Research article

On existence results for a class of biharmonic elliptic problems without (AR) condition

  • Received: 28 January 2024 Revised: 03 March 2024 Accepted: 07 March 2024 Published: 05 June 2024
  • MSC : 35J35, 35B38, 35J91

  • In this paper, we study the following biharmonic elliptic equation in RN:

    Δ2ψΔψ+P(x)ψ=g(x,ψ),  xRN,

    where g and P are periodic in x1,,xN, g(x,ψ) is subcritical and odd in ψ. Without assuming the Ambrosetti-Rabinowitz condition, we prove the existence of infinitely many geometrically distinct solutions for this equation, and the existence of ground state solutions is established as well.

    Citation: Dengfeng Lu, Shuwei Dai. On existence results for a class of biharmonic elliptic problems without (AR) condition[J]. AIMS Mathematics, 2024, 9(7): 18897-18909. doi: 10.3934/math.2024919

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  • In this paper, we study the following biharmonic elliptic equation in RN:

    Δ2ψΔψ+P(x)ψ=g(x,ψ),  xRN,

    where g and P are periodic in x1,,xN, g(x,ψ) is subcritical and odd in ψ. Without assuming the Ambrosetti-Rabinowitz condition, we prove the existence of infinitely many geometrically distinct solutions for this equation, and the existence of ground state solutions is established as well.



    In the present paper, we consider the following biharmonic elliptic equation with potential:

    {Δ2ψΔψ+P(x)ψ=g(x,ψ)   in RN,ψ(x)H2(RN), (1.1)

    where Δ2 is the biharmonic operator. We assume that P(x) and g(x,ψ) satisfy the hypotheses below:

    (P)P(x)C(RN,R) is 1-periodic in each of xi,1iN, and infxRNP(x)a0>0.

    (g1)g(x,t)C(RN×R,R) is 1-periodic in each of xi,1iN, and there exists a1>0 such that |g(x,t)|a1(1+|t|q1) for 2<q<2, where 2=2NN4 if N>4, 2=+ if N4.

    (g2)lim|t|0g(x,t)t=0 uniformly for xRN.

    (g3)lim|t|+G(x,t)t2=+ uniformly for xRN, where G(x,ψ)=ψ0g(x,t)dt.

    (g4)g(x,t)t is strictly increasing on (,0) and on (0,+).

    Problem (1.1) is usually used to describe some phenomena appearing in different physical, engineering and other sciences. Over the course of the last decades, plenty of results for the biharmonic elliptic equations have been presented. When ΩRN(N>4) is a smooth bounded domain, the problem

    {Δ2ψ+aΔψ=g(x,ψ),xΩ,ψ=Δψ=0,xΩ, (1.2)

    arising in the study of traveling waves in suspension bridges (see for instance, [2,8,9,15]) and the study of the static deflection of an elastic plate in a fluid, has drawn a great deal of attention, see for example, [1,3,7,11,12,13,14,23] and the references therein. Furthermore, biharmonic elliptic problems on the whole space RN also attract a lot of attention, see [4,5,10,21,22,24,25]. It is worth noticing that in the paper by Yin and Wu [22], a sequence of high energy solutions to problem (1.1) has been established by using variational methods. Later, based on Rabinowitz's symmetric mountain pass theorem, Ye and Tang extended the results in [21] to a more generic conditions, and obtained similar results. Subsequently, Zhang et al. [24] obtained the existence of infinitely many solutions by applying the genus properties in critical point theory.

    Resting on the different assumptions (g1)(g4) from those applied previously, our paper states some new existence results of problem (1.1) and, meanwhile, we do not assume the Ambrosetti-Rabinowitz condition ((AR) in short):

    (AR) there is μ>2 such that 0<μG(x,ψ)g(x,ψ)ψ for ψ0 and xRN.

    It is noticeable that the (AR) condition is to ensure the boundedness of the Palais-Smale sequences of the corresponding functional, which is very essential in applying the critical point theory. It would be more complicated for this problem without (AR) condition. However, there are numerous functions superlinear at infinity not satisfying the (AR) condition for any μ>2. Virtually, the (AR) condition implies that G(x,ψ)c1|ψ|μc2 for some c1,c2>0. Thus, for example the superlinear function

    g(x,ψ)=a(x)ψln(1+|ψ|), where a(x)>0 is 1-periodic in xi,1iN,

    does not satisfy (AR) condition. But it satisfies our conditions (g1)(g4).

