Research article

Normalized solutions for Kirchhoff-Carrier type equation

  • Received: 21 May 2023 Revised: 18 June 2023 Accepted: 26 June 2023 Published: 07 July 2023
  • MSC : 49J35

  • In this paper, we study the following Kirchhoff-Carrier type equation

    (a+bM(|u|2,|u|τ))Δuλu=|u|p2u, in R3,

    where a, b>0 are constants, λR, p(2,6). By using a minimax procedure, we obtain infinitely solutions (vbn,λn) with vbn having a prescribed L2-norm. Moreover, we give a convergence property of vbn as b0+.

    Citation: Jie Yang, Haibo Chen. Normalized solutions for Kirchhoff-Carrier type equation[J]. AIMS Mathematics, 2023, 8(9): 21622-21635. doi: 10.3934/math.20231102

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  • In this paper, we study the following Kirchhoff-Carrier type equation

    (a+bM(|u|2,|u|τ))Δuλu=|u|p2u, in R3,

    where a, b>0 are constants, λR, p(2,6). By using a minimax procedure, we obtain infinitely solutions (vbn,λn) with vbn having a prescribed L2-norm. Moreover, we give a convergence property of vbn as b0+.



    In this paper, we study the following Kirchhoff-Carrier type equation

    (a+bM(|u|2,|u|τ))Δuλu=|u|p2u, in R3, (1.1)

    where a, b>0 are constants, λR, 2<p<τ<6. M satisfies the following condition

    (H) M:R+×R+R+ is continuous and

    0M(ξ1,ξ2)C0(ξd11+ξd22+1), (1.2)

    for some C0, d1, d2>0. Moreover, for each σ[0,Sτ], t0,

    t0sM(s,σs)ds23(p2)t2M(t,σt), (1.3)

    where Sτ is the best constant of H1(R3) embedding into Lτ(R3).

    It is well known that problem (1.1) comes from two classes of typical nonlocal problem which are Kirchhoff type and Carrier type problems. In 1883, Kirchhoff [10] firstly proposed the Kirchhoff type problem

    ρ2ut2(P0h+E2LL0|ux|2dx)2ux2=0,

    which extends the claasical D'Alembert wave equation by considering the effects of changes in the length of the strings during the vibrations. In [12], Lions proposed a functional analysis method to solve the following Kirchhoff problem

    {(a+bΩ|u(x)|2dx)Δu=f(x,u),xΩ,u(x)=0,xΩ. (1.4)

    Since then, problem (1.4) has attracted the attention of researchers, see [4,5,14,15,19,20] and the references therein.

    Carrier in [7] proposed the following problem

    2ut2(1+α22ππ0|u(x,t)|2dx)2ux2=0in (0,π)×(0,+). (1.5)

    The vibration of elastic string is described by (1.5) when the tension change is not very small. We note that problem (1.5) is similar to the Kirchhoff equation which has many research results. However, there is very little work on the problem (1.5) or its generalization. The appearance of nonlocal terms leads to some difficulties. It lacks a variational structure so we cannot study the problem (1.5) with variational method. Some authors are concerned about the existence of positive solutions or some generalized cases of problem (1.5) by the topological theory and pseudo montone operator theory, see [1,6]. It is worth mentioning that in [9], Jin and Yan studied the following Carrier type problem

    {(a+bΩ|u(x)|γdx)Δu=f(u),xΩ,u(x)=0,xΩ, (1.6)

    where a,b>0 and ΩRN is a bounded open set with smooth boundary and γ1. They obtained the existence of sign-changing solutions of problem (1.6) by using the fixed point index method. Xu and Qin [21] considered the Kirchhoff-Carrier type equation

    {(a+bα(u,|u|γ))Δu=f(u),xΩ,u(x)=0,xΩ, (1.7)

    where a,b>0 and ΩRN is a bounded domain with smooth boundary and 1<γ<2, α:R+×R+R+. By applying the mountain pass theorem and the Ekeland theorem, they obtained some existence results for problem (1.7). Let us emphasize that all the previous existence results for Kirchhoff-Carrier type equation are obtained in a bounded domain. Obviously, the approaches adopted in [9,21] do not work when the Kirchhoff-Carrier type problem defined on the whole space R3.

