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≤
Q−p
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 Ω⊂Hn,
u=0, on∂Ω, (1.2)
where 1< p < Q=2n+ 2, λ ∈R, f(ξ)∈Ᏺp := {f : Ω→R+|lim
d(ξ)→0(dp(ξ)f(ξ)/
(ψp(ξ))=0, f(ξ)∈Lloc∞(Ω\ {0})},Ωis a bounded domain in the Heisenberg group, and
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 theCα 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
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|2y−xt
, ∇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
(p−1)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 domainΩ⊂Hn: 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
dξ, (2.1)
choice of a suitable vector field and an elementary integral inequality. Note that we also require that 0∈Ω.
Theorem2.1. Letu∈D01,p(Ω/{0}). Then
(1)ifp=Qand there exists a positive constantM0such thatsupξ∈Ωd(ξ)e1/M0:=R0<∞,
then for anyR≥R0,
Ω
∇Hup
dξ≥Q−p p
pΩψp|u| p
dp dξ+ p−1
2p
Q−p pp−2Ωψp|u| p
dp
ln
R
d −2
dξ;
(2.2)
moreover, if2≤p < Q, then choosesupξ∈Ωd(ξ)=R0;
(2)ifp=Qand there exitsM0such thatsupξ∈Ωd(ξ)e1/M0< R, then
Ω
∇Hup dξ≥
p−1
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,pΓ(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,
0≤Ꮾd(ξ) R
≤M, sup
ξ∈Ωd(ξ)< R, ξ∈Ω. (2.5)
Furthermore,
∇HᏮγd R
=γᏮ
γ+1(d/R)∇Hd
d ,
dᏮγ(ρ/R)
dρ =γ
Ꮾγ+1(ρ/R)
ρ ∀γ∈R, (2.6) b
a
Ꮾγ+1(s)
s ds=
1 γ
Ꮾγ(b)−Ꮾγ(a). (2.7)
Proof. Let T be aC1vector field onΩand let it be specified later. For anyu∈C∞
0(Ω\ {0}),
we use H¨older’s inequality and Young’s inequality to get
Ω
divHT
|u|pdξ= −p
Ω
T,∇Hu|u|p−2udξ
≤p
Ω
∇Hup dξ
1/ p
Ω|T|
p/(p−1)|u|pdξ
(p−1)/ p
≤
Ω
∇Hup
dξ+ (p−1)
Ω|T|
p/(p−1)|u|pdξ.
Thus, the following elementary integral inequality:
Ω
∇Hup dξ≥
Ω
divHT−(p−1)|T|p/(p−1)
|u|pdξ (2.9)
holds.
(1) Letabe a free parameter to be chosen later. Denote
I1(Ꮾ)=1 +p−1
pA Ꮾ
d
R
+aᏮ2 d
R
, I2(Ꮾ)= p−1
pA Ꮾ
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−(p−1)∇Hd
p
dp
=A|A|p−2(Q−1−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
p−1
pA Ꮾ
2d
R
+ 2aᏮ3 d
R
=p|A|p∇Hd p
dp I1+A|A|p−
2∇Hd
p
dp I2,
divHT−(p−1)|T|p/(p−1)=p|A|p
∇Hdp
dp I1+A|A|p−
2∇Hd
p
dp I2
−(p−1)|A|p∇Hd p
dp I p/(p−1) 1
= |A|p∇Hd p
dp
pI1+ 1
AI2−(p−1)I p/(p−1) 1
.
(2.12)
We claim
divHT−(p−1)|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)−(p−1)I p/(p−1)
andM=M(R) :=supξ∈ΩᏮ(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, 0≤s≤M, (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)Ꮾp−1(d/R). It gives
divHT=
p−1
p
p−1 Q−1−(p−1)∇Hdp
dp Ꮾp−
1d
R
+ (p−1)Ꮾp d
R
∇Hdp dp
=p p−1
p p
Ꮾp
d R
∇Hdp dp ,
(2.19)
and hence
divHT−(p−1)|T|p/(p−1)=
p−1
p p
Ꮾp
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−(p−1)|T|p/(p−1)=p|A|p
∇Hdp
dp −(p−1)|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.
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| pdξ≤
Ω
∇Hup
dξ, (2.22)
wherec=((p−1)/ 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| pdξ≤
Ω
∇Hup
dξ, (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| pdξ=
BH(δ)
f|u|pdξ+ Ω\BH(δ)
f|u|pdξ
≤
BH(δ) ψp|u|
p
dp dξ+Cδ
Ω\BH(δ)
|u|pdξ≤C−1
f,Q,p
Ω
∇Hup dξ,
(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∞(Ω\BH(δ)). From (2.1),
Ωf um
j−u p
dξ=
BH(δ)
fumj−u p
dξ+
Ω\BH(δ)
fumj−u p
dξ
≤
BH(δ) ψp
um j−u
p
dp dξ+Cδ
Ω\BH(δ) um
j−u p
dξ
≤C−1
Q,p
Ω
∇H
umj−u p
dξ+Cδ
Ω um
j−u p
dξ.