    To state our results, we need to present some notations. For y=(y1,,yN)ZN, the action of ZN on H2(RN) is given by

    τyψ(x)=ψ(xy),  yZN. (1.3)

    It follows from (P) and (g1) that if ψ0 is a solution of (1.1), then so is τyψ0 for all yZN. Set O(ψ0)={τyψ0:yZN}. Two solutions ψ1,ψ2 of (1.1) are regarded as geometrically distinct if O(ψ1)O(ψ2).

    The main results of this paper are the following:

    Theorem 1.1. Assume that P(x) satisfies (P), and g(x,ψ) satisfies (g1)(g4). Then the Eq (1.1) has at least one ground state solution.

    In our next result, we verify the existence of infinitely many solutions for (1.1) if g(x,t) is an odd function about t. More specifically, we suppose

    (g5)g(x,t)=g(x,t) for all (x,t)RN×R.

    Theorem 1.2. Assume that P(x) satisfies (P), and g(x,ψ) satisfies (g1)(g5). Then problem (1.1) has infinitely many pairs ±ψ of geometrically distinct solutions.

    The paper is organized as follows. In Section 2, some preliminary results for proving our main results are presented, and the fact that problem (1.1) has a ground state solution is proved. Section 3 is devoted to the proof of Theorem 1.2.

    First, let us set some notations to be used in this paper. Lr(RN)(1r<+) denotes Lebesgue space, the usual norm of Lr(RN) is denoted by r for 1r<+. Let

    E={ψH2(RN):RN(|Δψ|2+|ψ|2+P(x)ψ2)dx<+},

    then E is a Hilbert space with the inner product

    ψ,vE=RN(ΔψΔv+ψv+P(x)ψv)dx,

    and the induced norm is denoted by ψ=ψ,ψE. Note that the following embedding is continuous:

    ELr(RN)  (2r<2),

    consequently, for each r[2,2), there exists a constant ar>0 such that

    ψrarψ,   ψE. (2.1)

    The dual space of a space E will be denoted by E1 and S is the unit sphere in E, that is

    S={ψE:ψ=1}.

    The corresponding energy functional of problem (1.1) is defined on E by

    E(ψ)=12RN(|Δψ|2+|ψ|2+P(x)ψ2)dxRNG(x,ψ)dx, (2.2)

    where G(x,ψ)=ψ0g(x,t)dt. Under the assumptions (g1)(g4) and (P), we can easily check that E(ψ)C1(E,R) and

    E(ψ),v=RN(ΔψΔv+ψv+P(x)ψv)dxRNg(x,ψ)vdx, (2.3)

    for all ψ,vE. Thus, solutions to problem (1.1) can be obtained as the critical points of the functional E(ψ). We consider the Nehari manifold

    M={ψE{0}:E(ψ),ψ=0},

    and let

    c=infψME(ψ).

    Note that M contains every nonzero solution of problem (1.1). For t>0, we consider the fibering maps ϕψ:tE(tψ) defined by

    ϕψ(t)=E(tψ)=t22RN(|Δψ|2+|ψ|2+P(x)ψ2)dxRNG(x,tψ)dx.

    Now we have the following lemma. Hereafter, we suppose that (P) and (g1)(g4) are satisfied.

    Lemma 2.1. (i) For each ψE{0}, there is a unique tψ>0 such that ϕψ(t)>0 for 0<t<tψ and ϕψ(t)<0 for t>tψ. Moreover, tψM if and only if t=tψ.

    (ii) There exists ρ>0 such that cinfψSρE(ψ)>0, where Sρ={ψE:ψ=ρ}.

    (iii) For all ψM, there holds ψ2c.

    (iv) For all ψM, there holds E(ψ) as ψ.

    Proof. (i) First, we claim that ϕψ(t)>0 for t>0 small. Indeed, the conditions (g1) and (g2) imply that for any ε>0, there exists Cε>0 such that, for all (x,ψ)RN×R, there hold

    |g(x,ψ)|ε|ψ|+Cε|ψ|q1,   |G(x,ψ)|ε|ψ|2+Cε|ψ|q. (2.4)

    Then, by (2.4) and the Sobolev embedding theorem, for ε>0 sufficiently small, we obtain

    ϕψ(t)t22RN(|Δψ|2+|ψ|2+P(x)ψ2)dxεt22RN|ψ|2dxCεtqqRN|ψ|qdxt22ψ2t22εC1ψ2tqqCεC2ψq,

    since q>2, ϕψ(t)>0 whenever t>0 is small enough.

