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Research article Special Issues

Qualitative properties of stable solutions to some supercritical problems

  • Received: 06 July 2021 Revised: 14 February 2022 Accepted: 15 February 2022 Published: 24 March 2022
  • In this paper, we study symmetry properties of stable solutions to the Lane-Emden equation

    Δu+|u|p1u=0inRn

    with n11, p in a suitable range and the Liouville equation

    Δu+eu=0inRn

    with n=10.

    Citation: Yong Liu, Kelei Wang, Juncheng Wei, Ke Wu. Qualitative properties of stable solutions to some supercritical problems[J]. Electronic Research Archive, 2022, 30(5): 1668-1690. doi: 10.3934/era.2022084

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  • In this paper, we study symmetry properties of stable solutions to the Lane-Emden equation

    Δu+|u|p1u=0inRn

    with n11, p in a suitable range and the Liouville equation

    Δu+eu=0inRn

    with n=10.



    In this paper, we consider the Lane-Emden equation

    Δu+|u|p1u=0inRn. (1.1)

    and the equation

    Δu+eu=0inRn. (1.2)

    The structures of the positive solutions of (1.1) and (1.2) have been studied intensively in the last several years. When n=3, (1.1) arises in the stellar structure in astrophysics. When n=4, (1.1) is relevant to the famous Yang-Mills equations. When n=2, (1.2) is an interesting problem in differential geometry and is known as the "Prescribing Gaussian Curvature" problem.

    For Eq (1.1), the Sobolev exponent

    ps(n)={+if1n2,n+2n2ifn3

    plays a central role in the solvability question. In the subcritical case 1<p<ps(n), it was established by Gidas and Spruck in their celebrated work [1] that (1.1) has no positive solution. If p=(n+2)/(n2), then (1.1) is a special case of the Yamabe problem in conformal geometry. In [2], using the asymptotic symmetry technique, Caffarelli, Gidas and Spruck were able to classify all the positive solutions of (1.1) for n3. They showed that any positive solutions of (1.1) can be written in the form

    ux0,λ(x)=(λn(n2)λ2+|xx0|2)n22,

    where λ>0 and x0 is some point in Rn. In [3], Chen and Li proved the same result for (1.1) by applying the moving plane method. In n=2, Eq (1.2) is also classified in [3] under the additional assumption that

    R2eudx<. (1.3)

    It is proved in [3] that if u is a solution of (1.2) such that (1.3) holds, then

    u=ln32λ2(4+λ2|xx0|2)2

    for some λ>0 and some point x0R2.

    In the supercritical case p>ps(n), it is more difficult to classify the positive solutions of (1.1). The first result in this direction was given by Zou in [4]. It was proved in [4] that if ps(n)<p<ps(n1) and if u is a positive solution of (1.1) with algebraic decay rate 2/(p1) at infinity, then u is radially symmetric about some point x0Rn. In [5], Guo generalized Zou's result to pps(n1) by assuming that

    lim|x|+|x|2p1u(x)[2p1(n22p1)]1p1. (1.4)

    Moreover, it is showed in [5] that (1.4) is a necessary and sufficient condition for a positive solution of (1.1) to be radially symmetric about some point. The analogous result for second order equation (1.2) is considered in [6]. It is proved in [6] that if n4 and if uC2(Rn) is an entire solution of (1.2), then u is radially symmetric about some point x0Rn if and only if

    lim|x|u(x)+2ln(|x|)ln(16)=0.

    If we focus on radial solutions, then the structure of positive solutions of (1.1) has been completely classified in [7]. They showed that for any a>0, (1.1) admits a unique positive radial solution u=ua(r) with ua(0)=a. Moreover, no two positive radial solutions of (1.1) can intersect each other when p>pJL(n), where pJL(n) is the exponent given by

    pJL(n)={if3n10,(n2)24n+8n1(n2)(n10)ifn11.

    Another important topic is the classification of stable solutions. In general, a solution of the semilinear equation

    Δu+f(u)=0inRn

    with f be a Lipschitz function is called stable if

    Rn|ψ|2dxRnf(u)ψ2dx0ψC0(Rn).

    One of the most interesting questions concerning stable solutions is the following De Giorgi's conjecture.

    Conjecture: Let u be a bounded solution of the equation

    Δu+uu3=0inRn

    such that uxn>0. Then the level sets of u are hyperplanes, at least if n8.

    De Giorgi's conjecture was proved in dimension n=2 by Ghoussoub and Gui in [8]. For n=3, this is proved by Ambrosio and Cabré in [9]. Savin proved in [10] that for 4n8, the above conjecture is true under the additional limit condition that

    u(x1,...,xn)±1asxn±. (1.5)

    For n>9, a counterexample is constructed in [11]. The conjecture is still open for dimensions 4n8 without the additional assumption (1.5).

    For Eq (1.1), there are also some results concerning stable solutions. In [12], Liouville type results for solutions with finite Morse index were established. By making a delicate use of the classical Moser iteration method, Farina was able to classify finite Morse index solutions in his seminal paper [13]. It was proved in [13] that if uC2(Rn) is a stable solution of (1.1) with 1<p<pJL(n), then u0. Moreover, (1.1) admits a smooth positive, bounded, stable and radial solution for n11,p>pJL(n). Actually, it was showed in [13] that the radial solutions considered in [7] are stable when n11,p>pJL(n). The results in [13] also have a lot of generalizations, we refer to [14,15,16,17,18,19]. As for the classification of the stable solutions of (1.2), it was proved in [20] that for 1n9, there is no stable solution uC2(Rn) of (1.2).

