• No results found

A Hardy Inequality with Remainder Terms in the Heisenberg Group and the Weighted Eigenvalue Problem

N/A
N/A
Protected

Academic year: 2020

Share "A Hardy Inequality with Remainder Terms in the Heisenberg Group and the Weighted Eigenvalue Problem"

Copied!
24
0
0

Loading.... (view fulltext now)

Full text

(1)

Volume 2007, Article ID 32585,24pages doi:10.1155/2007/32585

Research Article

A Hardy Inequality with Remainder Terms in the Heisenberg

Group and the Weighted Eigenvalue Problem

Jingbo Dou, Pengcheng Niu, and Zixia Yuan

Received 22 March 2007; Revised 26 May 2007; Accepted 20 October 2007

Recommended by L´aszl ´o Losonczi

Based on properties of vector fields, we prove Hardy inequalities with remainder terms in the Heisenberg group and a compact embedding in weighted Sobolev spaces. The best constants in Hardy inequalities are determined. Then we discuss the existence of solutions for the nonlinear eigenvalue problems in the Heisenberg group with weights for the p-sub-Laplacian. The asymptotic behaviour, simplicity, and isolation of the first eigenvalue are also considered.

Copyright © 2007 Jingbo Dou et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Let

Lp,μu= −ΔH,pu−μψp|u| p−2u

dp , 0≤μ≤

Qp

p p

, (1.1)

be the Hardy operator on the Heisenberg group. We consider the following weighted eigenvalue problem with a singular weight:

Lp,μu=λ f(ξ)|u|p−2u, in ΩHn,

u=0, onΩ, (1.2)

where 1< p < Q=2n+ 2, λ R, f(ξ)Ᏺp := {f : Ω→R+|lim

d(ξ)0(dp(ξ)f(ξ)/

p(ξ))=0, f(ξ)∈Lloc\ {0})},Ωis a bounded domain in the Heisenberg group, and

(2)

well as the eigenfunctions corresponding to other eigenvalues change sign. Our proof is mainly based on a Hardy inequality with remainder terms. It is established by the vec-tor field method and an elementary integral inequality. In addition, we show that the constants appearing in Hardy inequality are the best. Then we conclude a compact em-bedding in the weighted Sobolev space.

The main difficulty to study the properties of the first eigenvalue is the lack of regu-larity of the weak solutions of thep-sub-Laplacian in the Heisenberg group. Let us note that the regularity for the weak solutions of the p-subelliptic operators formed by the vector field satisfying H¨ormander’s condition was given in [1] and theC1,αregularity of the weak solutions of thep-sub-LaplacianΔH,pin the Heisenberg group for pnear 2 was proved in [2]. To obtain results here, we employ the Picone identity and Harnack inequality to avoid effectively the use of the regularity.

The eigenvalue problems in the Euclidean space have been studied by many authors. We refer to [3–11]. These results depend usually on Hardy inequalities or improved Hardy inequalities (see [4,12–14]).

Let us recall some elementary facts on the Heisenberg group (e.g., see [15]). LetHnbe a Heisenberg group endowed with the group law

ξ◦ξ=

x+x,y+y,t+t+ 2 n

i=1

xiyi−xiyi

, (1.3)

whereξ=(z,t)=(x,y,t)=(x1,x2,. . .,xn,y1,. . .,yn,t),z=(x,y),x∈Rn, y∈Rn,t∈R, n≥1;ξ=(x,y,t)R2n+1. This group multiplication endowsHnwith a structure of nilpotent Lie group. A family of dilations onHnis defined as

δτ(x,y,t)=

τx,τ y,τ2t, τ >0. (1.4)

The homogeneous dimension with respect to dilations isQ=2n+ 2.The left invariant vector fields on the Heisenberg group have the form

Xi=∂x i

+ 2yi∂t, Yi=∂y i−

2xi∂t, i=1, 2,. . .,n. (1.5)

We denote the horizontal gradient by ∇H = (X1,. . .,Xn,Y1,. . .,Yn), and write divH(v1,v2,. . .,v2n)= ni=1(Xivi+Yivn+i). Hence, the sub-LaplacianΔH and the p-sub-LaplacianΔH,pare expressed by

ΔH= n

i=1

X2

i +Yi2= ∇H·∇H,

ΔH,pu= ∇H∇Hup−2∇Hu=divH∇Hup−2∇Hu, p >1,

(1.6)

respectively.

The distance function is

(3)

Ifξ=0, we denote

d(ξ)=d(ξ, 0)=|z|4+t21/4, with|z| =x2+y21/2. (1.8)

Note thatd(ξ) is usually called the homogeneous norm. Ford=d(ξ), it is easy to calculate

∇Hd= 1 d3

|z|2x+yt

|z|2yxt

, ∇Hdp=|z|p

dp =ψp, ΔH,pd=ψp Q−1

d . (1.9)

Denote byBH(R)= {ξ∈Hn|d(ξ)< R}the ball of radiusRcentered at the origin. Let Ω1=BH(R2)\BH(R1) with 0≤R1< R2≤ ∞andu(ξ)=v(d(ξ))∈C2(Ω1) be a radial

function with respect tod(ξ). Then

ΔH,pu=ψpvp−2

(p1)v+Q−1

d v

. (1.10)

Let us recall the change of polar coordinates (x,y,t)→(ρ,θ,θ1,. . .,θ2n−1) in [16]. If

u(ξ)=ψp(ξ)v(d(ξ)), then

Ω1u(ξ)dξ =sH R2

R1

ρQ−1v(ρ)dρ, (1.11)

wheresH=ωn

π

0(sinθ)n− 1+p/2

dθ,ωnis the 2n-Lebesgue measure of the unitary Euclidean sphere inR2n.

The Sobolev space inHnis written byD1,p(Ω)={u:Ω→R;u,|∇Hu|∈Lp(Ω)}.D1,p

0 (Ω)

is the closure ofC∞0(Ω) with respect to the normuD1,0p(Ω)=(

Ω|∇Hu|pdξ)1/ p.

In the sequel, we denote byc,c1,C, and so forth some positive constants usually except

special narrating.

This paper is organized as follows. InSection 2, we prove the Hardy inequality with remainder terms by the vector field method in the Heisenberg group. InSection 3, we discuss the optimality of the constants in the inequalities which is of its independent interest. InSection 4, we show some useful properties concerning the Hardy operator (1.1), and then check the existence of solutions of the eigenvalue problem (1.2) (1< p < Q) and the asymptotic behavior of the first eigenvalue asμincreases to ((Q−p)/ p)p. In

Section 5, we study the simplicity and isolation of the first eigenvalue.

