A NONLINEAR LIOUVILLE THEOREM FOR
FRACTIONAL EQUATIONS IN THE HEISENBERG
GROUP
ELEONORA CINTI AND JINGGANG TAN
Abstract. We establish a Liouville-type theorem for a subcritical nonlinear problem, involving a fractional power of the subLaplacian in the Heisenberg group. To prove our result we will use
the local realization of fractional CR covariant operators, which
can be constructed as the Dirichlet-to-Neumann operator of a degenerate elliptic equation in the spirit of Caffarelli and Silvestre
[4], as established in [9]. The main tools in our proof are the CR
inversion and the moving plane method, applied to the solution of
the lifted problem in the half-space Hn × R+ .
1. Introduction and Main Results
In this paper we establish a Liouville-type result for the following
fractional nonlinear problem in the Heisenberg group:
P 1 u = up
2
in Hn .
(1.1)
Here P 1 denotes the CR square root of the Heisenberg Laplacian −∆H ,
2
which is a CR covariant operator of order 1/2 in Hn . In [9] Frank, Gonzalez, Monticelli and one of the author, study CR covariant operators
of fractional orders on orientable and strictly pseudoconvex CR manifolds. In this context, the Heisenberg group Hn plays the same role as
Rn in conformal geometry.
Given a Kähler-Einstein manifold X , CR covariant operators of fractional order γ are pseudodifferential operators whose principal symbol
agrees with the pure fractional powers of the CR sub-Laplacian on the
boundary M = ∂X . They can be defined using scattering theory, as
done in [7, 15, 14, 13].
One of the main results in [9] establishes that it is possible to characterize fractional CR covariant operators on some CR manifold M =
2010 Mathematics Subject Classification. Primary: 35A01,35B50,35J70. Secondary: 35B53,35J50.
Key words and phrases. Fractional sublaplacian, Heisenberg group, Louville theorem, moving plane method .
1
2
ELEONORA CINTI AND JINGGANG TAN
∂X , as the Dirichlet-to-Neumann map of a degenerate elliptic equation
in the interior of X , in the spirit of Caffarelli-Silvestre result [4].
In [9], in order to construct fractional CR covariants operators in
the specific case of the Heisenberg group, Hn is identified with the
boundary of the Siegel domain in R2n+2 (see section 2 for the precise
definition) and it is crucial to use its underlying complex hyperbolic
structure.
Another possible approach in the construction of fractional powers
of the sub-Laplacian consists in using purely functional analytic tools
as done in [8]. In [8] Ferrari and Franchi proved an extension result
for fractional operators defined by using the spectral resolution of the
sub-Laplacian in general Carnot groups. The operators considered in
[8] are different in nature from the ones in [9], since they are not CR
covariant.
In this paper, the CR covariance of P 1 will be crucial, this is the
2
reason why the operator that we consider is the one constructed in [9].
Since it will be of utmost importance in the sequel, we recall here the
extension result proven in [9].
Theorem 1.1 (see Theorem 1.1 in [9]). Let γ ∈ (0, 1), a = 1 − 2γ. For
each u ∈ C ∞ (Hn ), there exists a unique solution U for the extension
problem
 2
2
a ∂U
∂ U
2∂ U
+
+
λ
+ ∆H U = 0 in Hn × R+ ,
(1.2)
∂λ2
λ ∂λ
∂t2
U = u
n
on H × {λ = 0}.
Moreover,
∂U
,
λ→0
∂λ
where cγ is a constant depending only on γ.
Pγ u = −cγ lim λa
Observe that, differently from the extension result established in
2
[8], here we have the additional term λ2 ∂∂tU2 which appears when one
considers CR sub-Laplacians.
Our Liouville-type theorem is the analogue, for the fractional operator P 1 , of a result by Birindelli and Prajapat [2], for the sublaplacian
2
∆H . In [2], the authors establish a nonexistence result for a class of
positive solution of the equation
− ∆H u = up ,
Q+2
,
Q−2
(1.3)
for p subcritical (i.e. 0 < p <
where Q = 2n + 2 denotes the
homogeneous dimension of Hn ). The technique they used is based on
the moving plane method (which goes back to Alexandrov and Serrin
NONLOCAL LIOUVILLE THEOREM
3
[20]), adapted to the Heisenberg group setting. This method requires
two basic tools: the maximum principle and invariance under reflection
with respect to a hyperplane. Since the operator −∆H is not invariant
under the usual reflection with respect to hyperplanes, Birindelli and
Prajapat needed to introduce a new reflection, called H-reflection, under which −∆H is invariant. Since it will be important in the sequel,
we recall here the definition of H-reflection.
Definition 1.2. For any ξ = (x, y, t) ∈ Hn , we consider the plane
Tµ := {ξ ∈ Hn : t = µ}. We define
ξµ := (y, x, 2µ − t),
to be the H-reflection of ξ with respect to the plane Tµ .
Due to the use of this reflection, the proof of the non existence result
in [2] requires the solution u of (1.3) to be cylindrical, that is u(x, y, t) =
1
u(r, t) must depend only on r and t where r = (x2 + y 2 ) 2 .
We can now state our main result, which is the analogue for the
operator P 1 of the Liouville result contained in [2].
2
Theorem 1.3. Let 0 < p < Q+1
, where Q = 2n+2 is the homogeneous
Q−1
n
dimension of H . Then there exists no bounded cylindrical solution
u ∈ C 2 (Hn ) of
(
P 1 u = up in Hn
2
.
(1.4)
u>0
in Hn
Using the local formulation (1.2) established in [9], the above theorem will follow as a corollary of the following Liouville-type result for
a nonlinear Neumann problem in the half-space Hn × R+ .
Q+1
Theorem 1.4. Let 0 < p < Q−1
and let U ∈ C 2 (Hn ×R+ )∩C(Hn × R+ )
be a nonnegative solution of
 2
2

 ∂ U + λ2 ∂ U + ∆H U = 0 in Hn × R+
∂λ2
∂t2
(1.5)
∂U

p
n
−
=U
on H × {λ = 0}.
