Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 32585, 24 pages
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 László 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 psub-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
L p,μ u = −ΔH,p u − μψ p
|u| p−2 u
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:
L p,μ u = λ f (ξ)|u| p−2 u,
u = 0,
on ∂Ω,
in Ω ⊂ Hn ,
(1.2)
where 1 < p < Q = 2n + 2, λ ∈ R, f (ξ) ∈ Ᏺ p := { f : Ω→R+ | limd(ξ)→0 (d p (ξ) f (ξ)/
(ψ p (ξ)) = 0, f (ξ) ∈ L∞
loc (Ω \ {0})}, Ω is a bounded domain in the Heisenberg group, and
the definitions of d(ξ) and ψ p (ξ); see below. We investigate the weak solution of (1.2) and
the asymptotic behavior of the first eigenvalue for different singular weights as μ increases
to ((Q − p)/ p) p . Furthermore, we show that the first eigenvalue is simple and isolated, as
2
Journal of Inequalities and Applications
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 vector 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 embedding in the weighted Sobolev space.
The main difficulty to study the properties of the first eigenvalue is the lack of regularity of the weak solutions of the p-sub-Laplacian in the Heisenberg group. Let us note
that the C α regularity for the weak solutions of the p-subelliptic operators formed by
the vector field satisfying Hörmander’s condition was given in [1] and the C 1,α regularity
of the weak solutions of the p-sub-Laplacian ΔH,p in the Heisenberg group for p near 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]). Let Hn be
a Heisenberg group endowed with the group law
ξ ◦ ξ = x + x , y + y ,t + t + 2
n
xi y i − xi y i
,
(1.3)
i=1
where ξ = (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 endows Hn with a structure of
nilpotent Lie group. A family of dilations on Hn is defined as
δτ (x, y,t) = τx,τ y,τ 2 t ,
τ > 0.
(1.4)
The homogeneous dimension with respect to dilations is Q = 2n + 2. The left invariant
vector fields on the Heisenberg group have the form
Xi =
∂
∂
+ 2yi ,
∂xi
∂t
Yi =
∂
∂
− 2xi ,
∂yi
∂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 (Xi vi + Yi vn+i ). Hence, the sub-Laplacian ΔH and the p-subLaplacian ΔH,p are expressed by
ΔH =
n
Xi2 + Yi2 = ∇H ·∇H ,
(1.6)
i=1
p −2
p −2
ΔH,p u = ∇H ∇H u
∇H u = divH ∇H u
∇H u ,
p > 1,
respectively.
The distance function is
d(ξ,ξ ) =
2
(x − x )2 + (y − y )2 + t − t − 2(x· y − x · y)
2 1/4
,
for ξ,ξ ∈ Hn .
(1.7)
Jingbo Dou et al. 3
If ξ = 0, we denote
d(ξ) = d(ξ,0) = |z|4 + t 2
1/4
with |z| = x2 + y 2
,
1/2
.
(1.8)
Note that d(ξ) is usually called the homogeneous norm.
For d = d(ξ), it is easy to calculate
∇H d =
p
∇H d p = |z| = ψ p ,
p
1 |z|2 x + yt
,
d3 |z|2 y − xt
d
ΔH,p d = ψ p
Q−1
.
d
(1.9)
Denote by BH (R) = {ξ ∈ Hn | d(ξ) < R} the ball of radius R centered at the origin. Let
Ω1 = BH (R2 )\ BH (R1 ) with 0 ≤ R1 < R2 ≤ ∞ and u(ξ) = v(d(ξ)) ∈ C 2 (Ω1 ) be a radial
function with respect to d(ξ). Then
p −2
ΔH,p u = ψ p v (p − 1)v +
Q−1 v .
d
(1.10)
Let us recall the change of polar coordinates (x, y,t)→(ρ,θ,θ1 ,...,θ2n−1 ) in [16]. If
u(ξ) = ψ p (ξ)v(d(ξ)), then
Ω1
u(ξ)dξ = sH
R2
R1
ρQ−1 v(ρ)dρ,
(1.11)
π
where sH = ωn 0 (sinθ)n−1+p/2 dθ, ωn is the 2n-Lebesgue measure of the unitary Euclidean
sphere in R2n .
1,p
The Sobolev space in Hn is written by D1,p (Ω)={u : Ω→R;u,|∇H u| ∈ L p (Ω)}. D0 (Ω)
1/ p
∞
p
is the closure of C0 (Ω) with respect to the norm uD01,p (Ω) = ( Ω |∇H u| dξ) .
In the sequel, we denote by c, c1 , C, and so forth some positive constants usually except
special narrating.
This paper is organized as follows. In Section 2, we prove the Hardy inequality with
remainder terms by the vector field method in the Heisenberg group. In Section 3, we
discuss the optimality of the constants in the inequalities which is of its independent
interest. In Section 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 as μ increases 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 domain Ω ⊂ Hn : let
1,p
p > 1 and p=Q. For any u ∈ D0 (Ω, |z| p /d2p ), it holds that
CQ,p
Ω
ψp
|u| p
dp
dξ ≤
Ω
∇H u
p dξ,
(2.1)
where CQ,p = |(Q − p)/ p| p . Moreover, if 0 ∈ Ω, then the constant CQ,p is best. In this
section, we give the Hardy inequality with remainder terms on Ω, based on the careful
4
Journal of Inequalities and Applications
choice of a suitable vector field and an elementary integral inequality. Note that we also
require that 0 ∈ Ω.
1,p
Theorem 2.1. Let u ∈ D0 (Ω / {0}). Then
(1) if p=Q and there exists a positive constant M0 such that supξ ∈Ω d(ξ)e1/M0 := R0 < ∞,
then for any R ≥ R0 ,
Ω
p
p −2 −2
p −1
Q− p
|u| p
|u| p
R
∇H u
p dξ ≥ Q − p ψ
dξ
+
ψ
ln
dξ;
p
p
p dp
2p p dp
d
Ω
Ω
(2.2)
moreover, if 2 ≤ p < Q, then choose supξ ∈Ω d(ξ) = R0 ;
(2) if p = Q and there exits M0 such that supξ ∈Ω d(ξ)e1/M0 < R, then
Ω
∇H u
p dξ ≥
p−1
p
p
|u| p
ψp p dξ.
Ω
d ln(R/d)
(2.3)
Before we prove the theorem, let us recall that
⎧
⎨d(ξ)(p−Q)/(p−1)
if p=Q,
if p = Q
Γ d(ξ) = ⎩
− ln d(ξ)
(2.4)
is the solution of ΔH,p at the origin, that is, ΔH,p Γ(d(ξ)) = 0 on Ω \ {0}. Equation (2.4)
is useful in our proof. For convenience, write Ꮾ(s) = −1/ ln(s), s ∈ (0,1), and A = (Q −
p)/ p. Thus, for some positive constant M > 0,
0≤Ꮾ
d(ξ)
≤ M,
R
sup d(ξ) < R,
ξ ∈Ω
ξ ∈ Ω.
