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Theory article Special Issues

Group invariant solutions for the planar Schrödinger-Poisson equations

  • This paper is concerned with the following planar Schrödinger-Poisson equations

    Δu+V(x)u+(ln|||u|p)|u|p2u=f(x,u),xR2,

    where p2 is a constant, and V(x) and f(x,u) are continuous, mirror symmetric or rotationally periodic functions. The nonlinear term f(x,u) satisfies a certain monotonicity condition and has critical exponential growth in the Trudinger-Moser sense. We adopted a version of mountain pass theorem by constructing a Cerami sequence, which in turn leads to a ground state solution. Our method has two new insights. First, we observed that the integral R2R2ln(|xy|)|u(x)|p|u(y)|pdxdy is always negative if u belongs to a suitable space. Second, we built a new Moser type function to ensure the boundedness of the Cerami sequence, which further guarantees its compactness. In particular, by replacing the monotonicity condition with the Ambrosetti–Rabinowitz condition, our approach works also for the subcritical growth case.

    Citation: Ganglong Zhou. Group invariant solutions for the planar Schrödinger-Poisson equations[J]. Electronic Research Archive, 2023, 31(11): 6763-6789. doi: 10.3934/era.2023341

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  • This paper is concerned with the following planar Schrödinger-Poisson equations

    Δu+V(x)u+(ln|||u|p)|u|p2u=f(x,u),xR2,

    where p2 is a constant, and V(x) and f(x,u) are continuous, mirror symmetric or rotationally periodic functions. The nonlinear term f(x,u) satisfies a certain monotonicity condition and has critical exponential growth in the Trudinger-Moser sense. We adopted a version of mountain pass theorem by constructing a Cerami sequence, which in turn leads to a ground state solution. Our method has two new insights. First, we observed that the integral R2R2ln(|xy|)|u(x)|p|u(y)|pdxdy is always negative if u belongs to a suitable space. Second, we built a new Moser type function to ensure the boundedness of the Cerami sequence, which further guarantees its compactness. In particular, by replacing the monotonicity condition with the Ambrosetti–Rabinowitz condition, our approach works also for the subcritical growth case.



    The present paper is concerned with the existence of solution to the planar Schrödinger-Poisson equations

    Δu+V(x)u+(ln|||u|p)|u|p2u=f(x,u),xR2, (1.1)

    where p2, V,f are continuous, mirror symmetric or rotationally periodic functions, and f(x,t) has exponential critical growth in the Trudinger-Moser sense ([1]).

    In last decades, considerable attention has been paid to the following Schrödinger-Poisson equations:

    {Δu+V(x)u+K(x)ϕ(x)|u|p2u=f(x,u),xR3,Δϕ=K(x)|u|p,xR3 (1.2)

    with various conditions on the parameters p,N and functions V,K,f. These kinds of equations arise in many contexts of physics, such as, in quantum mechanics [2,3,4] and semiconductor theory[5,6,7,8]. In [5], Eq (1.2) was introduced as a model describing solitary waves for nonlinear stationary equations of Schrödinger type interacting with an electrostatic field where the unknown functions u and ϕ wave function for particles and potential, respectively. Let p(1,6] and KL(R3). For each uH1(R3), the second equation in (1.2) determines the Newton potential ϕu in D1,2(R3), i.e.,

    ϕu(x):=14πR3K(y)|u(y)|p|xy|dy.

    Many minimization techniques, such as minimizing on a constraint set [9,10] and the Mountain Pass Theorem [11,12,13,14,15], were used in the Eq (1.2).

    When K(x)0, Eq (1.2) becomes the Schrödinger equation. In this case, there are many results to Eq (1.2) with the dual method if V and f satisfy some certain conditions, such as a positive lower bound on V or a monotonicity condition on f (see [16,17,18,19,20] and references therein).

    In the following, let us focus on the two-dimensional case. Stubbe [21] considered the equations

    {Δu+λu+ϕ(x)u=0,xR2,Δϕ=u2,xR2, (1.3)

    where λR is a constant. They set up a variational framework for Eq (1.3) with a subspace Z of H1(R2):

    Z:={uH1(R2):R2ln(1+|x|)u2dx<}.

    They proved that there exists a unique radial ground state solution for any λ0. In addition, they proved that there exists a negative number λ, such that for any λ(λ,0) there are two radial ground states with different L2 norms. Cigolani and Weth [22] considered Eq (1.1) with p=2 and f(x,u)=b|u|σ2u. Specifically, VC(R2,(0,)) is Z2 periodic. Using the concentration-compactness theory, they proved that Eq (1.1) has a ground state uX2 and a solution sequence {un}nX2, such that limnJ(un)=. Here,

    X2:={uH1(R2):R2[|u|2+V(x)u2+ln(1+|x|)u2]dx<}

    and J are the energy functionals associated with Eq (1.1).

    Chen and Tang [23] considered Eq (1.1) with p=2, i.e.,

    {Δu+V(x)u+ϕ(x)u=f(x,u),xR2,Δϕ=u2,xR2, (1.4)

    where VC(R2,[0,)) is axially symmetrical and fC(R2×R) is of subcritical or critical exponential growth in the sense of Trudinger-Moser. More precisely, we say that f(x,t) has subcritical exponential growth at t=± if it verifies

    (F1') For every A>0,

    supxR2,|s|A|f(x,s)|<+ (1.5)

    and

    lim|t||f(x,t)|eαt2=0,uniformly inR2, (1.6)

    for any α>0;

    and the function f(x,t) is said to have the critical exponential growth at t=± if it verifies

    (F1) The nonlinearity f satisfies (1.5) and there exists α0>0 such that, for any α>α0, (1.6) holds but

    lim|t||f(x,t)|eαt2=+,uniformly inR2for allα<α0.

    This notion of criticality can be referred to [24].