∂λ
Suppose that U (x, y, t, λ) = U (r, t, λ) depends only on r, t, λ, where
1
r = (|x|2 + |y|2 ) 2 . Then U ≡ 0.
In the Euclidean case, classical nonexistence results for subcritical
nonlinear problems in the all space Rn are contained in two works by
Gidas and Spruck [12] and by Chen and Li [5]. Analogue results for
nonlinear Neumann problems in the half-space Rn+ where established
in [17, 18], using the methods of moving planes and moving spheres.
4
ELEONORA CINTI AND JINGGANG TAN
In the Heisenberg group setting there are several papers concerning
nonexistence results for problem (1.3). In [10] Garofalo and Lanconelli
proved some nonexistence results for positive solutions of (1.3) when p
is subcritical, but their results require some integrability conditions on
u and ∇u. In [16, 21] similar nonexistence results for positive solutions
of (1.3) in the half-space are established for the critical exponent p =
Q+2
. In [3], a Liouville-type result for solution of (1.3) is proved without
Q−2
Q
requiring any decay condition on u, but only for 0 < p < Q−2
. As
explained before, in [2] Birindelli and Prajapat extends this last result
but only in the class of cylindrical solution. A
to any 0 < p < Q+2
Q−2
last more recent result in this context was proven by Xu in [22], who
established that there are no positive solution of (1.3) for 0 < p <
Q(Q+2)
. This result uses a different technique, based on the vector filed
(Q−1)2
method, and improves the results contained in [10] and [3], since it does
not require any decay on the solution u and it improves the exponent p.
Nevertheless it does not let to reach the optimal exponent Q+2
(observe
Q−2
Q
< Q(Q+2)
< Q+2
).
that Q−2
(Q−1)2
Q−2
In this paper we aim to establish a first Liouville-type result for
a CR fractional power of −∆H ; this is, to our knowledge, the first
nonexistence result in this fractional setting.
Let us comment now on the basic tools in the proof of our main
result. Following [2], in order to get a nonexistence result, we combine
the method of moving planes with the CR inversion of the solution u.
The CR inversion was introduced by Jerison and Lee in [19], and it is
the analogue of the Kelvin transform, in the Heisenberg group context.
In Section 3 we will give the precise definition of CR inversion and we
will show which problem is satisfied by the CR inversion of a solution
of (1.5).
As said before, the moving plane method is based on several version
of the maximum principles. More precisely we will need to prove that
our operator satisfies a weak maximum principle (see Proposition 4.2)
and two versions of the Hopf’s Lemma (see Propositions 4.4 and 4.6).
The paper is organized as follows:
• in Section 2 we recall some basic facts on the Heisenberg group
and we will introduce the fractional CR operator P 1 ;
2
• in Section 3 we will introduce the CR inversion of a function u
and prove a lemma concerning the CR inversion of a solution
of our problem (1.5);
NONLOCAL LIOUVILLE THEOREM
5
• in Section 4 we establish Maximum principles and Hopf’s Lemma
for our operator, which will be basic tools in the method of moving plane;
• in Section 5 we will prove our main result (Theorem 1.4).
2. Preliminary facts on the Heisenberg group
In this section we recall some basic notions and properties concerning
the Heisenberg group.
We will denote the points in Hn using the notation ξ = (x, y, t) =
(x1 , ..., xn , y1 , ..., yn , t) ∈ Rn × Rn × R. The Heisenberg group Hn is
the space R2n+1 endowed with the group law ◦ defined in the following
way:
n
X
ˆ
ξ ◦ ξ := (x̂ + x, ŷ + y, t̂ + t + 2
(xi ŷi − yi x̂i )).
i=1
The natural dilation of the group is given by δ` (ξ) := (`x, `y, `2 t), and
ˆ ◦ δ` (ξ).
it satisfies δ` (ξˆ ◦ ξ) = δ` (ξ)
n
In H we will consider the gauge norm defined as

 41
!2
n
X
|ξ|H := 
(x2i + yi2 ) + t2  ,
i=1
which is homogeneous of degree one with respect to δ` . Using this
norm, one can define the distance between two points in the natural
way:
ˆ ξ) = |ξˆ−1 ◦ ξ|H ,
dH (ξ,
where ξˆ−1 denotes the inverse of ξˆ with respect to the group action.
We denote the ball associated to the gauge distance by
BH (ξ0 , R) := {ξ ∈ Hn : dH (ξ, ξ0 ) < R}.
Denoting by |A| the Lebesgue measure of the set A, we have that
|BH (ξ0 , R)| = |BH (0, R)| = RQ |BH (0, 1)|.
Here Q = 2n + 2 denotes the homogeneous dimension of Hn .
For every i = 1, · · · , n, we denote by Xi , Yi , and T the following
vector fields:
∂
∂
∂
∂
∂
Xi =
+ 2yi , Yi =
− 2xi , T = .
∂xi
∂t
∂yi
∂t
∂t
They form a basis of the Lie Algebra of left invariant vector fields.
Moreover, an easy computions shows that [Xi , Yj ] = −4δij T . The
6
ELEONORA CINTI AND JINGGANG TAN
Heisenberg gradient of a function f is given by
∇H f = (X1 f, ..., Xn f, Y1 f, ..., Yn f ).
Finally, we define the sublaplacian as
∆H :=
n
X
(Xi2 + Yi2 ).
i=1
It can be written also in the form ∆H = div(A∇T ), where A = aij is the
(2n+1)×(2n+1) symmetric matrix given by aij = δij for i, j = 1, ..., 2n,
aj(2n+1) = (2n + 1)j = 2yj for j = 1, ..., n, aj(2n+1) = (2n + 1)j = −2xj
for j = n + 1, ..., 2n and a(2n+2)(2n+2) = 4(|x|2 + |y|2 ). It is easy to
observe that A is positive semidefinite for any (x, y, t) ∈ Hn . This
operator is degenerate elliptic, and it is hypoelliptic since it satisfies
the Hormander condition.
We pass now to describe CR covariant operators of fractional orders
in Hn . For a rigourous geometric description of CR covariant operators
on CR manifold, we remind to [9] and references therein. Here we just
consider the case of the Heisenberg group, since it is the one of interest.