2. The Hardy inequality with remainder terms

D’Ambrosio in [17] has proved a Hardy inequality in the bounded domainΩHn: let p >1 andp=Q. For anyu∈D1,0p(Ω,|z|p/d2p), it holds that

CQ,p

Ωψp |u|p

dp dξ≤

Ω

∇Hup

, (2.1)

(4)

choice of a suitable vector field and an elementary integral inequality. Note that we also require that 0Ω.

Theorem2.1. Letu∈D01,p/{0}). Then

(1)ifp=Qand there exists a positive constantM0such thatsupξΩd(ξ)e1/M0:=R0<∞,

then for anyR≥R0,

Ω

∇Hup

dξ≥Q−p p

pΩψp|u| p

dp + p−1

2p

Q−p pp−2Ωψp|u| p

dp

ln

R

d 2

dξ;

(2.2)

moreover, if2≤p < Q, then choosesupξΩd(ξ)=R0;

(2)ifp=Qand there exitsM0such thatsupξΩd(ξ)e1/M0< R, then

Ω

∇Hup dξ≥

p1

p p

Ωψp

|u|p

dln(R/d)pdξ. (2.3)

Before we prove the theorem, let us recall that

Γd(ξ)=

⎧ ⎨ ⎩

d(ξ)(p−Q)/(p−1) if p=Q,

lnd(ξ) if p=Q (2.4)

is the solution ofΔH,pat the origin, that is,ΔH,(d(ξ))=0 onΩ\ {0}. Equation (2.4)

is useful in our proof. For convenience, writeᏮ(s)= −1/ln(s),s∈(0, 1), andA=(Q p)/ p. Thus, for some positive constantM >0,

0d(ξ) R

≤M, sup

ξ∈Ωd(ξ)< R, ξ∈Ω. (2.5)

Furthermore,

∇Hγd R

γ+1(d/R)∇Hd

d ,

dγ(ρ/R)

γ+1(ρ/R)

ρ ∀γ∈R, (2.6) b

a

γ+1(s)

s ds=

1 γ

γ(b)γ(a). (2.7)

Proof. Let T be aC1vector field onΩand let it be specified later. For anyuC

0(Ω\ {0}),

we use H¨older’s inequality and Young’s inequality to get

Ω

divHT

|u|p= −p

Ω

T,∇Hu|u|p−2udξ

≤p

Ω

∇Hup

1/ p

Ω|T|

p/(p−1)|u|p

(p−1)/ p

Ω

∇Hup

+ (p−1)

Ω|T|

p/(p−1)|u|pdξ.

(5)

Thus, the following elementary integral inequality:

Ω

∇Hup dξ≥

Ω

divHT(p1)|T|p/(p−1)

|u|p (2.9)

holds.

(1) Letabe a free parameter to be chosen later. Denote

I1(Ꮾ)=1 +p−1

pA

d

R

+aᏮ2 d

R

, I2(Ꮾ)= p−1

pA

2d

R

+ 2aᏮ3 d

R

,

(2.10)

and pick T(d)=A|A|p−2(|∇Hd|p−2∇Hd/dp−1)I

1. An immediate computation shows

divH

A|A|p−2∇Hd

p−2

∇Hd dp−1

=A|A|p−2dΔH,pd−(p1)∇Hd

p

dp

=A|A|p−2(Q1−p+ 1)∇Hd

p

dp =p|A|

p∇Hd p

dp .

(2.11)

By (2.6),

divHT=p|A|p

∇Hdp

dp I1+A|A|

p−2∇Hd

p−2

∇Hd dp−1

×∇Hd d

p1

pA

2d

R

+ 2aᏮ3 d

R

=p|A|p∇Hd p

dp I1+A|A|p−

2∇Hd

p

dp I2,

divHT(p1)|T|p/(p−1)=p|A|p

∇Hdp

dp I1+A|A|p−

2∇Hd

p

dp I2

(p1)|A|p∇Hd p

dp I p/(p−1) 1

= |A|p∇Hd p

dp

pI1+ 1

AI2(p1)I p/(p−1) 1

.

(2.12)

We claim

divHT(p1)|T|p/(p−1)≥ |A|p

∇Hdp dp

1 + p−1 2pA2Ꮾ

2d

R

. (2.13)

In fact, arguing as in the proof of [13, Theorem 4.1], we set

f(s) :=pI1(s) + 1

AI2(s)(p1)I p/(p−1)

(6)

andM=M(R) :=supξΩᏮ(d(ξ)/R), and distinguish three cases (i)

1< p <2< Q, a >(2−p)(p−1)

6p2A2 , (2.15)

(ii)

2≤p < Q, a=0, (2.16)

(iii)

p > Q, a <(2−p)(p−1)

6p2A2 <0. (2.17)

It yields that

f(s)1 + p−1 2pA2s

2, 0sM, (2.18)

(see [13]) and then follows (2.13). Hence (2.2) is proved.

(2) If p=Q, then we choose the vector field T(d)=((p−1)/ p)p−1(|∇Hd|p−2∇Hd/

dp−1)p−1(d/R). It gives

divHT=

p1

p

p−1 Q1(p1)∇Hdp

dpp−

1d

R

+ (p1)Ꮾp d

R

∇Hdp dp

=p p1

p p

p

d R

∇Hdp dp ,

(2.19)

and hence

divHT(p1)|T|p/(p−1)=

p1

p p

p

d

R

∇Hdp

dp . (2.20)

Combining (2.20) with (2.9) follows (2.3).

Remark 2.2. The domainΩin (2.9) may be bounded or unbounded. In addition, if we select that T(d)=A|A|p−2(|∇Hd|p−2∇Hd/dp−1), then

divHT(p1)|T|p/(p−1)=p|A|p

∇Hdp

dp (p1)|A|p

∇Hdp dp = |A|p

∇Hdp dp .

(2.21)

Hence, from (2.9) we conclude (2.1) on the bounded domainΩand onHn(see [15]), respectively.

(7)

Lemma2.3. LetΩbe a subset ofHnbounded inx

1direction, that is, there existsR >0 such

that 0< r= |x1| ≤R for ξ=(x1,x2,. . .,xn,y1,. . .,yn,t)Ω. Then for any u∈D

1,p

0 (Ω),

then

c

Ω|u| p

Ω

∇Hup

, (2.22)

wherec=((p1)/ pR)p.

Using (2.9) by choosing T= −((p−1)/ p)p−1(∇Hr/rp−1) immediately provides a

dif-ferent proof to (2.22).

In the following, we describe a compactness result by using (2.1) and (2.22).

Theorem2.4. Supposep=Qand f(ξ)Ᏺp. Then there exists a positive constantCf,Q,p

such that

Cf,Q,p

Ωf(ξ)|u| p

Ω

∇Hup

, (2.23)

and the embeddingD01,p(Ω)Lp(Ω,f dξ)is compact.

