Table Of ContentA Liouville Theorem for the Fractional
4
Laplacian
1
0
2
n Ran Zhuo Wenxiong Chen Xuewei Cui Zixia Yuan
a
J
9
2
Abstract
]
P Inthispaper,weconsiderthefollowingfractionalLaplaceequation
A
. (−∆)α/2u(x) = 0, in Rn,
h (1)
t ( u(x) ≥0, in Rn,
a
m
where n ≥ 2 and α is any real number between 0 and 2. We prove
[
that the only solution for (1) is constant. Or equivalently,
1
Every α-harmonic function bounded either above or below in all of
v
2 Rn must be constant.
0 Thisextends theclassical Liouville TheoremfromLaplacian tothe
4
fractional Laplacian.
7
. As an immediate application, we use it to obtain an equivalence
1
between a semi-linear pseudo-differential
0
4
1 (−∆)α/2u= up(x), x ∈ Rn (2)
:
v
i and the corresponding integral equation
X
r 1
a u(x) = dy, x ∈ Rn.
ZRn |x−y|n−α
Combining this with the existing results on the integral equation, one
can obtained much more general results on the qualitative properties
of the solutions for (2).
1 Introduction
The well-known Liouville’s Theorem states that
1
Any harmonic function bounded either above or below in all of Rn is
constant.
One of its important applications is the proof of the Fundamental Theo-
rem of Algebra. It is also a key ingredient in deriving a priori estimates for
solutions in PDE analysis.
The main purpose of this article is to extend this classical theorem to the
fractional Laplacian.
Essentially different from the Laplacian, the fractional Laplacian in Rn is
a nonlocal operator, taking the form
u(x)−u(z)
(−∆)α/2u(x) = C PV dz (3)
n,α
ZRn |x−z|n+α
where α is any real number between 0 and 2 and PV stands for the Cauchy
principal value. This operator is well defined in S, the Schwartz space of
rapidly decreasing C∞ functions in Rn. In this space, it can also be defined
equivalently in terms of the Fourier transform
(−∆)α/2u(ξ) = |ξ|αuˆ(ξ)
d
where uˆ is the Fourier transform of u. One can extend this operator to a
wider space of distributions as the following.
Let
|u(x)|
L = {u : Rn → R | < ∞}.
α ZRn (1+|x|n+α)
For u ∈ L , we define (−∆)α/2u as a distribution:
α
< (−∆)α/2u(x),φ >=< u,(−∆)α/2φ >, ∀φ ∈ C∞(Rn).
0
This defines the operator in a weak sense. In this paper, we consider the
class of functions where the fractional Laplacian is defined in a little bit
more stronger sense, that is
u ∈ L , such that the right hand side of (3) is well defined for every
α
x ∈ Rn.
One can verify that, all the above definitions coincides when u is in S.
We say that u is an α-harmonic function if u ∈ L , such that the right
α
hand side of (3) is well defined for every x ∈ Rn and equals zero. In this
sense, we have
2
Theorem 1 Every α-harmonic function bounded either above or below in all
of Rn for n ≥ 2 must be constant.
This is the main result of the paper. To prove it, we study
(−∆)α/2u(x) = 0, in Rn,
(4)
u(x) ≥ 0, in Rn.
(
We say that u ≥ 0 is a strong solution of (4), if u ∈ L , such that the right
α
hand side of (3) is well defined for every x ∈ Rn and equals zero.
Apparently, Theorem 1 is equivalent to the following
Theorem 2 Assume that n ≥ 2. Let u be a strong solution of (4), then
u ≡ C.
As an immediate application of the Liouville theorem for α harmonic
functions, we prove an equivalence between a psedodifferential equation and
an integral equation.
Theorem 3 Assume that n ≥ 2 and u ∈ L is a nonnegative strong solution
α
of
(−∆)α/2u(x) = up(x), x ∈ Rn, (5)
then u also satisfies
c
u(x) = n up(y)dy,
ZRn |x−y|n−α
and vice versa.