Introducing complex coordinate ζ = x + iy ∈ Cn , we can identify the
Heisenberg group Hn with the boundary of the Siegel domain Ωn+1 ⊂
Cn+1 , which is given by
Ωn+1 := {(ζ1 , . . . , ζn+1 ) = (ζ, ζn+1 ) ∈ Cn × C | q(ζ, ζn+1 ) > 0} ,
with
q(ζ, ζn+1 ) = Im ζn+1 −
n
X
|ζj |2 ,
j=1
through the map (ζ, t) ∈ Hn → (ζ, t + i|ζ|2 ) ∈ ∂Ωn+1 . It is possible
to see that X = Ωn+1 is a Kähler-Einstein manifold, endowed with a
Kähler metric g+ , which can be identified with the complex hyperbolic
space. The boundary manifold M = ∂Ωn+1 inherits a natural CR
structure from the complex structure of the ambient manifold.
Scattering theory tells us that for s ∈ C, Re(s) ≥ m2 , and except
for a set of exceptional values, given f smooth on M, the eigenvalue
equation
−∆g+ u − s(m − s)u = 0, in X
has a solution u with the expansion
(
u = q (m−s) F + q s G for some F, G ∈ C ∞ (X ),
F |M = f.
NONLOCAL LIOUVILLE THEOREM
7
The scattering operator is defined as
S(s) : C ∞ (M) → C ∞ (M)
by
S(s)f := G|M .
For γ ∈ (0, m)\N, we set s = m+γ
. The conformal fractional sub2
Laplacian on Hn is defined in the following way:
Pγ f = cγ S(s)f,
(2.1)
for a constant
cγ = 2γ
Γ(γ)
.
Γ(−γ)
For γ = 1 and γ = 2 we have:
P1 = ∆H
and P2 = ∆2H + T 2 .
A crucial property of Pγ is its conformal covariance.
As explained in the introduction, one of the main result in [9], is the
characterization of these fractional operators via the extension problem
(1.2). Since, throughout this paper, we will work on this lifted problem
in the halfspace, let us introduce some notations in Hn × R+ .
We will denote by dλ the distance between two points in Hn × R+ .
More precisely, for any z = (x1 , ..., xn , y1 , ..., yn , t, λ) and any ẑ =
(x̂1 , · · · , x̂n , ŷ1 , ..., ŷn , t̂, λ̂), let us define
n
h X
i 41
2
dλ (z, ẑ) =
(|xi − x̂i |2 + |yi − ŷi |2 ) + |λ − λ̂|2 + |t − t̂|2 .
i=1
Observe that when λ = λ̂ = 0, that is z and ẑ lies on Hn , dλ (z, ẑ) =
dH (z, ẑ). Moreover, given z0 ∈ Hn × {0}, we set
B + (z0 , R) = {z ∈ Hn × R+ | dλ (z, z0 ) < R , λ > 0}.
The operator of our interest will be
2
∂2
2 ∂
+
λ
,
∂λ2
∂t2
which, writing explicitly all the terms, becomes
n 2
X
∂2
∂
∂2
∂2
∂2
L=
+
+ 2 + 4yj
− 4xj
∂λ2 j=1 ∂x2j
∂yj
∂xj ∂t
∂yj ∂t
L = ∆H +
n
X
∂2
(x2j + yj2 )) 2 .
+ 4(λ +
∂t
j=1
2
8
ELEONORA CINTI AND JINGGANG TAN
Also in this case, we can write L = div(A∇T ), where now A is the
(2n+2)×(2n+2) symmetric matrix given by aij = δij if i, j = 1, · · · , 2n,
aj(2n+1) = a(2n+1)j = 2yj if j = 1, · · · , n, aj(2n+1) = a(2n+1)j = −2xj if
j = n + 1, · · · , 2n, a(2n+1)(2n+1) = 4(|x|2 + |y|2 + λ2 ), a(2n+2)(2n+2) = 1,
aj(2n+2) = a(2n+2)j = 0 if i = 1, · · · , n.
In the sequel it will be useful to express L for cylindrical and radial
functions. Let r = (|x|2 + |y|2 + λ2 )1/2 , ρ = (r4 + t2 )1/4 . Suppose
that Ψ is a radial function, that is, Ψ depends only on ρ; then a direct
computations gives:
∂Ψ ∂ρ
∂Ψ
∂Ψ
=
= ρ−3 r2 xj
,
∂xj
∂ρ ∂xj
∂ρ
∂Ψ
∂Ψ ∂ρ
∂Ψ
=
= ρ−3 r2 λ
.
∂λ
∂ρ ∂λ
∂ρ
∂Ψ
∂Ψ ∂ρ
∂Ψ
=
= ρ−3 r2 yj
,
∂yj
∂ρ ∂yj
∂ρ
Then we deduce that
r4 x2j ∂ 2 Ψ ρ4 r2 + 2ρ2 x2j − 3r2 x2j ∂Ψ
∂ 2Ψ
+
,
= 6
∂x2j
ρ ∂ρ2
ρ7
∂ρ
r4 yj2 ∂ 2 Ψ ρ4 r2 + 2ρ2 yj2 − 3r2 yj2 ∂Ψ
∂ 2Ψ
+
,
= 6
∂yj2
ρ ∂ρ2
ρ7
∂ρ
t2 ∂ 2 Ψ 2ρ4 − 3t2 ∂Ψ
∂ 2Ψ
=
+
.
∂t2
4ρ6 ∂ρ2
4ρ7
∂ρ
Hence, by using that
n
1 X 4 2
r2
4 2
4 2
2 2
(
(r
x
+
r
y
)
+
r
λ
+
r
t
)
=
j
j
ρ6 j=1
ρ2
and
X 2
1
4 2
4
4
4
2
(2n
+
1)ρ
r
+
2ρ
−
3r
+
(2ρ
−
3t
)
( (xj + yj2 ) + λ2 )
7
ρ
j=1
=
Qr2
1
4 2
4 2
6
2
4
2
[(2n
+
1)ρ
r
+
2ρ
r
−
3r
+
r
(2ρ
−
3t
)]
=
,
ρ7
ρ3
we conclude that
r2
LΨ(ρ) = 2
ρ
∂ 2 Ψ(ρ) Q ∂Ψ(ρ)
+
∂ρ2
ρ ∂ρ
.
