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On isolated singularities of Kirchhoff equations

  • Huyuan Chen , Mouhamed Moustapha Fall and Binling Zhang EMAIL logo
Published/Copyright: May 27, 2020

Abstract

In this note, we study isolated singular positive solutions of Kirchhoff equation

Mθ(u)(Δ)u=upinΩ{0},u=0onΩ,

where p > 1, θ ∈ ℝ, Mθ(u) = θ + ∫Ω |∇ u| dx, Ω is a bounded smooth domain containing the origin in ℝN with N ≥ 2.

In the subcritical case: 1 < p < NN2 if N ≥ 3, 1 < p < + ∞ if N = 2, we employee the Schauder fixed point theorem to derive a sequence of positive isolated singular solutions for the above equation such that Mθ(u) > 0. To estimate Mθ(u), we make use of the rearrangement argument. Furthermore, we obtain a sequence of isolated singular solutions such that Mθ(u) < 0, by analyzing relationship between the parameter λ and the unique solution uλ of

Δu+λup=kδ0inB1(0),u=0onB1(0).

In the supercritical case: NN2 p < N+2N2 with N ≥ 3, we obtain two isolated singular solutions ui with i = 1, 2 such that Mθ(ui) > 0 under other assumptions.

MSC 2010: 35J75; 35B40; 35A01

1 Introduction and main results

A model with small variation of tension due to the changes of the length of a string is described by D’Alembert wave equation, it is also well-known as the Kirchhoff equation, see [15], which states as follows

m2ut2τ0+κ2L0αβux2dx2ux2=0,

where τ0 is the tension, L0 = βα is the length of the string at rest, m is the mass density, κ is the Young’s modulus. The Kirchhoff—type problems have been attracted great attentions in the analysis of different nonlinear term due to the gradient term, see [9, 11, 24, 37].

Observe that in the prototype of Kirchhoff model, the tension, for small deformations of the string, takes the linear form as follows:

M(u)=a+bΩ1+|u|2dx, (1.1)

where a > 0, b > 0. When the displacement gradient is small, i.e. |∇ u| ≪ 1, M(u) ∽ a + b|Ω| + b2 Ω |∇ u|2dx. The advantage for this approximation makes the problem have variational structure and the approximating solution could be constructed by variational methods. For example, the stationary analogue and qualitative properties of solutions to the Kirchhoff—type equation

a+bΩ|u|2dxΔu+V(x)u=f(x,u)inΩ

has been extensively studied in [9, 10, 14, 16, 19, 30] and extended into the fractional setting in [27, 28, 33] and the references therein. In this case, M(u) = a + bΩ |∇ u|2dx is often called Kirchhoff function. In fact, the Kirchhoff function has been greatly extended for recent years. For example, the case a = 0, b > 0, which is called degenerate, has been intensely investigated recently, we refer to [34] for a physical explanation and [8, 25, 26, 40, 41] for related results in this direction.

Our interest of this paper is to study a new Kirchhoff—type problem by taking into account that |∇ u| is not small in a bounded smooth domain Ω and the tension could be vector in a proper coordinate axis. In this situation, the Kirchhoff function (1.1) may be taken as

Mθ(u)=θ+Ω|u|dx, (1.2)

where θ is assumed to be real number. Given a sequence of extra pressures {σm} with the support in B1m (0) and the total force F = ∫Ωσm dx = 1 keeps invariant. The limit of {σm} as m → + ∞ in the distributional sense is Dirac mass. As we know that the corresponding solutions may blow up at the origin or blow up in the whole domain. Our aim is to clarify this limit phenomena of the solutions to some elliptic problems involving the Kirchhoff—type function (1.2).

More precisely, in this article we are interested in nonnegative singular solutions of the following Kirchhoff—type equation:

=Mθ(u)Δu=upinΩ{0},M(u)Δu=0onΩ, (1.3)

where p > 1, Mθ is defined by (1.2) with θ ∈ ℝ and Ω is a bounded, smooth domain containing the origin in ℝN with N ≥ 2. The following parameter plays an important role in obtaining the solutions of (1.3):

ap=supxΩw1w0, (1.4)

where w0 = 𝔾Ω[δ0] and w1=GΩ[w0p],GΩ is Green operator defined as

GΩ[u](x)=ΩGΩ(x,y)u(y)dy,

here GΩ is the Green kernel of − Δ in Ω × Ω with zero Dirichlet boundary condition. Note that ap is well-defined when p is subcritical, that is, p < p*, where

p=NN2ifN3,+ifN=2. (1.5)

Our first existence result about isolated singular solutions with Mθ(u) > 0 is stated as follows.

Theorem 1.1

Assume that N ≥ 2, Mθ is defined by (1.2) with θ ∈ ℝ, ap is given by (1.4), p* is given by (1.5) and Ω is a bounded smooth domain containing the origin such that

B1(0)Ωand|Br0(0)|=|Ω|

where 1 ≤ r0 < +∞.

