, b, c>0, 6+8s3< q< 2s, p2s (2s=632s), μ>0 is a real parameter, and ()s denotes the fractional Laplacian operator.

    The operator ()s can be seen as the infinitesimal generators of Lévy stable diffusion processes, see [1,2] for example. This operator appears in several areas such as biology, chemistry and physics (see [3,4,5,6]). Problem (1.1) is viewed as being nonlocal because of the appearance of the term bR3|()s2u|2, which implies that Eq (1.1) is no longer a pointwise identity. This also results in lack of weak sequential continuity of the energy function associated to (1.1), so it make the study of (1.1) particularly interesting. Over the last decade, many mathematicians were particularly keen on the study of nonlinear equations involving nonlocal operators, we can look it up in [7,8,9,10,11,12,13,14] and the references therein.

    It is well known that problem (1.1) arises from looking for the standing wave type solutions φ(x,t)=eiλtu(x), λR for the following time-dependent nonlinear fractional Kirchhoff equation

    iφt=(a+bR3|()s2φ|2dx)()sφf(|φ|)φ, xR3, (1.2)

    where 0<s<1, i denotes the imaginary unit. The stationary case of (1.2) is the following equation

    (a+bR3|()s2φ|2dx)()sφ=f(|φ|)φ, xR3. (1.3)

    Clearly, φ solves (1.2) if and only if the stand wave u(x) satisfies (1.1) with f(u)=|u|q2+μ|u|p2. Alternatively one can consider the existence of normalized solutions to (1.1), that is, solutions with prescribed L2-norm. Since solutions φC([0,1], Hs(R3)) to (1.2) maintain their mass along time (In fact, multiplying (1.2) by the conjugate φ¯ of φ, integrating over R3, and taking the imaginary part, we get ddt|φ(t)|22=0, t[0,T].), it is natural and interesting, from a physical point view, to search for such solutions.

    When s=1, Problem (1.3) becomes the Kirchhoff equation. In the past several years, the Kirchhoff type equations has been studied extensively by many researchers(see [15,16,17,18,19,20,21,22,23]). For all we know, the existence results to problem (1.1) have been mostly available for the case where p,q(2,2s) and λ is fixed and assigned. When a=1, b=0, s=1 and μ=0, i.e., for the Laplacian operator, Jeanjean's [24] was the first paper to prove existence of normalized solutions in purely L2-supercritical case. Li and Ye in [25] considered problem (1.1) with s=1,μ=0,N=3,λ=1,q(3,6) and proved that (1.1) has at least one least energy solution by dealing with a constrained minimization problem on a manifold of H1(R3), which is obtained by combining the Nehari manifold and the corresponding Pohozaev identity. Liu, Chen and Yang in [26] considered problem (1.1) with 2<q<p<2s and proved some existence results about the normalized solutions. However, there is few literature concerned about the normalized solutions for fractional Kirchhoff equation with critical or supercritical nonlinearity. With regard to the point, we attempt to study this kind of problem in this paper.

    It is well known that the fractional order Sobolev space H:=Hs(R3) can be defined as follows

    Hs(R3)={uL2(R3):R6|u(x)u(y)|2|xy|3+2sdxdy<+},

    endowed with the norm

    uH=(R6|u(x)u(y)|2|xy|3+2sdxdy+R3|u|2dx)12,

    and the inner product is

    (u,v)H=R6(u(x)u(y))(v(x)v(y))|xy|3+2sdxdy+R3uvdx.

    According to [26], we know that

    u=(R3(|()s2u|2dx+R3|u|2dx)12

    is also a norm on Hs(R3) which is equivalent to uH. Moreover, we define Hrs(R3):={uHs(R3):u(x)=u(|x|),xR3}.

    Let H=H×R with the scalar product

    (,)H=(,)H+(,)R

    and the corresponding norm

    H2=H2+||R2.