(2.2)
In a similar way, we deduce that for cylindrical symmetric functions
φ = φ(r, t), we have
Lφ =
2
∂ 2 φ Q − 2 ∂φ
2∂ φ
+
+
4r
.
∂r2
r ∂r
∂t2
NONLOCAL LIOUVILLE THEOREM
9
Using the radial form (2.2) for L, an easy computation yelds the following lemma.
1
1
Lemma 2.1. For ρ 6= 0, let ψ(ρ) = ρQ−1
= ρ2n+1
. Then we have that
(
Lψ(ρ) = 0
∈ Hn × R+ \ {0}
(2.3)
∂
ψ(ρ) = 0 on Hn \ {0} × {λ = 0}.
− ∂λ
3. CR inversion
Following [2] and [19], we define the CR inversion in the half-space
Hn × R+ .
1
For any (x, y, t, λ) ∈ Hn × R+ , let as before r = (|x|2 + |y|2 + λ2 ) 2
1
and ρ = (r4 + t2 ) 4 . We set
x
ei =
x i t + yi r 2
y i t − xi r 2 e
t e
λ
,
y
e
=
, t = − 4, λ
= 2.
i
4
4
ρ
ρ
ρ
ρ
The CR inversion of a function U defined on Hn × R+ , is given by
1
v(x, y, t, λ) = Q−1 U (x̃, ỹ, t̃, λ̃).
ρ
The following lemma shows which equation is satisfied by the CR inversion of a solution of problem (1.5).
Lemma 3.1. Suppose that U ∈ C 2 (Hn × R+ \ {0}) ∩ C(Hn × R+ \ {0})
is a solution of (1.5). Then the CR inversion v of U satisfies

Lv = 0
in Hn × R+ \ {0}
(3.1)
∂v
−
= ρp(Q−1)−(Q+1) v on Hn \ {0} × {λ = 0}.
∂λ
Proof. Since it is just a long computation, we will give the details of
the proof for cylindrical solutions U (r, t), but the statement holds true
for any solution.
It is clear that
r̃2 =
|x̃|2 + |ỹ|2 + λ̃2
t2 (|x|2 + |y|2 + λ2 ) r4 (|x|2 + |y|2 + λ2 )
r2
=
+
=
.
ρ8
ρ8
ρ8
ρ4
The following relations are useful
∂r̃
t2 − r6 ∂r̃
−rt
=
,
= 6 ,
6
∂r
ρ
∂t
ρ
3
2
∂ t̃
4tr ∂ t̃
2t − t2 − r4
t2 − r 4
= 8 ,
=
=
.
∂r
ρ
∂t
ρ8
ρ8
10
ELEONORA CINTI AND JINGGANG TAN
Now we have
∂v
(1 − Q)r3
4tr3 ∂U
1 t2 − r4 ∂U
=
+
]
U
+
[
∂r
ρQ+3
ρQ−1 ρ6 ∂r̃
ρ8 ∂ t̃
and
∂ 2v
2(1 − Q)r3 t2 − r4 ∂U
4r3 t ∂U
∂ (1 − Q)r3 U+
+( 8 )
]
=
[(
)
∂r2
∂r
ρQ+3
ρQ+3
ρ6
∂r̃
ρ
∂ t̃
1 ∂ t2 − r4 ∂U
∂ 4r3 t ∂U
]
+ Q−1 [ (
+
(
)
)
ρ
∂r
ρ6
∂r̃
∂r ρ8 ∂ t̃
1 h t2 − r4 2 ∂ 2 U
4r3 t 2 ∂U i
t2 − r4 4r3 t ∂ 2 U
+ Q−1 (
+
(
)
+
(
)(
)
)
ρ
ρ6
∂r̃2
ρ6
ρ8 ∂r̃∂ t̃
ρ8 ∂ t̃2
1 ∂ t2 − r4 ∂U
∂ 4r3 t ∂U
+ Q−1 [ (
)
+
(
)
].
ρ
∂r
ρ6
∂r̃
∂r ρ6 ∂ t̃
The derivative with respect to t are given by
(1 − Q)t
1 −rt ∂U
t2 − r4 ∂U
∂v
=
U
+
[
+
]
∂t
2ρQ+3
ρQ−1 ρ6 ∂r̃
ρ8 ∂ t̃
and
∂ (1 − Q)t t2 − r4 ∂U
2(1 − Q)t −rt ∂U
∂ 2v
=
[(
)
+
(
)
]
U
+
+
∂t2
∂r 2ρQ+3
2ρQ+3
ρ6 ∂r̃
ρ8
∂ t̃
1 ∂ −rt ∂U
∂ t2 − r4 ∂U
+ Q−1 [ ( 6 )
+ (
)
]
ρ
∂r ρ ∂r̃
∂r
ρ8
∂ t̃
1 h −rt 2 ∂ 2 U
−rt t2 − r4 ∂ 2 U
t2 − r4 2 ∂U i
+ Q−1 ( 6 )
+ ( 6 )(
)
+(
) 2
ρ
ρ
∂r̃2
ρ
ρ8
ρ8
∂r̃∂ t̃
∂ t̃
2
4
1 ∂ −rt ∂U
∂ t − r ∂U
+ Q−2 [ ( 6 )
+ (
)
].
ρ
∂t ρ ∂r̃
∂t
ρ6
∂ t̃
Let us denote
∂ 2U
∂ 2U
∂ 2U
∂U
∂U
+ b2
LU = a1 2 + a2
+ a3 2 + b 1
.
∂r̃
∂r̃
∂r̃∂ t̃
∂ t̃
∂ t̃
Then
a1 =
a2 =
a3 =
1
1
ρ
[(t2 − r4 )2 + 4r2 (rt)2 ] =
Q+11
1
ρQ+13
1
ρ
ρQ+3
,
[2(t2 − r4 )(4r3 t) − 8r2 (rt)(t2 − r4 )2 ] = 0,
[16r6 t3 + 4r2 (t2 − r4 )2 ] =
Q+13
1 4r2
.