    For the critical growth case, Chen and Tang [23] established the existence of a ground state solution for Eq (1.4) by assuming following conditions on V and f:

    (V0) VC(R,[0,)) and lim inf|x|V(x)>0;

    (CF1) V(x):=V(x1,x2)=V(|x1|,|x2|) for all xR2, f(x,t):=f(x1,x2,t)=f(|x1|,|x2|,t) for all (x,t)R2×R;

    (CF2) f(x,t)t>0 for all (x,t)R2×R{0}, and there exists M0>0 and t0>0 such that

    F(x,t)M0|f(x,t)|,xR2,|t|t0,

    where F(x,t):=t0f(x,s)ds;

    (CF3) lim inf|t|t2F(x,t)eα0t2>2α20ρ2 uniformly on xR2, where ρ(0,1/2) satisfying ρ2max|x|ρV(x)1;

    (CF4) f(x,t)V(x)t|t|3 is non-decreasing on tR{0}.

    Recently, Cao et al. [25] considered the equations

    Δu+V(x)u+(ln|||u|p)|u|p2u=b|u|σ2u,xR2, (1.7)

    where σ2p,b0 and VC(R2,(0,)) are Z2 periodic. With a similar method in [22], they obtained the existence of a positive ground state solution of Eq (1.7) in Xp, where

    Xp:={uH1(R2):R2[|u|2+V(x)u2+ln(1+|x|)|u|p]dx<}.

    Here, we will prove the existence of a nontrivial solution to Eq (1.1), not only for all p2, but also for general nonlinearities f and potentials V.

    To describe our main results, we introduce the following notations: Let us view R2 as C, let kN,k2 and we say that vPk if v(ze2πi/k)=v(z) over C. We define

    Ek,p:=XpPk,Vk,1:=C(C)Pk
    Fk,1:={fC(C×R):f(,t)Pk,tR}.

    We say that v is mirror symmetric denoted by vM if v(¯z)=v(z) in C. Let

    Tk,p:=Ek,pM,Vk,2:=Vk,1M,
    Fk,2:={fFk,1:f(,t)M,tR}.

    Finally, we write the associated functional of Eq (1.1) in the following form

    Φ(u)=12u2+14pπR2R2ln(|xy|)|u(x)|p|u(y)|pdxdyR2F(x,u)dx, (1.8)

    and the associated Nehari manifold of the functional (1.8) is

    N1:={uEk,p{0}:Φ(u),u=0},N2:={uTk,p{0}:Φ(u),u=0}. (1.9)

    Our main result is stated as follows.

    Theorem 1.1. Let p2, V and f satisfy (V0),(F1) and the following conditions

    (VF)VVk,1andfFk,1 with k4 or VVk,2andfFk,2 with k2;

    (F2)f(x,t)t0 for all (x,t)R2×R{0}, and there exists M0>0 and t0>0 such that F(x,t)M0|f(x,t)| for xR2, |t|t0;

    (F3) There exists qR such that lim inf|t||t|qF(x,t)eα0t2=+;

    (F4)gp(x,t) is non-decreasing on t(,0) and t(0,), where

    gp(x,t):={f(x,t)V(x)t|t|2p1,p=2,f(x,t)μV(x)t|t|2p1,p>2

    for some μ<1.

    (F5) If p=2, f(x,t)=o(t) as t0 uniformly on R2; and if p>2, f(x,t)=O(ts0) with s0>1 as t0 uniformly on R2.

    Then, Eq (1.1) has a nontrivial solution ˉu. Moreover, if VVk,1 and fFk,1, then ˉuEk,p satisfies

    Φ(ˉu)=minN1Φ;

    if VVk,2 and fFk,2, then ˉuTk,p satisfies

    Φ(ˉu)=minN2Φ.

    Remark 1.2. Comparing to [23, Theorem 4], we have weakened the assumptions (CF1)(CF3) to (VF) and (F2)(F3), respectively. More precisely,

    (CF1) means VV2,2 and fF2,2, hence, it is a special case of (VF);

    The condition (F3) is less restrictive than (CF3) for the behavior of f at infinity;

    (F2) improves slightly (CF2) where f(x,t)t>0 is replaced by f(x,t)t0;

    Here is an example of f, which satisfies (VF) and (F1)(F5), but not (CF3). Let θ>0,p0p,q0>2 and

    f0(x,t)={θeα0t2(2α0t2q0)tq0+1,tp0+q0α0,θep0+q0(2p0+q0)αp0+q0/20(p0+q0)p0+q0/2t2p01,0t<p0+q0α0

    with odd extension to t<0. Finally, it seems that [23] used implicitly f(t)=o(t) as t0 with p=2 in (F5) (see the proof of Lemma 2.6 there).

    Our approach works also for the subcritical case.

    Theorem 1.3. Let p2, V and f satisfy (V0), (VF), (F1'), (F5) and the following condition:

    (F4') f(x,t)t>0 for all (x,t)R2×(R{0}) and there exists ν(2,),t1(0,) such that

    f(x,t)tνF(x,t),xR2,|t|t1;

    Furthermore, if p>2, we assume that

    Mt1<(121μ)γ2,

    where

    Mt1=sup(x,t)R2×[t1,t1]{0}F(x,t)t2andγ=infuXpuuH1(R2)>0.

    Then, Eq (1.1) has a nontrivial solution ˉu. Moreover, ˉuEk,p if VVk,1 and fFk,1, and ˉuTk,p if VVk,2 and fFk,2.

    This paper is organized as follows: In Section 2, we present some basic results; in particular we show that the energy functional corresponding to the nonlocal term is non positive, which is our key observation and different from the available results, see Lemma 2.3. In Section 3, we prove a mountain pass type theorem using a new test function, see Lemma 3.2 below. In Sections 4 and 5, we give the proof of Theorems 1.1 and 1.3, respectively.

    In this section, we will give some preliminary definitions and basic facts about inequalities, such as the Moser-Trudinger inequality, the energy estimate of the nonlocal term. In the following, the letter C denotes generic positive constants and q denotes the standard norm in Lq(R2).