    When b=0, adding a repulsive nonlocal Coulombic potential, problem (1.1) reduces to the Schrödinger-Poisson-Slater equation

    Δuλu+(|x|1|u|2)u=|u|p2u, in R3. (1.8)

    Luo [13] proved a multiplicity result of solutions for problem (1.8) if p(103,6). They were interested in normalized solutions, that is, solutions to (1.8) satisfying

    R3|u|2dx=c.

    The normalized solutions associated to Schrödinger or Kirchhoff problems have been extensively studied in recent years, see [11,17,22]. Because people are particularly interested in normalized solutions, we can search for normalized solutions of (1.1). Precisely, for given c>0 we look for

    (u,λ)H1r(R3)×Rwith|u|22=c.

    For this reason, we construct the variational structure of problem (1.1). Define Jb:H1r(R3)R is a functional with its Fréchet derivative operator given by

    Jb(u)h=(a+bM(|u|2,|u|τ))R3uhdxR3|u|p2uhdx,u, hH1r(R3). (1.9)

    For any uH1r(R3){0} set βu=u|u|2 and v(t)=tβu for all t0. It follows from (1.9) that

    d(Jb(v(s)))ds=Jb(v(s))v(s)=s(a+bM(s,s|βu|τ))sp1R3|βu|pdx.

    Integrating over [0,|u|2], using Fubini's theorem and Jb(0)=0 we obtain

    Jb(u)=a2|u|22+b|u|20sM(s,s|βu|τ)ds1pR3|u|pdx (1.10)

    for uH1r(R3). Since M is continuous, JbC1(H1r(R3),R) and the couple of weak solution as above can be viewed as a critical point of Jb restricted on the constraint

    Xr(c)={uH1r(R3):R3|uc|2dx=c},c>0.

    As far as we know, there is no result on the existence and asymptotic behaviour of normalized solutions of problem (1.1). In this paper, we focus on the existence of high energy normalized solutions to (1.1). Let us state our main result.

    Theorem 1.1. Let max{23(d1+2)+2, 23(d2+2)+2}<p<6. Then, for any fixed c>0 (1.1) has a sequence of couples of weak solutions {(vn,λn)}H1r(R3)×R with |vn|22=c for each nN+,

    vn2+,andJb(vn)+,as n+.

    Theorem 1.1 shows (vn,λn) depends on the parameter b. Therefore, we next focus on whether some converence phenomena appear when b0.

    Theorem 1.2. Let max{23(d1+2)+2, 23(d2+2)+2}<p<6. {(vbn,λbn)}Xr(c)×R are obtained in Theorem 1.1. Then, for any sequence {bm}0+(m+) there exists a subsequence still denoted by {bm} such that for any nN+ vbmnv0n in H1r(R3) and λbmnλ0n in R as m+ where {(v0n,λ0n)}Xr(c)×R is a sequence of couples of weak solutions to the following equation

    aΔuλu=|u|p2u,in R3. (1.11)

    Remark 1.1. A typical example of M satisfying condition (H) is M(s,t)=a+2bt2 with a,b>0.

    The rest of this paper is organized as follows. In Section 2, we introduce some preliminaries results. In Section 3, we prove Theorems 1.1 and 1.2.

    Throughout the paper, let u=(R3(|u|2+|u|2)dx)12 denote the usual norm of H1(R3), ||τ denote the norm of Lτ(R3) for 2<τ<6. H1(R3) is the dual space of H1(R3) and H1r(R3) is the subspace of radially symmetric functions in H1(R3).

    Proposition 2.1. [3,Lemma 2.1] Assume that p(2,6). Then, there holds

    μn:=infuVn1R3(|u|2+|u|2)dx(R3|u|pdx)2p+,as n+

    where {Vn}H1r(R3) be a strictly increasing sequence of finite-dimensional linear subspace in H1r(R3) such that nVn is dense in H1r(R3) and Vn denotes the orthogonal complementary space of Vn in H1r(R3).

    This lemma is useful to show the compactness of the Palais-Smale sequence.