    In spite of the above mentioned results, there are some intriguing problems which are still open. In [21], the authors proposed the following conjecture, which is a natural extension of De Giorgi type conjecture:

    Conjecture: Let n11,pJL(n)<p<pJL(n1), then all stable solutions to (1.1) must be radially symmetric around some point.

    Remark 1.1. When p>pJL(n1), (1.1) has a positive stable solution which is not radially symmetric. Indeed, let u be a positive radial stable solution of the equation

    Δu+|u|p1u=0inRn1

    for p>pJL(n1) (see [13]), then u can also be viewed as a stable solution of the equation

    Δu+|u|p1u=0inRn.

    But it is obvious that this solution is not radially symmetric in Rn.

    Our first objective in this paper is to give some partial results toward the above conjecture. To state our results in a more precise way, let us first introduce some new exponents. Let γ be a constant such that

    1γ<2p+2p(p1)1,(p+γ)/(p1)>(n1)/2,

    we define

    α(p,γ,n)={[(n2)2γ(γ+1)2β4pγ(γ+1)2]p+γp1(n1)πn12n2(p+γ)}22(p+γ)(p1)(n1). (1.6)

    Let

    β(p,n)=(2n+βp)1p1 (1.7)

    and let

    cs(p,n)=ωn1((n2)24βp1)2p1ωn24(2π)n2(β(p,n)β1p1)n+32(α(p,n))p(n1)2. (1.8)

    We consider

    p1=((n2)242pp1(n22p1))cs(p,n) (1.9)

    as a function with respect to the variable p. Let S be the set

    S={p:pisasolutionof(1.9)suchthatp>pJL(n)}

    and let p be the infimum of S. We define

    pcs(n)={pJL(n1)ifppJL(n1),pifp<pJL(n1). (1.10)

    With the help of these numbers, we can give the statement of our first result.

    Theorem 1.2. Let n11,pJL(n)<p<pcs(n), where pcs(n) is defined in (1.10). Let u be a positive stable solution of (1.1) such that u is even symmetric with respect to the planes {xi=0},i=1,2,n, then u is radially symmetric with respect to the origin.

    In Theorem 1.2, we need the assumption that u is a positive solution of (1.1). But under suitable conditions, we can show that stable solutions of (1.1) do not change sign. For our purpose, we use p1(n)p2(n)p3(n) to denote the three numbers such that

    (n1)(p1)=(n2)242pp1(n22p1). (1.11)

    We define

    psi(n)={pJL(n1)ifp3(n)pJL(n),pJL(n1)ifp2(n)pJL(n1),p2(n)ifp3(n)>pJL(n)andp2(n)<pJL(n1). (1.12)

    Theorem 1.3. Let n11,pJL(n)<p<psi(n), where psi(n) is defined in (1.12). Let u be a stable solution of Eq (1.1), then u does not change sign. If u is a axially symmetric stable solution of (1.1), then u does not change sign when pJL(n)<p<pJL(n1).

    Remark 1.4. By using MATLAB, we can give some examples for pcs(n) and psi(n).

    n pJL(n) pcs(n) psi(n) pJL(n1)
    n = 12 3.9266499161 4.1229824119 6.9220245868 6.9220245868
    n = 13 2.9306913006 3.0772258656 3.9266499161 3.9266499161
    n = 14 2.4342585459 2.5559714732 2.9306913006 2.9306913006
    n = 15 2.1374347552 2.2443064930 2.4342585459 2.4342585459

     | Show Table
    DownLoad: CSV

    For Eq (1.2), we have the following result.

    Theorem 1.5. Let u be a smooth stable solution of Eq (1.2) for n=10, then u is radially symmetric with respect to some point in Rn.

    Remark 1.6. If n11, then Eq (1.2) has a smooth stable solution which is not radially symmetric. For more discussions, we refer to [22].

    The rest of the paper will be organized as follows. In section 2, we consider rigidity results on the unit sphere for some second order equations. Rigidity results on compact manifolds have been considered by several authors, see for instance [1,23,24,25,26,27]. We point out that in the above papers, the proof of the rigidity results depends heavily on the classical Bochner formula. In section 3, we first use a monotonicity formula to study the qualitative properties of solutions. Then, by combing the rigidity results and the qualitative properties of solutions, we can verify the assumption of Theorem 1.1 in [5] and obtain the symmetry properties of stable solutions. In section 4, we give the prove of Theorem 1.5.

    Notation. In some situations, we will write a point xRn as x=(r,θ), where (r,θ) is the spherical coordinates and Sn1Rn is the unit sphere. In the rest of the paper, c will denote a positive constant which may vary from line to line.

    In this section, we consider the equation

    ΔSn1ϕβϕ+|ϕ|p1ϕ=0, (2.1)

    where

    β=2p1(n22p1)

    and ΔSn1 is the Laplace-Beltrimi operator on the unit sphere. In the rest of this section, we will always assume that n11. The main result in this section is the following.

    Theorem 2.1. Let γ be a constant such that

    1γ<2p+2p(p1)1,(p+γ)/(p1)>(n1)/2.

    If pJL(n)<p<pcs(n) with pcs(n) be the number defined in (1.9) and if ϕ is a positive solution of (2.1) such that (2.2) holds. Suppose

    Sn1ϕΦidθ=0,i=1,2,n

    where Φi,i=1,2,,n are the eigenfunctions of the operator ΔSn1 corresponding to the eigenvalue n1, then ϕ is a constant solution of Eq (2.1). If {ϕβ1p1} has at least three connected components, then the same result holds.