    The function space Xp is a Banach space equipped with the norm

    uXp:=u+u, (2.1)

    where

    u:=(R2ln(1+|x|)|u|pdx)1p; (2.2)

    while

    u:=(R2[|u|2+V(x)u2]dx)12 (2.3)

    is induced by the scalar product

    u,v:=R2(uv+V(x)uv)dx. (2.4)

    We will use the following bilinear functionals (see [21]):

    A1(u,v):=R2R2ln(1+|xy|)u(x)v(y)dxdy; (2.5)
    A2(u,v):=R2R2ln(1+1|xy|)u(x)v(y)dxdy; (2.6)
    A0(u,v):=A1(u,v)A2(u,v)=R2R2ln(|xy|)u(x)v(y)dxdy. (2.7)

    By the Hardy-Littlewood-Sobolev inequality (see [26]), there exists C>0 such that for any u,vL4/3(R2),

    |A2(u,v)|R2R21|xy|u(x)v(y)dxdyCu43v43. (2.8)

    Corresponding to (2.5)–(2.7), we define

    Ii(u):=Ai(|u|p,|u|p),i=0,1,2. (2.9)

    The following bound for I2(u) is a direct consequence of (2.8):

    |I2(u)|Cu2p4p3,uL4p3(R2),p1. (2.10)

    We can rewrite the associated functional of Eq (1.1) in the following form

    Φ(u)=12u2+14pπI0(u)R2F(x,u)dx. (2.11)

    Next, we state several lemmas.

    Lemma 2.1. (i) Let uH1(R2), then for any α>0,

    R2(eαu21)dx<.

    (ii) Given M>0, α(0,4π), there exists a constant C(M,α) such that for all uH1(R2) satisfying u21, u2M, there holds

    R2(eαu21)dx<C(M,α).

    The statements (i) and (ii) of the above lemma were first established by [27, Lemma 1] and [1, Lemma 2.1], respectively (see also [28,29]).

    Lemma 2.2. Assume that V and f satisfy (V0),(F1)(or (F1)),(F5). Then, Ii,ΦC1(Xp,R) and

    Ii(u),v=2pAi(|u|p,|u|p2uv),i=1,2Φ(u),v=u,v+12πA0(|u|p,|u|p2uv)R2f(x,u)vdx. (2.12)

    For the sake of completeness, we present a proof of Lemma 2.2 in the appendix. The following lemma is our first key observation.

    Lemma 2.3. For any uXp, we have I0(u)0.

    Proof. First, let uC0(R2) with supp(u)B12(0). Then

    I0(u)=B12(0)B12(0)ln(|xy|)|u(x)|p|u(y)|pdxdy0.

    Consider now uC0(R2,R). Take R>0 such that supp(u)BR(0) and let w(x)=u(2Rx), so supp(w)B12(0) and ϕw(x):=14R2ϕu(2Rx). Hence,

    12πI0(u)=BR(0)BR(0)ln(|xy|)|u(x)|p|u(y)|pdxdy=16R4B12(0)B12(0)ln(|xy|)|w(x)|p|w(y)|pdxdy0.

    We conclude by the density argument. For any R>0, let φR(r) be a C0 cut-off function such that 0φ1, φR1 on [0,R] and φR0 on [R+1,). Let η be the standard mollifier and ηδ(x):=1δ2η(xδ), where δ>0. Given ϵ>0, since VuL2(R2),[ln(1+||)]1/puLp(R2), [30, Pages 264 and 714], we can choose δ small enough such that

    φ1δ(||)uuXp<ϵ,ηδ[φ1δ(||)u]φ1δ(||)uXp<ϵ.

    Therefore, for any uXp, there exists {un}nC0(R2) such that limnunuXp=0. By the fact I0=I1I2 and Lemma 2.2, we conclude that I0(u)0.

    Corollary 2.4. Assume that V and f satisfy (V0) and (F1)(or (F4)). Then

    limtΦ(tω)=,ωXp{0}.

    Proof. For any ωXp{0}, there exists δ>0 such that m{|ω(x)|δ}>0. For a critical case, by Lemma 2.3 and (F1), one has

    Φ(tω)=t22ω2+t44pπI0(ω)R2F(x,tω)dxt22ω2C{|ω(x)|δ}eα0δ2t2/2dx,ast.

    For a subcritical case, we choose large enough R>0 such that m(G)>0, where

    G:={|ω(x)|δ}BR(0).

    By (F4') and choosing M:=ω2m(G)>0, there exists tM>0 such that

    |F(x,tw)|Mt2,xG,|t|tM,

    which together with Lemma 2.3 implies

    Φ(tω)=t22ω2+t44pπI0(ω)R2F(x,tω)dxt22ω2t2Mm(G)=t22ω2

    as t, and we complete the proof.

    The following lemma is inspired by [31, Lemma 2.2].

    Lemma 2.5. Assume that V and f satisfy (V0) and (VF). Then there exists Ck>0 such that

    A1(|u|p,|v|p)Ckupvpp,u,vEk,p. (2.13)

    In particular, since Tk,pEk,p, (2.13) holds for u,vTk,p.

    Proof. Let Ω1:={(x1,x2)R2:xi0}, Ω2=Ω1. For any xΩ1 and yΩ2, one has

    |xy|2=|x|2+|y|22xy|x|2+|y|2.

    Then, it follows from the definition of Ek,p and k4 that

    A1(|u|p,|v|p)=R2R2ln(1+|xy|)|u(x)|p|v(y)|pdxdyΩ2|v(y)|pdyΩ1ln(1+|xy|)|u(x)|pdxΩ2|v(y)|pdyΩ1ln(1+|x|)|u(x)|pdx1k2R2|v(y)|pdyR2ln(1+|x|)|u(x)|pdxCkupvpp,u,vEk,p,

    so we obtain (2.13).

    In this section, we will quote a version of Mountain Pass Theorem and prepare the proof of Theorems 1.1 and 1.3.

    Lemma 3.1. Let Y be a real Banach space and IC1(Y,R). Let S be a closed subset of Y, which disconnects Y into distinct connected Y1 and Y2. Suppose further that I(0)=0 and

    (i) 0Y1, and there exists α>0 such that I|Sα,

    (ii) There is eY2 such that I(e)0.