Let k > r0 θ with θ := min{0, θ} be such that

kp1θ+r01k1appp1pp1. (1.6)

Then for p ∈ (1, p*), problem (1.3) has a nonnegative solution uk satisfying that

Mθ(uk)θ+r01k>0 (1.7)

and uk has following asymptotic behaviors at the origin

lim|x|0+uk(x)Φ1(x)=cNk, (1.8)

where cN > 0 is the normalized constant and

Φ(x)=|x|2NifN3,ln|x|ifN=2.

Furthermore, uk is a distributional solution of

Δu=upMθ(u)+kδ0inΩ,Δu=0onΩ, (1.9)

where δ0 is Dirac mass concentrated at the origin.

Remark 1.1

Note that ap depends on p and Ω and the value p = 2 is critical for assumption (1.6) for N = 2, 3. Indeed, p* > 2 occurs only for N = 2 and N = 3. Due to the parameter θ, (1.6) gives a rich structure of isolated singular solutions for problem (1.3). Moreover, a discussion is put in Proposition 2.2 in Section 2.

Involving Kirhchoff function Mθ(u), the classical method of Lions’ iteration argument in [17] does not work due to the lack of monotonicity of nonlinearity Mθ1 (u)up, and also the variational method in [29] fails, since (1.3) has no variational structure. Furthermore, it is difficult to calculate precise value for ∫Ω |∇ uk| to express Mθ(uk), especially, when Ω is a general bounded domain. To overcome these difficulties, we make use of the rearrangement argument to estimate the value of Mθ(u) and employ the Schauder fixed-point theorem to obtain the existence of isolated singular solutions in the class set of Mθ(u) > 0.

When θ < 0, we can derive a branch of singular solutions such that Mθ(u) < 0.

Theorem 1.2

  1. Let N ≥ 2, p ∈ (1, p*), θ < 0 and Ω = B1(0). For k ∈ (0, − θ), problem (1.3) has a nonnegative solution uk, which is a distributional solution of (1.9) with Ω = B1(0), satisfying that

    θ<Mθ(uk)<k+θ<0

    and uk has the asymptotic behavior (1.8).

  2. Let N ≥ 2, p ∈ ( N+1N1 , p*), θ < 0 and Ω is a bounded smooth domain containing the origin. Then problem (1.3) has a nonnegative solution up, which is not a distributional solution of (1.9), satisfying that

    Mθ(up)<0

    and up has the asymptotic behavior

    lim|x|0+up(x)|x|2p1=cp(Mθ(up))1p1,

    where cp=2p12p1+2N1p1.

For Mθ(u) < 0, problem (1.3) could be written as

Δu+λup=0inB1(0){0},u=0onB1(0), (1.10)

where λ = − Mθ1 (u) > 0. For λ = 1, the nonlinearity in problem (1.10) is an absorption and Lions showed in [17] that it is always studied by considering the very weak solutions of

Δu+λup=kδ0inB1(0). (1.11)

Véron in [39] gave a survey on the isolated singularities of (1.10), in which B1(0) is replaced by general bounded domain containing the origin. With a general Radon measure and a more general absorption nonlinearity g : ℝ → ℝ satisfies the subcritical assumption:

1+(g(s)g(s))s1pds<+,

problem (1.11) has been studied by Benilan-Brézis [1], Brézis [3], by approximating the measure by a sequence of regular functions, and find classical solutions which converges to a weak solution. For this approach to work, uniform bounds for the sequence of classical solutions are necessary to be established. The uniqueness is then derived by Kato’s inequality. Such a method has been applied to solve equations with boundary measure data in [13, 20, 21, 22] and other extensions in [2, 5].

In the case λ = − Mθ1 (u), depending on the unknown function u, a different approach has to be taken into account to study problem (1.11). A branch of solutions such that Mθ(u) < 0 are derived from the observations that the function F(λ) = − Mθ1 (u) − λ is continuous and it has a zero, because we will find two values λ1, λ2 such that F(λ1)F(λ2) < 0, where vλ is the unique solution of problem (1.11). This zero indicates a solution of problem (1.3).

For the singularity as |x|−2/(p−1), the diffusion and the nonlinear terms play the predominant roles in (1.3), so we just consider λ up, where up is the solution of − Δ u + up = 0 in Ω ∖ {0}. By scaling λ to meet the Kirchhoff function and then a solution with this type singularity is derived in Theorem 1.2. This scaling technique could be extended to obtain solutions in the supercritical case in Theorem 5.1 in Section 5.

It is worth pointing out that the method of searching solutions with the weak singularities as Φ in Theorem 1.1 could be extended into dealing with general nonlinearity f(u) when 0 ≤ f(u) ≤ c|u|p with p ∈ (1, p*). This method to prove Theorem 1.2 is based on the homogeneous property of nonlinearity and when the nonlinearity is not a power function, it is open but challenging to obtain solutions with such isolated singularity.