    On the other hand, we have

    ϕψ(t)=t22RN(|Δψ|2+|ψ|2+P(x)ψ2)dxRNG(x,tψ)dx=t2(12ψ2RNG(x,tψ)(tψ)2ψ2dx).

    By (g3) and Fatou's lemma, one has

    RNG(x,tψ)(tψ)2ψ2dx+ (t+).

    Hence ϕψ(t) as t+ and maxt>0ϕψ(t) is achieved at a t=tψ>0. In addition, the condition ϕψ(t)=0 is equivalent to

    RN(|Δψ|2+|ψ|2+P(x)ψ2)dx=RNg(x,tψ)tψψ2dx.

    By (g4), the function g(x,t)t is strictly increasing for t>0, so there exists a unique tψ>0 such that ϕψ(tψ)=0. On the other hand, we note that

    ϕψ(t)=t1E(tψ),tψ.

    Therefore, tψM if and only if t=tψ.

    (ii) For ψE, we have

    E(ψ)=12ψ2RNG(x,ψ)dx,

    and thus by (2.3) there holds

    RNG(x,ψ)dx=o(ψ2)  as  ψ0,

    hence infψSρE(ψ)>0 if ρ>0 is sufficiently small. The inequality infψME(ψ)infψSρE(ψ) is a consequence of (i), since for every ψM there exists t>0 such that tψSρ, and E(tψψ)E(tψ).

    (iii) Note that by using (g2) and (g4), we can get

    G(x,ψ)0,   g(x,ψ)ψ2G(x,ψ),   ψ0. (2.5)

    Then by the definition of c and (2.5), for ψM one has

    c12ψ2RNG(x,ψ)dx12ψ2,

    hence ψ2c.

    (iv) Arguing by contradiction, suppose there exists a sequence {ψm}M such that ψm and E(ψm)d for some d>0. Let vm=ψmψm. Then {vm} is bounded (vm=1) in E, after passing to a subsequence, if necessary, we may assume that vmv in E and vm(x)v(x) a.e. in RN. Choose ymRN to satisfy

    B1(ym)v2mdx=maxyRNB1(y)v2mdx.

    Since E and M are invariant with respect to the action of ZN given by (1.3), we may assume translating vm, if necessary, that {ym} is bounded in RN. If

    limmB1(ym)v2mdx=0, (2.6)

    according to P. L. Lions' vanishing lemma (see [20], Lemma 1.21), we get vm0 in Lr(RN) for 2<r<2. By (2.4), fixing an s>2c and using the Lebesgue dominated convergence theorem, we have

    limmRNG(x,svm)dx=RNlimmG(x,svm)dx=0.

    Note that {ψm}M, and then by Lemma 2.1, we obtain that

    dE(ψm)E(svm)=s22vm2RNG(x,svm)dxs22,

    which is a contradiction for s>2d. Hence (2.6) cannot hold, and then vmv0 in L2loc(RN). Since |ψm(x)| if v(x)0, then by (g3) and Fatou's lemma, we have

    RNG(x,ψm)ψ2mv2mdx+ (m),

    and therefore

    0E(ψm)ψm2=12vm2RNG(x,ψm)ψ2mv2mdx,

    as m, which is a contradiction. This completes the proof.

    Lemma 2.2. Let V be a compact subset of E{0}, then there exists R>0 such that E(ψ)<0 on (R+V)BR(0) for all ψV, where R+V={tψ:tR+,ψV}.

    Proof. Without loss of generality, we may assume that VS, i.e., ψ=1 for every ψV. Arguing by contradiction, suppose there exists ψmV and wm=tmψm such that E(wm)0 and tm+ as m. Passing to a subsequence, we may assume that ψmψS. Note that |wm(x)| if ψ(x)0, then by (g3) and Fatou's lemma we have

    RNG(x,wm)t2mdx=RNG(x,wm)w2mψ2mdx+ (m),

    and therefore

    0E(wm)t2m=12RNG(x,wm)t2mdx,

    which is a contradiction. This completes the proof.