    In order to prove Theorem 2.1, we need several lemmas.

    Lemma 2.2. Let pJL(n)<p<pJL(n1) and let ϕH1(Sn1) be a weak solution of (2.1) such that

    Sn1|Sn1ψ|2dθ+(n2)24Sn1ψ2dθpSn1|ϕ|p1ψ2dθ (2.2)

    for every ψH1(Sn1), then ϕC2(Sn1).

    Proof. We take ψ=|ϕ|γ12ϕ into (2.2), where γ is a positive constant which will be chosen later. Then

    pSn1|ϕ|p+γdθ(n2)24Sn1|ϕ|γ+1dθ+Sn1|Sn1(|ϕ|γ12ϕ)|2dθ. (2.3)

    Multiplying the both sides of (2.1) by |ϕ|γ1ϕ and integrating over Sn1, we can get that

    Sn1Sn1ϕSn1(|ϕ|γ1ϕ)dθ+βSn1|ϕ|γ+1dθ=Sn1|ϕ|p+γdθ. (2.4)

    (2.4) is equivalent to

    4γ(γ+1)2Sn1|Sn1(|ϕ|γ12ϕ)|2dθ+βSn1|ϕ|γ+1dθ=Sn1|ϕ|p+γdθ. (2.5)

    By combining (2.3) and (2.5) together, we can obtain that

    [p(γ+1)24γ]Sn1|ϕ|p+γdθ[(n2)24(γ+1)2β4γ]Sn1|ϕ|γ+1dθ. (2.6)

    It is easy to check that

    p(γ+1)24γ>0

    when 1γ<2p+2p(p1)1. By applying Hölder's inequality, we have

    [p(γ+1)24γ]Sn1|ϕ|p+γdθc(Sn1|ϕ|p+γdθ)γ+1p+γ. (2.7)

    It follows from (2.7) that Sn1|ϕ|p+γdθ is finite. By formula (5.10) in [13], we know that there exists a constant γ such that (p+γ)/(p1)>(n1)/2. Therefore, |ϕ|p1Lq(Sn1) for some q>(n1)/2. The standard regularity results in [28] imply that ϕC2(Sn1).

    Lemma 2.3. Let pJL(n)<p<pJL(n1) and let ϕC2(Sn1) be a positive solution of (2.1) such that (2.2) holds, then

    ϕL(Sn1)α(p,γ,n), (2.8)

    where γ is the constant used in the proof of Lemma 2.2 and α(p,γ,n) is the constant defined in (1.6).

    Proof. Let θ0Sn1 be a point such that ϕ(θ0)=ϕL(Sn1)=η. By taking suitable orthogonal transformation, we may assume that θ0 is the south pole. Let us introduce the following coordinates on Sn1,

    {θ1=sinξsinξn2sinξ2sinξ1,θ2=sinξsinξn2sinξ2cosξ1,θ3=sinξsinξn2cosξ2,,θn1=cosξ,

    where ξ[0,π),ξ1[0,2π),ξk[0,π) for k=2,3,n2. The coordinate of the point θ0 is given by (0,0,,0). By (2.1), we know that ϕ satisfies the equation

    1sinn2ξddξ(sinn2ξdϕdξ(ξ))+1sin2ξΔSn2ϕβϕ+ϕp=0, (2.9)

    where Sn2 is the unit sphere in Rn1 and ΔSn2 is the Laplace -Beltrami operator on Sn2. We define

    ˆϕ(ξ)=1ωn2Sn2ϕ(ξ,θ)dθ,

    where ωn2 is the area of Sn2. It follows from (2.9) that ˆϕ satisfies

    1sinn2ξddξ(sinn2ξdˆϕdξ(ξ))βˆϕ+1ωn2Sn2ϕp(ξ,θ)dθ=0. (2.10)

    By the Jensen's inequality, we can get that

    1sinn2ξddξ(sinn2ξdˆϕdξ(ξ))βˆϕ+ˆϕp0,in(0,π). (2.11)

    Let ξ1 be the first point such that ˆϕ(ξ1)=η2. It follows from (2.11) that ˆϕ is strictly decreasing in (0,ξ1). We will focus on the case ξ1<π2 since the case ξ1π2 can be dealt with similarly. Let γ be the constant used in the proof of Lemma 2.2. By (2.10), we can obtain that

    ˆϕ(ξ1)ˆϕ(0)=ξ101sinn2ξξ0sinn2τ[βˆϕ(τ)1ωn2Sn2ϕp(τ,θ)dθ]dτdξηpξ101sinn2ξξ0sinn2τdτdξξ212ηp.

    This implies ξ1η1p2. By the above analysis, we can get that

    {ξξ1}ϕp+γdθ=ξ10Sn2sinn2ξϕp+γ(ξ,θ)dθdξωn2ξ10sinn2ξˆϕp+γ(ξ)dξωn22n2(p+γ)(n1)πn2ηp+γ+(1p)(n1)2. (2.12)

    By Lemma 2.2, we know that

    Sn1ϕp+γdθ[(n2)24(γ+1)24γβp(γ+1)24γ]p+γp1ωn1. (2.13)

    We get from (2.12) and (2.13) that

    η{[(n2)24(γ+1)24γβp(γ+1)24γ]p+γp1ωn1(n1)πn2ωn22n2(p+γ)}1p+γ+(1p)(n1)2{[(n2)2γ(γ+1)2β4pγ(γ+1)2]p+γp1(n1)πn12n2(p+γ)}22(p+γ)(p1)(n1). (2.14)

    Hence (2.8) holds.