Since{um} ⊆D01,p(Ω) is bounded, we have
Ω f um
j−u p
dξ≤M+Cδ
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+Ꮾ−κ
d
R
, 1 p < κ <
2 p,
Jγ()=
Ωϕ
p(ξ)∇Hd p
dQ−p Ꮾ−γ
d
R
dξ, γ∈R.
(3.2)
Lemma3.1. For>0small, it holds
(i)c−1−γ≤J
γ()≤C−1−γ,γ >−1,
(ii)Jγ()=(p/(γ+ 1))Jγ+1() +O(1),γ >−1, (iii)Jγ()=O(1),γ <−1.
Proof. By the change of polar coordinates (1.11) and 0≤ϕ≤1, we have
Jγ()≤
BH(δ)
∇Hdp dQ−p Ꮾ−γ
d
R
dξ=sH
ρ<δρ
−Q+pᏮ−γρ R
ρQ−1dρ
=sH
ρ<δρ
−1+pᏮ−γρ R
dρ.
By (2.7) we know that forγ <−1 the right-hand side of (3.3) has a finite limit, hence (iii) follows from→0.
To show (i), we setρ=Rτ1/. Thus,dρ=(1/)Rτ1/−1dτ,Ꮾ−γ(τ1/)=−γᏮ−γ(τ), and
Jγ()≤sH
ρ<δρ
−1+pᏮ−γ
ρ
R
dρ=sH
(δ/R)
0
Rτ1/−1+p Ꮾ−γ
Rτ1/
R
1
Rτ1/−1dτ
=sHRp−1−γ
(δ/R)
0 τ
p−1Ꮾ−γ(τ)dτ.
(3.4)
It follows the right-hand side of (i). Using the fact thatϕ=1 inBH(δ/2),
Jγ()≥
BH(δ/2)
∇Hdp dQ−p Ꮾ−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
Ꮾ−γ−1
d R
∇Hd·ndS−→0 asη−→0. (3.6)
From (2.6),
ΩdivH
ϕp∇Hdp−2∇Hd
dQ−1−p
Ꮾ−γ−1d
R dξ = − Ω
ϕp∇Hdp−2 dQ−1−p
∇Hd,∇HᏮ−γ−1 d
R
dξ
=(γ+ 1)
Ω
ϕp∇Hdp dQ−p Ꮾ−γ
d R
dξ=(γ+ 1)Jγ().
(3.7)
On the other hand,
ΩdivH
ϕp∇Hdp−2∇Hd
dQ−1−p
Ꮾ−γ−1d
R dξ =p Ωϕ
p−1∇Hd
p−2
∇Hd,∇Hϕ dQ−1−p Ꮾ−γ−1
d R
dξ
+ (1−Q+p+Q−1)
Ωϕ
p∇Hd p
dQ−p Ꮾ−γ−1
d R dξ =p Ωϕ
p−1∇Hd
p−2
∇Hd,∇Hϕ dQ−1−p Ꮾ−γ−1
d
R
dξ+pJγ+1().
We claim thatpΩϕp−1(|∇Hd|p−2∇Hd,∇Hϕ/dQ−1−p)Ꮾ−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 Ꮾ−γ−1
d
R
dξ
≤2
BH(δ)
∇Hdp dQ−p Ꮾ−γ−1
d
R
dξ
≤2sH
BH(δ)
ρ−Q+pᏮ−γ−1 ρ
R
ρQ−1dρ
=2sH
BH(δ)
ρp−1Ꮾ−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(p−1)/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,b∈R2n,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
Since∇H= −d−A+−1Ꮾ−κ(d/R)(A−+κᏮ(d/R))∇Hd, it follows
Ω
∇HVp dξ≤
BH(δ)
∇Hdp
ϕpd−Q+pᏮ−pκd R
A−−κᏮ d
R
pdξ
+ 2cp
BH(δ)
∇Hdp
ϕp−1d−Q+pᏮ−pκ
d R A−
−κᏮ
d R
p−1dξ
+ 2pc p
BH(δ)
∇Hdp
d−Q+pᏮ−pκ
d
R
dξ:=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+pᏮ−pκ
d
R
A−
−κᏮ
d
R
p−1dξ
≤C
BH(δ)
∇Hdp
d−Q+pᏮ−pκd R
dξ,
Π2≤C
BH(δ)
∇Hdp
d−Q+pᏮ−pκ d
R
dξ.
(3.16)
By (i) ofLemma 3.1, it derivesΠ1,Π2=O(1), as→0.