    Let t be the usual norm of space Lt(R3) where 2t. H is continuously embedding into Lt(R3) for t[2,2s] and there exists a best constant Ss such that

    Ss=infuH,u0R3|()s2u|2dxu2s2. (1.4)

    The normalized weak solution for the problem (1.1) is obtained by looking for critical points of the following C1 functional

    Jμ(u)=a2R3|()s2u|2+b4(R3|()s2u|2)21qR3|u|qμpR3|u|p

    constrained on the L2-spheres in H:

    S(c)={uH| u2=c>0}.

    uc is called a ground state of (1.1) on S(c) if

    dJμ|S(c)(uc)=0 and  Jμ(uc)=infuS(c){Jμ(u): dJμ|S(c)(u)=0}.

    Since p2s, the functional Jμ is not well defined on Hs(R3) unless p=2s. Moreover, we need to overcome the lack of compactness in studying critical and supercritical growth. Hence, we cannot directly use variational methods to prove the existence of normalized solutions. To overcome these difficulties, we use a new method, which came from [14,18]. The main idea of this method is to reduce the supercritical problem into a subcritical one. In comparison with previous works, this paper has several new features. Firstly, we consider the nonlinear term with supcritical growth. Secondly, we give the existence of normalized solution for the appropriate truncation problem of (1.1). Finally, the existence of a normalized ground state solution is obtained by Moser iteration method. The results in this paper extend the results in paper [4,24,26]. There have been no previous studies considering the existence of normalized ground state solutions for problem (1.1) involving supcritical growth to the best of our knowledge.

    Our main result is the following:

    Theorem 1.1. For any c>0, there exists a μ>0 such that, problem (1.1) has a couple of solutions (uc, λc)Hs(R3)×R for any μ(0,μ]. Moreover, uc is a positive ground state, radially symmetric function and λc<0.

    Remark 1.2. When 6+8s3<q<2s, Jμ is not bounded from below on S(c), i.e., infuS(c)Jμ(u)=. So, the minimization problem constrained on S(c) does not work. We try to look for a critical point with a minimax characterization. Although Jμ has a mountain-pass geometry on S(c), the boundedness of the obtained Palais-Smale sequence is not yet clear. Motivated by [4], we try to construct an auxiliary map Iμ, which on S(c)×R has the same type of geometric structure as Jμ on S(c). Besides, the Palais-Smale sequence of Iμ satisfies the additional condition, which is the key point to obtain the boundedness of the Palais-Smale sequence.

    In this section, we give a truncation argument in order to overcome the lack of compactness in studying critical and supercritical growth. Let M>0 be a constant. For fixed c>0, μ>0, M>0, we investigate the existence of ground state for the following truncation problem

    {(a+bR3|()s2u|2dx)()su=λu+|u|q2u+μdM(u), xR3,R3u2=c2, (2.1)

    where s(34, 1), a,b>0, 6+8s3< q< 2s, p2s (2s=632s),  and

    dM(t)={|t|p2t,|t|M,Mpq|t|q2t,|t|>M,

    To investigate (2.1), we define the the energy functional Eμ:HR by

    Eμ(u)=a2R3|()s2u|2dx+b4(R3|()s2u|2dx)21qR3|u|qdxμR3DM(u)dx, (2.2)

    where DM(t)0tdM(τ)dτ. It is easy to obtain that EμC1(H,R) and

    Eμ(u),v=(a+bR3|()s2u|2dx)R3()s2u()s2vdxR3|u|q2uvdxμR3dM(u)vdx (2.3)

    for all u,vH.

    Theorem 2.1. For any c>0  and M>0, there exists a μ1>0, such that, problem (2.1) has a couple of solutions (uc, λc)Hrs(R3)×R for any μ(0,μ1]. Moreover, uc is a positive ground state, λc<0 and Eμ(uc)=mc,μ, where

    mc,μ:=infuV(c)Eμ(u)

    and V(c) is the Pohozaev manifold defined in lemma 2.4.

    Next, we give some useful preliminary lemmas to prove Theorem 2.1.

    Lemma 2.1. [8] If α(2, 2s), there exists an optimal constant C(s,α) such that for any uH,

    R3|u|αC(s,α)(R3|()s2u|2)αβαs(R3|u|2)ααβαs, (2.4)

    where βα:=3(α2)2α.