(2.5)
Furthermore,
∇H Ꮾγ
Ꮾγ+1 (d/R)∇H d
d
,
=γ
R
d
b
Ꮾγ+1 (ρ/R)
dᏮγ (ρ/R)
=γ
dρ
ρ
∀γ ∈ R,
Ꮾγ+1 (s)
1
ds = Ꮾγ (b) − Ꮾγ (a) .
s
γ
a
(2.6)
(2.7)
Proof. Let T be a C 1 vector field on Ω and let it be specified later. For any u ∈ C0∞ (Ω \ {0}),
we use Hölder’s inequality and Young’s inequality to get
Ω
divH T |u| p dξ = − p
≤p
Ω
Ω
≤
Ω
T, ∇H u |u| p−2 udξ
∇H u
p dξ
1/ p ∇H u
p dξ + (p − 1)
(p−1)/ p
Ω
|T| p/(p−1) |u| p dξ
Ω
|T| p/(p−1) |u| p dξ.
(2.8)
Jingbo Dou et al. 5
Thus, the following elementary integral inequality:
Ω
∇H u
p dξ ≥
Ω
divH T − (p − 1)|T| p/(p−1) |u| p dξ
(2.9)
holds.
(1) Let a be a free parameter to be chosen later. Denote
I1 (Ꮾ) = 1 +
p−1
d
d
+ aᏮ2
,
Ꮾ
pA
R
R
I2 (Ꮾ) =
p−1 2 d
d
+ 2aᏮ3
,
Ꮾ
pA
R
R
(2.10)
and pick T(d) = A|A| p−2 (|∇H d| p−2 ∇H d/d p−1 )I1 . An immediate computation shows
divH A|A|
p −2
∇ H d p −2 ∇ H d d p −1
p
p−2 dΔH,p d − (p − 1) ∇H d
= A|A|
dp
p
p
(Q − 1 − p + 1)
∇H d
p ∇H d
= A|A| p−2
=
p
|
A
|
.
p
p
d
d
(2.11)
By (2.6),
divH T = p|A| p
∇H d p
I1 + A|A| p−2
∇ H d p −2 ∇ H d
dp
d p −1
∇H d p − 1 2 d
d
×
+ 2aᏮ3
Ꮾ
d
pA
R
R
= p|A|
divH T − (p − 1)|T| p/(p−1) = p|A|
p
∇H d p
dp
p
p ∇H d
dp
I1 + A|A|
p −2
dp
I1 + A|A| p−2
∇H d p
∇H d p
∇H d p
dp
I2 ,
I2
p/(p−1)
− (p − 1)|A| p
I1
dp
p ∇H d 1
p/(p−1)
= |A| p
pI
+
−
(p
−
1)I
.
I
1
2
1
p
d
A
(2.12)
We claim
divH T − (p − 1)|T| p/(p−1) ≥ |A| p
∇H d p dp
1+
p−1 2 d
Ꮾ
2pA2
R
.
(2.13)
In fact, arguing as in the proof of [13, Theorem 4.1], we set
f (s) := pI1 (s) +
1
p/(p−1)
(s)
I2 (s) − (p − 1)I1
A
(2.14)
6
Journal of Inequalities and Applications
and M = M(R) := supξ ∈Ω Ꮾ(d(ξ)/R), and distinguish three cases
(i)
1 < p < 2 < Q,
a>
(2 − p)(p − 1)
,
6p2 A2
(2.15)
(ii)
a = 0,
(2.16)
(2 − p)(p − 1)
< 0.
6p2 A2
(2.17)
2 ≤ p < Q,
(iii)
p > Q,
a<
It yields that
f (s) ≥ 1 +
p−1 2
s,
2pA2
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 (|∇H d| p−2 ∇H d/
d p−1 )Ꮾ p−1 (d/R). It gives
divH T =
p−1
p
p −1 p−1
=p
p
p
Ꮾ
p
p
∇ H d p Q − 1 − (p − 1) ∇H d
p−1 d
p d
Ꮾ
+
(p
−
1)Ꮾ
dp
R
R
dp
p
d ∇ H d dp
R
,
(2.19)
and hence
divH
T − (p − 1)|T| p/(p−1)
p−1
=
p
p
Ꮾp
d
R
p
∇H d dp
(2.20)
.
Combining (2.20) with (2.9) follows (2.3).
Remark 2.2. The domain Ω in (2.9) may be bounded or unbounded. In addition, if we
select that T(d) = A|A| p−2 (|∇H d| p−2 ∇H d/d p−1 ), then
divH
T − (p − 1)|T| p/(p−1)
=
p|A| p
∇H d p
dp
− (p − 1)|A| p
∇H d p
dp
= |A| p
∇H d p
dp
.
(2.21)
Hence, from (2.9) we conclude (2.1) on the bounded domain Ω and on Hn (see [15]),
respectively.
We will prove in next section that the constants in (2.2) and (2.3) are best.
Now, we state the Poincaré inequality proved in [17].
Jingbo Dou et al. 7
Lemma 2.3. Let Ω be a subset of Hn bounded in x1 direction, that is, there exists R > 0 such
1,p
that 0 < r = |x1 | ≤ R for ξ = (x1 ,x2 ,...,xn , y1 ,..., yn ,t) ∈ Ω. Then for any u ∈ D0 (Ω),
then
c
Ω
|u| p dξ ≤
Ω
∇H u
p dξ,
(2.22)
where c = ((p − 1)/ pR) p .
Using (2.9) by choosing T = −((p − 1)/ p) p−1 (∇H r/r p−1 ) immediately provides a different proof to (2.22).
In the following, we describe a compactness result by using (2.1) and (2.22).
Theorem 2.4. Suppose p=Q and f (ξ) ∈ Ᏺ p . Then there exists a positive constant C f ,Q,p
such that
p
C f ,Q,p
Ω
f (ξ)|u| dξ ≤
Ω
∇H u
p dξ,
(2.23)
1,p
and the embedding D0 (Ω) L p (Ω, f dξ) is compact.
Proof. Since f (ξ) ∈ Ᏺ p , we have that for any > 0, there exist δ > 0 and Cδ > 0 such that
dp
f (ξ) ≤ ,
BH (δ)⊆Ω ψ p
f (ξ)|Ω\BH (δ) ≤ Cδ .
sup
(2.24)
By (2.1) and (2.22), it follows
p
Ω
p
f (ξ)|u| dξ =
BH (δ)
f |u| dξ +
≤
BH (δ)
ψp
|u| p
dp
Ω\BH (δ)
f |u| p dξ
dξ + Cδ
Ω\BH (δ)
1
|u| p dξ ≤ C −f ,Q,p
Ω
∇H u
p dξ,
(2.25)
then (2.23) is obtained.