    Then, I possesses a Cerami sequence with cα>0 given by

    c=infγΓmaxt[0,1]I(γ(t)),

    where

    Γ={γC([0,1],X):γ(0)=0,γ(1)=e},

    and a Cerami sequence means a sequence {un}X such that

    I(un)c,I(un)Y(1+unY)0.

    The proof of the above lemma can be found in [32, Theorem 3]. We state another result that serves as a bridge between the mountain pass structure (see Lemma 3.3) and Theorem 1.1.

    Lemma 3.2. Assume that V and f satisfy (V0), (F1) and (F3)(F5). Then there exists n0N such that

    maxt0Φ(tωn0)<2πα0, (3.1)

    where

    ωn(x)={lnn2πqln(lnn)22πlnn,0|x|(lnn)q/2/n;ln(1/|x|)2πlnn,(lnn)q/2/n|x|1;0,|x|1.

    Proof. Without loss of generality, we can fix q2. Direct computation yields

    ωn2B1|ωn|2dx+V1B1ω2ndx=1qln(lnn)2lnn+δn, (3.2)

    where δn=O(1lnn) as n. By (F3), there exists t0>0 such that

    |t|qF(x,t)eα0t21,|t|t0. (3.3)

    There are three cases for the value of t.

    Case (i): 0t3πα0. For large n, then it follows from (3.2) and Lemma 2.3 that

    Φ(tωn)=t22ωn2+t2p2pI0(ωn)R2F(x,tωn)dx1+δn2t27π4α0. (3.4)

    Case (ii): 3πα0t8πα0. For large n, we have tωn(x)t0 for xB(lnn)q/2/n. Then it follows from (3.2), (3.3) and Lemma 2.3 that

    Φ(tωn)=t22ωn2+t2p2pI0(ωn)R2F(x,tωn)dx1+δn2t2qln(lnn)4lnnt22q/2π1+q/2(lnn)qn2tqTq/2neα02πt2Tn1+δn2t2qln(lnn)4lnnt2αq/20π(lnn)q2qn2Tq/2neα02πt2Tn=:φn(t), (3.5)

    where

    Tn:=lnnqln(lnn)+q2ln2(lnn)4lnn.

    Let tn>0 be the unique maximum of φn in R+, then (as n)

    t2n=4πα0[1+(q1)ln(lnn)2lnn+O(1lnn)] (3.6)

    and

    φn(t)φn(tn)=1+δn2t2nqln(lnn)4lnnt2n+O(1lnn). (3.7)

    Combining (3.5)–(3.7), one has

    Φ(tωn)φn(tn)=2πα0ln(lnn)2lnn+O(1lnn). (3.8)

    Case (iii): t8πα0. As in the above case (ii), we have

    Φ(tωn)1+δn2t22q/2π1+q/2(lnn)qn2tqTq/2neα02πt2Tn1+δn2t22q/2π1+q/2(lnn)qtqTq/2nexp[2(α04πt21)Tn]4π(1+δn)α0αq/20π(lnn)q/22qn20 (3.9)

    for large n. To get the third inequality, we used the fact that the function

    1+δn2t22q/2π1+q/2(lnn)qtqTq/2nexp[2(α04πt21)Tn]

    is decreasing on t8πα0 when n is large enough. Combining the conclusions for cases (i)–(iii), the proof is completed.

    Now we show the existence of the Cerami sequence.

    Lemma 3.3. Assume that V and f satisfy (V0), (VF), (F1)(or (F1) and (F4)) and (F5). Then there exists a constant ˜c(0,supt0Φ(tωn0)] and a Cerami sequence {un}Ek,p such that

    Φ(un)˜c,Φ(un)Xp(1+unXp)0. (3.10)

    Proof. Applying the Sobolev embedding theorem for given s[2,), there exists γs>0 such that

    usγsu,uXp. (3.11)

    By (F1) (or (F1')) and (F5) for any ϵ>0, there exists some constant Cϵ>0 such that

    |F(x,t)|ϵt2+Cϵ(e3α0t2/21)|t|3,(x,t)R2×R. (3.12)

    On the other hand, in view of Lemma 2.1, one has

    R2(e3α0u21)dxC,uπα0. (3.13)

    Let ϵ=14γ22 from (3.11)–(3.13), and there holds

    R2F(x,u)dx14u2+C3u3,uπα0. (3.14)

    Hence, it follows from (2.11) and (3.14) that if uπα0,

    Φ(u)=12u2+14pπ(I1(u)I2(u))R2F(x,u)dx14u2C3u3C4u2p. (3.15)

    Therefore, there exists κ0>0 and 0<ρ<πα0 such that

    Φ(u)κ0,uS:={uEk,p:u=ρ}. (3.16)

    By (V0), (F1) (or (F4')) and Corollary 2.4, we have limtΦ(tωn0)=, and then we can choose t>0 such that e=tωn0Y2:={uEk,p:u>ρ} and Φ(e)<0. Let Y1:={uEk,p:uρ}, then in view of Lemma 3.1, one deduces that there exists ˜c[κ0,supt0Φ(tωn0)] and a Cerami sequence {un}Ek,p satisfying (3.10).

    The proof of Theorem 1.1 is based on the following lemmas. As in Lemma 2.5, we only consider the Ek,p case.

    Lemma 4.1. Assume that V and f satisfy (V0),(VF),(F1),(F4) and (F5), then we have

    (i) Let m1:=infN1Φ(u), then there exists a constant c(0,m1] and a sequence {un}Ek,p satisfying

    Φ(un)c,Φ(un)Xp(1+unXp)0. (4.1)

    (ii) For any uEk,p{0}, there exists a unique tu>0 such that tuuN1. Moreover, we have

    m1=infEk,p{0}maxt0Φ(tu).

    Proof. We will prove that

    Φ(u)=maxt0Φ(tu),uN1 (4.2)

    and then we can get the statement (i). Indeed, if (4.2) holds the same as [33, Lemma 3.2], we can choose vkN1 such that

    m1Φ(vk)m1+1k,kN.