Remark 1 i) Actually, the right hand side of equation (5) can be a much
more general function f(x,u), such that, for any constant c > 0,
1
f(y,c)dy = ∞. (6)
ZRn |x−y|n−α
ii) The idea of proof can be extended to establish the equivalence between
a general system of m equations in Rn
(−∆)α/2u (x) = f (x,u (x),···u (x)), i = 1,···,m,
i i 1 m
u ≥ 0, i = 1,···,m,
( i
and the corresponding integral system
u (x) = cn f (y,u (y),···,u (y)), i = 1,···,m,
i Rn |x−y|n−α i 1 m
( ui(x) ≥ R0, i = 1,···,m.
3
Combining Theorem 3 with the qualitative properties established for the
integral equations in [CLO] and [CLO1], one obtain immediately that
Theorem 4 Assume that n ≥ 2 and u is a nonnegative strong solution of
(5) for 0 < α < 2. Then
i) In the critical case when p = n+α, it must assume the form
n−α
t
u(x) = c( )(n−α)/2
t2 +|x−x |2
o
for some t > 0, x ∈ Rn.
o
ii) In the subcritical case when 1 < p < n+α, we must have u ≡ 0.
n−α
Remark 2 i) In [CLO] and [CLO1], in order the results in Theorem 4 to
hold, one requires u to be in Hα/2(Rn). Here we only requires u ∈ L , a
α
much weaker restriction.
ii) In [BCPS], by using the extension method to obtain the same results
as in Theorem 4, the authors require that 1 ≤ α < 2 and u be bounded.
Obviously, our condition here is much weaker.
In Section 2, we prove the Liouville Theorem 2 and hence Theorem 1. In
Section 3, we establish the equivalence and hence prove Theorem 3 and 4.
2 The proof of the Liouville Theorem
In this section, we prove Theorem 2.
Proof. First, we define
u(x), |x| ≤ k,
u (x) = (7)
k P (y,x)u(y)dy, |x| > k,
( Bk k
R
where P (y,x) is a Poisson kernel in the exterior of the ball B with radius
k k
k and centered at the origin:
n n πα (|x|2 −k2)α/2 1
Pk(y,x) = Γ( )π−2−1sin( ) , |y| < k, |x| > k.
2 2 (k2 −|y|2)α/2 |x−y|n
(8)
4
Obviously, for each x,
lim u (x) = u(x).
k
k→∞
One can also verify (see [L]) that
(−∆)α/2u (x) = 0, for |x| > k, and u (x) ≤ u(x), ∀x ∈ Rn. (9)
k k
Moreover, by Taylor expansion, it is easy to derive that
c 1
1
u (x) = +O( ). (10)
k
|x|n−α |x|n−α+1
In order to prove that u is constant, it suffice to show that for any ψ ∈
C∞(Rn), satisfying the condition
0
ψ(x)dx = 0, (11)
ZRn
we have
u(x)ψ(x)dx = 0.
ZRn
Actually, we only need to prove
lim u (x)ψ(x)dx = 0.
k
k→∞ZRn
We divided the proof into two steps.
Step 1. Let
ψ(y)
ϕ(x) = .
ZRn |x−y|n−α
Combining Taylor expansion with (11), we deduce that
1
ϕ(x) = O( ), as|x| → ∞. (12)
|x|n−α+1
It follows that ϕ(x) ∈ L2(Rn) for n ≥ 2 , and
(−∆)α/2ϕ(x) = ψ(x), x ∈ Rn. (13)
In this step, we will show that
u (x)(−∆)α/2ϕ(x)dx = (−∆)α/2u (x)ϕ(x)dx. (14)
k k
ZRn ZRn
5
Let
c
1
v (x) = u (x)− ,
k k
|x|n−α
then v (x) ∈ L2(Rn) due to (10).
k
Applying the Parseval’s formula to one part of the left hand side of (14),
we derive that
c
u (x)(−∆)α/2ϕ(x)dx = (v (x)+ 1 )(−∆)α/2ϕ(x)dx
ZRn k ZRn k |x|n−α
c
= v (x)(−∆)α/2ϕ(x)dx+ 1 (−∆)α/2ϕ(x)dx
ZRn k ZRn |x|n−α
= v (ξ)|ξ|αϕ(ξ)+cϕ(0). (15)
k
ZRn
Herewehaveusedaresultin[L]thact,inthesbenseofdistributions, theFourier
transform of c1 is a constant multiple of |ξ|−α, and by the definition of
|x|n−α
Fourier transform on distributions (see [L]), we have
c
1 (−∆)α/2ϕ(x)dx = |ξ|−α|ξ|αϕ(ξ)dξ = cϕ(0).