The rest of this paper is organized as follows. In Section 2, we introduce the very weak solution of equation (1.3) involving Dirac mass and give a discussion of (1.6). Section 3 is devoted to show the existence of a solution to (1.3) with Mθ(u) > 0 in Theorem 1.1. In Section 4, we search the solutions of (1.3) with Mθ(u) < 0 in Theorem 1.2. The supercritical case: N/(N − 2) ≤ p < (N + 2)/(N − 2) with N ≥ 3, is considered in Section 5, and we obtain there multiple isolated singular solutions of (1.3) such that Mθ(ui) > 0.

2 Preliminary

2.1 Kirchhoff-type problem with Dirac mass

In order to drive solutions of (1.3) with singularity (1.8), it is always transformed into finding solutions of (1.9). A function u is said to be a super (resp. sub) distributional solution of (1.9), if uL1(Ω), |∇ u| ∈ L(Ω), upL1(Ω, ρ dx) and

Ωu(Δ)ξupMθ(u)ξdμ(resp.)kξ(0),ξC01.1(Ω),ξ0, (2.1)

where ρ(x) = dist(x, Ω). A function u is a distributional solution of (1.9) if u is both super and sub distributional solutions of of (1.9).

Next we build the connection between the singular solutions of (1.3) and the distributional solutions of (1.9).

Theorem 2.1

Assume that N ≥ 2, p > 1 and uL1(Ω) is a nonnegative classical solution of problem (1.3) satisfying that Mθ(u) ≠ 0 and upL1(Ω, ρ dx). Then u is a very weak solution of problem (1.9) for some k ≥ 0. Furthermore,

  1. Mθ(u) < 0.

    1. For N ≥ 3, pp*, problem (1.3) only has zero solution and θ < 0.

    2. For N ≥ 2, 1 < p < p*, we have that k > 0 and

      lim|x|0u(x)Φ1(x)=cNk. (2.2)

  2. Mθ(u) > 0.

    1. For N ≥ 3, pp*, we have that k = 0 and

      lim|x|0u(x)|x|N2=0;
    2. Assume more that 1 < p < p*. If k = 0, then u is removable at the origin, and if k > 0, then u satisfies (1.8).

In order to prove Theorem 2.1, we need the following lemmas.

Lemma 2.1

Let τ ∈ (0, N), then for xB1/2(0) ∖ {0},

GΩ[||τ](x)c2|x|τ+2ifτ>2,c2log(|x|)ifτ=2,c2ifτ<2. (2.3)

For N ≥ 3, p ∈ (1, p*), there holds

GΩ[GΩp[δ0]]c2|x|p(2N)+2ifp(2N2,p),c2log(|x|)ifp=2N2,c2ifp<2N2. (2.4)

Proof

We follow the idea of Lemma 2.3 in [6]. In fact, from [2, Propsition 2.1] it follows that the Green kernel verifies that

GΩ(x,y)cNΦ(xy),

By direct computation, we get (2.3). Since lim|x|→0+ 𝔾Ω[δ0](x)Φ−1(x) → cN, (2.3) with τ = (2 − N)p implies (2.4). □

Proposition 2.1

([36] or [7, Propostion 5.1]) Let hLs(Ω) with s ≥ 1, then there exists c3 > 0 such that

  1. GΩ[h]L(Ω)c3hLs(Ω)if1s<2N; (2.5)
  2. GΩ[h]Lr(Ω)c3hLs(Ω)if1s1r+2Nands>1; (2.6)
  3. GΩ[h]Lr(Ω)c3hL1(Ω)if1<1r+2N. (2.7)

Proof of Theorem 2.1

For Mθ(u) ≠ 0, we rewrite (1.3) as

Δu=upMθ(u)inΩ{0},Δu=0onΩ. (2.8)

Since upL1(Ω, ρ dx) and uL1(Ω), we may define the operator L by the following

L(ξ):=Ωu(Δ)ξupMθ(u)ξdx,ξCc(RN). (2.9)

First we claim that for any ξ Cc (Ω) with the support in Ω ∖ {0},

L(ξ)=0.

In fact, since ξ Cc (Ω) has the support in Ω ∖ {0}, then there exists r ∈ (0, 1) such that ξ = 0 in Br(0) and then

L(ξ)=ΩBr(0)u(Δ)ξupMθ(u)ξdx=ΩBr(0)ΔuupMθ(u)ξdx=0.

From Theorem 1.1 in [4], it implies that

L=kδ0forsomek0, (2.10)

that is,

L(ξ)=Ωu(Δ)ξupMθ(u)ξdx=kξ(0),ξCc(RN). (2.11)

Then u is a weak solution of (1.9) for some k ≥ 0.

  1. Mθ(u) < 0. We observe that

    u=kGΩ[δ0]1Mθ(u)GΩ[uq]kGΩ[δ0],

    then

    kGΩ[δ0]kpMθ(u)GΩ[GΩ[δ0]p]ukGΩ[δ0]inΩ{0}.