ρQ+3 ρ4
NONLOCAL LIOUVILLE THEOREM
11
By using that
∂ t2 − r4
−4r3 (t2 + r4 ) −6(t2 − r4 )r3
(
)
=
−
,
∂r
ρ6
ρ10
ρ10
∂ −rt
r(t2 + r4 ) 3rt2
(
)=−
+ 10 ,
∂t ρ6
ρ10
ρ
we can see that
2(1 − Q)r4 (t2 − r4 )
1 ∂ t2 − r4
(
+
)
rρQ+9
ρQ−1 ∂r
ρ6
4r2 ∂ −rt
Q − 2 (t2 − r4 )(t2 + r4 ) 4r2 (1 − Q)(−r2 t2 )
+
+
(
)
+
r
ρQ+9
rρQ+9
ρQ−1 ∂t ρ6
2(1 − Q)r4 (t2 − r4 )
−4r4 (t2 + r4 ) −6(t2 − r4 )r4
=
+
(
−
)
rρQ+9
rρQ+9
rρQ+9
Q − 2 (t2 − r4 )(t2 + r4 ) 4(1 − Q)(−r4 t2 )
+
+
r
ρQ+9
rρQ+9
4 2
4
4 2
4r (t + r ) 12r t
−
+ Q+9
rρQ+9
rρ
2
1 (Q − 2)
1 (Q − 2)ρ
= Q+3
.
= Q+3
ρ
r
ρ
r̃
b1 =
We have that
12r2 t(t2 + r4 ) 32r6 t
∂ 4r3 t
( 8 )=
− 12 ,
∂r ρ
ρ12
ρ
∂ t2 − r4
2t(t2 + r4 ) 4t(t2 − r4 )
(
)
=
−
.
∂t
ρ8
ρ12
ρ12
Then we see that
8(1 − Q)r6 t
1 ∂ 4r3 t
4(Q − 2)r3 t(t2 + r4 )
+
(
)
+
ρQ+11
ρQ−1 ∂r ρ8
rρQ+11
4r2 ∂ t2 − r4
4(1 − Q)r2 t(t2 − r4 )
+
+
(
)
ρQ+11
ρQ−1 ∂t
ρ8
8(1 − Q)r6 t
12r2 t(t2 + r4 )
32r3 t
4(Q − 2)r2 t(t2 + r4 )
=
+
(
−
)
+
ρQ+11
ρQ+11
ρQ+11
ρQ+11
4(1 − Q)r2 t(t2 − r4 )
8r2 t(t2 + r4 ) 16r2 t(t2 − r4 )
+
+
(
−
) = 0.
ρQ+11
ρQ+11
ρQ+11
b2 =
12
ELEONORA CINTI AND JINGGANG TAN
1
e = 0. Finally, the
This shows that Lv(x, y, t, λ) = ρ(Q+3)
LU (e
x, ye, e
t, λ)
Neumann data becomes:
1
∂U
∂v
e
(x, y, t, λ) = − lim Q−1 ρ2
(e
x, ye, e
t, λ)
− lim
e
λ→0 ∂λ
e
ρ
λ→0
∂λ
1
= (Q+1) U p ((e
x, ye, e
t, 0) = ρp(Q−1)−(Q+1) v(x, y, t, 0).
ρ
4. Maximum principles and Hopf’s Lemma
Basic tools in the method of moving planes are the maximum principle and Hopf’s Lemma. In this section we will prove that our operator
L satisfies both of them.
We start by stating the maximum principle for the classical solution
of the problem related to L.
Proposition 4.1. Let D be a bounded domain in Cn+1 , and let b ∈
C(D) be a nonnegative function. Assume that v ∈ C 2 (D) ∩ C 1 (D) is a
solution of
(
−Lv + bv ≥ 0 in D,
(4.1)
v≥0
on D.
Then v ≥ 0 in D.
Proof. Suppose that ξ0 ∈ D ⊂ Cn+1 is a local minimum point of v
satisfying v(ξ0 ) < 0 then we have Dv(ξ0 ) = 0 and the Hessian matrix
D2 v(ξ0 ) > 0. Writing L = div(A∇T ) and using the semipositivity of
A, we reach a contradiction.
We also give the maximum principle for weak solutions of the equations. For D ⊂ Cn+1 , we say that U ∈ H 1 (D) if it satisfies
Z X
n
∂U 2
∂U 2 (|Xi U |2 + |Yi U |2 ) + + λ2 + |U |2 dz < ∞. (4.2)
∂λ
∂t
D
i=1
Proposition 4.2. Assume that D is a subset of Cn+1 with D0 := D ∩
{λ = 0} ⊂ ∂D. Let c ∈ L∞ (Ω) be a nonnegative function. Assume
that v ∈ H 1 (D) ∩ C 2 (D) ∩ C 1 (D) is a solution of


Lv = 0
in D,


∂v
(4.3)
−
+ c(ξ)v ≥ 0 on D0 = D ∩ {λ = 0},

∂λ


v≥0
on ∂D ∩ {λ 6= 0}.
Then v ≥ 0 in D and v(ξ, 0) ≥ 0 in Ω.
NONLOCAL LIOUVILLE THEOREM
13
Proof. Let v − denotes the negative part of the function v. We multiply
the first equation in (4.3) by v − and integrate by parts to get
− 2
− 2 !
Z
Z
n
X
∂v
− 2
− 2
2 ∂v −
(|Xi v | + |Yi v | ) + c(ξ)|v − |2 dξ = 0,
+ λ ∂t dz−
∂λ
D
D0
i=1
which implies v − ≡ 0 in D0 = D ∩ {λ = 0}.
We prove now two Hopf’s Lemmas. The first one is Hopf’s Lemma
for the operator L in a subset D of Cn+1 . We first define the interior
ball condition in this setting.