ZRn |x|n−α ZRn
b
Also note that we are not able to apply the Parseval’s formula directly to
u (x)(−∆)α/2ϕ(x)dx because u may not be in L2(Rn).
Rn k k
For the right hand side of (14), we have
R
c
(−∆)α/2u (x) ϕ(x)dx = (−∆)α/2(v (x)+ 1 )ϕ(x)dx
ZRn k ZRn k |x|n−α
c
= (−∆)α/2v (x) ϕ(x)dx+ (−∆)α/2( 1 ) ϕ(x)dx
k
ZRn ZRn |x|n−α
= |ξ|αv (ξ)ϕ(ξ)dξ +cϕ(0). (16)
k
ZRn
Here we have used a well-known fact thcat b1 is a constant multiple of the
|x|n−α
fundamental solution of (−∆)α/2.
Now from (15) and (16), we arrive at (14).
Step 2. We prove
(−∆)α/2u (x) ϕ(x)dx → 0, ask → ∞. (17)
k
ZRn
6
By elementary calculation, we separate the integral in(17) into two parts,
u (x)−u (y)
(−∆)α/2u (x)ϕ(x)dx = c k k dyϕ(x)dx
ZRn k ZRnZRn |x−y|n+α
u (x)−u (y)
k k
= c dyϕ(x)dx
ZBr(0)ZRn |x−y|n+α
u (x)−u (y)
k k
+ c dyϕ(x)dx
ZRn\Br(0)ZRn |x−y|n+α
= I +I ,
1 2
where r < k.
First, we consider I .
1
From the first equation of (4), we have
u(x)−u(y)
0 = c dy
ZRn |x−y|n+α
u(x)−u(y) u(x)−u(y)
= c dy +c dy.
ZBk(0) |x−y|n+α ZRn\Bk(0) |x−y|n+α
It follows that
u (x)−u (y)
k k
I = c dyϕ(x)dx
1 ZBr(0)ZRn |x−y|n+α
u(x)−u (y)
k
= c dyϕ(x)dx
|x−y|n+α
ZBr(0)ZBk(0)
u(x)−u (y)
k
+ c dyϕ(x)dx
ZBr(0)ZRn\Bk(0) |x−y|n+α
u(y)−u (y)
k
= c dyϕ(x)dx. (18)
ZBr(0)ZRn\Bk(0) |x−y|n+α
Applying (18), we obtain, for each fixed r,
|u(y)|
|I | ≤ c dy|ϕ(x)|dx
1 ZBr(0)ZRn\Bk(0) |x−y|n+α
|u(y)|
≤ c dy|ϕ(x)|dx → 0, ask → ∞. (19)
ZBr(0)ZRn\Bk(0) (1+|y|)n+α
Here we have used the facts that u ∈ L and 0 ≤ u ≤ u.
α k
7
Next we will show that
I = c (−∆)α/2u ϕ(x)dx → 0, asr → ∞, (20)
2 k
ZRn\Br(0)
uniformly in k.
Let
f (x) = (−∆)α/2u (x).
k k
We first show that
f (y)
k
dy = u (x). (21)
ZRn |x−y|n−α k
To this end, denote
f (y)
k
g (x) = dy.
k
ZRn |x−y|n−α
For any ψ(x) ∈ C∞(Rn), we show that
0
g (x)ψ(x)dx = u (x)ψ(x)dx. (22)
k k
ZRn ZRn
Actually,
(−∆)α/2u (y)
k
g (x)ψ(x)dx = dyψ(x)dx
ZRn k ZRn ZRn |x−y|n−α
(−∆)α/2(v (y)+c /|y|n−α)
k 1
= dyψ(x)dx
ZRn ZRn |x−y|n−α
(−∆)α/2v (y) c
k 1
= dyψ(x)dx+ ψ(x)dx
ZRn ZRn |x−y|n−α ZRn |x|n−α
= I +I .