    Lemma 2.4. Let ϕ be a positive solution of (2.1) such that

    Sn1ϕΦidθ=0fori=1,2,,n, (2.15)

    where Φi,i=1,2,,n are the eigenfunctions of the operator ΔSn1 corresponding to the eigenvalue n1, then

    ϕL(Sn1)(2n+βp)1p1. (2.16)

    Proof. We define

    ˜ϕ=ϕ¯ϕ,

    where ¯ϕ is given by

    ¯ϕ=1ωn1Sn1ϕdθ.

    Then ˜ϕ satisfies the equation

    ΔSn1˜ϕβϕ+ϕp=0. (2.17)

    Multiplying the both sides of (2.17) by ˜ϕ and using integration by part, we can get that

    Sn1|Sn1˜ϕ|2dθ+βSn1˜ϕ2dθSn1(ϕp¯ϕp)(ϕ¯ϕ)dθ=0. (2.18)

    By (2.15) and the definition of ˜ϕ, we know that

    Sn1˜ϕdθ=0,Sn1˜ϕΦidθ=0,i=1,2,n.

    By (2.18) and the Poincaré's inequality, we have

    2nSn1˜ϕ2dθ+βSn1˜ϕ2dθpϕp1L(Sn1)Sn1˜ϕ2dθ0. (2.19)

    If ˜ϕ0, then

    2n+βpϕp1L(Sn1)0.

    It follows that

    ϕL(Sn1)(2n+βp)1p1, (2.20)

    Hence (2.16) holds.

    Lemma 2.5. If ϕ is a positive solution of (2.1) such that {ϕβ1p10} has at least three connected components, then

    ϕL(Sn1)(2n+βp)1p1. (2.21)

    Proof. The equation (2.1) can be written as

    ΔSn1ϕβ(ϕβ1p1)+ϕpβpp1=0. (2.22)

    Since {ϕβ1p10} has at least three connected components, then there is a connected component Ω0 of {ϕβ1p10} such that the area of Ω0 is less than 13ωn1. let 1Ω0 be the function defined by

    1Ω0={1inΩ0,0onSn1Ω0.

    Multiplying the both sides of (2.22) by (ϕβ1p1)1Ω0 and using integration by part, we can get that

    Ω1|Sn1ϕ|2dθβΩ1(ϕβ1p1)2dθ+Ω1(ϕpβpp1)(ϕβ1p1)dθ=0. (2.23)

    Let λ1(Ω0) be the first eigenvalue of the eigenvalue problem

    {ΔSn1Φ+λΦ=0inΩ0,Φ=0onΩ0.

    By (2.23) and the mean value theorem, we can get that

    (λ1(Ω0)β+pϕp1L(Sn1))Ω1(ϕβ1p1)2dθ0. (2.24)

    It follows from (2.24) that

    λ1(Ω0)β+pϕp1L(Sn1)0. (2.25)

    Since the area of Ω0 is less than 13ωn1, where ωn1 is the area of the unit sphere in Rn. By using Schwartz symmetrization, we can get that

    λ1(Ω0)2n. (2.26)

    It follows from (2.25) and (2.26) that

    ϕL(Sn1)(2n+βp)1p1, (2.27)

    Hence (2.21) holds.

    Remark 2.6. We notice that

    (p1)β=2(n22p1)<2n,

    then

    ϕL(Sn1)(2n+βp)1p1>β1p1.

    Lemma 2.7. Let ¯p be a constant such that pJL(n)<¯p<pJL(n1). There exists a positive constant c such that if ϕC2(Sn1) is a nonconstant solution of (2.1) for pJL(n)<p<¯p, then

    Sn1ϕ2dθcSn1|Sn1ϕ|2dθ. (2.28)

    Proof. Suppose (2.28) does not hold, then there exists a sequence {ϕm} such that ϕm satisfies

    ΔSn1ϕm2pm1(n22pm1)ϕm+|ϕ|pm1ϕm=0 (2.29)

    and

    Sn1ϕ2mdθmSn1|Sn1ϕm|2dθ. (2.30)

    Since ϕm is also a solution of (2.29), without loss of generality, we can assume that

    ϕm(θm)=maxθSn1ϕm(θ)>0. (2.31)

    It follows from the proof of Lemma 2.2 that Sn1ϕ2mdθ remains bounded. So (2.30) implies

    limm+Sn1|Sn1ϕm|2dθ=0. (2.32)

    By (2.8) and (2.32), we can get that there exist two constants p0 and c0 such that

    limm+pm=p0,limm+ϕm=c0.

    Moreover, c0 is a constant solution of (2.1) for p=p0. Therefore,

    c0=0orc0=[1p01(n22p01)]1p01.