Proof. Since f(ξ)Ᏺp, we have that for any>0, there existδ >0 andCδ>0 such that

sup BH(δ)Ω

dp

ψp f(ξ), f(ξ)|Ω\BH(δ)≤Cδ. (2.24)

By (2.1) and (2.22), it follows

Ωf(ξ)|u| p=

BH(δ)

f|u|p+ Ω\BH(δ)

f|u|p

BH(δ) ψp|u|

p

dp +

Ω\BH(δ)

|u|pC1

f,Q,p

Ω

∇Hup ,

(2.25)

then (2.23) is obtained.

Now, we prove the compactness. Let{um} ⊆D1,0p(Ω) be a bounded sequence. By

re-flexivity of the spaceD1,0p(Ω) and the Sobolev embedding for vector fields (see [18]), it

yields

umjuweakly inD

1,p

0 (Ω),

umj−→ustrongly in Lp(Ω)

(2.26)

for a subsequence{umj}of{um}as j→∞. WriteCδ= fL∞\BH(δ)). From (2.1),

Ωf um

j−u p

dξ=

BH(δ)

fumj−u p

+

Ω\BH(δ)

fumj−u p

BH(δ) ψp

um j−u

p

dp +

Ω\BH(δ) um

j−u p

≤C−1

Q,p

Ω

∇H

umj−u p

+

Ω um

j−u p

dξ.

(8)

Since{um} ⊆D01,p(Ω) is bounded, we have

Ω f um

j−u p

dξ≤M+

Ω um

j−u p

dξ, (2.28)

whereM >0 is a constant depending onQandp. By (2.26),

lim j→∞

Ω f um

j−u p

dξ≤M. (2.29)

Asis arbitrary, limj→∞Ωf|umj−u|pdξ=0.HenceD

1,p

0 (Ω)Lp(Ω,f dξ) is compact.

Remark 2.5. The class of the functionsf(ξ)Ᏺphas lower-order singularity thand−p(ξ) at the origin. The examples of such functions are

(a) any bounded function,

(b) in a small neighborhood of 0, f(ξ)=ψp(ξ)/dβ(ξ), 0< β < p, (c) f(ξ)=ψp(ξ)/dp(ξ)(ln (1/d(ξ)))2in a small neighborhood of 0.

3. Proof of best constants in (2.2) and (2.3)

In this section, we prove that the constants appearing inTheorem 2.1are the best. To do this, we need two lemmas. First we introduce some notations.

For some fixed smallδ >0, let the test functionϕ(ξ)∈C∞0(Ω) satisfy 0≤ϕ≤1 and

ϕ(ξ)= ⎧ ⎪ ⎨ ⎪ ⎩

1 ifξ∈BH

0,δ

2

,

0 ifξ∈Ω\BH(0,δ),

(3.1)

with|∇Hϕ|<2|∇Hd|/d.Let>0 small enough, and define

V(ξ)=ϕ(ξ), with=d−A+−κ

d

R

, 1 p < κ <

2 p,

()=

Ωϕ

p(ξ)∇Hd p

dQ−p−γ

d

R

, γ∈R.

(3.2)

Lemma3.1. For>0small, it holds

(i)c−1−γJ

γ()≤C−1−γ,γ >−1,

(ii)()=(p/(γ+ 1))Jγ+1() +O(1),γ >−1, (iii)()=O(1),γ <−1.

Proof. By the change of polar coordinates (1.11) and 0≤ϕ≤1, we have

()

BH(δ)

∇Hdp dQ−p−γ

d

R

dξ=sH

ρ<δρ

−Q+p−γρ R

ρQ−1

=sH

ρ<δρ

1+p−γρ R

dρ.

(9)

By (2.7) we know that forγ <−1 the right-hand side of (3.3) has a finite limit, hence (iii) follows from0.

To show (i), we setρ=Rτ1/. Thus,=(1/)Rτ1/−1dτ,−γ1/)=−γ−γ(τ), and

()≤sH

ρ<δρ

1+p−γ

ρ

R

dρ=sH

(δ/R)

0

1/−1+p−γ

1/

R

1

1/−1

=sHRp−1−γ

(δ/R)

0 τ

p−1−γ(τ)dτ.

(3.4)

It follows the right-hand side of (i). Using the fact thatϕ=1 inBH(δ/2),

()

BH(δ/2)

∇Hdp dQ−p−γ

d R

dξ=sHRp−1−γ

(δ/2R)

0 τp−1Ꮾ−γ(τ)dτ, (3.5)

and the left-hand side of (i) is proved.

Now we prove (ii). LetΩη:= {ξ∈Ω|d(ξ)> η},η >0, be small and note the boundary term

d=η

ϕp∇Hdp−2∇Hd

dQ−1−p

−γ−1

d R

∇Hd·ndS−→0 asη−→0. (3.6)

From (2.6),

ΩdivH

ϕp∇Hdp−2∇Hd

dQ−1−p

−γ−1d

R = − Ω

ϕp∇Hdp−2 dQ−1−p

∇Hd,∇H−γ−1 d

R

=(γ+ 1)

Ω

ϕp∇Hdp dQ−p−γ

d R

dξ=(γ+ 1)Jγ().

(3.7)

On the other hand,

ΩdivH

ϕp∇Hdp−2∇Hd

dQ−1−p

−γ−1d

R =p Ωϕ

p−1∇Hd

p−2

∇Hd,∇Hϕ dQ−1−p−γ−1

d R

+ (1−Q+p+Q−1)

Ωϕ

p∇Hd p

dQ−p−γ−1

d R =p Ωϕ

p−1∇Hd

p−2

∇Hd,∇Hϕ dQ−1−p−γ−1

d

R

+pJγ+1().

(10)

We claim thatpΩϕp−1(|∇Hd|p−2∇Hd,∇Hϕ/dQ−1−p)−γ−1(d/R)dξ=O

(1). In fact, by (3.1) and (1.11),

Ωϕ

p−1∇Hd

p−2

∇Hd,∇Hϕ dQ−1−p−γ−1

d

R

2

BH(δ)

∇Hdp dQ−p−γ−1

d

R

2sH

BH(δ)

ρ−Q+p−γ−1 ρ

R

ρQ−1

=2sH

BH(δ)

ρp−1−γ−1 ρ

R

dρ.

(3.9)

Using the estimate (i) follows thatBH(δ)ρp−1Ꮾ−γ−1(ρ/R)dρ=O(1).Combining (3.7)

with (3.8) gives

(γ+ 1)Jγ()=pJγ+1() +O(1). (3.10)

This allows us to conclude (ii).