    So if k = 0, we obtain that u ≡ 0, which implies Mθ(u) = θ < 0; and if k > 0

    lim|x|0+u(x)Φ1(x)=cNk.

    We prove that k = 0 if pp* with N ≥ 3. By contradiction, if k > 0, then

    u(k/2)ΦinBr0(0){0},

    which implies that

    up(x)(k/2)p|x|(2N)p,xBr0(0){0},

    where (2 − N)p ≤ − N and r0 > 0 is such that B2r0(0) ⊂ Ω. A contradiction is obtained that upL1(Ω). Therefore, when pp*, there is no nontrivial nonnegative solution (1.3) such that Mθ(u) < 0.

  2. Mθ(u) > 0. We refer to [17] for the proof. For the reader’s convenience, we give the details. When p ∈ (1, N/(N − 2)) and k = 0, then

    u=1Mθ(u)GΩ[up].

    We infer from upLt0(Ω) with t0=12(1+1pNN2)>1 and Proposition 2.1 that uLt1p(Ω) and upLt1(Ω) with

    t1=1pNN2t0t0>t0.

If t1 > Np/2, by Proposition 2.1, uL(Ω ) and then it could be improved that u is a classical solution of

Δu=1Mθ(u)upinΩ. (2.12)

If t1 < Np/2, we proceed as above. By Proposition 2.1, uLt2p(Ω), where

t2=1pNt1N2t1>1pNN2t0t1=1pNN2t02t0.

Inductively, let us define

tm=1pNtm1N2tm1>1pNN2t0mt0+asm+.

Then there exists m0 ∈ ℕ such that

tm0>12Np

and by part (i) in Proposition 2.1,

uL(Ω).

It then follows that u is a classical solution of (2.12).

When p ∈ (1, N/(N − 2)) and k ≠ 0, we observe that

limx0GΩ[δ0](x)|x|N2=cN,α

and

u=1Mθ(u)GΩ[up]+kGΩ[δ0]. (2.13)

We let

u1=1Mθ(u)GΩ[up]andΓ0=kGΩ[δ0].

Then by Young’s inequality,

up2pu1p+Γ0p. (2.14)

By the definition of u1 and (2.14), we obtain

u12pGΩ[u1p]+Γ1, (2.15)

where u1Ls(Ω) for any s ∈ (1, N/(N − 2)) and

Γ1=2pGΩ[Γ0p].

Denoting μ1 = 2 + (2 − N)p, then for 0 < |x| < 1/2,

Γ1(x)c1|x|μ1ifμ1<0,c1log|x|ifμ1=0,c1ifμ1>0.

If μ1 ≤ 0, letting

u2=2pGΩ[u1p],

then u2Ls(Ω) with s ∈ [1, NN2 ), u1u2 + Γ1 and

u22pGΩ[u2p]+GΩ[Γ1p].

Let μ2 = μ1 p + 2, then μ2 > μ1 and for 0 < |x| < 12 ,

Γ2(x):=2pGΩ[Γ1p](x)c2|x|μ2ifμ2<0,c2log|x|ifμ2=0,c2ifμ2>0.

Inductively, we assume that

un12pGΩ[un1p]+2pGΩ[Γn2p],

where un−1Ls(Ω) for s ∈ [1, N/(N − 2)), Γn−2(x) ≤ |x|μn−2 for μn−2 < 0.

Let

un=2pGΩ[un1p],Γn1=2pGΩ[Γn2p],

and

μn1=μn2p+2.

Then unLs(Ω) for s ∈ [1, N/(N − 2)) and for 0 < |x| < 1/2,

Γn1(x):=GΩ[Γn2p](x)cn|x|μn1ifμn1<0,cnlog|x|ifμn1=0,cnifμn1>0.

We observe that

μn1μn2=p(μn2μn3)=pn3(μ2μ1)+asn+.

Then there exists n2 ≥ 1 such that

μn21>0andμn220

and

uun2+i=1n21Γi+Γ0, (2.16)

where Γic|x|μi and

un22p(GΩ[un2p]+1).

Next, we claim that un2L(Ω). Since un2Ls(Ω) for s ∈ [1, N/(N − 2)), letting

t0=121+1pNN21,NN2,

then 1pNN2t0>1 and by Proposition 2.1, we have that un2Lt1(Ω) with

t1=1pNt0N2t0.

Inductively, it implies by un2Ltn−1(Ω) that un2Ltn(Ω) with

tn=1pNtn1N2tn1>1pNN2t0nt0+asn.

Then there exists n3 ∈ ℕ such that

sn3>Np2

and by part (i) in Proposition 2.1, it infers that

un2L(Ω).

Therefore, it implies by uΓ0 and (2.16) that

limx0u(x)|x|N2=cN,αk.