    Recall that S is the unit sphere in E, and define the mapping φ:SM by setting

    φ(w)=tww,

    where tw is the same as in Lemma 2.1 (i). Note that φ(w)=tw.

    Lemma 2.3. (i) The mapping φ is a homeomorphism between S and M, and the inverse of φ is given by φ1(ψ)=ψψ.

    (ii) The mapping φ1 is Lipschitz continuous.

    Proof. (i) See [19], Proposition 8.

    (ii) For ψ,uM, by Lemma 2.1(iii), we have

    φ1(ψ)φ1(u)=ψψuu=ψuψ+(uψ)uψu2ψψu2cψu,

    this implies that the mapping φ1 is Lipschitz continuous.

    Now we consider the functional I:SR given by I(w)=E(φ(w)). Then we have

    Lemma 2.4. (i) IC1(S,R) and

    I(w),z=φ(w)E(φ(w)),z  for all  zTw(S)={uE:w,u=0}.

    (ii) If {wm} is a (PS) sequence for I, then {φ(wm)} is a (PS) sequence for E. If {ψm}M is a bounded (PS) sequence for E, then {φ1(ψm)} is a (PS) sequence for I.

    (iii)

    infψSI(ψ)=infψME(ψ)=c.

    Moreover, w is a critical point of I if and only if φ(w) is a nontrivial critical point of E(ψ), and the corresponding critical values coincide.

    (iv) If E(ψ) is even, then I(ψ) is also even.

    Proof. The proof is entirely analogous to that of Corollary 10 in [19]. By Lemmas 2.1 and 2.3, it can be concluded that the hypotheses in [19] are satisfied. Indeed, if ϕw(t)=E(tw) and wS, then ϕw(t)>0 for 0<t<tw and ϕw(t)<0 for t>tw by Lemma 2.1(i), twδ>0 by Lemma 2.1 (ii) and twR for wVS by Lemma 2.2.

    Now we give the proof of Theorem 1.1.

    Proof of Theorem 1.1. From the conclusion (ii) of Lemma 2.1, we know that c>0. Moreover, if ψ0M satisfies E(ψ0)=c, then φ1(ψ0)S is a minimizer of I, and therefore a critical point of I, so that ψ0 is a critical point of E by Lemma 2.4. It remains to show that there exists a minimizer ψ of E|M. By Ekeland's variational principle [20], there exists a sequence {wm}S such that

    I(wm)c  and  I(wm)0  as  m.

    Set ψm=φ(wm)M for all mN. Then E(ψm)c and E(ψm)0 as m. By Lemma 2.1(iv), we know that {ψm} is bounded and hence ψmψ after passing to a subsequence. Choose ymRN to satisfy

    B1(ym)ψ2mdx=maxyRNB1(y)ψ2mdx. (2.7)

    Since E and M are invariant with respect to the action of ZN given by (1.3), we may assume that {ym} is bounded in RN. If

    limmB1(ym)ψ2mdx=0, (2.8)

    then by P. L. Lions' vanishing lemma, we have ψm0 in Lr(RN) for 2<r<2. From (2.4) and the Sobolev embedding theorem, we infer that

    RNg(x,ψm)ψmdx=o(ψm)  as  m.

    Hence

    o(ψm)=E(ψm),ψm=ψm2RNg(x,ψm)ψmdx=ψm2o(ψm),

    and therefore ψm0, contrary to Lemma 2.1(iii). It follows that (2.8) cannot hold, and thus ψmψ0, E(ψ)=0.

    In the following we claim that E(ψ)=c. Notice that {ψm} is bounded, by (2.5) and Fatou's lemma we get that

    c=lim infm(E(ψm)12E(ψm),ψm)=lim infm(RN(12g(x,ψm)ψmG(x,ψm))dx)RN(12g(x,ψ)ψG(x,ψ))dx=E(ψ)12E(ψ),ψ=E(ψ).

    Hence E(ψ)c. On the other hand, by the definition of c and note that ψM, we have cE(ψ), so we obtain that E(ψ)=c. This completes the proof of Theorem 1.1.

    We begin with the following lemma (see Lemma 2.13 in [18]).

    Lemma 3.1. Let K={ψS:I(ψ)=0}, then α:=inf{ψw:ψ,wK,ψw}>0.