1,p
Now, we prove the compactness. Let {um } ⊆ D0 (Ω) be a bounded sequence. By re1,p
flexivity of the space D0 (Ω) and the Sobolev embedding for vector fields (see [18]), it
yields
1,p
um j u weakly in D0 (Ω),
(2.26)
um j −→ u strongly in L p (Ω)
for a subsequence {um j } of {um } as j →∞. Write Cδ = f L∞ (Ω\BH (δ)) . From (2.1),
Ω
p
f um j − u
dξ =
BH (δ)
≤
BH (δ)
−1
≤ CQ,p
p
f um j − u
dξ +
ψp
Ω
u m − u p
j
dp
Ω\BH (δ)
p
f um j − u
dξ
dξ + Cδ
Ω\BH (δ)
∇H um − u p dξ + Cδ
j
Ω
um − u
p dξ
j
um − u
p dξ.
j
(2.27)
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Journal of Inequalities and Applications
1,p
Since {um } ⊆ D0 (Ω) is bounded, we have
p
f um j − u
dξ ≤ M + Cδ
Ω
Ω
um − u
p dξ,
j
(2.28)
where M > 0 is a constant depending on Q and p. By (2.26),
lim
j →∞
As is arbitrary, lim j →∞
p
f um j − u
dξ ≤ M.
Ω
(2.29)
1,p
− u| p dξ = 0. Hence D0 (Ω) L p (Ω, f dξ) is compact.
Ω f |um j
Remark 2.5. The class of the functions f (ξ) ∈ Ᏺ p has lower-order singularity than d− p (ξ)
at the origin. The examples of such functions are
(a) any bounded function,
(b) in a small neighborhood of 0, f (ξ) = ψ p (ξ)/dβ (ξ), 0 < β < p,
(c) f (ξ) = ψ p (ξ)/d p (ξ)(ln(1/d(ξ)))2 in 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 in Theorem 2.1 are the best. To do
this, we need two lemmas. First we introduce some notations.
For some fixed small δ > 0, let the test function ϕ(ξ) ∈ C0∞ (Ω) satisfy 0 ≤ ϕ ≤ 1 and
⎧
⎪
⎨1
δ
,
2
if ξ ∈ Ω \ BH (0,δ),
if ξ ∈ BH 0,
ϕ(ξ) = ⎪
⎩0
(3.1)
with |∇H ϕ| < 2|∇H d|/d. Let > 0 small enough, and define
with = d−A+ Ꮾ−κ
V (ξ) = ϕ(ξ) ,
Jγ () =
Ω
ϕ p (ξ)
∇H d p
d Q− p Ꮾ−γ
d
dξ,
R
d
,
R
1
2
<κ< ,
p
p
(3.2)
γ ∈ R.
Lemma 3.1. For > 0 small, it holds
(i) c−1−γ ≤ Jγ () ≤ 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
∇H d p
Jγ () ≤
BH (δ)
= sH
ρ<δ
d Q− p ρ
−1+p
Ꮾ−γ
−γ
Ꮾ
d
dξ = sH
R
ρ
dρ.
R
ρ<δ
ρ−Q+p Ꮾ−γ
ρ Q−1
ρ dρ
R
(3.3)
Jingbo Dou et al. 9
By (2.7) we know that for γ < −1 the right-hand side of (3.3) has a finite limit, hence (iii)
follows from →0.
To show (i), we set ρ = Rτ 1/ . Thus, dρ = (1/ )Rτ 1/−1 dτ, Ꮾ−γ (τ 1/ ) = −γ Ꮾ−γ (τ), and
Jγ () ≤ sH
ρ<δ
ρ−1+p Ꮾ−γ
= sH R p −1−γ
(δ/R)
0
ρ
dρ = sH
R
(δ/R)
0
Rτ 1/
−1+p
Ꮾ−γ
Rτ 1/ 1 1/−1
Rτ
dτ
R
τ p−1 Ꮾ−γ (τ)dτ.
(3.4)
It follows the right-hand side of (i). Using the fact that ϕ = 1 in BH (δ/2),
Jγ () ≥
∇H d p
BH (δ/2)
d Q− p Ꮾ−γ
(δ/2R) p−1 −γ
d
dξ = sH R p −1−γ 0
τ Ꮾ (τ)dτ,
R
(3.5)
and the left-hand side of (i) is proved.
Now we prove (ii). Let Ωη : = {ξ ∈ Ω | d(ξ) > η}, η > 0, be small and note the boundary
term
−
d =η
p −2
ϕ p ∇H d ∇ H d
d
Ꮾ−γ−1
∇H d ·
ndS −→ 0 as η −→ 0.
Q
−
1
−
p
d
R
(3.6)
From (2.6),
Ω
divH
p −2
ϕ p ∇H d ∇ H d
d
Ꮾ−γ−1
dξ
Q
−
1
−
p
d
R
p −2 ϕ p ∇H d d
dξ
∇H d, ∇H Ꮾ−γ−1
dQ−1− p
R
Ω
p
ϕ p ∇H d
−γ d
= (γ + 1)
Ꮾ
dξ = (γ + 1)Jγ ().
d Q− p R
Ω
=−
(3.7)
On the other hand,
Ω
divH
p −2
ϕ p ∇H d ∇ H d
d
Ꮾ−γ−1
dξ
dQ−1− p
R
∇H d p−2 ∇H d, ∇H ϕ
d
dξ
R
Ω
∇H d p
d
+ (1 − Q + p + Q − 1) ϕ p Q− p Ꮾ−γ−1
dξ
d
R
Ω
=p
ϕ
=p
Ω
p −1
ϕ p −1
dQ−1− p
∇H d p−2 ∇H d, ∇H ϕ
dQ−1− p
Ꮾ−γ−1
Ꮾ−γ−1
d
dξ + pJγ+1 ().
R
(3.8)
10
Journal of Inequalities and Applications
We claim that p Ω ϕ p−1 (|∇H d| p−2 ∇H d, ∇H ϕ/dQ−1− p )Ꮾ−γ−1 (d/R)dξ = O (1). In fact,
by (3.1) and (1.11),
Ω
∇H d p−2 ∇H d, ∇H ϕ
d
dξ
dQ−1− p
R
∇H d p
−γ −1 d
≤2
Ꮾ
dξ
Q− p R
BH (δ) d
ρ Q −1
≤ 2sH
ρ−Q+p Ꮾ−γ−1
ρ dρ
R
BH (δ)
ρ
= 2sH
ρ p−1 Ꮾ−γ−1
dρ.
R
BH (δ)
ϕ p −1
Using the estimate (i) follows that
with (3.8) gives
BH (δ) ρ
Ꮾ−γ−1
p−1 Ꮾ−γ−1 (ρ/R)dρ
(3.9)
= O (1). Combining (3.7)
(γ + 1)Jγ () = pJγ+1 () + O (1).
(3.10)
This allows us to conclude (ii).
We next estimate the quantity
I[V ] =
Ω
∇H V p dξ − |A| p
Ω
p
∇H d p V dξ.
dp
(3.11)
Lemma 3.2. As →0, it holds
1)/2)|A| p−2 J pκ−2 () + O (1);
(i) I(V
) ≤ (κ(p −
p
(ii) BH (δ) |∇H V | dξ ≤ |A| p J pκ () + O (1− pκ ).