    Lemma 2.2. [19] Hrs(R3) is compactly embedding into Lt(R3) for t(2,2s).

    As in [4], we introduce the useful fiber map preserving the L2-norm, that is,

    (τu)(x):=e32τu(eτx),  for a.e. xR3. (2.5)

    Define the auxiliary functional I:HR by

    Iμ(u,τ):=Eμ(τu)=a2e2sτR3|()s2u|2+b4e4sτ(R3|()s2u|2)21qeqβqτR3|u|qdxμerβrτR3DM(u)dx, (2.6)

    where

    r={p,|u|M,q,|u|>M, (2.7)

    then we can obtain that Iμ is a C1-functional.

    Lemma 2.3. [13] The map (u,τ)HτuH is continuous.

    Similar to Lemma 2.1 in [4], we can easily get the following lemma.

    Lemma 2.4. Let (u,λ)S(c)×R be a weak solution of Eq (2.2). Then u belongs to the set

    V(c):={uS(c): Pμ(u)=0}

    where

    Pμ(u)=aR3|()s2u|2dx+b(R3|()s2u|2dx)2βqsR3|u|qdxμβrrsR3DM(u)dx. (2.8)

    Lemma 2.5. For any uS(c), τR is a critical point for Φu(τ):=Iμ(u,τ) if and only if τuV(c).

    Proof. For any uS(c) and τR, we have

    (Φu)(τ)=ase2sτR3|()s2u|2+bse4sτ(R3|()s2u|2)2βqeβqqτR3|u|qμβrreβrrτR3DM(u)=asR3|()s2(τu)|2+bs(R3|()s2(τu)|2)2βqR3|(τu)|qμβrrR3DM(τu)=sPμ(τu). (2.9)

    It is easy to see that Lemma 2.5 holds.

    Lemma 2.6. Let uS(c) be arbitrary fixed, then

    (1) R3|()s2(τu)|20 and Iμ(u,τ)0 as τ;

    (2) R3|()s2(τu)|2+ and Iμ(u,τ) as τ+.

    Proof. For fixed uS(c), we can easily get the conclusions (1) and (2) from the facts

    R3|()s2(τu)|2=e2sτR3|()s2u|2,
    Iμ(u,τ)=a2e2sτR3|()s2u|2+b4e4sτ(R3|()s2u|2)2eβqqτqR3|u|qμeβrrτR3DM(u)dx

    and βrrβqq>4s.

    Lemma 2.7. For every uS(c), there exists a unique τuR such that τuuV(c), where τu is a strict maximum point for Φu(τ) and Φu(τu)>0.

    Proof. For uS(c) and τR, by (2.9) we have

    (Φu)(τ)=2as2R3|()s2(τu)|2+4bs2(R3|()s2(τu)|2)2qβq2R3|(τu)|qμ(βrr)2R3DM(τu). (2.10)

    Since rβrqβq>4s, it is easy to see that (Φu)(τ)>0 as τ, and (Φu)(τ)<0 as τ. So, there exists τuR such that (Φu)(τu)=0. From Lemma 2.5, τuuV(c).

    Combining with (Φu)(τu)=0, (2.9) and (2.10), we have

    (Φu)(τu)=2as2R3|()s2(τu)|2βq(qβq4s)R3|(τu)|qμrβr(rβr4s)R3DM(τu)<0,

    which together with Lemma 2.6 implies that τu is unique and it is a strict global maximum point for Φu(τ) and Φu(τu)>0.

    As in [4], firstly, we prove that Eμ(u) has the mountain pass geometry on S(c)×R in the following lemma.

    Lemma 3.1. There exists kc>0 such that

    Pμ(u), Eμ(u)>0 for all  uAc,and0<supuAcEμ(u)<infuBcEμ(u)

    with

    Ac={uS(c):R3|()s2u|2kc}, Bc={uS(c):R3|()s2u|2=2kc}.

    Proof. Let k>0 be arbitrary fixed and suppose u,vS(c) are such that

    R3|()s2u|2k, and R3|()s2v|2=2k.