Definition 4.3. Let D ⊂ Cn+1 . We say that D satisfies the interior
dλ -ball condition at P ∈ ∂D if there exists a constant R > 0 and a
point z0 ∈ D, such that the ball B(z0 , R) ⊂ D and P ∈ ∂B(z0 , R),
where B(z0 , R) = {z ∈ Cn+1 | dλ (z, z0 ) < R}.
Lemma 4.4. Let D ⊂ Cn+1 satisfy the interior dλ -ball condition at the
point P0 ∈ ∂D and let U ∈ C 2 (D) ∩ C 1 (D), be a solution of
− LU ≥ c(z)U in D
(4.4)
with c ∈ L∞ (D). Suppose that U (z) > U (P0 ) = 0 for every z ∈ D.
Then
U (P0 ) − U (P0 − sν)
< 0.
lim
s→0
s
where ν is the outer normal to ∂D in P0 .
Proof. By assumption, there exist a point z0 = (x̂1 , ..., x̂n , ŷ1 , ..., ŷn , t̂, λ̂)
and a radius R > 0 such that the ball B(z0 , R) ⊂ D and P0 ∈ ∂B(z0 , R).
We consider the function
2
ψ = U e−K(x1 −x̂1 ) , for K > 0.
An easy computations yields
∂U
∂ 2ψ
∂ 2U
−K(x1 −x̂1 )2
2
2
=
e
[4K
(x
−
x̂
)
U
−
2KU
−
4K(x
−
x̂
)
+
],
1
1
1
1
∂x21
∂x1
∂x21
and
2
∂ 2ψ
−K(x1 −x̂1 )2 ∂ U
=
e
,
∂x2j
∂x2j
2
∂ 2ψ
−K(x1 −x̂1 )2 ∂ U
=
e
,
∂yk2
∂yk2
where j = 2, · · · , n, k = 1, · · · , n.
2
∂ 2ψ
−K(x1 −x̂1 )2 ∂ U
=
e
,
∂λ2
∂λ2
14
ELEONORA CINTI AND JINGGANG TAN
Moreover,
∂ 2ψ
∂U
∂ 2U
2
= e−K(x1 −x̂1 ) [−2K(x1 − x̂1 )
+
],
∂x1 ∂t
∂t
∂xj ∂t
2
∂ 2ψ
2 ∂ U
= e−K(x1 −x̂1 )
,
∂xj ∂t
∂xj ∂t
Therefore, we have
2
Lψ + 4K(x1 − x̂1 )X1 ψ = e−K(x1 −x̂1 ) [−4K 2 (x1 − x̂1 )2 U + LU − 2KU ].
and hence for K sufficiently large, we deduce
−Lψ − 4K(x1 − x̂1 )X1 ψ ≥ 0.
2
2
We introduce now the function φ = e−αR − e−αρ , where ρ =
dλ (z, z0 ), and 0 < ρ < R. We can choose α sufficiently large such
that
−
Lφ − 4K(x1 − x̂i )X1 φ
2
=[
r
2
(4α2 ρ2 − 2Qα) − 8Kα(x1 − x̂1 )X1 ρ]e−αρ ≥ 0.
2
ρ
Let A := B(z0 , R) \ B(z0 , R1 ) for 0 < R1 < R. For ε small enough
ψ(z) + εφ(z) ≥ 0 in ∂A := ∂B(z0 , R) ∪ ∂B(z0 , R1 )
Then, by Lemma 4.2 we obtain that ψ(z) + εφ ≥ 0 in A. Therefore,
using that ψ(P0 ) = φ(P0 ) = 0, we deduce that
ψ(P0 ) − ψ(P0 − sν) + ε(φ(P0 ) − φ(P0 − sν)) ≤ 0.
Using that φ is strictly increasing in ρ, we deduce
ψ(P0 ) − ψ(P0 − sν)
< 0,
s→0
s
which, in turns, implies that
lim
U (P0 − sν) − U (P0 )
< 0.
s→0
s
lim
For any Ω ⊂ R2n+1 , we denote by C the infinite cylinder
C = Ω × (0, +∞).
Before proving our second Hopf’s Lemma, let us recall the notion of
interior ball condition in the Heisenberg group.
NONLOCAL LIOUVILLE THEOREM
15
Definition 4.5. Let Ω ⊂ R2n+1 . We say that Ω satisfies the interior
Heisenberg ball condition at ξ ∈ ∂Ω if there exists a constant R > 0
and a point ξ0 ∈ Ω, such that the Heisenberg ball BH (ξ0 , R) ⊂ Ω and
ξ ∈ ∂BH (ξ0 , R).
Lemma 4.6. Let Ω ⊂ Hn satisfy the interior Heisenberg ball condition
at the point P ∈ ∂Ω and let w ∈ C 2 (C) ∩ C 1 (C), be a nonnegative
solution of
(
−Lw ≥ c(z)w in C
(4.5)
−∂λ w ≥ d(z)w on Ω,
with c ∈ L∞ (C) and d ∈ L∞ (Ω). Suppose that w((P, 0)) = 0
Then
∂ν w(P, 0) < 0
where ν is the outer normal to ∂Ω in P .
(4.6)
Proof. We follow the proof of Lemma 2.4 in [6]. Without loss of generality, we can assume that Ω intersects the positive ξ1 half-axes and
that P = (b, 0, ..., 0) for some b > 0. We have to show that
∂ξ1 w(P, 0) < 0.
(4.7)
By the maximum principle and the Lemma 4.4 we have that
w > 0 on C ∪ Ω.
(4.8)
We start by proving the lemma in the case c(z) = d(z) ≡ 0.
Since Ω satisfies the interior Heisenberg ball condition at P , there
exist z0 ∈ Ω × {0} and R > 0, such that the ball B + (z0 , R) is contained
in the cylinder C and (∂C ∩ {λ > 0}) ∩ ∂B + (z0 , R) = {(P, 0)}. We may
assume that z0 = (a, 0, ..., 0) and we consider the set
+
+
A = B (z0 , b − a) \ B (z0 , (b − a)/2) ∩ {λ > 0}.
We observe that {(P, 0)} = ∂A ∩ ∂C.