Proof. By the definition of V (ξ), we see ∇H V (ξ) = ϕ(ξ)∇H + ∇H ϕ. Using the elementary inequality
|a + b| p ≤ |a| p + c p |a| p−1 |b| + |b| p ,
a,b ∈ R2n , p > 1,
(3.12)
one has
∇H V p ≤ ϕ p ∇H p + c p ∇H ϕ
ϕ p−1 ∇H p−1 + ∇H ϕ
p p
2 ∇H d 2 ∇H d p p p
p −1
.
ϕ p−1 ∇H +
≤ ϕ p ∇H + c p
d
d
(3.13)
Jingbo Dou et al.
11
Since ∇H = −d−A+−1 Ꮾ−κ (d/R)(A − + κᏮ(d/R))∇H d, it follows
Ω
∇H V p dξ ≤
p
∇H d p ϕ p d −Q+p Ꮾ− pκ d A − − κᏮ d
dξ
R R BH (δ)
p−1
∇H d p ϕ p−1 d −Q+p Ꮾ− pκ d A − − κᏮ d
dξ
+ 2c p
R
R
BH (δ)
∇H d p d −Q+p Ꮾ− pκ d dξ := ΠA + Π1 + Π2 .
+ 2pcp
R
BH (δ)
(3.14)
We claim that
Π1 ,Π2 = O (1).
(3.15)
Indeed, since |A − ( − κᏮ(d/R))| is bounded, using (3.1) we get
p−1
∇H d p ϕ p−1 d −Q+p Ꮾ− pκ d A − − κᏮ d dξ
R
R
BH (δ)
∇H d p d −Q+p Ꮾ− pκ d dξ,
≤C
Π1 ≤ C
R
p −Q+p − pκ d
∇H d d
Π2 ≤ C
Ꮾ
dξ.
R
BH (δ)
(3.16)
BH (δ)
By (i) of Lemma 3.1, it derives Π1 , Π2 = O (1), as →0.
From (3.14), (3.15) and the definition of Jγ (), it clearly shows
I V =
BH (δ)
∇H V p dξ − |A| p J pκ () ≤ ΠA − |A| p J pκ () + O (1) = Π3 + O (1),
(3.17)
where Π3 = BH (δ) |∇H d| p ϕ p d−Q+p Ꮾ− pκ (d/R)(|A − ( − κᏮ(d/R))| p − |A| p )dξ. For simplicity, denote ζ = − κᏮ(d/R). Since ζ is small compared to A, we use Taylor’s expansion
to yield
|A − ζ | p − |A| p ≤ − pA|A| p−2 ζ +
p(p − 1)
|A| p−2 ζ 2 + C |ζ |3 .
2
(3.18)
Thus, we can estimate Π3 by
Π3 ≤ Π31 + Π32 + Π33 ,
where
Π31 = − pA|A| p−2
Π32 =
BH (δ)
p(p − 1)
|A| p−2
2
Π33 = C
BH (δ)
(3.19)
∇H d p ϕ p d −Q+p Ꮾ− pκ d − κᏮ d dξ,
BH (δ)
R
R
2
∇H d p ϕ p d −Q+p Ꮾ− pκ d − κᏮ d
dξ,
R
3
∇H d p ϕ p d −Q+p Ꮾ− pκ d − κᏮ d dξ.
R
R R
(3.20)
12
Journal of Inequalities and Applications
We will show that
Π31 , Π33 = O (1),
as −→ 0.
(3.21)
In fact, using (ii) of Lemma 3.1 with γ = −1 + pκ yields
Π31 = − pA|A| p−2 J pκ () − κJ pκ−1 ()
= − pA|A| p−2 J pκ () − J pκ () + O (1) = O (1).
(3.22)
Recalling (a − b)3 ≤ (|a| + |b|)3 ≤ c(|a|3 + |b|3 ), we obtain
Π33 ≤ c3 J pκ () + cJ pκ−3 ()
for > 0.
(3.23)
From (i) and (iii) in Lemma 3.1 and the fact that 1 < pκ < 2 it follows Π33 = O (1). Using
(ii) of Lemma 3.1 twice (pick γ = pκ − 1 > −1 and γ = pκ − 2 > −1, resp.), we conclude
that
p(p − 1)
∇H d p ϕ p d −Q+p Ꮾ− pκ d 2−2κᏮ d +κ2 Ꮾ2 d
|A| p−2
dξ
2
R
R
R
BH (δ)
p(p − 1)
pκ − 1 1
=
|A| p−2 2 J pκ () − 2κJ pκ−1 () + κ2
J pκ−2 ()
+
2
pκ
pκ
p
p(p − 1)
|A| p−2 2 J pκ () − κ J pκ () − κJ pκ−1 ()
=
2
pκ
κ2 (pκ − 1) p
κ
+
J pκ−1 () + J pκ−2 () + O (1)
pκ
pκ − 1
p
κ(p − 1)
|A| p−2 J pκ−2 () + O (1).
=
2
(3.24)
Π32 =
In virtue of (3.17), (3.19), (3.21), and (3.24) we deduce (i) of Lemma 3.2. By (3.17),
(3.24), and (i) of Lemma 3.2,
BH (δ)
∇H V p dξ = I V + |A| p J pκ () ≤ |A| p J pκ () + κ(p − 1) |A| p−2 J pκ−2 () + O (1).
2
(3.25)
Hence (ii) of Lemma 3.2 follows from (i) in Lemma 3.1. It completes the proof.
We are now ready to give the proof of the best constants in Theorem 2.1.
Theorem 3.3. Let 0 ∈ Ω be a bounded domain in Hn and p=Q. Suppose that for some
constants B > 0, D ≥ 0, and ι > 0, the following inequality holds for any u(ξ) ∈ C0∞ (Ω \ {0}):
Ω
∇H u
p dξ ≥ B
Ω
ψp
|u| p
dp
dξ + D
Ω
ψp
|u| p
dp
Ꮾι
d
dξ.
R
(3.26)
Jingbo Dou et al.
13
Then,
(i) B ≤ |A| p ;
(ii) if B = |A| p and D > 0, then ι ≥ 2;
(iii) if B = |A| p and ι = 2, then D ≤ ((p − 1)/2p)|A| p−2 .
Proof. Choose u(ξ) = V (ξ).
(i) By (ii) of Lemma 3.2, we have
B≤ =
BH (δ)
∇H V p dξ
|A| p J pκ () + O 1− pκ
≤
p p
−A+ Ꮾ−κ (d/R)| p /d p )dξ
BH (δ) ψ p (|V | /d )dξ
BH (δ) ψ p (|ϕd
|A| p 1 + c2 J pκ () + O (1)
(3.27)
.
J pκ ()
Note that J pκ ()→∞, as →0, so B ≤ |A| p .