We next estimate the quantity

I[V]=

Ω

∇HVp

dξ− |A|p

Ω

∇HdpVp

dp dξ. (3.11)

Lemma3.2. As→0, it holds

(i)I(V)(κ(p1)/2)|A|p−2J

pκ−2() +O(1); (ii)BH(δ)|∇HV|pdξ≤ |A|pJpκ() +O(1−pκ).

Proof. By the definition ofV(ξ), we see∇HV(ξ)=ϕ(ξ)∇H+∇Hϕ.Using the ele-mentary inequality

|a+b|p≤ |a|p+c p

|a|p−1|b|+|b|p, a,bR2n,p >1, (3.12)

one has

∇HVp

≤ϕp∇H

p+cp∇Hϕϕp−1∇Hp−1+∇Hϕpp

≤ϕp∇H

p+cp 2∇Hd

d ϕ

p−1∇H

p−1+ 2∇Hd

d

p p

(11)

Since∇H= −d−A+1Ꮾ−κ(d/R)(A+κᏮ(d/R))∇Hd, it follows

Ω

∇HVp dξ≤

BH(δ)

∇Hdp

ϕpd−Q+p−pκd R

A−−κd

R

pdξ

+ 2cp

BH(δ)

∇Hdp

ϕp−1d−Q+p−pκ

d R A−

−κ

d R

p−1

+ 2pc p

BH(δ)

∇Hdp

d−Q+p−pκ

d

R

:=ΠA+Π1+Π2.

(3.14)

We claim that

Π1,Π2=O(1). (3.15)

Indeed, since|A−(−κᏮ(d/R))|is bounded, using (3.1) we get

Π1≤C

BH(δ)

∇Hdp

ϕp−1d−Q+p−pκ

d

R

A−

−κ

d

R

p−1

≤C

BH(δ)

∇Hdp

d−Q+p−pκd R

,

Π2≤C

BH(δ)

∇Hdp

d−Q+p−pκ d

R

dξ.

(3.16)

By (i) ofLemma 3.1, it derivesΠ1,Π2=O(1), as0.

From (3.14), (3.15) and the definition of(), it clearly shows

IV=

BH(δ)

∇HVp

dξ− |A|pJ

()ΠA− |A|pJpκ() +O(1)=Π3+O(1),

(3.17)

whereΠ3=

BH(δ)|∇Hd|

pϕpd−Q+p−pκ(d/R)(|A(κ(d/R))|p− |A|p)dξ.For

sim-plicity, denoteζ=−κᏮ(d/R).Sinceζis small compared toA, we use Taylor’s expansion to yield

|A−ζ|p− |A|p≤ −pA|A|p−2ζ+ p(p−1)

2 |A|

p−2ζ2+C|ζ|3. (3.18)

Thus, we can estimateΠ3by

Π3Π31+Π32+Π33, (3.19)

where

Π31= −pA|A|p−2

BH(δ)

∇Hdp

ϕpd−Q+p−pκd R

−κ

d

R

,

Π32= p(p−1)

2 |A| p−2

BH(δ)

∇Hdp

ϕpd−Q+p−pκ d

R

−κ

d

R

2

dξ,

Π33=C

BH(δ)

∇Hdp

ϕpd−Q+p−pκd R

−κ

d

R

3dξ.

(12)

We will show that

Π31,Π33=O(1), as −→0. (3.21)

In fact, using (ii) ofLemma 3.1withγ= −1 +yields

Π31= −pA|A|p−2

Jpκ()−κJpκ−1()

= −pA|A|p−2J

()−Jpκ() +O(1)=O(1). (3.22)

Recalling (a−b)3(|a|+|b|)3≤c(|a|3+|b|3), we obtain

Π33≤c3Jpκ() +cJpκ−3() for>0. (3.23)

From (i) and (iii) inLemma 3.1and the fact that 1< pκ <2 it followsΠ33=O(1). Using (ii) ofLemma 3.1twice (pickγ=pκ−1>−1 andγ=pκ−2>−1, resp.), we conclude that

Π32= p(p−1)

2 |A| p−2

BH(δ)

∇Hdp

ϕpd−Q+p−pκd R

22κ

d

R

+κ22 d

R

= p(p−1) 2 |A|

p−2

2J

()2κJpκ−1() +κ2 1

+

1

Jpκ−2()

= p(p−1) 2 |A|

p−2

2J

()−κp

pκJpκ()−κJpκ−1()

+κ

2(pκ1)

p

pκ−1Jpκ−1() + κ

pJpκ−2() +O(1)

=κ(p−1) 2 |A|

p−2J

pκ−2() +O(1).

(3.24)

In virtue of (3.17), (3.19), (3.21), and (3.24) we deduce (i) ofLemma 3.2. By (3.17), (3.24), and (i) ofLemma 3.2,

BH(δ)

∇HVp

dξ=IV+|A|pJpκ()≤ |A|pJpκ() +κ(p−1) 2 |A|

p−2J

pκ−2() +O(1).

(3.25)

Hence (ii) ofLemma 3.2follows from (i) inLemma 3.1. It completes the proof.

We are now ready to give the proof of the best constants inTheorem 2.1.

Theorem3.3. Let0Ωbe a bounded domain inHnand p=Q. Suppose that for some

constantsB >0,D≥0, andι >0, the following inequality holds for anyu(ξ)∈C∞0(Ω\ {0}):

Ω

∇Hup dξ≥B

Ωψp |u|p

dp +D

Ωψp |u|p

dpι

d

R

(13)

Then,

(i)B≤ |A|p;

(ii)ifB= |A|pandD >0, thenι2;

(iii)ifB= |A|pandι=2, thenD((p1)/2p)|A|p−2.

Proof. Chooseu(ξ)=V(ξ).

(i) By (ii) ofLemma 3.2, we have

B≤

BH(δ)∇HV p

BH(δ)ψp(|V|

p/dp)dξ

|A|pJ

() +O1−pκ

BH(δ)ψp(|ϕd−A

+−κ(d/R)|p/dp)dξ

=|A|p

1 +c2J

() +O(1) Jpκ()

.

(3.27)

Note thatJpκ()→∞, as0, soB≤ |A|p.

(ii) SetB= |A|pand assume by contradiction thatι <2. Sinceι >1, using (i) ofLemma 3.2and (i) ofLemma 3.1leads to

0< D≤ I

V

Ωψp(|V|p/dp)Ꮾι(d/R)dξ = IV Jpκ−ι()

κ(p−1)/2|A|p−2J

pκ−2() +O(1) Jpκ−ι()

C1−pκ c−1−pκ+ι=C

2−ι−→0, as −→0,

(3.28)

which is a contradiction. Henceι≥2.