    We get from (2.31) that

    Δϕm(θm)=(βmϕpm1m(θm))ϕm(θm)0. (2.33)

    Therefore,

    ϕm(θm)(βm)1pm1. (2.34)

    It follows from (2.34) that c0 is not zero. Let

    ϕm=β1pm1m+ψm,

    then limm+ψm=0 and ψm satisfies the equation

    ΔSn1ψm+(pm1)βmψm+(ψm+β1pm1m)pmβpmpm1mpmβmψm=0. (2.35)

    It is easy to verify that

    (ψm+β1pm1m)pmβpmpm1mpmβmψmcψm2L(Sn1)

    for some positive constant c independent of m. We define

    vm=ψmψmL(Sn1),

    then vm satisfies

    ΔSn1vm+(pm1)βmvm+(ψm+β1pm1m)pmβpmpm1mpmβmψmψmL(Sn1)=0. (2.36)

    Since

    vmL(Sn1)=1

    and

    limm+(ψm+β1pm1m)pmβpmpm1mpmβmψmψmL(Sn1)L(Sn1)=0.

    By standard elliptic estimates, we know that there exists a nontrivial function v such that vmv in H1(Sn1). Moreover, v satisfies the equation

    ΔSn1v+(p01)β0v=0. (2.37)

    Then we deduce that v is a nontrivial eigenfunction of ΔSn1 corresponding to the eigenvalue (p01)β0. On the other hand, it is easy to see that

    (p01)β0=2(n22p01)<2n

    and pJL(n)>(n+1)/(n3) when n11. Therefore, (p01)β0 can not be an eigenvalue of ΔSn1. By combining these two facts together, we obtain a contradiction. Next, we can give some estimates about the constant c in Lemma 2.7.

    Lemma 2.8. Let pJL(n)<p<pJL(n1) and let ϕ be a positive solution of (2.1) such that

    ϕL(Sn1)(2n+βp)1p1,

    then the constant c in Lemma 2.7 can be estimated by cs(p,n), where cs(p,n) is defined by (1.8).

    Proof. Multiplying the both sides of (2.1) by ϕ and integrating over Sn1, we can get that

    Sn1|Sn1ϕ|2dθ+βSn1ϕ2dθ=Sn1|ϕ|p+1dθ. (2.38)

    We take ψ=ϕ into (2.2), then

    Sn1|Sn1ϕ|2dθ+(n2)24Sn1ϕ2dθpSn1|ϕ|p+1dθ. (2.39)

    By (2.38) and (2.39), we can get that

    Sn1ϕp+1dθ(n2)24βp1Sn1ϕ2dθ. (2.40)

    By the Poincaré's inequality, we know that

    Sn1ϕ2dθωp1p+1n1(Sn1ϕp+1dθ)2p+1 (2.41)

    It follows from (2.40) and (2.41) that

    Sn1ϕ2dθωn1((n2)24βp1)2p1. (2.42)

    In order to estimate the constant c in Lemma 2.7, we need to give a lower bound for Sn1|Sn1ϕ|2dθ. Since we have assumed that

    ϕL(Sn1)(2n+βp)1p1=β(p,n),

    then there exists a point θ0 such that ϕ(θ0)=β(p,n). By taking suitable orthogonal transformation, we may assume that θ0 is the south pole. We use the coordinates used in the proof of Lemma 2.2. By (2.1), we know that ϕ satisfies Eq (2.9). We define

    ˆϕ(ξ)=1ωn2Sn2ϕ(ξ,θ)dθ,

    then ˆϕ satisfies (2.10) and (2.11). Let ξ1 be the first point such that

    ˆϕ(ξ1)=β(p,n)+β1p12.

    We know from (2.11) that

    ˆϕ(ξ)>β(p,n)+β1p12in(0,ξ1).

    We will assume that ξ1<π2 since the case ξ1<π2 can be dealt with similarly. By (2.10), we can get that

    ˆϕ(ξ1)ˆϕ(0)=ξ101sinn2ξξ0sinn2τ[βˆϕ(τ)1ωn2Sn2ϕp(τ,θ)dθ]dτdξ(α(p,n))pξ101sinn2ξξ0sinn2τdτdξξ212(α(p,n))p.

    We deduce that

    ξ1>(β(p,n)β1p1)12(α(p,n))p2. (2.43)

    Let

    ¯ϕ=1ωn1Sn1ϕdθ.

    By (2.1) and the Jensen's inequality, we can get that ¯ϕβ1p1. Therefore,

    Sn1(ϕ¯ϕ)2dθ=π0Sn2sinn2ξ(ϕ¯ϕ)2dθdξωn2ξ10sinn2ξ(ˆϕ¯ϕ)2dξωn24(n1)(2π)n2(β(p,n)β1p1)n+32(α(p,n))p(n1)2. (2.44)

    It follows from the Poincaré's inequality that

    Sn1|Sn1ϕ|2dθωn24(2π)n2(β(p,n)β1p1)n+32(α(p,n))p(n1)2.

    Therefore,

    Sn1|Sn1ϕ|2dθSn1ϕ2dθωn1((n2)24βp1)2p1ωn24(2π)n2(β(p,n)β1p1)n+32(α(p,n))p(n1)2.

    By the above analysis, we know that (1.8) holds.

    Lemma 2.9. Let ϕ be a positive solution of (2.1) such that (2.2) holds. If

    ϕL(Sn1)(2n+βp)1p1,

    then ϕ is a constant when pJL(n)<p<pcs(n), where pcs(n) is defined by (1.9).