    Then for k small enough, by (2.4) and 3(r2)2s3(q2)2s>4, there exist constants C1 and C2 such that

    Pμ(u)aR3|()s2u|2+b(R3|()s2u|2)2C1(R3|()s2u|2)3(q2)2sC2(R3|()s2u|2)3(r2)2s,
    Eμ(u)a2R3|()s2u|2+b4(R3|()s2u|2)2C1(R3|()s2u|2)3(q2)2sC2(R3|()s2u|2)3(r2)2s

    and

    Eμ(v)Eμ(u)a2R3|()s2v|2+b4(R3|()s2v|2)21qR3|v|qμR3DM(v)a2R3|()s2u|2b4(R3|()s2u|2)2ak+bk2a2kb4k2C1(R3|()s2v|2)3(q2)2sC2(R3|()s2v|2)3(r2)2sak2+ak22C1k3(q2)2sC2k3(r2)2s.

    By the above inequalities, we can obtain that there exists kc>0 sufficiently small such that Lemma 3.2 holds.

    Next, we need to construct the minimax characterization of Iμ and Eμ.

    Lemma 3.2. Let

    γ~c,μ :=infh~Γ~cmaxt[0,1]Iμ(h~(t))

    with

    Γ~c={h~C([0,1],S(c)×R):h~(0)(Ac,0),h~(1)(Eμ0,0)}, Eμ0:={uS(c): Eμ(u)0}

    and

    γc,μ :=infhΓcmaxt[0,1]Eμ(h(t))

    with

    Γc={hC([0,1],S(c)):h(0)Ac, h(1)Eμ0},

    then we have

    γ~c,μ=γc,μ=mc,μ>0.

    Proof. Firstly, we prove that γ~c,μ=γc,μ.

    For any h~Γ~c, we can write it into

    h~(t)=(h~1(t),h~2(t))S(c)×R.

    We set h(t)=h~2(t)h~1(t), then h(t)Γc, and

    maxt[0,1]Iμ(h~(t))=maxt[0,1]Eμ(h~2(t)h~1(t))=maxt[0,1]Eμ(h(t)),

    which implies γ~c,μγc,μ. On the other hand, for any hΓc, if we set h~(t)=(h(t),0), then we get h~Γ~c and

    maxt[0,1]Iμ(h~(t))=maxt[0,1]Eμ(h(t)).

    This infers that γ~c,μγc,μ. So, γ~c,μ=γc,μ.

    Secondly, we claim that for uS(c), Eμ(u)0 implies Pμ(u)<0.

    For uS(c), if Eμ(u)0, then Φu(0)0. By the proof of Lemma 2.7 and Lemma 2.6, we easily see that τu<0, so

    Pμ(u)=Pμ(0u)=1s(Φu)(0)<1s(Φu)(τu)=0.

    That is

    Eμ(u)0Pμ(u)<0. (3.1)

    Next, we prove that mc,μ=γc,μ.

    For any uV(c), by Lemma 2.6 and Lemma 2.3, there exists t1 and t+1 such that

    hu:τ[0,1]((1τ)t+τt+)u andhuΓc.

    By Lemma 2.7, we have maxτ[0,1]Eμ(hu(τ))=Eμ(u). So we have mc,μγc,μ. On the other hand, for any h~(τ)=(h~1(τ),h~2(τ))Γ~c, we know that h2~(0)h1~(0)=h1~(0)Ac, h~2(1)h~1(1)=h~1(1)Eμ0. Hence by Lemma 3.1, we can deduce that

    Pμ(h~2(0)h~1(0))>0,

    and using (3.1),

    Pμ(h~2(1)h~1(1))<0.

    From Lemma 2.3, the function P~μ(τ):=Pμ(h~2(τ)h~1(τ)) is continuous in [0,1]. Therefore, there exists τ¯(0,1) such that P~μ(τ¯)=0, which implies that h~2(τ¯)h~1(τ¯)V(c), and

    maxτ[0,1]I(h~(τ))=maxτ[0,1]Eμ(h~2(τ)h~1(τ))infuV(c)Eμ(u),

    which implies that γc,μ=γ~c,μmc,μ. So, mc,μ=γc,μ.