(iii) IfB= |A|pandι=2, then by (i) ofLemma 3.2,

D≤ I

V Jpκ−2()

κ(p−1)/2|A|p−2J

pκ−2() +O(1)

Jpκ−2() . (3.29)

The assumptionκ >1/ p impliesJpκ−2()→∞, as0. Hence, by (i) ofLemma 3.1we

conclude thatD≤(κ(p1)/2)|A|p−2, as0. Then lettingκ1/ p, the proof is finished.

Theorem3.4. Set0Ωand p=Q. Suppose that there exist some constantsD≥0and

ι >0such that the following inequality holds for allu(ξ)∈C0\ {0}):

Ω

∇Hup dξ≥D

Ωψp |u|p

dpι

d

R

dξ. (3.30)

Then,

(i)ifD >0, thenι≥p;

(ii)ifι=p, thenD≤((p−1)/ p)p.

(14)

Using (3.12) yields

Ω

∇HVp dξ=

BH(δ)

ϕ∇H+∇Hϕp

BH(δ)

ϕp∇H

pdξ+cp

BH(δ) ϕp−1

∇Hp−1∇Hϕdξ

+cp

BH(δ)

p∇Hϕpdξ:=Π4+Π5+Π6.

(3.31)

Arguing as in the proof of previous theorem and letting0, we have

Π5,Π6=O(1). (3.32)

Denoting bycipthe coefficients of binormial expansion, we get

∇Hp

=∇Hdpdp−p−pκd R

−κ

d

R

p

≤∇Hdpdp−p−pκ d

R

+κ

d

R

p

=∇Hdpdp−p−pκ

d R

Σp i=0c

p

ip−iκii

d R

.

(3.33)

Hence,

Π4Σip=0c

p

ip−iκiJpκ−i(), (3.34)

where()=

Ω|∇Hd|pϕpdp−p−γ(d/R). By (ii) ofLemma 3.1and the induction ar-gument it holds

p−iJ

pκ−i()=

κ− i

p

κ−i+ 1 p

···

κ−p−1 p

Jpκ−p()+O(1), i=0, 1,. . .,p−1.

(3.35)

Now (i) ofLemma 3.1and the assumptionκ >(p1)/ pimply thatJpκ−p()→∞, as0, and

D≤lim sup 0

Ω∇HVpdξ

ΩψpVp/dpp(d/R)dξ

lim sup 0

κp+Σp−1 i=0c

p iκi

κ−i/ pκ−(i+ 1)/ p···κ−(p1)/ pJpκ−p() +O(1) Jpκ−p()

=

κp+Σp−1 i=0c

p iκi

κ− i

p

κ−i+ 1 p

···

κ−p−1 p

.

(3.36)

(15)

4. The weighted eigenvalue problem

This section is devoted to the problem (1.2) by using the Hardy inequality with remainder terms.

We begin with some properties concerning the Hardy operator (1.1).

Lemma4.1. Suppose thatu(ξ)∈D01,p(Ω)andp=Q. Then

(1)Lp,μ is a positive operator if μ≤CQ,p; in particular, if μ=CQ,p, then v(ξ)= d(p−Q)/ p(ξ)is a solution ofL

p,μu=0; (2)Lp,μis unbounded from below ifμ > CQ,p.

Proof. (1) It is obvious from (2.1) thatLp,μis a positive operator.

We now suppose thatμ=CQ,pand verify thatv=d(p−Q)/ psatisfiesLp,μu=0. For the purpose, setv=d(p−Q)/ p+∈D01,p(Ω) andA=(Q−p)/ p. Since

v =(−A)d−A−1+, v

=(−A)(−A−1)d−A−2+, (4.1)

it yields from (1.10) and (4.1) that

−ΔH,pv= −ψpvp−2

(p1)v+Q−1

d v

= −ψp(−A)d−A−1+p−2

(p1)(−A)(−A−1)d−A−2+

+Q−1

d (−A)d− A−1+

= −ψp(−A)|−A|p−2d(−A−1+)(p−2)+(−A−2+)[(p−1)(−A−1) +Q−1] = −ψp(−A)(p−1) +Q−p(−A)|−A|p−2d(−A+)(p−1)−p

= −(−A)(p−1) +pA(−A)|−A|p−2ψ

pv p−1

dp .

(4.2)

Letting0, the conclusion follows.

(2) By the density argument, we selectφ(ξ)∈C∞0(Ω),φLp=1, such thatCQ,p=

Ω|∇Hφ|p/(

Ω(|z|p/dp)(|φ|p/dp)). Using the best constant of the inequality (2.1), one has

Lp,μφ,φ

=

Ω

∇Hφp −μ

Ω |z|p

dp |φ|p

dp <

Ω

∇Hφp −CQ,p

Ω |z|p

dp |φ|p

dp =0. (4.3)

Denote(x,y,t)=τQ/ pφ(δτ(x,y,t)) andξ=(x,y,t). Thus,

uτ(x,y,t)p Lp=

Ω

τQ/ pφδ

τ(x,y,t)pdξ=

Ω φ

δτ(x,y,t)pτQdξ=1, (4.4)

andLp,μuτ,uτ =τpLp,μφ,φ<0. This concludes the result.

(16)

Lemma4.2. Let{gm} ⊂Lp(Ω)(1p <)be such that asm→∞,

gmg weakly inLp(Ω),

gm−→g a.e. inΩ. (4.5)

Then,

lim m→∞gm

p

Lp(Ω)−gm−gLpp(Ω)

= gLpp(Ω). (4.6)

The proof is similar to one in the Euclidean space (see [19, Chapter 1, Section 4]. We omit it here.

Lemma4.3. Suppose that{um} ⊂D01,p(Ω)(1≤p <∞)satisfies

umu weakly inD01,p(Ω),

um−→u strongly inLlocp (Ω),

(4.7)

asm→∞, and

−ΔH,pum=fm+gm, in(Ω), (4.8)

where fm→f strongly inD−1,p

(Ω)(p=p/(p−1)),gm is bounded in M(Ω)(the space of

Radon measures), that is,

gmCKϕL (4.9)

for allϕ∈Ᏸ(Ω)withsupp(ϕ)⊂K,whereCK is a constant which depends on the compact

setK. Then there exists a subsequence{umj}of{um}such that

umj−→u strongly inD

1,q

0 (Ω),∀q < p. (4.10)

Its proof is similar to one of [20, Theorem 2.1].