    Proof. By (2.38) and (2.39), we have

    (p1)Sn1|Sn1ϕ|2dθSn1((n2)24pβ)ϕ2dθ. (2.45)

    Let ϕ be a nonconstant solution of (2.1) satisfying (2.2), we know from Lemma 2.7 that ϕ satisfies (2.28). By combining (2.28) and (2.45) together, we can get that

    (p1)Sn1|Sn1ϕ|2dθ((n2)24pβ)cs(p,n)Sn1|Sn1ϕ|2dθ. (2.46)

    It follows from (2.46) that

    Sn1|Sn1ϕ|2dθ=0

    when pJL(n)<p<pcs(n). Since we have assumed that ϕ is a nonconstant solution of (2.1), this is a contradiction.

    Proof of Theorem 2.1:. This result follows from Lemma 2.4, Lemma 2.5 and Lemma 2.9.

    Remark 2.10. In this section, we always assume that ϕ is positive. But we can prove that if ϕ is a solution of (2.1) depends only on the variable ξ, then ϕ does not change sign. The proof of this fact will be given in the appendix.

    Remark 2.11. It is proved in [29] that if n4 and (n+1)/(n3)<p<pJL(n1), then (2.1) has a nonconstant positive solution.

    Remark 2.12. By Lemma 1 in [30], we have the following Hardy type inequality,

    Sn1|Sn1ϕ|2dθ+(n2)24Sn1ϕ2dθ(n3)24Sn1ϕ2sin2ξdθ. (2.47)

    The equation (2.1) has a singular solution which is given by

    ϕ(ξ)=[2p1(n32p1)]1p1(sinξ)2p1=β(sinξ)2p1.

    Suppose ϕ satisfies (2.2), then

    Sn1|Sn1ϕ|2dθ+(n2)24Sn1ϕ2dθpβp1Sn1ϕ2sin2ξdθ. (2.48)

    If p=pJL(n1), then

    2pp1(n32p1)=(n3)24.

    Let us define

    g(p)=2pp1(n32p1),

    then

    g(p)=2(p1)2(n54p1).

    If p>(n1)/(n5), then g(p)<0. Therefore, the singular solution ϕ satisfies (2.2) if ppJL(n1).

    In this section, we give the proof of Theorem 1.2 and Theorem 1.3.

    Lemma 3.1. Let n11,pJL(n)<p<psi(n), where psi(n) is defined in (1.12). If ϕ is a nontrivial solution of (2.1) such that (2.2) holds, then ϕ does not change sign.

    Proof. We assume that ϕ change sign. Without loss of generality, we can assume that there exists a connected component Ω1 of {ϕ>0} such that λ1(Ω1)n1, where λ1(Ω1) is the first eigenvalue of the eigenvalue problem

    {ΔSn1Φ+λΦ=0inΩ1,Φ=0onΩ1.

    Multiplying the both sides of (2.1) by ϕ and integrating over Ω1, we can get that

    Ω1|Sn1ϕ|2dθ+βΩ1ϕ2dθ=Ω1|ϕ|p+1dθ. (3.1)

    We take ψ=u1Ω1 into (2.2), where 1Ω1 is the function defined by

    1Ω1={1inΩ10onSn1Ω1.

    Then

    Ω1|Sn1ϕ|2dθ+(n2)24Ω1ϕ2dθpΩ1|ϕ|p+1dθ. (3.2)

    By (3.1) and (3.2), we know that

    (p1)Ω1|Sn1ϕ|2dθ1λ1(Ω1)Ω1((n2)24pβ)|Sn1ϕ|2dθ. (3.3)

    It follows that if pJL(n)<p<psi(n), then ϕ vanishes identically on Ω1. Since we have assumed that ϕ>0 on Ω1, this is a contradiction.

    Proof of Theorem 1.3: We consider the transform

    u(r,θ)=r2p1w(t,θ),t=lnr.

    Since u satisfies (1.1), then w is a bounded solution of the equation

    ttw+(n24p1)tw+ΔSn1w2p1(n22p1)w+|w|p1w=0. (3.4)

    We set

    A=n24p1,B=2p1(n22p1),E(w)=Sn112|Sn1w|2B2w21p+1|w|p+1dθ. (3.5)

    By (3.4), we get that

    ASn1(tw)2dθ=ddt[E(w)(t)12Sn1(tw)2dθ]. (3.6)

    By the estimates in [19], we can get that tw,ttw,|Sn1| are uniformly bounded. Integrating (3.6) from s to s, we find

    AssSn1(tw)2dθdt<c (3.7)

    for some constant c independent of s. Let s tend to + in (3.7), then

    A+Sn1(tw)2dθdt=0.

    Similar to the proof of Theorem 1.4 in [1], we can obtain that

    limt+Sn1(tw)2dθ=0. (3.8)

    For any sequence {tk} such that tk as k, we consider the translation of w defined by wk(t,θ)=w(t+tk,θ). Then there exist a subsequence {wlk(t,θ)} and a function w(t,θ) such that wlk(t,θ)w(t,θ) in C2([1,1]×Sn1). By (3.8) and the dominated convergence theorem, we know that there exists a function ϕ(θ) such that w(t,θ)=ϕ(θ). Moreover, ϕ is a solution of (2.1) such that (2.2) holds. If ϕ=0, then limt+E(w)(t)=0. But we also have limtE(w)(t)=0 since u is regular at the origin. It follows easily that w0. Since we have assumed that u is a nontrivial solution, this is a contradiction. Therefore ϕ is not zero. If ϕ0, we know from Lemma 3.1 that ϕ does not change sign. Suppose there exist two sequences {tk} and {˜tk} such that

    limkw(tk,θ)<0

    and

    limkw(˜tk,θ)>0,

    then {u0} has a bounded connected component. Without loss of generality, we can assume there exists a bounded connected component Ω such that u<0 on Ω. Then u satisfies the equation

    {Δu+|u|p1u=0inΩ,u=0onΩ. (3.9)

    Since u is a stable solution of (1.1), then L=Δ+p|u|p1 satisfies the refined maximum principle (see [31]). Since

    {Lu=(p1)|u|p1u0inΩ,u=0onΩ, (3.10)

    we get from the refined maximum principle that u0 on Ω. In view of the definition of Ω, we get a contradiction. By the above arguments, we know that there exits a positive constant R0 such that u doesn't change sign on RnBR0. By applying the refined maximum principle again, we know that u does not change sign.