(ii) Set B = |A| p and assume by contradiction that ι < 2. Since pκ − ι > −1, using (i)
of Lemma 3.2 and (i) of Lemma 3.1 leads to
κ(p − 1)/2 |A| p−2 J pκ−2 () + O (1)
I V
I V
=
≤
0<D≤ p p
ι
J pκ−ι ()
J pκ−ι ()
Ω ψ p (|V | /d )Ꮾ (d/R)dξ
1
− pκ
C
≤ −1− pκ+ι = C 2−ι −→ 0, as −→ 0,
c
(3.28)
which is a contradiction. Hence ι ≥ 2.
(iii) If B = |A| p and ι = 2, then by (i) of Lemma 3.2,
κ(p − 1)/2 |A| p−2 J pκ−2 () + O (1)
I V
≤
.
D≤
J pκ−2 ()
J pκ−2 ()
(3.29)
The assumption κ > 1/ p implies J pκ−2 ()→∞, as →0. Hence, by (i) of Lemma 3.1 we
conclude that D ≤ (κ(p − 1)/2)|A| p−2 , as →0. Then letting κ→1/ p, the proof is finished.
Theorem 3.4. Set 0 ∈ Ω and p = Q. Suppose that there exist some constants D ≥ 0 and
ι > 0 such that the following inequality holds for all u(ξ) ∈ C0∞ (Ω \ {0}):
Ω
∇H u
p dξ ≥ D
Ω
ψp
|u| p
dp
Ꮾι
d
dξ.
R
(3.30)
Then,
(i) if D > 0, then ι ≥ p;
(ii) if ι = p, then D ≤ ((p − 1)/ p) p .
Proof. The proof is essentially similar to one of Theorem 3.3. Let the test function ϕ be
as before (see (3.1)). For > 0, κ > (p − 1)/ p, define V = ϕ with = d (− ln(d/R))κ .
14
Journal of Inequalities and Applications
Using (3.12) yields
Ω
∇H V p dξ =
≤
BH (δ)
BH (δ)
ϕ∇H + ∇H ϕ
p dξ
ϕ
p
p
∇H dξ + c p
+ cp
p
BH (δ)
p
BH (δ)
p −1 ∇H ϕ
dξ (3.31)
ϕ p−1 ∇H ∇H ϕ
dξ := Π4 + Π5 + Π6 .
Arguing as in the proof of previous theorem and letting →0, we have
Π5 , Π6 = O (1).
(3.32)
p
Denoting by ci the coefficients of binormial expansion, we get
p
∇H p = ∇H d p d p− p Ꮾ− pκ d − κᏮ d R
R p
p
d
d
≤ ∇H d d p− p Ꮾ− pκ
+ κᏮ
R
R
p
d p p p −i i i d
= ∇H d d p− p Ꮾ− pκ
Σi=0 ci κ Ꮾ
.
R
(3.33)
R
Hence,
p
p
Π4 ≤ Σi=0 ci p−i κi J pκ−i (),
(3.34)
where Jγ () = Ω |∇H d| p ϕ p d p− p Ꮾ−γ (d/R). By (ii) of Lemma 3.1 and the induction argument it holds
p−1
i
i+1
p−i J pκ−i () = κ −
κ−
J pκ− p ()+O (1),
· · · κ−
p
p
p
i = 0,1,..., p − 1.
(3.35)
Now (i) of Lemma 3.1 and the assumption κ > (p − 1)/ p imply that J pκ− p ()→∞, as →0,
and
p
Ω ∇H V dξ
D ≤ limsup p p p
→0
Ω ψ p V /d Ꮾ (d/R)dξ
p
p −1 p κ + Σi=0 ci κi κ − i/ p κ − (i + 1)/ p · · · κ − (p − 1)/ p J pκ− p () + O (1)
≤ limsup
J pκ− p ()
→0
p−1
i
i
+
1
p −1 p
= κ p + Σi=0 ci κi κ −
κ−
.
··· κ −
p
p
p
(3.36)
The last expression converges to ((p − 1)/ p) p as κ→(p − 1)/ p. The proof is over.
Jingbo Dou et al.
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).
1,p
Lemma 4.1. Suppose that u(ξ) ∈ D0 (Ω) and p=Q. Then
(1) L p,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 of L p,μ u = 0;
(2) L p,μ is unbounded from below if μ > CQ,p .
Proof. (1) It is obvious from (2.1) that L p,μ is a positive operator.
We now suppose that μ = CQ,p and verify that v = d(p−Q)/ p satisfies L p,μ u = 0. For the
1,p
purpose, set v = d(p−Q)/ p+ ∈ D0 (Ω) and A = (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
p −2
Q−1 −ΔH,p v = −ψ p v
(p − 1)v +
v
d p −2
(p − 1)( − A)( − A − 1)d−A−2+
= −ψ p ( − A)d −A−1+ Q−1
+
( − A)d−A−1+
d
p−2 (−A−1+)(p−2)+(−A−2+)
[(p − 1)( − A − 1) + Q − 1]
= −ψ p ( − A)| − A| d
p−2 (−A+)(p−1)− p
= −ψ p ( − A)(p − 1) + Q − p ( − A)| − A| d
p −1
v
= − ( − A)(p − 1) + pA ( − A)| − A| p−2 ψ p p .
d
(4.2)
Letting →0, the conclusion follows.
argument, we select φ(ξ) ∈ C0∞ (Ω), φL p = 1, such that CQ,p =
(2) Bypthe
density
p p
p p
Ω |∇H φ| /( Ω (|z | /d )(|φ| /d )). Using the best constant of the inequality (2.1), one
has
L p,μ φ,φ =
Ω
∇H φ p − μ
|z| p |φ| p
Ω
dp
dp
<
Ω
∇H φ
p − CQ,p
|z| p |φ| p
Ω
dp dp
= 0.
(4.3)
Denote uτ (x, y,t) = τ Q/ p φ(δτ (x, y,t)) and ξ = (x, y,t). Thus,
uτ (x, y,t) pp =
L
Ω
Q/ p τ φ δτ (x, y,t) p dξ =
Ω
φ δτ (x, y,t) p τ Q dξ = 1,
and L p,μ uτ ,uτ = τ p L p,μ φ,φ < 0. This concludes the result.
(4.4)
In order to prove the main result (Theorem 4.6 below) we need the following two
preliminary lemmas.
16
Journal of Inequalities and Applications
Lemma 4.2. Let {gm } ⊂ L p (Ω)(1 ≤ p < ∞) be such that as m→∞,
gm g
weakly in L p (Ω),
gm −→ g
(4.5)
a.e. in Ω.
Then,
p
p
p
lim gm L p (Ω) − gm − g L p (Ω) = g L p (Ω) .
(4.6)
m→∞
The proof is similar to one in the Euclidean space (see [19, Chapter 1, Section 4]. We
omit it here.
1,p
Lemma 4.3. Suppose that {um } ⊂ D0 (Ω)(1 ≤ p < ∞) satisfies
1,p
um u weakly in D0 (Ω),
(4.7)
p
um −→ u strongly in Lloc (Ω),
as m→∞, and
in Ᏸ (Ω),
−ΔH,p um = fm + gm ,
(4.8)
where fm → f strongly in D−1,p (Ω)(p = p/(p − 1)), gm is bounded in M(Ω) (the space of
Radon measures), that is,
gm ,ϕ ≤ CK ϕL∞
(4.9)
for all ϕ ∈ Ᏸ(Ω) with supp(ϕ) ⊂ K, where CK is a constant which depends on the compact
set K. Then there exists a subsequence {um j } of {um } such that
1,q
um j −→ u strongly in D0 (Ω), ∀q < p.