Lemma4.4. Letp=Qand

Ip:=

f−→R+| lim

d(ξ)0

dp(ξ) ψp(ξ)f(ξ)

ln 1

d(ξ) 2

<∞, f(ξ)∈L∞loc

Ω\ {0}

. (4.11)

(i)If f(ξ)Ip, then there existsλ(f)>0 such that for allu∈D1,0p\ {0}), the

following holds:

Ω

∇Hup

dξ≥Q−p p

p Ωψp(ξ)|u| p

dp +λ(f)

Ω|u|

pf(ξ)dξ. (4.12)

(ii)If f(ξ)Ipand(dp(ξ)/ψp(ξ))f(ξ)(ln 1/d(ξ))2→∞asd(ξ)→0, then (4.12) is not

(17)

Proof. (i) If f(ξ)Ip, then

lim 0ξsupB

H() dp)

ψp f(ξ)

ln 1

d(ξ) 2

<∞. (4.13)

Without any loss of generality we assume thatR=1 in (2.2). For the sufficiently small

>0, we have

f(ξ)< Cψp

dp(ln 1/d)2, in BH(). (4.14)

OutsideBH(), f(ξ) is also bounded. Hence there existsC(f)>0 such that

Ω|u|

pf)dξ C(f)

Ωψp |u|p

dp(ln 1/d)2, onΩ. (4.15)

Takingλ(f)=C(f)1((p−1)/2p)|A|p−2>0, (4.12) follows from (2.2).

(ii) We write f(ξ)=ψph(ξ)/dp(ξ)(ln 1/d(ξ))2, whereh(ξ)→∞asd(ξ)→0. Then, for the sufficiently small>0, we selectu(ξ)=V(ξ)=ϕ(ξ)d−A+(ξ)Ꮾ−κ(d(ξ)/R) and get from (i) ofLemma 3.2,

0< λ(f) I

V

BH(δ)V p

f(ξ)dξ

IV

BH(δ)ψp(|V|

ph(ξ)/dp(ln 1/d)2)dξ

I

V

BH(δ)ψpV p

/dp(ln 1/d)2

κ(p−1)/2|A|p−2J

pκ−2() +O(1) KδJpκ−2()

c1−pκ cKδ1−pκ =

C

−→0, as δ,−→0,

(4.16)

whereKδ=infξ∈BH(δ)h(ξ). The impossibility shows that (4.12) cannot hold for f(ξ)

Ip.

Definition 4.5. Letλ∈R,u∈D01,p(Ω), andu≡0. Call that (λ,u) is a weak solution of

(1.2) if

Ω

∇Hup−2

∇Hu,∇Hϕdξ−μ

Ω ψp dp|u|

p−2uϕ dξ=λ

Ωf(ξ)|u|

p−2uϕ dξ (4.17)

for any ϕ∈C∞0(Ω). In this case, we call that u is the eigenfunction of problem (1.2)

associated to the eigenvalueλ.

Theorem4.6. Suppose that1< p < Q, 0≤μ <((Q−p)/ p)p, andf(ξ)Ᏺp. The problem (1.2) admits a positive weak solutionu∈D1,0p(Ω), corresponding to the first eigenvalueλ=

λ1

μ(f)>0. Moreover, asμincreases to((Q−p)/ p)p,λμ1(f)→λ(f)0for f(ξ)Ᏺp and

(18)

Proof. We define

(u) : =

Ω

∇Hup −μ

Ωψp |u|p

dp . (4.18)

Obviously,is continuous and Gˆateaux differentiable onD1,0p(Ω). By (2.1),

(u)

Ω

∇Hup

μ

CQ,p

Ω

∇Hup =

1 μ

CQ,p

Ω

∇Hup

(4.19)

for 0≤μ <((Q−p)/ p)p. Note thatCQ,p=((Q−p)/ p)p for 1< p < Q. HenceJμis coer-cive inD01,p(Ω). We minimize the function(u) over the mainfoldᏹ= {u∈D1,0p(Ω)|

Ω|u|pf(ξ)dξ=1}and let λ1μ be the infimum. It is clear thatλ1μ>0 fromLemma 4.1. Now, we choose a special minimizing sequence{um} ⊂ᏹwithΩ|um|pf)dξ=1, and (um)→λ1μ and(um)0 strongly inD−1,p

0 (Ω), when the component of(um) is re-stricted toᏹ. The coercivity ofimplies that{um}is bounded and then there exists a subsequence, still denoted by{um}, such that

umu weakly in D01,p(Ω),

umu weakly in LpΩ,ψpd−p, um−→u strongly inLp(Ω),

(4.20)

asm→∞. ByTheorem 2.4 inSection 2 we know thatD01,p(Ω) is compactly embedded

inLp(Ω,f dξ), and it follows that is weakly closed and henceu. Moreover,u m satisfies

−ΔH,pum−ψp μ dpum

p−2

um=λmump−

2

umf +fm, inᏰ(Ω), (4.21)

wherefm→0 strongly inD−1,p

(Ω) andλm→λ, asm→∞. Lettinggm=ψp(μ/dp)|um|p−2umm|um|p−2umf, we check easily thatgm is bounded in M(Ω) and conclude almost ev-erywhere convergence of∇Humto∇HuinΩbyLemma 4.3, and

(um)=∇HumLpp(Ω)−μumLpp(Ω,ψpd−p)

=∇Hum−uLpp(Ω)−μum−uLpp(Ω,ψpd−p)+∇Hu p

Lp(Ω)−μu p Lp(Ω,ψ

pd−p)+o(1) ≥CQ,p−μum−uLpp(Ω,ψ

pd−p)+λ

1

μ+o(1),

(4.22)

by applyingLemma 4.2toumand∇Hum, whereo(1)→0 asm→∞. ThusCQ,p> μ,um− uLpp(Ω,ψpd−p)0, and∇H(um−u)Lpp(Ω)0 asm→∞. It shows that(u)1μandλ=λ1μ. Since(|u|)=Jμ(u), we can takeu >0 inΩ. ByLemma 4.3,uis a distribution solution of (1.2) and sinceu∈D1,0p(Ω), it is a weak solution to eigenvalue problem (1.2)

corre-sponding toλ=λ1

μ. Moreover, if f(ξ)Ip, then byLemma 4.4,

λ1

μ(f)−→λ(f)=uD1,infp(Ω\{0})

Ω∇Hup−CQ,pψp

|u|p/dp

(19)

asμincreases to ((Q−p)/ p)p. When f(ξ)Ip, usingLemma 4.4again, it follows that (4.12) is not true and henceλ(f)=0. This completes the proof.

Remark 4.7. The setᏹis aC1manifold inD1,p

0 (Ω). By Ljusternik-Schnirelman critical

point theory onC1manifold, there exists a sequence{λ

m}of eigenvalues of (1.2), that is, writingΓm= {A⊂|Ais symmetric, compact, andγ(A)≥m}, whereγ(A) is the Krasnoselski’s genus ofA(see [19]), then for any integerm >0,

λm= inf

A∈Γm supuA(u) (4.24)

is an eigenvalue of (1.2). Moreover, limm→∞λm→∞.