2
2
For z ∈ A we consider the function η(z) = e−αρ − e−α(b−a) , where
ρ = dλ (z, z0 ). Writing L in radial coordinate as in (2.2), we have that
r2
2
Lη(ρ) = 2 4α2 ρ2 − 2Qα e−αρ ,
ρ
and therefore, for α sufficiently large, we have that
− Lη ≤ 0.
(4.9)
By (4.8) we deduce that w > 0 on ∂B + (z0 , (b − a)/2) ∩ {λ ≥ 0}. Hence,
we may choose ε > 0 such that
w − εη ≥ 0 on ∂B + (z0 , (b − a)/2) ∩ {λ ≥ 0}.
16
ELEONORA CINTI AND JINGGANG TAN
Claim: w − εη ≥ 0 in A.
Indeed, using (4.9), we deduce that −L(w − εη) ≥ 0 in A. Hence,
by the maximum principle, we have that the minimum of w − εη is
attained on ∂A. Now, on one side we have that
w − εη ≥ 0 on ∂A ∩ ∂B + (z0 , b − a) ∪ ∂B + (z0 , (b − a)/2) .
On the other side, since ∂λ η = 0 on {λ = 0}, we deduce that −∂λ (w −
εη) ≥ 0 on ∂A∩{λ = 0}. Thus, using Lemma
4.4, we conclude that the
+
minimum of w−εη cannot be achieved on B (z0 , b − a) \ B + (z0 , (b − a)/2) ∩
{λ = 0}. This conclude the proof of the claim.
Finally, since (w−εη)((P, 0)) = 0, we deduce that ∂ξ1 (w−εη)((P, 0)) ≤
0, which in turns implies that ∂ξ1 w((P, 0)) < 0 using that ∂ξ1 η((P, 0)) <
0. This concludes the proof of the theorem in the case c(z) = d(z) ≡ 0.
In the general case, we introduce the function v = e−βλ w. An easy
computation yields, for β large enough:
−Lv = −e−βλ Lw + β 2 w ≥ (c(z) + β 2 )v ≥ 0 in C,
and
−∂λ v ≥ (β + d(z))v ≥ 0.
The result finally follows after applying the first part of the proof to
the function v.
5. Proof of Theorems 1.3 and 1.4
In this last section we give the proof of our Liouville-type result.
We recall that U = U (|(x, y)|, t, λ) is a solution of


in Hn × R+ ,
LU = 0
(5.1)
−∂λ U = U p on Hn ,

U > 0.
First of all, we consider the CR inversion of U . For z = (x, y, t, λ) ∈
R
, we consider the function
1
z ),
w(z) = Q−1 U (e
ρ
hP
i1
n
1
2
2
2
2
2
2 2
2 4
where ze = ρ4 (xt + yr , yt − xr , −t, λρ ) and ρ = ( i=1 (xi + yi ) + λ ) + t .
We have seen in Lemma 3.1 that w satisfies
(
Lw = 0
in Hn × R+ \ {0},
(5.2)
−∂λ w = ρp(Q−1)−(Q+1) wp on Hn .
2n+2
NONLOCAL LIOUVILLE THEOREM
17
Observe that the function w could be singular at the origin and it
satisfies limρ→∞ ρQ−1 w(z) = U (0).
We start now applying the moving plane method. We will move
an hyperplane orthogonal to the t-direction and use the H-reflection.
More precisely, for any µ ≤ 0, let
Tµ = {z ∈ R2n+2 | t = µ}, and Σµ = {z ∈ R2n+2 : t < µ}. For
z ∈ Σµ , we define zµ = (y, x, 2µ − t, λ). To avoid the singular point, we
consider
e µ = Σµ \ {eµ },
Σ
where eµ = (0, 0, 2µ, 0) is the reflection of the origin. Let
wµ (z) = wµ (|(x, y)|, t, λ) := w(|(x, y)|, 2µ−t, λ) = w(y, x, 2µ−t, λ) = w(zµ ),
and
Wµ (z) := wµ (z) − w(z) := w(zµ ) − w(z) z ∈ Σµ .
By using the invariance of the operator under the CR transform as in
Lemma 3.1 and the fact that |zµ | ≤ |z|, we have that
(
LWµ = 0
in R2n+2
,
+
n
∂λ Wµ ≥ c(z, µ)Wµ on H ,
Ψp−1
µ
where c(z, µ) = ρ(Q+1)−p(Q−1)
and Ψµ (z) is between w(z) and wµ (z). By
the definition of wµ and w, we have that c(z, µ) ≈ C/ρ2 at infinity.
The following lemma will let us to start moving the hyperplane Tµ .
Proposition 5.1. Let Bρ+ = B + (0, ρ) be the half ball in R2n+2
. Assume
+
that w ∈ C 2 (R2n+2
)\{0} satisfies (5.2). Then (i) For µ < 0
)∩C 1 (R2n+2
+
+
with |µ| large enough, we have Wµ ≥ 0. (ii) For µ < 0 with |µ| large
enough, if inf Σµ Wµ < 0, then the infimum is attained at some point
z0 ∈ Σµ \ {eµ } such that |z0 | is bounded.
Proof. Define for z ∈ B1+ \ Bρ+0 and 0 < ρ0 < 1,
ε ελ
ε ρQ−1
φ0 (z) = − 0 Q−1 + .
4 |z|
2
Let ψ = w − φ0 . An easy computation implies that
(
Lψ ≤ 0
in B1+ \ Bρ+0 ,
∂λ ψ = wp − 2ε on ∂(B1+ \ Bρ+0 ) ∩ ∂Rn+1
+ .
It is clear that ψ = w − (ε/4 − ε + ελ/2) > 0 on ∂Bρ+0 ∩ ∂Bρ0 and that
ψ = w − [ε/4 −
ρQ−1
ε
0
|z|Q−1
+ ελ/2] > w − ε > 0 on ∂B1+ ∩ ∂B1 .
We claim that ψ ≥ 0 in B1+ \ Bρ+0 .