    For any vk, similarly to Lemma 3.3, we can obtain a Cerami sequence {uk,n}nEk,p such that

    Φ(uk,n)ck,Φ(uk,n)Xp(1+unXp)0,kN

    with ck(0,supt0Φ(tvk)]. By (4.2) and the diagonal rule, we can verify (4.1), and now we prove (4.2). By (2.11) and (2.12), one has

    Φ(u)Φ(tu)=1t22u2+1t2p4pπI0(u)+R2[F(x,tu)F(x,u)]dx=1t2p2pΦ(u),u+t2ppt2+p12pu2+R2[1t2p2pf(x,u)u+F(x,tu)F(x,u)]dx=1t2p2pΦ(u),u+t2ppt2+p12pu2+R21t[f(x,u)V(x)u|u|2p1f(x,su)V(x)su|su|2p1]s2p1|u|2p1udsdx1t2p2pΦ(u),u+t2ppt2+p12pu2. (4.3)

    According to the fact uN1 and mint0(t2ppt2+p1) attained at t=1, then (4.2) holds.

    Next, we consider statement (ii). Let uEk,p{0} be fixed and ζ(t):=Φ(tu) on [0,). By the definition (2.11),

    ζ(t)=0t2u2+t2p2πI0(u)R2f(x,tu)tudx=0tuN1.

    Using (3.15), (F1) and Lemma 2.3, one has ζ(0)=0, ζ(t)>0 for t>0 small and ζ(t)<0 for t large. Therefore maxt(0,)ζ(t) is achieved at some tu>0 so that ζ(tu)=0 and tuuN1. Now, we claim that tu is unique. In fact, for any given uEk,p{0}, let t1,t2>0 such that ζ(t1)=ζ(t2)=0. By (4.3), taking t=t2t1 and t=t1t2 respectively, it implies

    Φ(t1u)Φ(t2u)+t21g(t2/t1)2pu2andΦ(t2u)Φ(t1u)+t22g(t1/t2)2pu2,

    where g(t):=t2ppt2+p1. Therefore, we must have t1=t2, since g(s)>0 for any s>0, s1.

    Lemma 4.2. Assume that V and f satisfy (V0),(VF),(F1),(F4) and (F5). Then any sequence satisfying (4.1) is bounded w.r.t. .

    Proof. We only consider the case p>2. The case p=2 is obtained by [23, Lemma 2.11]. First, we prove that

    12pf(x,t)tF(x,t)μ(1p)2pV(x)t2,tR. (4.4)

    Indeed, by (F4), there holds

    F(x,t)μ2V(x)t2=t0[f(x,τ)μV(x)τ]dτt0f(x,t)μV(x)t|t|2p1|τ|2p2τdτ=f(x,t)tμV(x)t22p.

    By (4.4), one has

    c+o(1)=Φ(un)12pΦ(un),un=(1212p)un2+R2(12pf(x,un)unF(x,un))dx(1212p)un2(1212p)μR2V(x)u2ndx(p1)(1μ)2pun2. (4.5)

    Here, we also used (2.11), (2.12) and (4.1). Therefore, we complete the proof.

    Proof of Theorem 1.1 completed. Applying Lemmas 4.1 and 4.2, we deduce that there exists a sequence {un}Ek,p satisfying (4.1) and unC<. Now, we prove

    R2f(x,un)undxC. (4.6)

    Indeed, let p2, and by (2.11), (2.12) and (4.1) there holds

    c+o(1)=Φ(un)p+μ(1p)2pΦ(un),un(p1)μ2pun2(p1)μ2pR2V(x)u2ndx+(1p)(1μ)4pπI0(un)+(p1)(1μ)2pR2f(x,un)undx(p1)(1μ)2pR2f(x,un)undx,

    hence (4.6) holds true. Next, we complete the proof of Theorem 1.1 in three steps.

    Step 1: {un} is bounded in Ek,p.

    We first prove that δ0:=lim supnunp>0. Suppose the contrary δ0=0, then from the Gagliardo-Nirenberg inequality (see [34, Page 125]):

    unssCsunθpun1θ2, (4.7)

    where 2p<t<,θ=pt. Hence, un0 in Lη(R2) for η(2,+). Given any ε(0,M0C10/t2), we choose Mε>M0C10/ε, then it follows from (F2) and (4.6) that

    |un|MεF(x,un)dxM0|un|Mε|f(x,un)|dxM0Mε|un|Mεf(x,un)undx<ε. (4.8)

    Applying (F5), one has

    |un|MεF(x,un)dx{Cεun22=o(1),p=2,Cεuns+1s+1=o(1),p>2 (4.9)

    and

    |un|1f(x,un)undx{Cun22=o(1),p=2,Cuns+1s+1=o(1),p>2. (4.10)

    By the arbitrariness of ε>0, we deduce from (F2), (4.8) and (4.9) that

    R2F(x,un)dx=o(1). (4.11)

    Hence, by (2.10) we have

    0I2(un)Cun2p4p3=o(1). (4.12)

    By Lemmas 3.2 and 3.3, we know that ˉε:=13(1α0˜c2π)>0, which together with (2.11), (3.10), (4.11), (4.12) and the fact I1(un)0 implies

    un2=2˜c12pπI1(un)+12pπI2(un)+o(1)2˜c+o(1)=4πα0(13ˉε)+o(1). (4.13)

    Now, let d(1,pp1) satisfy

    (1+ˉε)(13ˉε)d1ˉε<1. (4.14)

    By (F1), there exists C>0 such that

    |f(x,t)|dC[eα0(1+ˉε)dt21],xR2,|t|1. (4.15)

    It follows from (4.13)–(4.15) and Lemma 2.1 that

    |un|1|f(x,un)|ddxCR2[eα0(1+ˉε)du2n1]dx=CR2[eα0(1+ˉε)dun2(un/un)21]dxC. (4.16)

    As d=dd1>p, using (4.16) there holds

    |un|1f(x,un)undx[|un|1|f(x,un)|qdx]1/dund=o(1). (4.17)

    Combining (2.10)–(2.12), (3.10), (4.10) and (4.17), we arrive at

    ˜c+o(1)=Φ(un)12Φ(un),un=(14π14pπ)I1(un)+(14π14pπ)I2(un)+R2[12f(x,un)unF(x,un)]dxo(1). (4.18)

    This contradiction shows that δ0>0. Now, from (2.10), (4.5) and Lemma 2.3, one has

    I1(un)I2(un)C,

    which, together with Lemma 2.5, implies that un is bounded and {un} is bounded in Ek,p.