    As a consequence of Lemma 2.4, we see as Remark 2.12 of [18] that since φ,φ1 are equivariant and E,I are invariant with respect to the action of ZN given by (1.3), there is a one-to-one correspondence between the critical orbits of E|M and I. Hence, the proof of Theorem 1.2 will be completed upon showing that I has infinitely many critical orbits. We shall proceed by contradiction. Namely let us suppose (to the contrary) that the set K only contains finitely many orbits.

    Note that by Theorem 1.1 and Lemma 2.3, the set K is nonempty. Choose a subset J of K such that J=J and each orbit O(ψ)K has a unique representative in J. So we assume by contradiction that

    J is a finite set. (3.1)

    From now on, we assume that the nonlinearity g(x,t) is odd in t. For a functional F we put

    Fd={ψ:F(ψ)d},   Fc={ψ:F(ψ)c},   Fdc={ψ:cF(ψ)d}.

    Lemma 3.2. Let dc. If {v1m}, {v2m}Id are two Palais-Smale sequences for I, then

    either  v1mv2m0 as m  or  lim supmv1mv2mρ(d)>0,

    where ρ(d) depends on d but not on the particular choice of Palais-Smale sequences.

    Proof. We put ψ1m:=φ(v1m) and ψ2m:=φ(v2m). By Lemma 2.4(ii), both sequences {ψ1m}, {ψ2m} are Palais-Smale sequences for E and since {ψ1m}, {ψ2m}Ed, {ψ1m}, {ψ2m} are bounded. We consider two cases.

    Case 1. For 2<q<2, ψ1mψ2mq0 as m. From (g1) and (g2), it follows that for each ε>0 and m large enough, we have that

    ψ1mψ2m2=E(ψ1m),ψ1mψ2mE(ψ2m),ψ1mψ2m+RN[g(x,ψ1m)g(x,ψ2m)](ψ1mψ2m)dxεψ1mψ2m+RN[ε(|ψ1m|+|ψ2m|)+Cε(|ψ1m|q1+|ψ2m|q1)](ψ1mψ2m)dx(1+C1)εψ1mψ2m+Dεψ1mψ2mq,

    where ε>0 is arbitrary, and C1 does not depend on the choice of ε. Notice that ψ1mψ2mq0, therefore ψ1mψ2m0 and Lemma 3.1 implies

    v1mv2m=φ1(ψ1m)φ1(ψ1m)0  as  m.

    Case 2. For 2<q<2, ψ1mψ2mq0 as m. It can be concluded from Lemma 1.21 in [20] that there exists ε>0 and ymRN such that after passing to a subsequence,

    B1(ym)(ψ1mψ2m)2dx=maxyRNB1(y)(ψ1mψ2m)2dxε   for all  m. (3.2)

    Since φ,φ1 and E,I are equivariant with respect to the action of ZN given by (1.3), we may assume that the sequence {ym} is bounded in RN. Passing to a subsequence once more, there exist ψ1,ψ2 and α1,α2 such that

    ψ1mψ1,   ψ2mψ2,   ψ1mα1,   ψ2mα2,

    and E(ψ1)=E(ψ2)=0. According to (3.2), ψ1ψ2 and by Lemma 2.1(iii),

    2cαiν(d)<+,(i=1,2),  where ν(d)=sup{ψ:ψEdM}.

    Suppose ψ1,ψ20. Then ψ1,ψ2M and v1:=φ1(ψ1)K, v2:=φ1(ψ2)K,v1v2. Hence

    lim infmv1mv2m=lim infmψ1mψ1mψ2mψ2mψ1mα1ψ2mα2=β1v1β2v2,

    where

    β1=ψ1α12cν(d),   β2=ψ2α22cν(d).

    Since v1=v2=1, it is easy to see from the inequalities above that

    lim infmv1mv2mβ1v1β2v2min{β1,β2}v1v2α2cν(d)>0, (3.3)

    where α is the constant in Lemma 3.1. Hence, (3.3) implies that

    lim infmv1mv2mρ(d)>0.

    Now the case where either ψ1=0 or ψ2=0 remains to be considered. If ψ2=0, then ψ10 and

    lim infmv1mv2m=lim infmψ1mψ1mψ2mψ2mψ1mα12cν(d)>0.

    The case ψ1=0 is similar.