    Lemma 2.2. [2,Lemma 3] Let F be a C1 functional on H1(R3), if {xk}X(c) is bounded in H1(R3). Then,

    F|X(c)(xk)0in H1(R3)F(xk)F(xk),xkxk0in H1(R3)as k,

    where X(c)={uH1(R3):R3|uc|2dx=c},c>0.

    Set

    A(u)=R3|u|2dx,B(u)=|u|20sM(s,s|βu|τ)ds,C(u)=R3|u|pdx,D(u)=R3|u|2dx.

    Now, we introduce a scaling. Define ut(x)=t32u(tx), for t>0, uXr(c). Then,

    A(ut)=t2A(u),B(ut)=t|u|20sM(s,s323τ|βu|τ)ds,C(ut)=t32(p2)C(u)andD(ut)=D(u).

    Lemma 2.3. Let p>max{23(d1+2)+2,23(d2+2)+2},c>0 and uXr(c). Then

    utXr(c), A(ut)+,andJb(ut),ast+.

    Proof. It follows from D(ut)=D(u) that utXr(c) for any uXr(c). By (1.2) and (1.10), we obtain

    Jb(ut)=at22R3|u|2dx+bt|u|20sM(s,s323τ|βu|τ)dst32(p2)pR3|u|pdxat22R3|u|2dx+bt|u|20C0(sd1+1+s(323τ)d2+1|βu|d2+1τ+s)dst32(p2)pR3|u|pdx=at22R3|u|2dx+C(td1+2|u|d1+22+t(323τ)d2+2|u|d2+22|βu|d2+2τ+t2|u|22)t32(p2)pR3|u|pdx,

    which implies that Jb(ut), as t+ since τ(2,6) and p>max{23(d1+2)+2,23(d2+2)+2}.

    For c>0 fixed, nN+ and n2 let

    θn:=L2p2a2p2μ2p2n,En:={uVn1Xr(c):|u|22=θn}

    and

    σn:=infuEnJb(u),

    where L=maxx>0(x2+c)p/2xp+cp/2, μn is given in Proposition 2.1.

    Lemma 2.4. For any p(2,6), then σn+ as n+. Particularly, we can suppose that σn1 for every nN+ without any restriction.

    Proof. For any uEn, we deduce that

    Jb(u)=a2R3|u|2dx+b|u|20sM(s,s|βu|τ)ds1pR3|u|pdxa2R3|u|2dx1pμn(|u|22+c)p/2a2R3|u|2dxLpμn(|u|p2+cp/2)=(121p)aθnLpμncp/2.

    Combining with Proposition 2.1, we obtain that σn+ as n+ since p>2.

    Define the continuous map

    η:R×H1r(R3)H1r(R3),η(t,v)(x)=e32tv(etx),for tR, vH1r(R3) and  xR3. (2.1)

    It follows from Lemma 2.3 that η(t,v)Xr(c) for any vXr(c), tR and

    {A(η(t,v))0,Jb(η(t,v))0,t,A(η(t,v))+,Jb(η(t,v)),t+.

    Recalling that Vn is finite dimensional, we obtain that for any nN+, there exists tn>0 such that

    ˉγn:[0,1]×(Xr(c)Vn)Xr(c),ˉγn(s,u)=η((2s1)tn,u)

    satisfies

    {A(ˉγn(0,u))<θn,A(ˉγn(1,u))>θn,Jb(ˉγn(0,u))<σn,Jb(ˉγn(1,u))<σn.

    Now, define

    Γn:={γ:[0,1]×(Xr(c)Vn)Xr(c)|γ is continuous, odd in uand such that for any u: γ(0,u)=ˉγn(0,u),γ(1,u)=ˉγn(1,u)}.

    Obviously, ˉγnΓn. According to [3,Lemma 2.3], we obtain the next key intersection result.

    Lemma 2.5. For each nN+,

    κbn(c):=infγΓnmax0s1,uXr(c)VnJb(γ(s,u))σn.

    Proof. The proof of the lemma can be easily obtained by [3,Lemma 2.3].