    Step 2: Φ(ˉu)=0 in Ek,p and Φ(ˉu)=m1.

    We can assume by [25, Lemma 2.3] and passing to a subsequence again if necessary, that unˉu in Ek,p, unˉu a.e. on R2 and

    unˉuinLs(R2),

    where s[2,) if p=2 and s(2,) if p>2. First, we need prove that

    limnR2F(x,un)dx=R2F(x,ˉu)dxandlimnR2f(x,un)ˉudx=R2f(x,ˉu)ˉudx. (4.19)

    Since (4.6) and the condition (F1), (F2) and (F5) hold the same as [23, Assertions 2 and 3], (4.19) still holds. Next, we prove that

    limnI2(un)=I2(ˉu). (4.20)

    Indeed, noting that unˉu in L4p3(R2) by [35, Lemma A.1], there exists w0L4p3(R2) such that

    |un(x)|w0(x)and|ˉu(x)|w0(x),

    a.e., for a subsequence if necessary, which together with the Lebesgue dominated convergence theorem and Hardy-Littlewood-Sobolev inequality implies

    |I2(un)I2(ˉu)||A2(|un|p,|un|p|ˉu|p)|+|A2(|un|p|ˉu|p,|ˉu|p)|=o(1) (4.21)

    as n and (4.20) is proved. Now, we claim that

    Φ(ˉu)=m1,Φ(ˉu),ˉuXp,Xp=0. (4.22)

    Indeed, similar as (4.20), we also have

    A2(|un|p,|un|punˉu)A2(|ˉu|p,|ˉu|p)=o(1). (4.23)

    By [25, Lemma 3.3], one has

    A1(|un|p,|un|p2ˉu(unˉu))=o(1),

    which together with (3.10), (4.19) and Fatou's Lemma implies

    o(1)=Φ(un),ˉuXp,Xp=un,ˉu+12πA1(|un|p,|un|p2unˉu)12πA2(|un|p,|un|p2ˉu2)R2f(x,un)ˉudx+o(1)Φ(ˉu),ˉuXp,Xp+o(1). (4.24)

    Hence, we can obtain

    Φ(ˉu),ˉuXp,Xp0. (4.25)

    Since ˉu0, by Lemma 4.1 there exists ˉt(0,1] such that ˉtˉuN1. By (3.10), (4.4) and (4.25), the weak lower semi-continuity of norm, Lemma 4.1, the condition (F4) and Fatou's Lemma, we have

    m1c=limn[Φ(un)12pΦ(un),unXp,Xp]=limn{(1212p)un22+R2[12pf(x,un)unF(x,un)+p12pV(x)u2n]dx}(1212p)ˉu22+R2[12pf(x,ˉu)ˉuF(x,ˉu)+p12pV(x)ˉu2]dx=Φ(ˉu)12pΦ(ˉu),ˉuXp,XpΦ(ˉtˉu)ˉt2p2pΦ(ˉu),ˉuXp,Xpm1ˉt2p2pΦ(ˉu),ˉuXp,Xpm1, (4.26)

    which implies (4.22) and

    limnΦ(un)=m1. (4.27)

    By (4.19), (4.20), (4.22) and (4.27) and the weak lower semi-continuity of norm and Fatou's Lemma, one has

    o(1)=Φ(un)Φ(ˉu)=un2ˉu2+14pπ[I1(un)I1(ˉu)]+o(1), (4.28)

    which implies

    limnunˉu=0andlimnI1(un)=I1(ˉu). (4.29)

    Hereafter, we claim that

    A1(|un|p,|vn|p)0,asn, (4.30)

    where |vn|p:=|un|p2|unˉu|2. Indeed, we have

    A1(|un|p,|vn|p)=I1(un)2A1(|un|p,|un|p2(unˉu)ˉu)A1(|un|p,|un|p2ˉu2).

    Then, we estimate

    |A1(|un|p,|un|p2(unˉu)ˉu)|unppR2|un|p2|unˉu||ˉu|dx+unpunp2punˉupˉupunppR2|un|p2|unˉu||ˉu|ln(1+|x|)dx+on(1). (4.31)

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

    h(Rϵ):=(R2BRϵ(0)|ˉu|pln(1+|x|)dx)1/p<ϵ.

    Now, we split

    R2|un|p2|unˉu||ˉu|ln(1+|x|)dx=dn(Rϵ)+en(Rϵ),

    where

    dn(Rϵ):=BRϵ(0)|un|p2|unˉu||ˉu|ln(1+|x|)dxRϵunp2punˉupˉupϵ

    for large enough n and

    en(Rϵ):=R2BRϵ(0)|un|p2|unˉu||ˉu|ln(1+|x|)dxunuunp2(R2BRϵ(0)|ˉu|pln(1+|x|)dx)1/pCh(Rϵ)Cϵ,

    which together with (4.31) implies

    limnA1(|un|p,|un|p2(unˉu)u)=0.

    Hence, by Fatou's Lemma, we have

    lim supnA1(|un|p,|vn|p)lim supnI1(un)lim infnA1(|un|p,|un|p2ˉu2)lim supnI1(un)I1(ˉu).

    Since I1(un)I1(ˉu) and A1(|un|p,|vn|p)0, we conclude with (4.30). Finally, by [25, Lemma 3.2] we obtain unˉu0, which together with (4.27) and (4.29) implies

    limnunˉuXp=0andΦ(ˉu)=m1.

    Step 3: Φ(ˉu)=0 in Xp.