(4.10)
Its proof is similar to one of [20, Theorem 2.1].
Lemma 4.4. Let p=Q and
I p :=
1
d p (ξ)
f : Ω −→ R | lim
f (ξ) ln
d(ξ)
d(ξ)→0 ψ p (ξ)
+
2
∞
< ∞, f (ξ) ∈ Lloc Ω \ {0}
. (4.11)
1,p
(i) If f (ξ) ∈ I p , then there exists λ( f ) > 0 such that for all u ∈ D0 (Ω \ {0}), the
following holds:
Ω
p |u| p
∇H u
p dξ ≥ Q − p ψ p (ξ) p dξ + λ( f ) |u| p f (ξ)dξ.
p d
Ω
Ω
(4.12)
(ii) If f (ξ) ∈ I p and (d p (ξ)/ψ p (ξ)) f (ξ)(ln1/d(ξ))2 →∞ as d(ξ)→0, then (4.12) is not
true.
Jingbo Dou et al.
17
Proof. (i) If f (ξ) ∈ I p , then
2
1
d p (ξ)
f (ξ) ln
lim sup
→0 ξ ∈B () ψ p
d(ξ)
H
< ∞.
(4.13)
Without any loss of generality we assume that R = 1 in (2.2). For the sufficiently small
> 0, we have
f (ξ) <
Cψ p
,
p
d (ln 1/d)2
in BH ().
(4.14)
Outside BH (), f (ξ) is also bounded. Hence there exists C( f ) > 0 such that
Ω
|u| p f (ξ)dξ ≤ C( f )
Ω
ψp
|u| p
d p (ln1/d)2
on Ω.
dξ,
(4.15)
Taking λ( f ) = C( f )−1 ((p − 1)/2p)|A| p−2 > 0, (4.12) follows from (2.2).
(ii) We write f (ξ) = ψ p h(ξ)/d p (ξ)(ln 1/d(ξ))2 , where h(ξ)→∞ as d(ξ)→0. Then, for
the sufficiently small > 0, we select u(ξ) = V (ξ) = ϕ(ξ)d−A+ (ξ)Ꮾ−κ (d(ξ)/R) and get
from (i) of Lemma 3.2,
I V
I V
p
0 < λ( f ) ≤ ≤
2
p
p
V f (ξ)dξ
BH (δ) ψ p (|V | h(ξ)/d (ln 1/d) )dξ
BH (δ)
κ(p − 1)/2 |A| p−2 J pκ−2 () + O (1)
I V
≤
p p
≤
2
Kδ J pκ−2 ()
Kδ BH (δ) ψ p V /d (ln 1/d) dξ
≤
c1− pκ
C
=
−→ 0,
cKδ 1− pκ Kδ
as δ, −→ 0,
(4.16)
where Kδ = inf ξ ∈BH (δ) h(ξ). The impossibility shows that (4.12) cannot hold for f (ξ) ∈
Ip.
1,p
Definition 4.5. Let λ ∈ R, u ∈ D0 (Ω), and u ≡ 0. Call that (λ,u) is a weak solution of
(1.2) if
Ω
∇H u
p−2 ∇H u, ∇H ϕ dξ − μ
ψ p p −2
|u| uϕ dξ = λ
p
Ωd
Ω
f (ξ)|u| p−2 uϕ dξ
(4.17)
for any ϕ ∈ C0∞ (Ω). In this case, we call that u is the eigenfunction of problem (1.2)
associated to the eigenvalue λ.
Theorem 4.6. Suppose that 1 < p < Q, 0 ≤ μ < ((Q − p)/ p) p , and f (ξ) ∈ Ᏺ p . The problem
1,p
(1.2) admits a positive weak solution u ∈ D0 (Ω), corresponding to the first eigenvalue λ =
λ1μ ( f ) > 0. Moreover, as μ increases to ((Q − p)/ p) p , λ1μ ( f )→λ( f ) ≥ 0 for f (ξ) ∈ Ᏺ p and
the limit λ( f ) > 0 for f (ξ) ∈ I p . If f (ξ) ∈ I p and ψ p−1 (ξ)d p (ξ) f (ξ)(ln 1/d(ξ))2 →∞, as
d(ξ)→0, then the limit λ( f ) = 0.
18
Journal of Inequalities and Applications
Proof. We define
Jμ (u) : =
Ω
∇H u p − μ
Ω
ψp
|u| p
dp
.
(4.18)
1,p
Obviously, Jμ is continuous and Gâteaux differentiable on D0 (Ω). By (2.1),
Jμ (u) ≥
Ω
∇H u p − μ
CQ,p
Ω
∇H u p = 1 − μ
∇H u
p
CQ,p
Ω
(4.19)
for 0 ≤ μ < ((Q − p)/ p) p . Note that CQ,p = ((Q − p)/ p) p for 1 < p < Q. Hence Jμ is coer1,p
1,p
cive
in D0 (Ω). We minimize the function Jμ (u) over the mainfold ᏹ = {u ∈ D0 (Ω) |
p
1
1 > 0 from Lemma 4.1.
Ω |u| f (ξ)dξ = 1} and let λμ be the infimum. It is clear that λ
μ
Now, we choose a special minimizing sequence {um } ⊂ ᏹ with Ω |um | p f (ξ)dξ = 1, and
−1,p
Jμ (um )→λ1μ and Jμ (um )→0 strongly in D0 (Ω), when the component of Jμ (um ) is restricted to ᏹ. The coercivity of Jμ implies that {um } is bounded and then there exists a
subsequence, still denoted by {um }, such that
1,p
um u weakly in D0 (Ω),
um u weakly in L p Ω,ψ p d− p ,
um −→ u
(4.20)
p
strongly in L (Ω),
1,p
as m→∞. By Theorem 2.4 in Section 2 we know that D0 (Ω) is compactly embedded
in L p (Ω, f dξ), and it follows that ᏹ is weakly closed and hence u ∈ ᏹ. Moreover, um
satisfies
p −2
μ p −2
−ΔH,p um − ψ p p um um = λm um um f + fm , in Ᏸ (Ω),
(4.21)
d
where fm →0 strongly in D−1,p (Ω) and λm →λ, as m→∞. Letting gm = ψ p (μ/d p )|um | p−2 um
+λm |um | p−2 um f , we check easily that gm is bounded in M(Ω) and conclude almost everywhere convergence of ∇H um to ∇H u in Ω by Lemma 4.3, and
p
p
Jμ (um ) = ∇H um L p (Ω) − μum L p (Ω,ψ p d− p )
p
p
p
p
= ∇H um − u L p (Ω) − μum − uL p (Ω,ψ p d− p ) + ∇H uL p (Ω) − μuL p (Ω,ψ p d− p ) +o(1)
p
1
≥ CQ,p − μ um − u p
− p + λ + o(1),
L (Ω,ψ p d
)
μ
(4.22)
by applying Lemma 4.2 to um and ∇H um , where o(1)→0 as m→∞. Thus CQ,p > μ, um −
p
p
uL p (Ω,ψ p d− p ) →0, and ∇H (um − u)L p (Ω) →0 as m→∞. It shows that Jμ (u) = λ1μ and λ = λ1μ .