From (3.14), (3.15) and the definition ofJγ(), it clearly shows
IV=
BH(δ)
∇HVp
dξ− |A|pJ
pκ()≤ΠA− |A|pJpκ() +O(1)=Π3+O(1),
(3.17)
whereΠ3=
BH(δ)|∇Hd|
pϕpd−Q+pᏮ−pκ(d/R)(|A−(−κᏮ(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
Π3≤Π31+Π32+Π33, (3.19)
where
Π31= −pA|A|p−2
BH(δ)
∇Hdp
ϕpd−Q+pᏮ−pκd R
−κᏮ
d
R
dξ,
Π32= p(p−1)
2 |A| p−2
BH(δ)
∇Hdp
ϕpd−Q+pᏮ−pκ d
R
−κᏮ
d
R
2
dξ,
Π33=C
BH(δ)
∇Hdp
ϕpd−Q+pᏮ−pκd R
−κᏮ
d
R
3dξ.
We will show that
Π31,Π33=O(1), as −→0. (3.21)
In fact, using (ii) ofLemma 3.1withγ= −1 +pκyields
Π31= −pA|A|p−2
Jpκ()−κJpκ−1()
= −pA|A|p−2J
pκ()−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+pᏮ−pκd R
2−2κᏮ
d
R
+κ2Ꮾ2 d
R
dξ
= p(p−1) 2 |A|
p−2
2J
pκ()−2κJpκ−1() +κ2 pκ−1
pκ +
1 pκ
Jpκ−2()
= p(p−1) 2 |A|
p−2
2J
pκ()−κp
pκJpκ()−κJpκ−1()
+κ
2(pκ−1)
pκ
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. Let0∈Ωbe 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 dξ+D
Ωψp |u|p
dp Ꮾι
d
R
Then,
(i)B≤ |A|p;
(ii)ifB= |A|pandD >0, thenι≥2;
(iii)ifB= |A|pandι=2, thenD≤((p−1)/2p)|A|p−2.
Proof. Chooseu(ξ)=V(ξ).
(i) By (ii) ofLemma 3.2, we have
B≤
BH(δ)∇HV p
dξ
BH(δ)ψp(|V|
p/dp)dξ ≤
|A|pJ
pκ() +O1−pκ
BH(δ)ψp(|ϕd−A
+Ꮾ−κ(d/R)|p/dp)dξ
=|A|p
1 +c2J
pκ() +O(1) Jpκ()
.
(3.27)
Note thatJpκ()→∞, as→0, soB≤ |A|p.
(ii) SetB= |A|pand assume by contradiction thatι <2. Sincepκ−ι >−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()→∞, as→0. Hence, by (i) ofLemma 3.1we
conclude thatD≤(κ(p−1)/2)|A|p−2, as→0. Then lettingκ→1/ p, the proof is finished.
Theorem3.4. Set0∈Ωand p=Q. Suppose that there exist some constantsD≥0and
ι >0such that the following inequality holds for allu(ξ)∈C0∞(Ω\ {0}):
Ω
∇Hup dξ≥D
Ωψp |u|p
dp Ꮾι
d
R
dξ. (3.30)
Then,
(i)ifD >0, thenι≥p;
(ii)ifι=p, thenD≤((p−1)/ p)p.
Using (3.12) yields
Ω
∇HVp dξ=
BH(δ)
ϕ∇H+∇Hϕp dξ
≤
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 letting→0, we have
Π5,Π6=O(1). (3.32)
Denoting bycipthe coefficients of binormial expansion, we get
∇Hp
=∇Hdpdp−pᏮ−pκd R
−κᏮ
d
R
p
≤∇Hdpdp−pᏮ−pκ d
R
+κᏮ
d
R
p
=∇Hdpdp−pᏮ−pκ
d R
Σp i=0c
p
ip−iκiᏮi
d R
.
(3.33)
Hence,
Π4≤Σip=0c
p
ip−iκiJpκ−i(), (3.34)
whereJγ()=
Ω|∇Hd|pϕpdp−pᏮ−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 >(p−1)/ pimply thatJpκ−p()→∞, as→0, and
D≤lim sup →0
Ω∇HVpdξ
ΩψpVp/dpᏮp(d/R)dξ
≤lim sup →0
κp+Σp−1 i=0c
p iκi
κ−i/ pκ−(i+ 1)/ p···κ−(p−1)/ 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)
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
(p−1)v+Q−1
d v
= −ψp(−A)d−A−1+p−2
(p−1)(−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)
Letting→0, 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)
Denoteuτ(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.