    It is well known that I admits a pseudo-gradient vector field, i.e., there exists a Lipschitz continuous map H:SKTS (see [17], p.86) such that

    H(w)<2I(w),  H(w),I(w)>12I(w)2,

    where TS denotes the tangent bundle of S. Moreover, seeing that I is even, we may assume H is odd. Let η:GSK be the flow defined by the following Cauchy problem:

    {ddtη(t,w)=H(η(t,w)),η(0,w)=w, (3.4)

    where

    G={(t,w):wSK,T(w)<t<T+(w)},

    and (T(w),T+(w)) is the maximal existence time for the trajectory tη(t,w). Note that η is odd in w because H is odd, and tI(η(t,w)) is strictly decreasing by the properties of a pseudo-gradient.

    Remark 3.1. We note that by the same argument as Lemma 2.15 of [18], we can get: For wS, the limit limtT+(w)η(t,w) exists and is a critical point of I.

    Let AS, δ>0 and define Uδ(A)={wS:dist(w,A)<δ}. Then we have

    Lemma 3.3. Let Kd={ψK:I(ψ)=d and I(ψ)=0} and dc. Then for every δ>0 there exists ε=ε(δ)>0 such that there hold

    (i) Id+εdεK=Kd.

    (ii) limtT+(w)I(η(t,w))<dε for all wId+εUδ(Kd).

    Proof. The proof is virtually identical to Lemma 2.16 in [18], and the details are omitted.

    Now, we will prove the Theorem 1.2. For this purpose, we should first introduce the definition of genus.

    Definition 3.1. For a closed symmetric set A that does not contain the origin, we define the Krasnoselskii genus of A, denoted γ(A), as the smallest integer k such that there exists an odd continuous mapping from A to Rk{0}. If there is no such mapping for any k, we define γ(A)=. Moreover, we set γ()=0.

    Proof of Theorem 1.2. Define

    dk=inf{dR:γ(Id)k}, (kN).

    Then the dk are the numbers at which the sets Id change genus. It is noticeable that dkdk+1. Let k1 and set d=dk. By Lemma 3.1, Kd is either empty set or discrete set, hence γ(Kd)=0 or 1. By the continuity property of the genus, there exists δ>0 such that γ(¯Uδ(Kd))=γ(Kd), where δ<α2. For the δ, we can choose ε>0 such that the conclusions of Lemma 3.3 hold. Then for each wId+εUδ(Kd) there exists t[0,T+(w)) such that I(η(t,w))<dε. Let ϱ=ϱ(w) be the infimum of the time for which I(η(t,w))dε, that is

    ϱ(w)=inf{t[0,T+(w)):I(η(t,w))dε}.

    Since dε is not a critical value of I by Lemma 3.3, it is apparent that by the implicit function theorem, ϱ(w) is a continuous mapping and since I is even, ϱ(w)=ϱ(w). Define a mapping h:Id+εUδ(Kd)Idε by setting h(w):=η(ϱ(w),w). Then h is odd and continuous, so it can be derived from the properties of the genus and the definition of dk that

    γ(Id+ε)γ(¯Uδ(Kd))+γ(Idε)γ(¯Uδ(Kd))+k1=γ(Kd)+k1.

    If γ(Kd)=0, then γ(Id+ε)k1, contrary to the definition of dk, so γ(Kd)=1 and Kd. If dk+1=dk=d, then γ(Kd)>1 (see [16], Proposition 8.5). However, this is impossible, so we have dk+1>dk and Kdk for all k1, hence there is an infinite sequence {±wk} of pairs of geometrically distinct critical points of I with I(wk)=dk, which is a contradiction to (3.1), and Theorem 1.2 is proved.

    In this paper, we are interested in studying a class of biharmonic elliptic equations with potential functions. Our problem is more complicated by the fact that the classical (AR)-type condition is not assumed. We establish the existence results of ground state solutions for the biharmonic elliptic equation (1.1) by using the Nehari manifold method and critical point theories. Moreover, the existence of infinitely many geometrically distinct solutions for this equation is also investigated. We believe that the proposed approach in the present paper can also be applied to study other related equations and systems. An interesting question is whether similar results still hold for a class of biharmonic elliptic systems under the same conditions.

    Dengfeng Lu: Investigation, Writing-original draft; Shuwei Dai: Writing-review & editing. The authors contributed equally to this paper. All authors have read and approved the final version of the manuscript for publication.

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

    The authors are grateful to the referees for their valuable comments and suggestions for improvement of the paper.

    This work is partially supported by the fund from NSFC(12326408).

    The authors declare that they have no competing interests.



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