This ends the proof. □

2.2 Discussion on (1.6)

The following two functions plays an important role in searching distributional solutions of problem (1.9)

w0=GΩ[δ0],w1=GΩ[w0p], (2.17)

which are the solutions respectively of

Δu=δ0inΩ,Δu=0onΩ (2.18)

and

Δu=w0pinΩ,Δu=0onΩ. (2.19)

Observe that ap > 0 defined in (1.4) is the smallest constant with p ∈ (1, NN2 ) such that

w1apw0inΩ{0}. (2.20)

Obviously, ap depends on the domain Ω.

Proposition 2.2

Let Ω = B1(0).

  1. If θ > 0 and 1 < p < min{2, p*}, there exists ap > 0 depending θ such that when 0 < ap ap , (1.6) holds for any k > 0; and when ap > ap , (1.6) holds for 0 < kk1 and k2k < + ∞, where 0 < k1 < k2 < + ∞.

    If θ > 0, p* > 2 and 2 < p < p*, there exists k3 > 0 such that for 0 < kk3, (1.6) holds.

    If θ > 0, p* > 2 and p = 2, then when a2 > 14 , (1.6) holds for 0 < k < θ4a21 ; and when a2 14 , (1.6) holds for any k > 0.

  2. If θ = 0 and 1 < p < min{2, p*}, then (1.6) is equivalent to

    k(p1)p1ppap12p;

    If θ = 0, p* > 2 and 2 < p < p*, then (1.6) is equivalent to

    0<k(p1)p1ppap1p2.

    If θ = 0, p* > 2 and p = 2, then when a2 > 14 , there is no k > 0 such that (1.6) holds; and when a2 14 , (1.6) holds for any k > 0.

  3. If θ < 0 and 1 < p < min{2, p*}, then (1.6) holds for kk4, where

    k4>(p1)p1ppap12p;

    If θ < 0, p* > 2 and 2 < p < p*, then ap = (− θ)2−p pp(p − 1)(p − 2)p−3 such that when 0 < ap ap , (1.6) holds for k5kk6, where 0 < k5 p12p θk6 < + ∞; and when ap > ap , there is no k > 0 such that (1.6) holds.

    If θ < 0, p* > 2 and p = 2, then when a2 < 14 , (1.6) holds for 0 < k < θ4a21 ; and when a2 14 , (1.6) holds for any k > 0.

Proof

When Ω = B1(0), we have that r0 = 1. Let

h(k)=kp1θ+k1appp1pp1,k(θ,+).

Note that

h(k)=(p1)kp2(θ+k)kp1(θ+k)2.

When p ≠ 2, h′(k0) = 0 implies that

k0=p12pθ.

When p = 2,

h(k)=kθ+k14a2,k(0,+).

The rest of the proof is simple and hence we omit it. □

When p = 2, note that 14 is a critical value for (1.6) and we show that a2 < 14 when Ω is a ball.

Lemma 2.2

Assume that Ω = B1(0), N = 2 or 3, p = 2 and a2 is given by (2.20). Then

α2<14.

Proof

When Ω = B1(0), take ξ(x) = 1 − |x| as a test function, we derive

B1(0)|w0|dx=B1(0)w0(1|x|)dx=1. (2.21)

Since w1 is radial symmetric and decreasing, then

(rN1w1(r))=rN1w02.

So for N = 3,

w1(r)=116π2r20r(1t)2dt

and

w1(r)=148π2r1s2[(1s)31]ds=148π23(r1)3lnrr212.

Then

w1(r)w0(r)=112π3r3rlnr1r+r+r22,

then rw1(r)w0(r) is increasing, so

a2=limr1w1(r)w0(r)=w1(1)w0(1).

So for N = 2,

w1(r)=14π2r10r(lnt)2tdt

and

w1(r)=18π2r1s(lns)2slnss2ds.

Then

w1(r)w0(r)=r2(lnr)22+r2lnr+1r2412πlnr,

then rw1(r)w0(r) is increasing, so

a2=limr1w1(rw0(r)=w1(1)w0(1).

We see that

w1(1)=148π2ifN=3,116π2ifN=2

and

w0(1)=14πifN=3,12πifN=2,

so

α2=112πifN=3,18πifN=2.

Therefore, we have that α2 < 1/4. The proof is thus complete. □

Corollary 2.1

Assume that N = 2 or 3, p = 2 Mθ is defined by (1.2) with θ ≥ 0, a2 is given by (1.4), Ω = B1(0). Then for any k > 0, problem (1.3) has a nonnegative solution uk satisfying (1.7) and (1.8).

3 Solutions with Mθ(u) > 0

In order to do estimates on Mθ(u), we introduce the following lemma.

Lemma 3.1

Let u, v be a radially symmetric, decreasing and nonnegative functions in C1(B1(0) ∖ {0}) ∩ W01,1 (B1(0)) such that

uL1(B1(0))vL1(B1(0))andlim inf|x|0+[u(x)v(x)]|x|N10. (3.1)

Then

B1(0)|u|dxB1(0)|v|dx.