    If u is axially symmetric, then the proof is essentially the same as the arguments above. The only difference is that we need to use remark 2.10 rather than Lemma 3.1 to show that there exits a positive constant R0 such that u doesn't change sign on RnBR0.

    Proof of Theorem 1.2. Let n11,pJL(n)<p<pcs(n) and let u be a positive stable solution of (1.1). Let us consider the transform

    u(r,θ)=r2p1w(t,θ),t=lnr.

    Since u satisfies (1.1), then w is a bounded solution of Eq (3.4). Let {tk} be a sequence such that tk as k. Similar to the arguments used in the proof of Theorem 1.3, we know that there exists a function ϕH1(Sn1) such that

    lim|k|w(tk,θ)=ϕ.

    Moreover, ϕ is a solution of (2.1) such that (2.2) holds. By Lemma 2.2, we have ϕC2(Sn1). If u is even symmetric with respect to the {xi=0},i=1,2,n, then

    Sn1ϕΦidθ=0fori=1,2,,n.

    Since pJL(n)<p<pcs(n), we know from Theorem 2.1 that ϕ is a constant function. In particular, we have

    ϕ=[2p1(n22p1)]1p1.

    Since the sequence {tk} can be arbitrary, we conclude that

    lim|x||x|2p1u(x)=[2p1(n22p1)]1p1.

    Since p>pJL(n), then p>n/(n4). By Theorem 4.4 in [5], we can get that

    u(x)=r2p1((B)1p1+ξ(r)+ν(r,θ)r), (3.11)

    where

    ξ(r)=r2p1¯u(r)(B)1p1 (3.12)

    and

    ¯u(r)=1ωn1Sn1u(r,θ)dθ.

    Moreover, for any integer τ0, we have ν(r,θ) satisfies

    ν(r,θ)V(θ)asr0 (3.13)

    uniformly in Cτ(Sn1), where V equals either zero or a first eigenfunctions of the operator ΔSn1. Since we have obtained the asymptotic expansion (3.11) which is good enough to apply the moving plane method, then the rest of the proof is essentially the same as the proof of Theorem 1.1 in [4].

    In this section, we give the proof of Theorem 1.5, the proof is mainly based on the following observation.

    Proposition 4.1. Let n=10 and let u be a smooth stable solution of Eq (1.2), then

    lim|x|u(x)+2ln(|x|)ln(16)=0. (4.1)

    In order to prove Proposition 4.1, we first recall a monotonicity formula.

    Lemma 4.2. If u is a solution of the equation (1.2), then

    dEdρ=ρ2nBρ(uρ+2ρ)2dθ, (4.2)

    where

    E(ρ,u)=ρ2nBρ(12|u|2eu)dx2ρ1nBρ(u+2ln(ρ))dθ.

    Moreover, if u is a smooth stable solution of (1.1), then

    limρ+E(ρ,u)<+. (4.3)

    Proof. The proof of (4.2) follows from a scaling argument which is similar to the proof Proposition 5.1 in [32]. The proof of (4.3) follows easily from the capacity estimates in [33].

    With the help of Lemma 4.2, we can give the proof of Proposition 4.1.

    proof of Proposition 4.1. The proof of Proposition 4.1 will consist of the following four steps.

    Step 1: Let {λk} be a sequence such that limk+λk=+. For any λk, we define uλk(x)=u(λkx)+2ln(λk). It is easy to check that uλk(x) is also a stable solution of (1.1). By the capacity estimates (see for instance [33]), we know that uλku for some function uH1loc(Rn). Moreover, u is a stable solution of (1.1).

    Step 2: For any 0<R1<R2<+, by Lemma 4.2,

    limk+E(λkR2;0,u)E(λkR1;0,u)=0. (4.4)

    By the scaling invariance of E, we have

    limk+E(R2;0,uλk)E(R1;0,uλk)=0. (4.5)

    We use Lemma 4.2 again, then

    0=limk+E(R2;0,uλk)E(R1;0,uλk)=limk+BR2BR1|x|2n(uλkr+2|x|)2dxBR2BR1|x|2n(uλr+2|x|)2dx. (4.6)

    Therefore,

    2r+ur=0a.e.inRN. (4.7)

    It follows that there exists a function ϕH1(Sn1) such that u=ϕ2ln(r). Moreover, ϕ satisfies the equation

    ΔSn1ϕ2(n2)+eϕ=0. (4.8)

    Step 3: For every δ>0, we choose a function ηδC0((δ2,2δ)) such that ηδ1 in (δ,1δ), and r|ηδ(r)|4. For every ψH1(Sn1), we define ψδ=rn22ψ(θ)ηδ(r). For every ψH1(Sn1), we define ψδ=rn22ψ(θ)ηδ(r). Since u is stable, we have

    Sn1eϕψ2dθ+0r1η2δdrSn1ψ2dθ0rn1(ηδrn22n22rn2ηδ)2dr+Sn1|Sn1ψ|2dθ0rn1(ηδrn2)2dr

    Therefore, ϕ satisfies

    Sn1|Sn1ψ|2dθ+(n2)24Sn1ψ2dθSn1eϕψ2dθ (4.9)

    for every ψH1(Sn1).