1 2
Since both (−∆)α/2v (x) and ψ(x) are compactly supported, one can
k
exchange the order of integration to derive
ψ(x)
I = (−∆)α/2v (y) dxdy
1 ZRn k ZRn |x−y|n−α
= |ξ|αv (ξ)ψ(ξ)|ξ|−αdξ
k
ZRn
= v (ξc)ψ(ξ)bdξ
k
ZRn
= vc(x)ψb(x)dx.
k
ZRn
8
Herewehaveusedaresult in[LL]thattheFouriertransformof ψ(x) dx
Rn |x−y|n−α
is a constant multiple of ψ(ξ)|ξ|−α. It follows that R
b
g (x)ψ(x)dx = I +I
k 1 2
ZRn
c
1
= v (x)+ ψ(x)dx
ZRn k |x|n−α!
= u (x)ψ(x)dx.
k
ZRn
This proves (22) and hence (21), and from which, we arrive immediately that
f (x)
k
dx ≤ u (0) ≤ u(0). (23)
ZRn\Br(0) |x|n−α k
Combining (12) with (23), we derive that
u (x)−u (y)
k k
I = c dy ϕ(x)dx
2 ZRn\Br(0)ZRn |x−y|n+α
= c f (x)ϕ(x)dx
k
ZRn\Br(0)
f (x)
k
≤ c dx
ZRn\Br(0) |x|n−α+1
c f (x)
k
≤ dx
r ZRn\Br(0) |x|n−α
c
≤ u(0) → 0, asr → ∞, uniformly in k. (24)
r
(19) and (24) imply that (17) holds. Hence we have
lim u (x)ψ(x)dx = lim u (x)(−∆)α/2ϕ(x)dx
k k
k→∞ZRn k→∞ZRn
= lim (−∆)α/2u (x)ϕ(x)dx = 0. (25)
k
k→∞ZRn
That is
u(x)ψ(x)dx = 0.
ZRn
Therefore we come to the conclusion that u ≡ C.
This complete the proof of Theorem 2.
9
The Proof of Theorem 1.
To see that Theorem 2 implies Theorem 1, let v be any α-harmonic func-
tion that is bounded from above by a constant M in Rn. Take u(x) =
M −v(x), then
(−∆)α/2u(x) = 0, x ∈ Rn,
u(x) ≥ 0, x ∈ Rn.
(
By Theorem 2, u must be constant, hence so does v. Similarly, if v is any α-
harmonic function that is bounded from below by a constant M in Rn, then
we let u(x) = v(x)−M to derive that v must be constant. This completes
the proof of Theorem 1.
3 Applications
The Proof of Theorem 3.
Assume u ∈ L is a nonnegative locally bounded strong solution of
α
(−∆)α/2u(x) = up(x), x ∈ Rn, (26)
Let
v (x) = G (x,y)up(y)dy, (27)
R R
ZBR
where G (x,y) is Green’s function on the ball B (0):
R R
(−∆)α/2G (x,y) = δ(x−y), x,y ∈ B (0),
R R
G (x,y) = 0, x or y ∈ Bc(0).
( R R
Thanks to Kulczycki [Ku], one can write
s (n−α)
A B (s−tb)
n,α n,α t 2
G (x,y) = 1− db , x,y ∈ B (0),
R s(n−2α) (s+t)(n−2α) Z0 bα2(1+b) R
wheres = |x−y|2,t = (1−|x|2)(1−|y|2). A andB areconstantsdepending
R2 R2 R2 n,α n,α
on n and α.
It is easy to verify that
(−∆)α/2v (x) = up(x), in B (0),
R R (28)
v = 0, in Bc(0).
( R R
10