    Next, we will show that {κbn(c)} is indeed a sequence of critical values of Jb restricted on Xr(c). First, we shall prove that there exists a bounded (PS) sequence at each level κbn(c). Now, we fix an nN+. We define the following auxiliary functional:

    ˜Jb:R×Xr(c)R,(t,u)Jb(η(t,u)),

    where η(t,u) is defined in (2.1). Set

    ˜Γn:={˜γ:[0,1]×(Xr(c)Vn)Xr(c)×R|˜γ is continuous, odd in uand such that η˜γΓn}.

    Obviously, ˜γ:=(0,γ)˜Γn for any γΓn. Define

    ˜κbn(c):=inf˜γ˜Γnmax0s1,uXr(c)Vn˜Jb(˜γ(s,u)).

    Applying the fact that the maps

    ϕ:Γn˜Γn,γϕ(γ):=(0,γ)

    and

    φ:˜ΓnΓn,˜γφ(˜γ):=η˜γ

    satisfy

    ˜Jb(ϕ(γ))=Jb(γ),Jb(φ(˜γ))=˜Jb(˜γ),

    we get ˜κbn(c)=κbn(c). Let us denote by E the space R×H1r(R3) endowed with the norm 2E=||2R+2 and by E its dual space. Similar to [8,Lemma 2.3], we get the following result.

    Lemma 2.6. For any ϵ>0, suppose that ˜γ0˜Γn satisfies

    max0s1,uXr(c)Vn˜Jb(˜γ0(s,u))˜κbn(c)+ϵ.

    Then, there exists a pair of (t0,u0)R×Xr(c) such that

    (i) ˜Jb(t0,u0)[˜κbn(c)ϵ,˜κbn(c)+ϵ];

    (ii) min0s1,uXr(c)Vn(t0,u0)˜γ0(s,u)Eϵ;

    (iii) (˜Jb|Xr(c)×R)(t0,u0)E2ϵ, i.e.,

    |˜Jb(t0,u0),zE×E|2ϵzE

    holds, for all z˜T(t0,u0):={(z1,z2)E,u0,z2L2=0}.

    Lemma 2.7. For any c>0 fixed and nN+, there exists a sequence {vnkXr(c)} satisfying

    Jb(vnk)κbn(c),Jb|Xr(c)(vnk)0,Gb(vnk)0, (2.2)

    where

    Gb(u)=aR3|u|2dx+bR3|u|2dxM(|u|2,|u|2|βu|τ)3(p2)2pR3|u|pdx.

    Particularly, {vnk} is bounded in Xr(c).

    Proof. By the definition of κbn(c) we have that for every kN+, there exists a γΓn such that

    max0s1,uXr(c)VnJb(γk(s,u))κbn(c)+1k.

    It follows from ˜κbn(c)=κbn(c) and ˜γk:=(0,γk)˜Γn that

    max0s1,uXr(c)Vn~Jb(˜γk(s,u))˜κbn(c)+1k.

    Using Lemma 2.6, we infer that there exists a sequence {(tnk,unk)}R×Xr(c) such that

    (i) ~Jb(tnk,unk)[κbn(c)1k,κbn(c)+1k];

    (ii) min0s1,uXr(c)Vn(tnk,unk)(0,γk)E1k;

    (iii) (˜Jb|Xr(c)×R)(tnk,unk)E4k, i.e.,

    |˜Jb(tnk,unk),zE×E|4kzE

    holds, for all z˜T(tnk,unk):={(z1,z2)E,unk,z2L2=0}.

    For every kN+, set vnk=η(tnk,unk). We shall show that {vnk}Xr(c) satisfies (2.2). It follows from (i) that Jb(vnk)κbn(c) as k, since Jb(vnk)=Jb(η(tnk,unk))=~Jb(tnk,unk). Noting that

    ~Jb(t,u),(r,v)=are2tR3|u|2dx+ae2tR3uvdx+bre2t|u|22M(et|u|2,e(323τ)t|u|323τ2|βu|τ)+be2tM(et|u|2,e(323τ)t|u|323τ2|βu|τ)R3uvdx3(p2)2pre3(p2)t2R3|u|pdxe3(p2)t2R3|u|p2uvdx,

    we obtain

    Gb(vnk)=aA(vnk)+b|vnk|22M(|vnk|2,|vnk|2|βvnk|τ)3(p2)2pC(vnk)=ae2tnkR3|unk|2dx+be2tnkR3|unk|2dxM(etnk|unk|,e(323τ)tnk|unk|2|βunk|τ)3(p2)2pe3(p2)tnk2R3|unk|pdx=~Jb(tnk,unk),(1,0).