    Step 4: We take ψ=eϕ2 into (4.9), then

    14Sn1eϕ|Sn1ϕ|2dθ+(n2)24Sn1eϕdθSn1e2ϕdθ. (4.10)

    Multiplying the both sides of (4.8) by eϕ and using integration by part, we have

    12Sn1eϕ|Sn1ϕ|2dθ+2(n2)Sn1eϕdθ=Sn1e2ϕdθ. (4.11)

    If n=10, then (n2)2/4=2(n2). By (4.10) and (4.11), we can get that

    Sn1eϕ|Sn1ϕ|2dθ0. (4.12)

    It follows from (4.12) that ϕ=ln(16) is a constant. Since {λk} can be arbitrary, we can obtain that proposition 4.1 holds.

    Proof of Theorem 1.5. It follows from proposition 4.1 and Theorem 1.3 in [6].

    In this appendix, we prove the claim in remark 2.10. The proof is based on the the following result.

    Proposition 4.3. Let pn+1n3 and (p1)μn1. If ϕ is a solution of the equation

    {(1+|x|22)n1div((21+|x|2)n3ϕ)μϕ+|ϕ|p1ϕ=0inBr,ϕ=0onBr, (4.13)

    where BrRn1 is a ball and 0<r<1, then ϕ=0.

    Proof. Multiplying the both sides of (4.13) by (21+|x|2)n1ϕ and using integration by part, we can get that

    Br|ϕ|2(21+|x|2)n3+μBrϕ2(21+|x|2)n1=Br|ϕ|p+1(21+|x|2)n1. (4.14)

    Multiplying the both sides of (4.13) by (21+|x|2)n1(xϕ) and using integration by part, we can get that

    h(r)Br|ϕ|2=Br(21+|x|2)n3ϕ(xϕ)+μBr(21+|x|2)n1ϕ(xϕ)Br(21+|x|2)n1|ϕ|p1ϕ(xϕ)=h(r)2Br|ϕ|2+3n2Br(21+|x|2)n3|ϕ|2(n1)μ2Br(21+|x|2)n1ϕ2+n1p+1Br(21+|x|2)n1|ϕ|p+112Brx(21+|x|2)n3|ϕ|2μ2Brx(21+|x|2)n1ϕ2+1p+1Brx(21+|x|2)n1|ϕ|p+1,

    where

    h(r)=r(21+r2)n3.

    It follows that

    3n2Br(21+|x|2)n3|ϕ|2(n1)μ2Br(21+|x|2)n1ϕ2+n1p+1Br(21+|x|2)n1|ϕ|p+112Brx(21+|x|2)n3|ϕ|2μ2Brx(21+|x|2)n1ϕ2+1p+1Brx(21+|x|2)n1|ϕ|p+1=h(r)2Br|ϕ|2. (4.15)

    Multiplying the both sides of (4.13) by x(21+|x|2)n1ϕ and using integration by part, we can get that

    0=(n1)Br(1+|x|22)n1div((21+|x|2)n3ϕ)(|x|2(21+|x|2)n)μBrx(21+|x|2)n1ϕ2+Brx(21+|x|2)n1|ϕ|p+1=(n1)Br2|x|21+|x|2ϕdiv((21+|x|2)n3ϕ)μBrx(21+|x|2)n1ϕ2+Brx(21+|x|2)n1|ϕ|p+1. (4.16)

    By some computations, we can get that

    (n1)Br2|x|21+|x|2ϕdiv((21+|x|2)n3ϕ)=(n1)Br(21+|x|2)n3((2|x|21+|x|2ϕ))ϕ=n1n3Brx(21+|x|2)n3|ϕ|2+n12(n2)BrΔ(21+|x|2)n2ϕ2, (4.17)

    By (4.16) and (4.17), we have

    0=n12Br[x(21+|x|2)n1+(n1)(21+|x|2)n1]ϕ2μBrx(21+|x|2)n1ϕ2+Brx(21+|x|2)n1|ϕ|p+1n1n3Brx(21+|x|2)n3|ϕ|2. (4.18)

    We combine (4.14), (4.15) and (4.18) in the following way:

    (4.14)×n1p+1+(4.15)1p+1×(4.18),

    then

    h(r)2Br|ϕ|2=(n1p+1n32)Br(21+|x|2)n31|x|21+|x|2|ϕ|2+n12(p+1)(n1(p1)μ)Br(21+|x|2)n11|x|21+|x|2ϕ2.

    If pn+1n3 and (p1)μ(n1), then the left hand side of the last identity will become non-positive, therefore, Eq (4.13) has only trivial solution.

    Corollary 4.4. If pn+1n3 and if ϕ is a nontrivial solution of Eq (2.1) depends only on the variable ξ, here we use the coordinates in the proof of Lemma 2.3, then ϕ does not change sign.

    Proof. If ϕ change sign, then there exists 0<r<1 such that (4.13) has a nontrivial solution, this is a contradiction.

    The research of J. Wei is partially supported by NSERC of Canada.

    We declare that we have no financial and personal relationships with other people or organizations that can inappropriately influence our work.



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