    Finally, we prove that mc,μ>0.

    For any uV(c), then Pμ(u)=0. By (2.4), we have

    aR3|()s2u|2dx+b(R3|()s2u|2dx)2C1(R3|()s2u|2dx)qβqs+C2(R3|()s2u|2dx)rβrs,

    noticing that rβrqβq>4s, there exists δ>0 such that infuV(c)R3|()s2u|2dxδ, and

    Eμ(u)=Eμ(u)14Pμ(u)=a4R3|()s2u|2dx+(βq4s1q)R3|u|qdx+μ(rβr4s1)R3DM(u)dxa4δ.

    Thus, mc,μ>0.

    Remark 3.3. Let

    ρc:=infhΓc0maxt[0,1]E0(h(t))

    with

    Γc0={hC([0,1],S(c)):h(0)Ac, h(1)E0}, E0:={uS(c): E0(u)0}.

    Obviously, Γc0Γc, E0(u)Eμ(u) for uS(c). Thus we can deduce that ρc is independent of positive numbers μ,M and ρcγc,μ for any μ>0.

    In the following lemma, we give the relationship between the Palais-Smale sequence for I and that of Eμ.

    Lemma 3.4. Let γ~c,μ and γc,μ be defined in Lemma 3.2. Then there exist a sequence {(vn,τn)}S(c)×R such that for n, we have

    (1) Iμ(vn,τn)γ~c,μ,

    (2) (Iμ)|S(c)×R(vn,τn)0, i.e., it holds that

    τIμ(vn,τn)0

    and

    uIμ(vn,τn),φ~0

    with

    φ~Tvn:={φH:R3vnφ=0}.

    In addition, setting un(x)=τnvn(x), then for n, we get

    (i) Eμ(un)γc,μ,

    (ii) Pμ(un)0,

    (iii) Eμ|S(c)(un)0, i.e., it holds that

    Eμ(un),φ0

    with

    φTun:={φH:R3unφ=0}.

    Proof. According to the construction of γc,μ~, we know that the conclusions (1) and (2) follow directly from the Ekeland's Variational Principle [8, Proposition 2.2]. Next we mainly show (i)–(iii).

    For (i), by Lemma 3.2, γ~c,μ=γc,μ. we notice that

    Eμ(un)=Eμ(τnvn)=Iμ(vn,τn),

    thus (i) holds.

    By (2.9), we can get that τIμ(vn,τn)=sPμ(τnvn). Thus, (ii) is a consequence of τIμ(vn,τn)0 as n.

    For the proof of (iii), by the definition of Iμ, we have

    uIμ(vn,τn),φ~=ae2sτnR6(vn(x)vn(y))(φ~(x)φ~(y))|xy|3+2sdxdy+be4sτnR6(vn(x)vn(y))2|xy|3+2sdxdyR6(vn(x)vn(y))(φ~(x)φ~(y))|xy|3+2sdxdyeβqqτnR3|vn|q2vnφ~μeβrrτR3dM(vn)φ~,

    where φ~Tvn.

    On the other hand, for any φ with satisfying φTun, by using (2.3), we have

      Eμ(un),φ=aR6(un(x)un(y))(φ(x)φ(y))|xy|3+2sdxdy+bR6(un(x)un(y))2|xy|3+2sdxdyR6(un(x)un(y))(φ(x)φ(y))|xy|3+2sdxdyR3|un|q2unφdxμR3dM(un)φdx=aR6e32τn(vn(eτnx)vn(eτny))(φ(x)φ(y))|xy|3+2sdxdy+bR6e3τn(vn(eτnx)vn(eτny))2|xy|3+2sdxdyR6e32τn(vn(eτnx)vn(eτny))(φ(x)φ(y))|xy|3+2sdxdyR3|e32τnvn(eτnx)|q2e32τnvn(eτnx)φ(x)dxμR3dM(e32τnvn(eτnx))φ(x)dx=ae2sτnR6(vn(x)vn(y))(e32τnφ(eτnx)e32τnφ(eτny))|xy|3+2sdxdy+be4sτnR6(vn(x)vn(y))2|xy|3+2sdxdyR6(vn(x)vn(y))(e32τnφ(eτnx)e32τnφ(eτny))|xy|3+2sdxdyeqβqτnR3|vn(x)|q2vn(x)e32τnφ(eτnx)dxμerβrτnR3dM(vn)e32τnφ(eτnx)dx