Proof

For radially symmetric decreasing function fC1(B1(0) ∖ {0}) ∩ W01,1 (B1(0)), we have that

ωNf(r)rN1+(N1)ωNr1f(s)sN2ds=ωNr1f(s)sN1ds,

then we have that

ωNlim|x|0+(uv)(x)|x|N1+(N1)B1(0)[u(x)v(x)]dx=B1(0)|u|dxB1(0)|v|dx.

From (3.1), we have that

B1(0)|u|dxB1(0)|v|dx.

This finishes the proof. □

Proof of Theorem 1.1

We search for distributional solutions of

Δu=1Mθ(u)up+kδ0inΩ,u=0onΩ (3.2)

by using the Schauder fixed-point theorem. Let w0, w1 be the solutions of (2.17) and denote

wt=tkpw1+kw0, (3.3)

where the parameter t > 0.

We claim that there exists kp > 0 independent of θ such that for k ∈ (0, kp], if θ + r01 k > 0 there exists tp > 0 such that

tpkpw1GΩ[wtpp]θ+r01k. (3.4)

We observe that if

(aptkp+k)pθ+r01ktkp, (3.5)

then wt verifies (3.4), since

GΩ[wtpp]θ+r01k(aptkp+k)pGΩ[w0p]θ+r01k=(aptkp+k)pθ+r01kw1tkpw1.

Now we discuss what condition on k guarantee that (3.5) holds for some t > 0. In fact, (3.5) is equivalent to

(aptkp1+1)pt(θ+r01k) (3.6)

or in the form

s=t(θ+r01k)andapkp1θ+r01ks+1ps.

For p > 1, since the function f(s)=(1p(p1p)p1s+1)p intersects the line g(s) = s at the unique point sp=pp1p, so k may be chosen such that

apkp1θ+r01k1pp1pp1. (3.7)

In fact, (1.6) implies (3.7). Therefore, for k > r0 θ satisfying (1.6) and taking tp=(θ+k)1pp1p, function wtp verifies (3.4).

Let

Dk={uW01.1(Ω):0utpkpw1}.

Denote

Tu=1Mθ(u+kw0)GΩ[(u+kw0)p],uDk.

We claim that

Mθ(u+kw0)θ+r01k>0foruDk. (3.8)

For u ∈ 𝓓k, we may let vn C01 (Ω) be a sequence of nonnegative functions converging to u in W01,1 (Ω). Let un = vn + kw0, and by the fact that w0C1(Ω ∖ {0}) ∩ W01,1 (Ω), then unC1(Ω ∖ {0}) ∩ W01,1 (Ω), unkw0 in Ω ∖ {0} and un converge to u + kw0 in W1,1(Ω). By the symmetric decreasing arrangement, we may denote un , the symmetric decreasing rearranged function of un in Br0(0), where r0 ≥ 1 such that |Br0(0)| = |Ω|. Observe that

lim inf|x|0+un(x)|x|N10=klim|x|0+w0(x)|x|N1

and

ΩundxkΩw0dx.

By Pólya-Szegő inequality, we have that

unL1(Ω)unL1(Br0(0))=r01wnL1(B1(0)).

where wn(x)=r0Nun(r0x) for xB1(0).

Let wB1(0) = k 𝔾B1(0)[δ0], since B1(0) ⊂ Ω, Kato’s inequality implies that

Ωw0dxB1(0)wB1(0)dx.

Thus,

B1(0)wndx=Br0(0)undxB1(0)wB1(0)dx. (3.9)

Thus, by Lemma 3.1, (3.9) and (2.21), we have

wnL1(B1(0))kwB1(0)L1(B1(0))=k

Therefore, passing to the limit as n → +∞ in the above inequality we get that

Mθ(u+kw0)θ+r01kwB1(0)L1(B1(0))=θ+r01k,

which implies (3.8).

Therefore, from (3.4) it follows that

Tu=GΩ[(kw0+u)p]Mθ(kw0+u)GΩ[(kw0+tpkpw1)p]θ+r01ktpkpw1,

then

TDkDk.

Note that for u ∈ 𝓓k, one has that (u + kw0)pLσ(Ω) with σ (1,1pNN2) , then 𝓣 𝓓kW2,σ(Ω), where σ (1,1pNN2) . Since the embeddings W2,σ(Ω) ↪ W1,1(Ω), L1(Ω) are compact and then 𝓣 is a compact operator.

Observing that 𝓓k is a closed and convex set in L1(Ω), we may apply the Schauder fixed-point theorem to derive that there exists vk ∈ 𝓓k such that

Tvk=vk.

Since 0 ≤ vktpkpw1, so vk is locally bounded in Ω ∖ {0}, then uk := vk + kw0 satisfies (1.8), and by interior regularity results, uk is a positive classical solution of (1.3). From Theorem 2.1 we deduce that uk is a distributional solution of (1.9). □

4 Solutions with Mθ(u) < 0

For θ < 0 and Mθ(u) < 0, equation (1.9) could be written as

Δu+1Mθ(u)up=kδ0inB1(0),u=0onB1(0). (4.1)

Lemma 4.1

Let p ∈ (1, p*) and λ > 0. For any k > 0, the problem

Δu+λup=kδ0inB1(0),u=0onB1(0) (4.2)

has a unique positive weak solution uλ,k verifying that

lim|x|0+uλ,k(x)|x|N2=cNk. (4.3)

Furthermore, uλ,k is radially symmetric and decreasing with to |x| and the map λuλ,k is decreasing.