    By using the group action on the space Xp, we will conclude Φ(ˉu)=0. Let GO(2) be a finite group of transforms acting on Xp, where O(2) denotes the group of orthogonal transformations in R2. The action of G on the space Xp is a continuous map (see [35, Definition 1.27]):

    G×XpXp:[τ,u]τ(u)=uτ.

    Assume that φC1(Xp,R) is invariant by G; that is, φ(wτ)=φ(w) for any τG, wXp. Let u be a critical point of φ in Xp,G, where

    Xp,G:={uXp:τu=u,τG}.

    Then φ(u)=0 in Xp. In fact, given any vXp, we define

    ˉv=1#(G)τGτv,

    where #(G) denotes the cardinal of G. For any τ0G, since GO(2), we have

    τ0ˉv=τ0[1#(G)τGτv]=1#(G)τGτ0τv=ˉv,

    which implies ˉvXp,G. Therefore, one has

    0=φ(u),ˉv=1#(G)τGφ(u),τv=1#(G)τGφ(u)τ1,v=1#(G)τGφ(uτ1)τ1,v=φ(u),v.

    For the second line, we used the fact uXp,G, so we have φ(u)=0 in Xp.

    The two cases in Theorem 1.1 are direct consequences of the above discussion. Indeed, let G1 be the subgroup of O(2) generated by zze2πi/k, then Xp,G1=Ek,p. If G2 is generated by zze2πi/k and zˉz, Xp,G2=Tk,p, so Φ(ˉu)=0 in Xp.

    We prove Theorem 1.3 in three steps for subcritical case and the Ambrosetti-Rabinowitz condition (F4'). As in the proof of Theorem 1.1, we only consider the function space Ek,p.

    Step 1: un is bounded.

    We only consider the case p>2. Same as critical case, the case p=2 can be obtained by [23, Lemma 2.11]. Applying Lemma 3.3 and (5.1), there exists a sequence {un}Ek,p satisfying (3.10). By (3.10) and (F4'), we can choose a constant λ0(1ν,12Mt1γ2) and then we have

    ˜c+o(1)=Φ(un)λ0Φ(un),un=(12λ0)un2+12π(12pλ0)I0(un)+R2(λ0f(x,un)unF(x,un))dx(12λ0)un2{|un|<t1}F(x,un)dx+(λ01ν)unt1f(x,un)undx(12Mt1γ2λ0)un2, (5.1)

    and then un is bounded.

    Step 2: {un} is bounded in Ek,p.

    As in the proof of the critical case, we first prove δ0:=lim supnunp>0. Suppose the contrary δ0=0. Denoting M:=supnun and M:=supnun2. By (F1') and (F5), choosing α(0,p1pM2), one has

    |f(x,t)|˜c2M2|t|+C(eαt21),(x,t)R2×R. (5.2)

    By (5.2) and Lemma 2.1, we have

    R2f(x,un)undx˜c2M2un22+CR2(eαu2n1)|un|dx˜c2+C[R2(epp1αu2n1)dx]p1punp=˜c2+C[R2(epp1αun2(u2n/un2)1)dx]p1punp˜c2+o(1). (5.3)

    Hence, by (5.3) and Lemma 3.3, we know

    ˜c+o(1)=Φ(un)12Φ(un),un=p14pπI1(un)+p14pπI2(un)+R2[12f(x,un)unF(x,un)]dx˜c2+o(1). (5.4)

    This contradiction shows that δ0>0. Now, from (2.10), (5.1) and Lemma 2.3, one has

    I1(un)I2(un)C,

    which, together with Lemma 2.5, implies that un is bounded and {un} is bounded in Ek,p.

    Step 3: Φ(ˉu)=0 in Xp.

    We may assume by [25, Lemma 2.3] and passing to a subsequence again if necessary, that unˉu in Ek,p, unˉu, a.e., on R2 and

    unˉuinLs(R2),

    where s[2,) if p=2 and s(2,) if p>2. Let M:=supnun2. By (F1'), we can choose α>0 small enough such that M2<4πα. Therefore, there is β>p big enough such that M2ββ1<4πα. Without loss of generality, we may assume that s0(1,2) in (F5). Then, it follows (F5) and Lemma 2.1 that

    R2|f(x,un)(unˉu)|dx{|un|<1}|f(x,un)(unˉu)|dx+{|un|1}|f(x,un)(unˉu)|dxCun2unˉu22s0+Cunˉuβ=o(1). (5.5)

    Similarly, one has

    R2|f(x,ˉu)(unˉu)|dx=o(1). (5.6)

    Furthermore, it follows from (2.9), (2.10) and the Hölder inequality that

    A2(|un|p,|un|p2un(unˉu))=o(1),A2(|ˉu|p,|ˉu|p2ˉu(unˉu))=o(1). (5.7)

    By [25, Lemma 3.3], we have

    A1(|un|p,|un|p2ˉu(unˉu))=o(1),A1(|ˉu|p,|ˉu|p2ˉu(unˉu))=o(1). (5.8)

    Combining (2.11), (2.12), (3.10), and (5.5)–(5.8), there holds

    o(1)=Φ(un)Φ(ˉu),unˉuXp,Xp=unˉu2+12πA1(|un|p,|un|p2(unˉu)2)+12πA1(|un|p,|un|p2ˉu(unˉu))12πA1(|ˉu|p,|ˉu|p2ˉu(unˉu))+12πA2(|ˉu|p,|ˉu|p2ˉu(unˉu))12πA2(|un|p,|un|p2un(unˉu))+R2f(x,ˉu)(unˉu)dxR2f(x,un)(unˉu)dx+o(1)unˉu2+o(1). (5.9)

    By (2.10), (5.9) and Lemma 2.1, we have

    A2(|un|p,|vn|p)=o(1),un,unˉu=o(1),R2f(x,un)(unˉu)dx=o(1), (5.10)

    where |vn|p:=|un|p2|unˉu|2 for every nN. By (3.10) and (5.10), one has

    o(1)=Φ(un),unˉuXp,Xp=12πA1(|un|p,|vn|p)12πA2(|un|p,|vn|p)+un,unˉu+R2f(x,un)(unˉu)dx=12πA1(|un|p,|vn|p)+o(1)

    which, together with Lemma 2.5, implies

    limn(vnp+vn)0. (5.11)

    From (5.11) and [25, Lemma 3.3], one has

    unˉup=R2ln(1+|x|)(|unˉu|p2|un|p2)|unˉu|2dx+o(1)12unˉup+CR2ln(1+|x|)|ˉu|p2|unˉu|2dx+o(1)=12unˉup+o(1), (5.12)

    where we used the following inequality

    ||a+b|p2|b|p2|12|b|p2+C|a|p2,

    and C is independent of a,bR. Combining with (5.9), we have unˉu in Ek,p. Hence, 0<˜c=limnΦ(un)=Φ(ˉu) and Φ(ˉu)=0 in Ek,p. We conclude that Φ(ˉu)=0 in Xp, as in the proof of step 3 of Theorem 1.1.