Since Jμ (|u|) = Jμ (u), we can take u > 0 in Ω. By Lemma 4.3, u is a distribution solution
1,p
of (1.2) and since u ∈ D0 (Ω), it is a weak solution to eigenvalue problem (1.2) corresponding to λ = λ1μ . Moreover, if f (ξ) ∈ I p , then by Lemma 4.4,
λ1μ ( f ) −→ λ( f ) =
p
p p dξ
Ω ∇H u − CQ,p ψ p |u| /d
inf
> 0,
p
1,p
u∈D (Ω\{0})
Ω |u| f dξ
(4.23)
Jingbo Dou et al.
19
as μ increases to ((Q − p)/ p) p . When f (ξ) ∈ I p , using Lemma 4.4 again, it follows that
(4.12) is not true and hence λ( f ) = 0. This completes the proof.
1,p
Remark 4.7. The set ᏹ is a C 1 manifold in D0 (Ω). By Ljusternik-Schnirelman critical
point theory on C 1 manifold, there exists a sequence {λm } of eigenvalues of (1.2), that
is, writing Γm = {A ⊂ Ᏻ | A is symmetric, compact, and γ(A) ≥ m}, where γ(A) is the
Krasnoselski’s genus of A (see [19]), then for any integer m > 0,
λm = inf sup J μ (u)
(4.24)
A∈Γm u∈A
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 in Theorem 4.6. From the previous results we know
clearly that the first eigenvalue is
1,p Ω \ {0 } ,
λ1μ = λ1μ ( f ) = inf Jμ (u) | u ∈ D0
Ω
|u| p f (ξ)dξ = 1 .
(5.1)
In what follows we need the Picone identity proved in [15].
Proposition 5.1 (Picone identity). For differentiable functions u ≥ 0, v > 0 on Ω ⊂ Hn ,
with Ω a bounded or unbounded domain in Hn , then
L(u,v) = R(u,v) ≥ 0,
(5.2)
with
p
p −1 up ∇H v p − p u
∇H v p−2 ∇H u·∇H v,
p
p
−
1
v
v
p p p −2
u
R(u,v) = ∇H u − ∇H v
∇H p−1 ·∇H v
v
L(u,v) = ∇H u
+ (p − 1)
(5.3)
for p > 1. Moreover, L(u,v) = 0 a.e. on Ω if and only if ∇H (u/v) = 0 a.e. on Ω.
Theorem 5.2. (i) λ1μ is simple, that is, the positive eigenfunction corresponding to λ1μ is
unique up to a constant multiple.
(ii) λ1μ is unique, that is, if v ≥ 0 is an eigenfunction associated with an eigenvalue λ with
p
1
Ω f (ξ)|v | dξ = 1, then λ = λμ .
(iii) Every eigenfunction corresponding to the eigenvalue λ (0 < λ=λ1μ ) changes sign in Ω.
Proof. (i) Let u > 0 and v > 0 be two eigenfunctions corresponding to λ1μ in ᏹ. For sufficiently small ε > 0, set φ = u p /(v + ε) p−1 ∈ D0 (Ω). Then
1,p
Ω
∇H v p−2 ∇H v, ∇H φ dξ = μ
Ω
ψp
u p −1
φdξ + λ1μ
dp
Ω
f (ξ)v p−1 φdξ.
(5.4)
20
Journal of Inequalities and Applications
Using Proposition 5.1 and (5.4),
0≤
=
Ω
Ω
L(u,v + ε) =
∇H u
p dξ −
=
Ω
Ω
R(u,v + ε)
Ω
∇ H v p −2 ∇ H
μ
=
Ω
ψp
+ λ1μ f (ξ) u p dξ −
dp
μ
Ω
up
, ∇H v dξ
(v + ε) p−1
μ
ψp
up
+ λ1μ f (ξ) v p−1
dξ
p
d
(v + ε) p−1
(5.5)
ψp
v p −1
1
p
+
λ
f
(ξ)
u
1
−
dξ.
μ
dp
(v + ε) p−1
The right-hand side of (5.5) tends to zero when ε→0. It follows that L(u,v) = 0 and by
Proposition 5.1 there exists a constant c such that u = cv.
(ii) Let u > 0 and v > 0 be eigenfunctions corresponding to λ1μ and λ, respectively. Similarly to (5.5), we have
Ω
μ
ψp
+ λ1μ f (ξ) u p dξ −
dp
μ
Ω
ψp
up
p −1
+
λ
f
(ξ)
v
dξ
dp
(v + ε) p−1
v p −1
ψp
=
μ p 1−
u p dξ +
Ω d
(v + ε) p−1
Ω
p
f (ξ)u
λ1μ − λ
v p −1
dξ ≥ 0.
(v + ε) p−1
(5.6)
Letting ε→0 shows that λ1μ − λ Ω f (ξ)u p dξ ≥ 0, which is impossible for λ > λ1μ . Hence
λ = λ1μ .
(iii) With the same treatment as in (5.6) we get
λ1μ − λ
Noting that
in Ω.
Ω f (ξ)u
p dξ
Ω
f (ξ)u p dξ ≥ 0.
(5.7)
> 0 and λ > λ1μ leads to a contradiction. So v must change sign
1,p
Lemma 5.3. If u ∈ D0 (Ω) is a nonnegative weak solution of (1.2), then either u(ξ) ≡ 0 or
u(ξ) > 0 for all ξ ∈ Ω.
1,p
Proof. For any R > r, BH (0,R) ⊃ BH (0,r), let u ∈ D0 (Ω) be a nonnegative weak solution
of (1.2). In virtue of Harnack’s inequality (see [1]), there exists a constant CR > 0 such that
sup u(ξ) ≤ CR inf
BH (0,R)
This implies u ≡ 0 or u > 0 in Ω.
BH (0,R)
u(ξ) .
(5.8)
Theorem 5.4. Every eigenfunction u1 corresponding to λ1μ does not change sign in Ω: either
u1 > 0 or u1 < 0.
Proof. From the proof of existence of the first eigenvalue we see that there exists a positive
eigenfunction, that is, if v is an eigenfunction, then u1 = |v| is a solution of the minimization problem and also an eigenfunction. Thus, from Lemma 5.3 it follows that |v| > 0 and
then u1 has constant sign.
Jingbo Dou et al.
21
1,p
Lemma 5.5. For u ∈ C(Ω \ {0}) ∩ D0 (Ω), let ᏺ be a component of {ξ ∈ Ω | u(ξ) > 0}.