    Hence, by (iii), we see that Gb(vnk)0 as k for (1,0)˜T(tnk,unk).

    Finally, we shall show that

    Jb|Xr(c)(vnk)0,as k.

    We claim that for kN large enough

    |Jb(vnk),w|4kw2

    holds for all wTvnk, where Tvnk:={wH1r(R),vnk,wL2=0}. Indeed, for wTvnk, taking ˜w=η(tnk,w) we have

    Jb(vnk),w=aR3vnkw+bM(|vnk|2,|vnk|2|evnk|τ)R3vnkwdxR3|vnk|p2vnkwdx=ae2tnkR3unk˜wdx+be2tnkR3unk˜wdxM(etnk|unk|2,e(323τ)tnk|unk|2|βunk|τ)e3(p2)tnk2R3|unk|p2unk˜wdx=~Jb(tnk,vnk),(0,˜w).

    From R3unk˜wdx=R3vnkwdx, we see that (0,˜w)˜T(tnk,unk) is equivalent to show wTvnk. It follows from (ii) that

    |tnk|=|tnk0|min0s1,uXr(c)Vn(tnk,unk)(0,γk(t,u))E1k.

    Hence, we conclude that

    (0,˜w)2E=˜w2=R3|w|2dx+e2tnkR3|w|2dx4w2.

    Therefore,

    |Jb(vnk),w|=~Jb(tnk,vnk),(0,˜w)4k(0,˜w)2E4kw2,

    which yields

    Jb|Xr(c)(vnk)=supwTvnk,w1|Jb(vnk),w|4k0,

    as k. Since (1.3) and p(103,6), we obtain

    Jb(u)23(p2)Gb(u)=(3p10)a6(p2)|u|22+b[|u|20sM(s,s|βu|τ)ds2|u|223(p2)M(|u|2,|u|2|βu|τ)](3p10)a6(p2)|u|22, (2.3)

    which completes the proof.

    Lemma 2.8. Let p(2,6), λR. If uH1(R3) is a weak solution of

    (a+bM(|u|2,|u|2|βu|τ))Δu|u|p2u=λu,

    then Gb(u)=0. Moreover if λ0, we obtain u=0.

    Proof. If uH1(R3) is a weak solution of (1.1) then it satisfies the Pohožaev identity

    a2R3|u|2dx+b2R3|u|2dxM(|u|2,|u|2|βu|τ)3pR3|u|pdx=32λR3|u|2dx

    and

    aR3|u|2dx+bR3|u|2dxM(|u|2,|u|2|βu|τ)R3|u|pdx=λR3|u|2dx. (2.4)

    Thus, we obtain

    Gb(u)=aR3|u|2dx+bR3|u|2dxM(|u|2,|u|2|βu|τ)3(p2)2pR3|u|pdx=0. (2.5)

    Combining (2.4) and (2.5), we can see that

    0a|u|22+b|u|22M(|u|2,|u|2|βu|τ)=3(p2)p6λR3|u|2dx.

    Since p(2,6), if λ>0, we obtain u0 immediately. If λ=0, we get A(u)=0. Together with Gb(u)=0 then u0.

    Lemma 2.9. Let vnkXr(c) be the Palais-Smale sequence obtained in Lemma 2.7. Then, there exist vnH1r(R3) and {λnk}R such that up to a subsequence as k+

    (i) vnkvn0 in H1r(R3);

    (ii) λnkλn<0 in R;

    (iii) (a+bM(|vnk|2,|vnk|2|βvnk|τ))Δvnkλnkvnk|vnk|p2vnk0 in H1r(R3);

    (iv) (a+bM(|vn|2,|vn|2|βvn|τ))Δvnλnvn|vn|p2vn=0 in H1r(R3).

    Moreover, if λn<0, then we obtain

    vnkvnin H1r(R3) as k.