Proof

The existence could be seen [38, theorem 3.7] and uniqueness follows by Kato’s inequality [38, theorem 2.4]. The radial symmetry of uλ,k and decreasing monotonicity with to |x| could be derived by the method of moving plane, see [12, 35] for the details. It follows from Kato’s inequality that the map λuλ,k is decreasing. This ends the proof. □

Proof of Theorem 1.2

  1. Observe that

    Mθ(kw0)=kB1(0)|w0|dx+θ=k+θ<0.

    From Lemma 4.1 with λ = λ1 := − Mθ1 (kw0), problem (4.2) with λ = λ1 has a unique solution vλ1 verifying that

    0<vλ1kw0,

    then it implies that

    kB1(0)|vλ1|dxB1(0)|kw0|dx

    and

    Mθ(vλ1)=kB1(0)|vλ1|dx+θkB1(0)|w0|dx+θ=k+θ,

    thus,

    θ<Mθ(vλ1)<k+θ,

    that is,

    1Mθ(vλ1)<λ1. (4.4)

    In terms of Lemma 4.1, let λ2 = − Mθ1 (vλ1) and {vλ2} be the solution of problem (4.2) with λ = λ2. Since λ2 > λ1, then

    vλ1<vλ2<kw0.

    So it follows by Lemma 3.1 that

    Mθ(vλ1)<Mθ(vλ2)<Mθ(kw0),

    that is,

    1Mθ(vλ2)>λ2. (4.5)

    We claim that the map λ ∈ [λ2, λ1] ↦ Mθ(uλ,k) is continuous.

    At this moment, we assume that the above argument is true. Let

    F(λ)=1Mθ(vλ)λ,

    where vλ is the solution of (4.2) with λ ∈ [λ2, λ1]. Since F is continuous in [λ2, λ1], by (4.4), (4.5) and the mean value theorem, there exists λ0 ∈ (λ2, λ1) such that F(λ0) = 0, that is, (4.1) has a solution uk with 1Mθ(uk) = λ0. From standard regularity, we have that uk is a classical solution of (1.3) and verifies the corresponding properties in the lemma.

    Now we prove that the map λ ∈ [λ2, λ1] ↦ Mθ(uλ,k) is continuous. Let λ2λ′ < λ″ ≤ λ1 and uλ′,k and uλ″,k be the solutions of (4.1) with λ = λ′ and λ = λ″ respectively. Then

    uλ,k<uλ,k

    and

    Mθ(uλ,k)<Mθ(uλ,k). (4.6)

    Let u¯=uλ,k+(λλλ2)1/pw0. Then

    Δu¯+λu¯pΔuλ,k+λλλ21p(Δ)w0+λuλ,kp+λλλλ2w0pΔuλ,k+λuλ,kp=kδ0.

    Therefore Kato’s inequality implies that

    uλ,kuλ,k+λλλ21pw0,

    which yields that

    Mθ(uλ,k)Mθ(uλ,k)+λλλ21pk.

    This together with (4.6), give

    |Mθ(uλ,k)Mθ(uλ,k)|λλλ21pk0as|λλ|0,

    thus, the map λ ∈ [λ2, λ1] ↦ Mθ(uλ,k) is continuous.

  2. It is well known that for p ∈ (1, p*), the problem

    Δu+up=0inΩ{0},u=0onΩ (4.7)

    has a positive solution vp verifying that

    lim|x|0+vp(x)|x|2p1=cp, (4.8)

    where cp=[2p1(2p1+2N)]1p1. Furthermore, vp is the unique solution of (4.7) such that

    lim inf|x|0+u(x)|x|2p2>0. (4.9)

We observe that

vλ:=λ1p1vp

is the unique solution of

Δu+λup=0inΩ{0},u=0onΩ (4.10)

in the set of functions satisfying (4.9).

For p ∈ ( N+1N1 , p*), we have that ∫Ω|∇ up| dx < +∞, so that

Mθ(vλ):=λ1p1m2+θ<0forλ(λ0,+),

where m2 = ∫Ω|∇ up| dx and λ0 = (m2/(− θ))p−1.

We define

F(λ):=1λ1p1m2+θ+λ,λ(λ0,+).

Observe that F is continuous, increasing and

limλλ0+F(λ)=,limλ+F(λ)=+.

Hence there exists a unique λ̄ such that

1λ¯1p1m2+θ=λ¯.