5. Simplicity and isolation for the first eigenvalue

This section is to consider the simplicity and isolation for the first eigenvalue. We always assume that f satisfies the conditions inTheorem 4.6. From the previous results we know clearly that the first eigenvalue is

λ1

μ=λ1μ(f)=inf

(u)|u∈D

1,p

0

Ω\ {0},

Ω|u|

pf)dξ=1

. (5.1)

In what follows we need the Picone identity proved in [15].

Proposition5.1 (Picone identity). For differentiable functionsu≥0,v >0onΩHn,

withΩa bounded or unbounded domain inHn, then

L(u,v)=R(u,v)≥0, (5.2)

with

L(u,v)=∇Hup+ (p−1)u p

vp∇Hv p

−pu p−1

vp−1∇Hv

p−2

∇Hu·∇Hv,

R(u,v)=∇Hup−∇Hvp−2∇H

up

vp−1

·∇Hv

(5.3)

forp >1. Moreover,L(u,v)=0a.e. onΩif and only if∇H(u/v)=0a.e. onΩ.

Theorem 5.2. (i) λ1

μ is simple, that is, the positive eigenfunction corresponding to λ1μ is

unique up to a constant multiple.

(ii)λ1

μis unique, that is, ifv≥0is an eigenfunction associated with an eigenvalueλwith

Ωf(ξ)|v|pdξ=1, thenλ=λ1μ.

(iii)Every eigenfunction corresponding to the eigenvalueλ(0< λ=λ1

μ)changes sign inΩ.

Proof. (i) Letu >0 andv >0 be two eigenfunctions corresponding toλ1

μinᏹ. For suffi -ciently smallε >0, setφ=up/(v+ε)p−1

∈D01,p(Ω). Then

Ω

∇Hvp−2

∇Hv,∇Hφdξ=μ

Ωψp up−1

dp φdξ+λ

1

μ

Ωf(ξ)v

(20)

UsingProposition 5.1and (5.4),

0

ΩL(u,v+ε)=

ΩR(u,v+ε)

=

Ω

∇Hup dξ−

Ω

∇Hvp−2

∇H

up

(v+ε)p−1

,∇Hv = Ω

μψp dp +λ

1

μf(ξ)

updξ−

Ω

μψp dp +λ

1

μf(ξ)

vp−1 u p

(v+ε)p−1

=

Ω

μψp dp +λ

1

μf(ξ)

up

1 vp−1 (v+ε)p−1

dξ.

(5.5)

The right-hand side of (5.5) tends to zero whenε→0. It follows thatL(u,v)=0 and by

Proposition 5.1there exists a constantcsuch thatu=cv. (ii) Letu >0 andv >0 be eigenfunctions corresponding toλ1

μandλ, respectively. Sim-ilarly to (5.5), we have

Ω

μψp dp +λ

1

μf(ξ)

updξ−

Ω

μψp

dp+λ f(ξ)

vp−1 up

(v+ε)p−1

= Ωμ ψp dp

1 vp−1 (v+ε)p−1

updξ+

Ωf(ξ)u pλ1

μ−λ vp−1

(v+ε)p−1

dξ≥0.

(5.6)

Lettingε→0 shows thatλ1

μ−λ

Ωf(ξ)updξ≥0, which is impossible forλ > λ1μ. Hence λ=λ1

μ.

(iii) With the same treatment as in (5.6) we get

λ1

μ−λ

Ωf(ξ)u

p0. (5.7)

Noting thatΩf(ξ)updξ >0 andλ > λ1

μleads to a contradiction. Sovmust change sign

inΩ.

Lemma5.3. Ifu∈D1,0p(Ω)is a nonnegative weak solution of (1.2), then eitheru(ξ)≡0or

u(ξ)>0for allξ∈Ω.

Proof. For anyR > r, BH(0,R)⊃BH(0,r), letu∈D01,p(Ω) be a nonnegative weak solution

of (1.2). In virtue of Harnack’s inequality (see [1]), there exists a constantCR>0 such that

sup BH(0,R)

u(ξ)≤CR inf BH(0,R)

u(ξ). (5.8)

This impliesu≡0 oru >0 inΩ.

Theorem5.4. Every eigenfunctionu1corresponding toλ1μdoes not change sign inΩ: either u1>0oru1<0.

Proof. From the proof of existence of the first eigenvalue we see that there exists a positive eigenfunction, that is, ifvis an eigenfunction, thenu1= |v|is a solution of the

minimiza-tion problem and also an eigenfuncminimiza-tion. Thus, fromLemma 5.3it follows that|v|>0 and

(21)

Lemma5.5. Foru∈C(Ω\ {0})∩D01,p(Ω), letbe a component of{ξ∈Ω|u(ξ)>0}.

Thenu|∈D1,0p(ᏺ).

Proof. Letum∈C(Ω\ {0})∩D01,p(Ω) be such thatum→uinD 1,p

0 (Ω). Therefore,u+m→u+ inD01,p(Ω).Setvm(ξ)=min{um(ξ),u(ξ)}, and letϕR(ξ)∈C(Ω) be a cutofffunction such that

ϕR(ξ)=

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

0 ifd(ξ)≤R 2,

1 ifd(ξ)≥R,

(5.9)

with|∇HϕR| ≤C|∇Hd|/d(ξ), for some positive constantC. Now, consider the sequence ωm(ξ)=ϕR(ξ)vm(ξ)|ᏺ. SinceϕR(ξ)vm(ξ)∈C(Ω), we claim thatωm∈C(ᏺ) andωm=0 on the boundaryᏺ. In fact, ifξ∈∂ᏺandξ=0, thenϕR=0, and soωm=0. Ifξ∈ Ωandξ=0, thenu=0 (sinceuis continuous except at{0}), and hencevm=0. If ξ∈∂Ω, thenum=0 and sovm=0. Therefore,ωm=0 onᏺ, andωm∈D

1,p

0 (ᏺ). Noting

Ω

∇Hωm− ∇H

ϕRupdξ=

ϕR∇Hvm+vm∇HϕRϕR∇Huu∇HϕRp

≤ϕR

∇Hvm− ∇HuLpp()+∇HϕR

vm−upLp(), (5.10)

it is obvious thatΩ|∇Hωm− ∇HRu)|pdξ→0, asm→∞. That isωm→ϕRu|ᏺinD1,0p(ᏺ).