Lemma4.2. Let{gm} ⊂Lp(Ω)(1≤p <∞)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, inᏰ(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,
gm,ϕ≤CKϕL∞ (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 :Ω−→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(Ω\ {0}), the
following holds:
Ω
∇Hup
dξ≥Q−p p
p Ωψp(ξ)|u| p
dp dξ+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
Proof. (i) If f(ξ)∈Ip, then
lim →0ξ∈supB
H() dp(ξ)
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(ξ)dξ ≤C(f)
Ωψp |u|p
dp(ln 1/d)2dξ, 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 Kδ
BH(δ)ψpV p
/dp(ln 1/d)2dξ ≤
κ(p−1)/2|A|p−2J
pκ−2() +O(1) KδJpκ−2()
≤ c1−pκ cKδ1−pκ =
C
Kδ −→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
Proof. We define
Jμ(u) : =
Ω
∇Hup −μ
Ωψp |u|p
dp . (4.18)
Obviously,Jμis continuous and Gˆateaux differentiable onD1,0p(Ω). By (2.1),
Jμ(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 functionJμ(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(ξ)dξ=1, and Jμ(um)→λ1μ andJμ(um)→0 strongly inD−1,p
0 (Ω), when the component ofJμ(um) is re-stricted toᏹ. The coercivity ofJμimplies 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 ᏹis weakly closed and henceu∈ᏹ. 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−2um +λm|um|p−2umf, we check easily thatgm is bounded in M(Ω) and conclude almost ev-erywhere convergence of∇Humto∇HuinΩbyLemma 4.3, and
Jμ(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 thatJμ(u)=λ1μandλ=λ1μ. SinceJμ(|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)=u∈D1,infp(Ω\{0})
Ω∇Hup−CQ,pψp
|u|p/dpdξ
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⊂Ᏻ|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 supu∈AJμ(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
Jμ(u)|u∈D
1,p
0
Ω\ {0},
Ω|u|
pf(ξ)dξ=1
. (5.1)
In what follows we need the Picone identity proved in [15].
Proposition5.1 (Picone identity). For differentiable functionsu≥0,v >0onΩ⊂Hn,
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
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 dξ = Ω
μψp dp +λ
1
μf(ξ)
updξ−
Ω
μψp dp +λ
1
μf(ξ)
vp−1 u p
(v+ε)p−1dξ
=
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−1dξ
= 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
pdξ≥0. (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
Lemma5.5. Foru∈C(Ω\ {0})∩D01,p(Ω), letᏺbe a component of{ξ∈Ω|u(ξ)>0}.
Thenu|ᏺ∈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 boundary∂ᏺ. In fact, ifξ∈∂ᏺandξ=0, thenϕR=0, and soωm=0. Ifξ∈ ∂ᏺ∩Ωandξ=0, thenu=0 (sinceuis continuous except at{0}), and hencevm=0. If ξ∈∂Ω, thenum=0 and sovm=0. Therefore,ωm=0 on∂ᏺ, andωm∈D
1,p
0 (ᏺ). Noting
Ω
∇Hωm− ∇H
ϕRupdξ=
ᏺ
ϕR∇Hvm+vm∇HϕR−ϕR∇Hu−u∇HϕRp dξ
≤ϕR
∇Hvm− ∇HuLpp(ᏺ)+∇HϕR
vm−upLp(ᏺ), (5.10)
it is obvious thatΩ|∇Hωm− ∇H(ϕRu)|pdξ→0, asm→∞. That isωm→ϕRu|ᏺinD1,0p(ᏺ).
By (2.1),
ᏺ
u∇HϕR+ϕR∇Hu− ∇Hup dξ
≤
ᏺ
ϕR∇Hu− ∇Hup dξ+
ᏺ∩{R/2<d<R}
u∇HϕRp dξ
≤
ᏺ
ϕR∇Hu− ∇Hup dξ+Cp
ᏺ∩{R/2<d<R}ψp
|u|p dp dξ
≤
ᏺ
ϕR∇Hu− ∇Hup dξ+C1
ᏺ∩{R/2<d<R}
∇Hup dξ,
(5.11)
which approaches 0, asR→0. Henceu|ᏺ∈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
μandᏺis a nodal domain ofv, then
CλfL∞−Q/ p≤ |ᏺ|, (5.12)
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
ξ∈Ω|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 dξ
≤
Ω−
∇Hump dξ−μ
Ω− ψp dpum
p dξ
=λm
Ω−f(ξ)
ump dξ
≤λmfL∞
Ω−
ump dξ
≤c1fL∞umDp1,pΩ−m p/Q
, Ω−
m≥
c2fL∞Q/ p, Ω−m=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|ᏺ∈D1,0p(ᏺ).Define the function
ω(ξ)=
⎧ ⎨ ⎩
v(ξ) ifξ∈ᏺ,
Clearly,ω(ξ)∈D1,0p(Ω). Takingωas a test function in (4.17) satisfied byvand arguing as
in (5.18), we have
1−Cμ Q,p
vDp1,p(ᏺ)≤λfL∞
ᏺ|v|
pdξ≤λCfL∞vp
D1,p(ᏺ)|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).
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