    Proof. Since {vnk}Xr(c) is bounded, we may assume that there exists vnH1r(R3) such that

    {vnkvn,in H1r(R3),vnkvn,in Lp(R3),vnkvn,a.e. in R3.

    We claim that vn0. In fact, we assume by contradiction that vn=0, then C(vkn)=o(1). By Gb(vnk)=o(1), we obtain A(vnk)=o(1). Consequently, Jb(vnk)=o(1) which contradicts Lemma 2.4. Hence, (i) is proved. It follows from Lemma 2.2 that

    Jb|Xr(c)(vnk)0 in H1r(R3)Jb(vnk)Jb(vnk),vnkvnk0 in H1r(R3),as k.

    Note that for any wH1r(R3),

    Jb(vnk)Jb(vnk),vnkvnk,w=(a+bM(|vnk|2,|vnk|2|βvnk|τ))R3vnkwdxR3|vnk|p2vnkwdxλnkR3vnkwdx,

    where

    λnk=1|vnk|22[(a+bM(|vnk|2,|vnk|2|βvnk|τ))R3|vnk|2dxR3|vnk|pdx]+o(1). (2.6)

    Therefore, (iii) is proved. By (2.6) and the fact {vnk}Xr(c) is bounded up to a subsequence, there exists λnR such that λnkλn as k+. Moreover, from 2<p<6 and (2.5), we deduce that

    λn=limk1c[(a+bM(|vnk|2,|vnk|2|βvnk|τ))|vnk|22R3|vnk|pdx]limk1c[(a+bM(|vnk|2,|vnk|2|βvnk|τ))|vnk|223(p2)2pR3|vnk|pdx]=0, (2.7)

    which yields λn0. Consequently, it follows from Lemma 2.8 that λn<0, that is, (ii) is true.

    Assume that limkR3|vnk|2dx=l20. By (ii) and (iii), we obtain

    (a+bM(l,|vn|τ))Δvnλnvn|vn|p2vn=0in H1r(R3). (2.8)

    From (2.8), we get

    (a+bM(l,|vn|τ))R3vn(vnkvn)dxλnR3vn(vnkvn)dx=R3|vn|p2vn(vnkvn)dx+o(1). (2.9)

    From (ii) and (iii), we obtain

    (a+bM(l,|vn|τ))R3vnk(vnkvn)dxλnR3vnk(vnkvn)dx=R3|vnk|p2vnk(vnkvn)dx+o(1). (2.10)

    By (2.9) and (2.10), we see that

    (a+bM(l,|vn|τ))R3|vnkvn|2dxλnR3|vnkvn|2dx+o(1)=R3(|vnk|p2vnk|vn|p2vn)(vnkvn)dx.

    Recalling that vnkvn in Lp(R3), we deduce that

    (a+bM(l,|vn|τ))R3|vnkvn|2dxλnR3|vnkvn|2dx=o(1),

    which yields that (a+bM(l,|vn|τ))R3|vnkvn|2dx=o(1) for a, b>0,l0,λn0. Hence,

    R3|vnk|2dxR3|vn|2dx.

    Together with (ii) and (iii), (iv) is easily obtained. Then, we obtain that R3|vnkvn|2dx=o(1). Thus, we obtain

    vnkvnin H1r(R3) as k.

    Proof of Theorem 1.1. From Lemma 2.9, we obtain that for any fixed c>0, {(vn,λn)}H1r(R3)×R with |vn|22=c is a sequence of couples of weak solutions to (1.1) for each nN+. Since Gb(vn)=0, (1.2) and (2.3) we obtain

    κbn(c)=Jb(vn)23(p2)Gb(vn)=(3p10)a6(p2)|vn|22+b[|vn|20sM(s,s|βvn|τ)ds2|vn|223(p2)M(|vn|2,|vn|2|βvn|τ)](3p10)a6(p2)vn2+C(vnd1+2+vnd2+2|βvn|d2+2τ+vn),

    which implies that {vn} is unbounded due to Lemmas 2.4 and 2.5.