    Setting

    φ~(x)=e32τnφ(eτnx),

    we get (iii) if we could show that φ~Tvn. In fact, φ~Tvn follows from the following equalities

    0=R3unφ=R3e32τnvn(eτnx)φ(x)=R3vn(x)e32τnφ(eτnx)=R3vnφ~.

    According to Lemma 3.4 and Lemma 3.2, there exist a Palais-Smale sequence {un}S(c) for Eμ|S(c) at level γc,μ>0, and it satisfies Pμ(un)0 as n. By applying the Lagrange multipliers rule there exists {λn}R such that

    (a+bR3|()s2un|2)()sun|un|q2unμdM(u)=λnun+o(1), as n. (4.1)

    (1). As Pμ(un)0, we have

    aR3|()s2un|2+b(R3|()s2un|2)2=βqsR3|un|q+μβrrsR3DM(un)+o(1)as  n. (4.2)

    Thus, by (4.2) we deduce that

    Eμ(un)+on(1)=a2R3|()s2un|2+b4(R3|()s2un|2)2μR3DM(un)1qsβq[aR3|()s2un|2+b(R3|()s2un|2)2μβrrsR3DM(un)]+o(1)=a(12sβqq)R3|()s2un|2+b(14sβqq)(R3|()s2un|2)2+μ(βrrβqq1)R3DM(un)γc,μ+1 (4.3)

    Since 6+8s3<q<2s and (2.7), it implies that sβqq<14 and βrrβqq. According to (4.3), we can deduce the boundedness of R3|()s2un|2, thus {un} is bounded in H.

    (2). According to Lemma 2.2, we know that the embedding Hrs(R3)Lt(R3) is compact for t(2, 2s), and we can deduce that there exists ucHrs(R3) such that, up to a subsequence, unuc weakly in H, unuc strongly in Lq(R3) for q(6+8s3,2s). since {un}S(c) is bounded in H. By (4.1), we obtain that

    λnc2=aR3|()s2un|2+b(R3|()s2un|2)2R3|un|qμrR3DM(un)+on(1). (4.4)

    Using the fact that the boundedness of {un} in H and (4.2), we can deduce that {λn} is bounded. Hence, up to a subsequence λnλcR.

    (3). We claim that uc0. We assume by contradiction that uc0, by (4.2) we deduce that R3|()s2un|20. Recalling that Pμ(un)0, according to (4.3), we have Eμ(un)0, which is a contradiction to the assumption that Eμ(un)γc,μ0. Now, since λnλc and unuc0 weakly in H, together with (4.1), we know (uc, λc) is a couple of solutions to (2.1). By the Pohozaev identity, we obtain

    32s2aR3|()s2uc|2+32s2b(R3|()s2uc|2)2=3(R312λc|uc|2+1q|uc|q+μDM(uc)).

    Combining with the (4.4) for uc, we get

    λcc2=λc|uc|22=(32s)(q2s)3(q2)[aR3|()s2uc|2+b(R3|()s2uc|2)2]+μ2(rq)(q2)R3DM(uc) (4.5)

    Since 6+8s3<q<2s and (4.5), there exists μ1>0 such that λc<0 for μ(0,μ1].

    (4). Testing (4.1) and (2.1) with unuc, we can obtain that

    Eμ(un)Eμ(uc),unucλcR3|unuc|2=on(1).