Meaning that − Mθ1 (vλ̄) = λ̄. We then conclude that (1.3) has a solution up := vλ̄ with Mθ(up) < 0. From (4.8) and the definition of vλ, we know that up is not a weak solution of problem (1.9). □

5 In the supercritical case

In the super critical case that p*p < 2* − 1, we have the following existence results.

Theorem 5.1

  1. Let N ≥ 3, p*p < 2* − 1, θ ∈ ℝ and Ω be a bounded smooth domain containing the origin. If

    1. p > 2, pp* and θ > 0;

    2. p = 2 ≥ p*, θ > 0 and m2 < 1;

    3. p*p < 2 and θ < 0;

    4. p*p < 2* − 1, p ≠ 2 and θ = 0,

      then problem (1.3) has two positive solutions ui with i = 1, 2 satisfying that

      Mθ(ui)>0,
      ifpp,N+2N2,lim|x|0+ui(x)|x|2p1=Mθ(ui)1p1cp (5.1)

      and

      ifp=p,lim|x|0+ui(x)|x|N2(ln|x|)N22=Mθ(ui)N22cp, (5.2)

      where cp=[2p1(N22p1)]1p1andcp=(N24)N2.

  2. Let N = 4, 5, p = 2 ∈ [p*, 2* − 1), θ = 0 and Ω be a bounded smooth domain containing the origin. If v is a solution of (5.3) such that Mθ(v) = 1, then for any λ > 0, u := λ v is a solution of problem (1.3) satisfying Mθ(u) = λ > 0 and (5.1)(5.2).

To prove Theorem 5.1, we need the following lemma.

Lemma 5.1

([31, 32]) Let N ≥ 3, p ∈ [p*, N+2N2 ) and Ω be a bounded smooth domain containing the origin. Then the following problem

Δu=upinΩ{0},u=0onΩ (5.3)

has two positive singular solution v1 and v2 verifying that

ifp(p,N+2N2),lim|x|0+vi(x)|x|2p1=cp (5.4)

and

ifp=p,lim|x|0+vi(x)|x|N2(ln|x|)N22=cp. (5.5)

Proof of Theorem 5.1

From Lemma 5.1, it is known that for p*p < 2* − 1, problem (5.3) has two positive solutions vi verifying that (5.4) and (5.5).

We observe that

vλ,i=λ1p1vi

is a solution of

Δu+λup=0inΩ{0},u=0onΩ. (5.6)

For p*p < 2* − 1, we have that ∫Ω|∇ up| dx < +∞, then

Mθ(vλ,i)=λ1p1mi+θ>0

for λ ∈ (0, λ+), where mi = ∫Ω|∇ vi| dx and

λ+=+ifθ0,(mi/θ)p1ifθ<0.

Denote

Fθ(λ)=1λ1p1mi+θλ,λ(0,λ+),

which is continuous and

limλλ+Fθ(λ)=ifθ>0,+ifθ<0.

  1. p > 2, pp* and θ > 0, then there exists t > 0 such that

    Fθ(λ)>0.
  2. p = 2 ≥ p*, θ > 0 and mi < 1, then there exists t > 0 such that

    Fθ(λ)>0.
  3. p*p < 2 and θ < 0, then there exists t > 0 such that

    Fθ(λ)<0.

    In the above three cases, there exists a unique λ̄i such that

    1λ¯i1p1mi+θ=λ¯i,

    that is, Mθ1 (vλ̄i,i) = λ̄i. Therefore, (1.3) has a solution ui := vλ̄i,i with Mθ(ui) > 0.

    When θ = 0,

    F0(λ)=λ1p1miλ,λ(0,+),

    When N ≥ 4, we have that 1/(p − 1) > 1 for p*p < 2* − 1, or when N = 3, p*p < 2* − 1, p ≠ 2, λ̄i = mi(p1)/(p2) , then (1.3) has a solution ui := vλi,i.

    When N = 4, 5 and p = 2 ∈ [p*, 2* − 1), if mi = 1, then for any λ > 0, u := λ1p1 vi is a solution (1.3) with Mθ( u) = λ > 0 and verifying (5.1)(5.2). □

Remark 5.1

Our method to prove Theorem 5.1 is based on the homogeneous property of the nonlinearity. When the nonlinearity is not a power function, this scaling method fails and so it is challenging to provide the existence results of isolated singular solutions.

Acknowledgments

H. Chen is supported by the National Natural Science Foundation of China (No. 11726614, No. 11661045) and the Alexander von Humboldt Foundation. M. Fall is supported by the Alexander von Humboldt Foundation. B. Zhang was supported by the National Natural Science Foundation of China (No. 11871199), the Heilongjiang Province Postdoctoral Startup Foundation (LBH-Q18109), and the Cultivation Project of Young and Innovative Talents in Universities of Shandong Province.

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Received: 2019-12-15
Accepted: 2020-03-14
Published Online: 2020-05-27

© 2021 Huyuan Chen et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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  70. Initial boundary value problems for the three-dimensional compressible elastic Navier-Stokes-Poisson equations
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