    Proof of Theorem 1.2. For every b>0, by Theorem 1.1 there exists a sequence of couples of weak solutions {(vbn,λbn)}H1r(R3)×R to (1.1). We show that for any sequence bm0+ as m+, {vbmn}mN+ is bounded in H1r(R3). It follows from Vn is finite-dimensional that for each nN+,

    κbmn(c):=infγΓnmax0s1,uSr(c)VnJbm(γ(s,u))infγΓnmax0s1,uSr(c)VnJ1(γ(s,u)):=αn<+.

    Since {(vbmn,λbmn)}H1r(R3)×R is a sequence of couples of weak solutions to (1.1) with b=bm and

    λbmn=1c[(a+bM(|vbmn|2,|βvbmn|τ))A(vbmn)C(vbmn)],

    then by Lemma 2.8, we conclude that Gbm(vbmn)=0. Together with (1.3) and (2.3), we deduce that

    κbmn(c)=Jbm(vbmn)23(p2)Gbm(vbmn)(3p10)a6(p2)|vbmn|22,

    which implies that {A(vbmn)}mN+ is bounded in R, {vbmn}mN+ is bounded in H1r(R3) and {λbmn}mN+ is bounded in R. Therefore, there exist a subsequence of {bm} (still denoted by {bm}) and λ0n0 such that λbmnλ0n as m+. Meanwhile,

    {vbmnv0nin H1r(R3),vbmnv0nin Lp(R3),vbmnv0n a.e. in R3. (3.1)

    Thus, for every nN+ (v0n,λ0n) is a couple of weak solution to (1.1) with b=0, that is, for any wH1r(R3),

    aR3v0nwdxλ0nR3v0nwdx=R3|v0n|p2v0nwdx. (3.2)

    Taking w=vbmnv0n in (3), we obtain

    aR3v0n(vbmnv0n)dxλ0nR3v0n(vbmnv0n)dx=R3|v0n|p2v0n(vbmnv0n)dx. (3.3)

    By {(vbmn,λbmn)}H1r(R3)×R is a sequence of couples of weak solutions to (1.1) with b=bm, λbmnλ0n and {vbmn} is bounded in H1r(R3), we obtain

    aR3vbmn(vbmnv0n)dxλ0nR3vbmn(vbmnv0n)dx=R3|vbmn|p2v0n(vbmnv0n)dx+o(1). (3.4)

    Combining (3.1), (3.3) and (3.4) we can see that

    aA(vbmnv0n)λ0nD(vbmnv0n)=o(1). (3.5)

    According to λ0n0, A(vbmnv0n)0 as m+. If λ0n=0, from (3.2) it follows that v0nH1r(R3) is a weak solution to aΔv=|v|p2v. This yields v0n=0. On the other hand, by Lemma 2.5 we obtain that

    1σnκbn(c):=Jb(vbmn)0,

    which yields a contradiction. Hence, λ0n<0 and D(vbmnv0n)0, as m+, due to (3.5). Thus, {(v0n,λ0n)}H1r(R3)×R is a sequence of couples of weak solutions to (1.11).

    The main purpose of this paper is to study the existence of high energy normalized solutions for the Kirchhoff-Carrier type equation. To prove Theorem 1.1, we first show that

    utXr(c), A(ut)+,andJb(ut),ast+.

    Combining with Proposition 2.1, we obtain that σn+ as n+. Similar to [8,Lemma 2.3], we get Lemma 2.6. We use Lemma 2.6 to obtain a Palais-Smale sequence {vnkXr(c)} of functional Jb satisfying Gb(vnk)0 where Gb(u) is given in Lemma 2.7. Finally, we conclude that

    vnkvnin H1r(R3) as k

    by using Lemma 2.2. Meanwhile, we find some converence phenomena appear when b0.

    In the proof, H1r(R3) is the subspace of radially symmetric functions in H1(R3) is very crucial, we do not know whether the solution can still exist in the more general space H1(R3). This is a question that we need to further consider.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    J. Yang is supported by Natural Science Foundation of Hunan Province of China (No. 2023JJ30482, 2022JJ30463), Research Foundation of Education Bureau of Hunan Province (No. 22A0540) and Huaihua University Double First-Class Initiative Applied Characteristic Discipline of Control Science and Engineering. H. B. Chen is supported by the National Natural Science Foundation of China (No. 12071486).

    The authors declare no conflict of interest.



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