    Using the strong Lp convergence of un, we infer that

    aR3|()s2(unuc)|2+b(R3|()s2(unuc)|2)2λcR3|unuc|2=on(1),

    which, being λc<0, implies unuc strongly in H. Therefore, Eμ(un)Eμ(uc), as n. From Lemma 2.4 and Lemma 3.2, we easily obtain that uc is a ground state of (2.1) and Eμ(uc)=mc,μ.

    In this section, we devote to complete the proof of Theorem 1.1. From the truncation argument in Sections 2–4, we can see that if the ground state uc of (2.1) satisfy ucM. Then ucH is a ground state of (1.1).

    Lemma 5.1. Let(uc, λc) be a couple of solutions of problem (2.1) for μ(0,μ1], then there exists a constant Kc>0 independent of μ,M>0 such that ucKc.

    Proof. By Theorem 2.1 and Lemma 2.4, it is easy to see that

    Eμ(uc)=γc,μ  and  Pμ(uc)=0, (5.1)

    It follows from (5.1) and Remark 3.3 that

    ρcγc,μEμ(uc)14Pμ(uc)a4uc2

    Consequently, there exists a constant Kc>0 independent of μ,M>0 such that ucKc.

    Lemma 5.2. If (uc, λc) be a couple of solutions of problem (2.1) for μ(0,μ1], then ucL(R3), and there exists a constant Bc>0 independent μ,M>0 such that

    ucBc(1+μ12sqMpq2sq).

    Proof. For convenience, we replace uc with u in the following. Let L>0 and β>1, we first define the following functions:

    Υ(u)=uuL2(β1)H,

    where uL=min{u,L}. Since Υ is an increasing function, we have

    (xy)[Υ(x)Υ(y]0,  x,yR.

    Let Φ(t)=12|t|2 and Ψ(t)=0t(Υ(τ))12dτ.Then, if x>y, by Cauchy-Schwarz inequality, we have

    Φ(xy)[Υ(x)Υ(y)]=(xy)[Υ(x)Υ(y)]=(xy)yxΥ(t)dt=(xy)yx(Ψ(t))2dt(yxΨ(t)dt)2=|Ψ(x)Ψ(y)|2.

    The same arguments hold for xy. Therefore,

    Φ(xy)[Υ(x)Υ(y)]|Ψ(x)Ψ(y)|2,  x,yR. (5.2)

    By the definition of uL, it is easy to see that |uuL2(β1)|L2(β1)u and Υ(u)H. Taking Υ(u) as a test function in Eq (2.1), and let gμ,M(x,t)=|t|q2t+μdM(t), we obtain

    (a+bR3|()s2u|2)R3×R3(u(x)u(y))(uuL2(β1)(x)uuL2(β1)(y))|xy|3+2sdxdy=λR3u(x)uuL2(β1)(x)dx+R3gμ,M(x,u(x))uuL2(β1)(x)dx. (5.3)

    Since Ψ(u)1βuuLβ1 and (5.2), we get

    aR3×R3(u(x)u(y))(uuL2(β1)(x)uuL2(β1)(y))|xy|3+2sdxdyaβ2R3×R3|uuLβ1(x)uuLβ1(y)|2|xy|3+2sdxdy. (5.4)

    For any ε>0, there exists Cε>0 such that

    |gμ,M(x,t)|ε|t|+Cε(1+μMpq)|t|q1. (5.5)

    Let ωL=uuLβ1. By employing Hölder's inequality and (5.3)–(5.5), we have

      aβ2R3×R3|ωL(x)ωL(y)|2|xy|3+2sdxdyεR3(ωL)2dx+Cε(1+μMpq)R3|u(x)|q2(ωL)2dxεR3(ωL)2dx+Cε(1+μMpq)(R3|u(x)|2sdx)q22s((ωL)2tdx)1t, (5.6)

    where q22s+1t=1 and 2t(2,2s). Moreover, it follows from (1.4) that

    Ssu2s2R3|()s2u|2. (5.7)

    Therefore, we deduce from (5.6) and (5.7) that

    ωL2s2Cβ2[ωL22+(1+μMpq)u2sq2ωL2t2],

    where C>0 is a constant. From the definition of uL, we have uLu in R3. Letting L+, using the Fatou's Lemma, one has

    uβ2s2βCβ2[u2β2β+(1+μMpq)u2sq2u2βt2β]. (5.8)