18
ELEONORA CINTI AND JINGGANG TAN
Suppose by contradiction that there exists P0 = (x0 , y0 , t0 , 0), ρ0 <
|P0 | < 1 such that ψ(P0 ) = minB + \Bρ+ ψ < 0. Then,
1
0
−∂λ ψ |P0 ≤ 0.
This inequality, together with the Neumann boundary condition for ψ,
imply that
ε p1
ε
≥ ,
w(P0 ) ≥
2
4
which in turns implies that
ε
ψ(P0 ) = w(P0 ) − φ0 (P0 ) > w(P0 ) − > 0.
4
This give a contradiction with w(P0 ) < 0. Thus, we have proven that
ψ ≥ 0 in B1+ \ Bρ+0 . By letting ρ0 → 0, we deduce that w(z) ≥ 4ε for
+
every z ∈ B 1 \ {0}.
We observe that
lim |z|Q−1 w(z) = lim U (z̃) = U (0).
|z|→∞
|z|→∞
1
For ε > 0, letting bε = max{U (z̃) | |z̃|
= |z| ≥ 1ε }, we see that w(z) ≤
bε
for all |z| ≥ 1ε .
|z|Q−1
e µ = Σµ \ {eµ }, if |µ| is sufficiently large.
Hence Wµ = wµ − w > 0 in Σ
This completes part (i).
To prove (ii), since by definition of w, w(z) → 0 as |z| → ∞, it follows
that for µ < 0 large in absolute value, we have w ≤ 8ε in B1+ (eµ ). For
such µ we therefore have that Wµ > 0 in B1+ (eµ ). Thus, combining
that Wµ vanishes on the plane Tµ , if inf Σe µ Wµ < 0, then Wµ attains its
+
infimum in the compact set B R1 ∩ (Σµ \ B1+ (eµ )).
2
Proof of Theorem 1.4. By Proposition 5.1, we know that for µ
negative and large in absolute value we have that Wµ ≥ 0 in Σµ . Let
us define µ0 = sup{µ < 0 | Wµ ≥ 0}. We only need to prove that
µ0 = 0. Suppose that µ0 6= 0 by contradiction. By continuity, Wµ0 ≥ 0
in Σµ0 . By the maximum principle, we deduce that Wµ0 ≡ 0 in Σµ0 or
Wµ0 > 0 on Σµ0 ∪ (∂Σµ0 ∩ {λ = 0} ∩ {t < µ0 }) \ {zµ0 }.
(5.3)
Using the Neumann condition satisfied by Wµ0 and the assumption
µ0 > 0, we see that Wµ0 ≡ 0 is impossible. Therefore (5.3) holds.
By the definition of µ0 there exists µk → µ0 , µk < µ0 such that
inf Σµk Wµk < 0.
We observe that for some positive d:
2n+2
min Wµ0 (z) : z ∈ ∂B + (zµ0 , |µ0 |/2) ∩ R+
= d.
NONLOCAL LIOUVILLE THEOREM
19
From this fact, using a similar argument to the one of point i) in Proposition 5.1, we deduce that
Wµ0 ≥ d in B + (zµ0 , |µ0 |/2) \ {zµ0 }.
Therefore, we have that
lim inf Wµk (z) : z ∈ B + (zµk , |µ0 |/2) \ {zµk } ≥ d.
k→∞
Using this bound and the fact that Wµk (z) → 0 as |z| → ∞, we deduce
that for k large enough, the negative infimum of Wµk is attained at
some point zk ∈ Σµk \ B + (zµk , |µ0 |/2).
By Proposition 5.1 we know that the sequence {zk } is bounded and
therefore, after passing to a subsequence, we may assume that zk → z0 .
By (5.3) we have that Wµ0 (z0 ) = 0 and z0 ∈ ∂Σµ0 ∩ {t = µ0 }.
If zk ∈ Σµk for an infinite number of k, then ∇Wµk (zk ) = 0, and
therefore, by continuity
∇Wµ0 (z0 ) = 0.
(5.4)
(z0 ) < 0,
If z0 ∈ ∂Σµ0 ∩ R2n+2
, then by Lemma 4.4, we have that ∂w
+
∂t
which give a contradiction. Analogously, using Lemma 4.6, we get a
contradiction if we assume that z0 ∈ ∂Σµ0 ∩ {λ = 0} ∩ {t = µ0 }.
In the case in which zk ∈ ∂Σµk ∩ {λ = 0} ∩ {t < µk }, we still
have that the derivatives of Wµk at zk in all directions except the λ
direction vanish. Passing to the limit and arguing as above, we get a
contradiction. Hence we have established that µ0 = 0. This implies
that v is even in t, but since the origin 0 on the t-axes is arbitrary, we
can perform the CR transform with respect to any point and then we
conclude that w is constant in the direction t.
This shows that U is actually a solution of the following problem
(
2n+1
∆U = 0
in R+
,
(5.5)
p
2n
−∂λ U = U on R .
Q+1
Since Q−1
= 2n+3
< 2n+1
, we conclude the proof by using the standard
2n+1
2n−1
Liouville type theorem for problem (5.5) (see [6, 17]).
2
Proof of Theorem 1.3. We observe that if u is a cilindrical function, that is u = u(|(x, y)|, t), then its extension U satisfying (1.2) is
also cilindrical in the all halfspace Hn × R+ , in the sense that U =
U (|(x, y)|, t, λ). This follows by the definition of the extension operator, which is constructed by taking the Fourier transform in Hn (see
for details Section 5.3, and in particular formula (5.9) in [9]). Using
this fact, the conclusion follows as a corallary of Theorem 1.4.
2
20
ELEONORA CINTI AND JINGGANG TAN
Acknowledgements: Both authors were supported by Spain Government grant MTM2011-27739-C04-01; J.T. was supported by Chile
Government grant Fondecyt 1120105, USM 121402; CMM in Universidad de Chile.
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E.C., Dipartimento di Matematica, Università degli Studi di Bologna,
Piazza di Porta San Donato 5, 40126 Bologna (Italy)
E-mail address: [email protected]
J.T., Departamento de Matemática, Universidad Técnica Federico
Santa Marı́a, Avda. España 1680, Valparaı́so (Chile)
E-mail address: [email protected]