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

    This work is partially supported by National Natural Science Foundation of China, No. 12071189. Thanks to the reviewers for their valuable suggestions and comments. The author would like to express the sincere gratitude to his tutors, professors Dong Ye and Feng Zhou, for their valuable guidance and insights that helped to improve the paper.

    The authors declare there are no conflicts of interest.

    In this section, we give the proof of Lemma 2.2. For any uXp, we denote Ψ(u):=R2F(x,u)dx. In fact, we just need to prove ΨC1(Xp,R), and the readers can refer to [25, Lemma 2.3] for the rest. First, given any u,vXp, for almost every xR2

    limt0F(x,u(x)+tv(x))F(x,u(x))t=f(x,u(x))v(x).

    On the other hand, we can choose a large enough number t1>0 such that

    |f(x,t)|e(α0+1)t21,|t|t1.

    By (F1), (F5) and Lemma 2.1, one has, for any uXp,

    R2|f(x,u)|2dx={|u|t1}|f(x,u)|2dx+{|u|t1}|f(x,u)|2dxCu22+R2(e(α0+1)u21)dxC.

    Then, for any uXp, f(x,u)L2(R2), it implies that the Gateaux derivative Ψg(u) exists and Ψg(u)Xp.

    Now let {un}Xp, unˉuXp0. Hence, unˉu in H1(R2). Let us prove Ψ(ˉu)=limnΨ(un). It suffices to prove

    limnsupvXp=1|R2[f(x,un)f(x,ˉu)]vdx|=0.

    Define that M:=supnun2, then we prove this lemma in two cases.

    Special case: Mπ2α0.

    For any given ε(0,1), we choose large enough Rε>0 such that

    vLp(BcR)vLp(BcRε)ε[ln(1+Rε)]1/pvLp(BcRε)εvXp,RRε

    and

    ˉuL2(BcR)ˉuL2(BcRε)ε,RRε.

    By Lemma 2.1 and α0π2M2, it is easy to verify that there is a C0>0 such that u4C0uXp for all uXp and

    BRε|f(x,un)|2dx+BRε|f(x,un)|2|un|dxC,n.

    Now, we claim that

    lim supn{f(x,un)f(x,ˉu)2+unˉu2}Cε. (A1)

    The proof of (A1) is in spirit of [36, Lemma 2.1]. As L2(BRε) is a Hilbert space, we need only to prove

    lim supnBRε(|f(x,un)|2|f(x,ˉu)|2)dxCε

    for the first part of (A1). Let M be large enough such that

    {|un|M}BRε|f(x,un)|2dx={|un|M}BRε|f(x,un)|2|un||un|dxC0Mε.

    By the dominated convergence theorem and Fatou's Lemma, one has

    |BRε(|f(x,un)|2|f(x,ˉu)|2)|{|un|M}BRε|f(x,un)|2|un|Mdx+{|u|M}BRε|f(x,un)|2|ˉu|Mdx+BRεhn(x)dx=2ε+on(1),

    where hn(x):=||f(x,un(x))|2χ{|un|<M}BRε|f(x,ˉu(x))|2χ{|un|<M}BRε|, and we use the fact

    |hn(x)|{|f(x,ˉu(x))|2,|un|M,sup{|f(x,t)|:x¯BRε,|t|<M}+|f(x,ˉu(x))|2,|un|<M.

    Therefore, we get (A1). By (A1), for large n, one has

    |R2[f(x,un)f(x,ˉu)]vdx|BRε|f(x,un)f(x,ˉu)||v|dx+BRcε|f(x,un)f(x,ˉu)||v|dx1γεvXp+Cγ(unˉu2+2ˉuL2(BRcε))vXp+22/pC(R2[exp(pαu2n)1+exp(pαˉu2)1]dx)1/pεvXpCεvXp,

    where 1p+1p=1 and γ:=infuXuuH1(R2)>0 (see [31, Lemma 2.1]).

    General case: M>0.

    For any R>0, let φR(r) be a C0 cut-off function such that 0φ1, φR1 on [0,R] and φR0 on [R+1,). Let δ>0 (to be determined later), we can choose large enough bounded domain BR(0) and its bounded open coverage {Ω}Nc which has a partition of unity w(1Nc) such that\\

    φR1(|x|)ˉu(x)ˉu(x)δ;
    BR(0)1NcΩ,Nc=1w(x)=1,xBR(0);
    wC1c(Ω),|w|C,;
    Ω|ψ|2dxδ,Ω|ψ,n|2dxδ,n,;
    Ω|ψ|2dxδ,Ω|ψ,n|2dxδ,n,;

    where

    ψ(x)=φR(|x|)wˉu,ψ,n(x)=φR(|x|)wun.

    Choosing δ>0 small enough and repeating now the proof of the special case, we can prove

    BcR(0)|[f(x,un)f(x,ˉu)]v|dxCεvX

    and

    Ω|[f(x,un)f(x,ˉu)]v|dxCεvX,.

    Therefore, one has

    |R2[f(x,un)f(x,ˉu)]vdx|BcR(0)|(f(x,un)f(x,ˉu))v|dx+Nc=1Ωi|(f(x,un)f(x,ˉu))v|dx(Nc+1)CεvX.

    So we obtain Lemma 2.2.



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