    By the interpolation inequality, we get u2βu21σu2βtσ, where σ(0,1) satisfies 12β=1σ2+σ2βt. Thus, σ=t(β1)tβ1, which shows that σ1 as β+. Since 2β(1σ)=2+2(1β)tβ1<2, we get

    u2β2βu22β(1σ)u2βt2βσ(1+u2)2u2βt2βσ,

    which together with (5.8) yields

    uβ2sC112ββ1β[1+u22+(1+μMpq)u2sq2]12βu2βtk. (5.9)

    where k{σ,1} and C1>0 is a constant. Let θ:=2s2t, then θ>1. Taking β=θ in (3.13), we deduce that

    uθ2sC112θθ1θ[1+u22+(1+μMpq)u2sq2]12θu2sk1, (5.10)

    where k1{σ1,1} and σ1=t(θ1)tθ1. Taking β=θ2 in (5.9), we get

    uθ22sC112θ2θ2θ2[1+u22+(1+μMpq)u2sq2]12θ2u2sθk2, (5.11)

    where k2{σ2,1} and σ2=t(θ21)tθ21. Combining (5.10) with (5.11), we have

    uθ22sC112θ+12θ2θ1θ+2θ2[1+u22+(1+μMpq)u2sq2]12θ+12θ2u2sk1k2.

    Taking β=θi,iN, one has

    uθi2sC1m=1i12θmθm=1imθm[1+u22+(1+μMpq)u2sq2]m=1i12θmu2sk1k2ki. (5.12)

    where ki{σi,1} and σi=t(θi1)tθi1.

    Next, we divide into two cases: u2s1 and u2s<1.

    (1) Assume that u2s1 is in force. In view of k1k2ki1, we have u2sk1k2kiu2s. Letting i+ in (5.12), we can know that

    uC112(θ1)θθ(θ1)2[1+u22+(1+μMpq)u2sq2]12(θ1)u2s.

    (2) Assume that u2s<1 is true. By σi=t(θi1)tθi1=1t1tθi1 and kiσi,1, we have 0<σ1σ2σik1k2ki1, which shows that m=1ilnσmm=1ilnkm0. From the fact that ln(1s)s1s for all s(0,1), one has

    m=1ilnkmm=1ilnσm=m=1iln(1t1tθm1)1ttm=1i1θm1:=A,

    which implies that

    k1k2kieA  for all  iN.

    By u2s<1, we have u2sk1k2kiu2seA. Similarly, letting i+ in (3.16), we reach

    uC112(θ1)θθ(θ1)2[1+u22+(1+μMpq)u2sq2]12(θ1)u2seA.

    Consequently, we have uL(R3) and

    uC112(θ1)θθ(θ1)2[1+c2+(1+μMpq)u2sq2]12(θ1)u2sτ, (5.13)

    where τ=1 or τ=eA1.

    Finally, by (1.4) and Lemma 5.1, there exists C2>0 such that u2sC2. Therefore, it follows from (5.13) and θ=2sq+22, there exists a constant Bc>0 independent μ, M>0 such that

    uBc(1+μ12sqMpq2sq).

    Proof of the Theorem 1.1. By Lemma 5.2, for any c>0, there exists a constant Bc>0 independent on μ and M such that

    ucBc(1+μ12sqMpq2sq).

    Thus, for large M>0, we can choose small μ>0 with μμ1 such that ucM for all μ(0,μ]. By Theorem 2.1, problem (1.1) has a couple of solutions (uc, λc)Hrs(R3)×R for any μ(0,μ]. Moreover, uc is a positive ground state, radially symmetric function and λc<0.

    The authors would like to thank the anonymous referees for carefully reading this paper and making valuable comments and suggestions. This research was funded by the National Natural Science Foundation of China (61803236) and Natural Science Foundation of Shandong Province (ZR2018MA022).

    All authors declare no conflicts of interest in this paper.



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