By (2.1),

u∇HϕR+ϕR∇Hu− ∇Hup

ϕR∇Hu− ∇Hup +

∩{R/2<d<R}

u∇HϕRp

ϕR∇Hu− ∇Hup +Cp

∩{R/2<d<R}ψp

|u|p dp

ϕR∇Hu− ∇Hup +C1

∩{R/2<d<R}

∇Hup dξ,

(5.11)

which approaches 0, asR→0. Henceu|∈D1,0p(ᏺ).

Theorem5.6. The eigenvalueλ1

μis isolated in the spectrum, that is, there existsδ >0such

that there is no other eigenvalues of (1.2) in the interval(λ1

μ,λ1μ+δ). Moreover, ifvis an

eigenfunction corresponding to the eigenvalueλ=λ1

μandis a nodal domain ofv, then

CλfL∞−Q/ p≤ |ᏺ|, (5.12)

(22)

Proof. Letu1 be the eigenfunction corresponding to the eigenvalueλ1μ. Let {λm}be a

sequence of eigenvalues such thatλm> λ1μandλmλ1μ, and the corresponding eigenfunc-tionsum→u1with

Ωf(ξ)|um|pdξ=1, that is,λmandumsatisfy

Lp,μum=λmf(ξ)ump−2um. (5.13)

Since

0<

Ω

∇Hump dξ−μ

Ω ψp dpum

p dξ=λm

Ωf(ξ) ump

dξ=λm, (5.14)

it follows thatum is bounded. ByLemma 4.3, there exists a subsequence (still denoted by{um}) of {um} such thatumu weakly inD

1,p

0 (Ω),um→ustrongly inLp(Ω) and

∇Hum→∇Hua.e inΩ. Lettingm→∞in (5.13) yields

Lp,μu=λ1μf(ξ)|u|p−2u. (5.15)

Therefore,u= ±u1. Using (iii) ofTheorem 5.2we see thatum changes sign. For conve-nience, we assume thatu=+u1. Then

ξΩ|um<0−→0. (5.16)

Now, we check (5.13) withum=u−m,

Ω ∇Hu

m p

dξ−μ

Ω ψp dpu−m

p dξ=λm

Ωf(ξ) u

m p

dξ. (5.17)

Using the Hardy inequality and Sobolev inequality yields

1 μ CQ,p

Ω

∇Hump

Ω

∇Hump dξ−μ

Ω ψp dpum

p

=λm

Ω−f(ξ)

ump

≤λmfL∞

Ω

ump

≤c1fL∞umDp1,pΩm p/Q

, Ω

m≥

c2fL∞Q/ p, Ωm=ξ∈Ω|um<0.

(5.18)

It contradicts with (5.16). Hence, there is no other eigenvalue of (1.2) in (λ1

μ,λ1μ+δ) for δ >0.

Next, we prove (5.12). Assumev >0 inᏺ(the casev <0 being treated similarly). In view ofLemma 5.5, we havev|∈D1,0p(ᏺ).Define the function

ω(ξ)=

⎧ ⎨ ⎩

v(ξ) ifξ∈ᏺ,

(23)

Clearly,ω(ξ)∈D1,0p(Ω). Takingωas a test function in (4.17) satisfied byvand arguing as

in (5.18), we have

1Cμ Q,p

vDp1,p()≤λfL∞

|v|

pλCfLvp

D1,p()|ᏺ|p/Q (5.20)

for some constantC=C(Q,p) and hence (5.12) holds.

Corollary5.7. Each eigenfunction has a finite number of nodal domains.

Proof. Letᏺjbe a nodal domain of an eigenfunction associated to some positive eigen-valueλ. It follows from (5.12) that

|Ω| ≥ j

ᏺj

CλfL∞−Q/ p

j

1. (5.21)

The result is proved.

Acknowledgments

The authors would like to thank the referee for their careful reading of the manuscript and many good suggestions. The project is supported by Natural Science Basic Research Plan in Shaanxi Province of China, Program no. 2006A09. Pengcheng Niu is the corresponding author (email address:pengchengniu@nwpu.edu.cn).

References

[1] L. Capogna, D. Danielli, and N. Garofalo, “An embedding theorem and the Harnack inequality for nonlinear subelliptic equations,”Communications in Partial Differential Equations, vol. 18, no. 9-10, pp. 1765–1794, 1993.

[2] A. Domokos and J. J. Manfredi, “C1,α-regularity forp-harmonic functions in the Heisenberg

group forpnear 2,” inThep-Harmonic Equation and Recent Advances in Analysis, vol. 370 of

Contemporary Mathematics, pp. 17–23, American Mathematical Society, Providence, RI, USA, 2005.

[3] W. Allegretto, “Principal eigenvalues for indefinite-weight elliptic problems inRN,”Proceedings of the American Mathematical Society, vol. 116, no. 3, pp. 701–706, 1992.

[4] Adimurthi, N. Chaudhuri, and M. Ramaswamy, “An improved Hardy-Sobolev inequality and its application,”Proceedings of the American Mathematical Society, vol. 130, no. 2, pp. 489–505, 2002.

[5] N. Chaudhuri and M. Ramaswamy, “Existence of positive solutions of some semilinear elliptic equations with singular coefficients,”Proceedings of the Royal Society of Edinburgh: Section A, vol. 131, no. 6, pp. 1275–1295, 2001.

[6] Adimurthi and M. J. Esteban, “An improved Hardy-Sobolev inequality inW1,pand its

applica-tion to Schr¨odinger operators,”Nonlinear Differential Equations and Applications, vol. 12, no. 2, pp. 243–263, 2005.

[7] J. P. Garc´ıa Azorero and I. Peral Alonso, “Hardy inequalities and some critical elliptic and para-bolic problems,”Journal of Differential Equations, vol. 144, no. 2, pp. 441–476, 1998.

References

Related documents

of a rotational grazing treatment on quantity and quality of available forage and amount ofground litter. The feed intake and arazina behavior of cattle grazing

This practice of investigating a crime’s space and actions – the chronotopian convergence or blending of place and plot – is best described as a narrative but also as a

Different analytical techniques including sequential injection analysis 2 , voltammetric 4 , capillary electrophoretic 5-7 , spectrofluorimetric method 8, 9 ,

The Table shows that heavy male out- migration from Murshidabad district is taking place exclusively for socio-economic or employment related reasons.. About 3.7

The FTIR and 1 H-NMR values of this compound were compared with reported literatures 12, 13, 14 and the mmp reading as well as Co-TLC data by comparison with

Batch B7 showed the lower value angle of repose and Carr’s index, resulting in better flow compare to other batches, it was observed the inlet temperature, atomization

Nigel Lockett Leeds University Business School, University of Leeds, UK Session 130 Limits to the use of debt David Hillier Strathclyde Business School,. University

What characteristics do modern concerns about taboo profanity found in film and television appealing to youth share with moral panics occurring between 1938 and 2010 that focused