1,p
Then u|ᏺ ∈ D0 (ᏺ).
1,p
1,p
Proof. Let um ∈ C(Ω \ {0}) ∩ D0 (Ω) be such that um →u in D0 (Ω). Therefore, u+m →u+
1,p
in D0 (Ω). Set vm (ξ) = min{um (ξ),u(ξ)}, and let ϕR (ξ) ∈ C(Ω) be a cutoff function such
that
⎧
⎪
⎪
⎨0
ϕR (ξ) = ⎪
⎪
⎩1
if d(ξ) ≤
R
,
2
(5.9)
if d(ξ) ≥ R,
with |∇H ϕR | ≤ C |∇H d|/d(ξ), for some positive constant C. Now, consider the sequence
ωm (ξ) = ϕR (ξ)vm (ξ)|ᏺ . Since ϕR (ξ)vm (ξ) ∈ C(Ω), we claim that ωm ∈ C(ᏺ) and ωm = 0
on the boundary ∂ᏺ. In fact, if ξ ∈ ∂ᏺ and ξ = 0, then ϕR = 0, and so ωm = 0. If ξ ∈
∂ᏺ ∩ Ω and ξ =0, then u = 0 (since u is continuous except at {0}), and hence vm = 0. If
1,p
ξ ∈ ∂Ω, then um = 0 and so vm = 0. Therefore, ωm = 0 on ∂ᏺ, and ωm ∈ D0 (ᏺ). Noting
Ω
∇H ωm − ∇H ϕR u p dξ =
ᏺ
ϕR ∇H vm + vm ∇H ϕR − ϕR ∇H u − u∇H ϕR p dξ
p
≤ ϕR ∇H vm − ∇H u p
L (ᏺ) +
∇H ϕR vm − u pp ,
L (ᏺ)
(5.10)
it is obvious that
By (2.1),
ᏺ
Ω |∇H ωm − ∇H (ϕR u)|
p
1,p
dξ →0, as m→∞. That is ωm →ϕR u|ᏺ in D0 (ᏺ).
u∇H ϕR + ϕR ∇H u − ∇H u
p dξ
≤
ᏺ
≤
ᏺ
≤
ᏺ
ϕR ∇H u − ∇H u
p dξ +
ᏺ∩{R/2<d<R}
ϕR ∇H u − ∇H u
p dξ + C p
ϕR ∇H u − ∇H u
p dξ + C1
u∇H ϕR p dξ
ᏺ∩{R/2<d<R}
ᏺ∩{R/2<d<R}
1,p
which approaches 0, as R→0. Hence u|ᏺ ∈ D0 (ᏺ).
ψp
|u| p
dp
(5.11)
dξ
∇H u
p dξ,
λ1μ
Theorem 5.6. The eigenvalue is isolated in the spectrum, that is, there exists δ > 0 such
that there is no other eigenvalues of (1.2) in the interval (λ1μ ,λ1μ + δ). Moreover, if v is an
eigenfunction corresponding to the eigenvalue λ=λ1μ and ᏺ is a nodal domain of v, then
Cλ f L∞
−Q/ p
where C is a constant depending only on Q and p.
≤ |ᏺ|,
(5.12)
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Journal of Inequalities and Applications
Proof. Let u1 be the eigenfunction corresponding to the eigenvalue λ1μ . Let {λm } be a
sequence of eigenvalues such that λm > λ1μ and λm λ1μ , and the corresponding eigenfunc
tions um →u1 with Ω f (ξ)|um | p dξ = 1, that is, λm and um satisfy
p −2
L p,μ um = λm f (ξ)
um Since
0<
Ω
∇H um p dξ − μ
ψp um p dξ = λm
p
Ωd
um .
(5.13)
Ω
p
f (ξ)
um dξ = λm ,
(5.14)
it follows that um is bounded. By Lemma 4.3, there exists a subsequence (still denoted
1,p
by {um }) of {um } such that um u weakly in D0 (Ω), um →u strongly in L p (Ω) and
∇H um →∇H u a.e in Ω. Letting m→∞ in (5.13) yields
L p,μ u = λ1μ f (ξ)|u| p−2 u.
(5.15)
Therefore, u = ±u1 . Using (iii) of Theorem 5.2 we see that um changes sign. For convenience, we assume that u = +u1 . Then
ξ ∈ Ω | um < 0 −→ 0.
Now, we check (5.13) with um = u−m ,
Ω
∇H u− p dξ − μ
m
ψp u− p dξ = λm
m
p
Ωd
(5.16)
Ω
p
f (ξ)
u−m dξ.
(5.17)
Using the Hardy inequality and Sobolev inequality yields
1−
μ
CQ,p
Ω−
≤
Ω−
= λm
∇H um p dξ
∇H um p dξ − μ
Ω−
p
Ω−
ψp um p dξ
dp
f (ξ)
um dξ
≤ λ m f L ∞
Ω−
(5.18)
p
um dξ
p p/Q
≤ c1 f L∞ um 1,p Ω− ,
−
Ω ≥ c2 f L∞ Q/ p ,
m
D
m
−
Ωm = ξ ∈ Ω | u m < 0 .
It contradicts with (5.16). Hence, there is no other eigenvalue of (1.2) in (λ1μ ,λ1μ + δ) for
δ > 0.
Next, we prove (5.12). Assume v > 0 in ᏺ (the case v < 0 being treated similarly). In
1,p
view of Lemma 5.5, we have v|ᏺ ∈ D0 (ᏺ). Define the function
⎧
⎨v(ξ)
ω(ξ) = ⎩
0
if ξ ∈ ᏺ,
if ξ ∈ Ω \ ᏺ.
(5.19)
Jingbo Dou et al.
23
1,p
Clearly, ω(ξ) ∈ D0 (Ω). Taking ω as a test function in (4.17) satisfied by v and arguing as
in (5.18), we have
μ
p
1−
v D1,p (ᏺ) ≤ λ f L∞
CQ,p
p
ᏺ
|v | p dξ ≤ λC f L∞ v D1,p (ᏺ) |ᏺ| p/Q
(5.20)
for some constant C = C(Q, p) and hence (5.12) holds.
Corollary 5.7. Each eigenfunction has a finite number of nodal domains.
Proof. Let ᏺ j be a nodal domain of an eigenfunction associated to some positive eigenvalue λ. It follows from (5.12) that
|Ω| ≥
ᏺ j ≥ Cλ f L∞ −Q/ p 1.
j
The result is proved.
(5.21)
j
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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Jingbo Dou: Department of Applied Mathematics, Northwestern Polytechnical University,
Xi’an, Shaanxi 710072, China
Email address: djbmn@126.com
Pengcheng Niu: Department of Applied Mathematics, Northwestern Polytechnical University,
Xi’an, Shaanxi 710072, China
Email address: pengchengniu@nwpu.edu.cn
Zixia Yuan: Department of Applied Mathematics, Northwestern Polytechnical University,
Xi’an, Shaanxi 710072, China
Email address: yzx8047@yahoo.com.cn