ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
MOSTAFA FAZLY, JUNCHENG WEI
Abstract. We derive a monotonicity formula for solutions of the fractional Hénon-Lane-Emden equation
(−∆)s u = |x|a |u|p−1 u
Rn
where 0 < s < 2, a > 0 and p > 1. Then we apply this formula to classify stable solutions of the above equation.
1. Introduction and Main Results
We study the classification stable solutions of the following equation
(−∆)s u = |x|a |u|p−1 u Rn
(1.1)
where (−∆)s is the fractional Laplacian operator for 0 < s < 2. Here is what we mean by stability.
Definition 1.1. We say that a solution u of (1.1) is stable if
Z Z
Z
(φ(x) − φ(y))2
(1.2)
dxdy − p
|x|a |u|p−1 φ2 ≥ 0
|x − y|n+2s
Rn Rn
Rn
for any φ ∈ Cc∞ (Rn ).
For the local cases s = 1 and s = 2, the classification of stable solutions is completely known for a ≥ 0. We
refer the interested readers to Farina [14] for the case of s = 1 and a = 0 and to Cowan-Fazly [6], Wang-Ye [31],
Dancer-Du-Guo [7], Du-Guo-Wang [11] for the case s = 1 and a > −2. Also, for the fourth order Lane-Emden
equation that is when s = 2 we refer to Davila-Dupaigne-Wang-Wei [10] where a = 0 and to Hu [20] where
a > 0. In this note, we focus on the case of fractional Laplacian operator.
It is by now standard that the fractional Laplacian can be seen as a Dirichlet-to-Neumann operator for a
degenerate but local diffusion operator in the higher-dimensional half-space Rn+1
+ . For the case of 0 < s < 1 this
in fact can be seen as the following theorem given by Caffarelli-Silvestre [2]. See also [27].
n+1
Theorem 1.1. Take s ∈ (0, 1), σ > s and u ∈ C 2σ (Rn ) ∩ L1 (Rn , (1 + |t|)n+2s dt). For X = (x, y) ∈ R+
, let
Z
ue (X) =
P (X, t)u(t) dt,
Rn
where
P (X, t) = pn,s t2s |X − t|−(n+2s)
and pn,s is chosen so that
R
Rn
n+1
1−2s
P (X, t) dt = 1. Then, ue ∈ C 2 (Rn+1
∂y ue ∈ C(Rn+1
+ ) ∩ C(R+ ), y
+ ) and

n+1
1−2s
∇ue ) = 0
in R+ ,

 ∇ · (y

ue = u
on ∂Rn+1
+ ,


 − lim y 1−2s ∂t ue = κs (−∆)s u
on ∂Rn+1
+ ,
y→0
The first author is pleased to acknowledge the support of a University of Alberta start-up grant. Both authors are supported by
NSERC grants.
1
2
MOSTAFA FAZLY, JUNCHENG WEI
where
(1.3)
κs =
Γ(1 − s)
.
22s−1 Γ(s)
From this theorem for a solution of the fractional Henon-Lane-Emden equation, we get the following equation
in the higher-dimensional half-space Rn+1
+ ,

 −∇ · (y 1−2s ∇ue ) = 0
in Rn+1
+
(1.4)
1−2s
a
p−1
∂t ue = κs |x| |ue | ue
in Rn
 − lim y
y→0
There are different ways of defining the fractional operator (−∆)s where 1 < s < 2, just like the case of 0 < s < 1.
Applying the Fourier transform one can define the fractional Laplacian by
\s u(ζ) = |ζ|2s û(ζ)
(−∆)
or equivalently define this operator inductively by (−∆)s = (−∆)s−1 o(−∆), see [26]. Recently, Yang in [29]
gave a characterization of the fractional Laplacian (−∆)s , where s is any positive, noninteger number as the
Dirichlet-to-Neumann map for a function ue satisfying a higher order elliptic equation in the upper half space
with one extra spatial dimension. This is a generalization of the work of Caffarelli and Silvestre in [2] for the
case of 0 < s < 1. We first fix the following notation then we present the Yang’s characterization. See also
Case-Chang [3] and Chang-Gonzales [4] for higher order fractional operators.
Notation 1.1. Throughout this note set b := 3 − 2s and define the operator
b
∆b w := ∆w + wy = y −b div(y b ∇w).
y
for a function w ∈ W 2,2 (Rn+1 , y b ).
As it is shown by Yang in [29], if u(x) is a solution
and y ∈ R+ satisfies

∆2b ue =

b
(1.5)
limy→0 y ∂y ue =

limy→0 y b ∂y ∆b ue =
of (1.1) then the extended function ue (x, y) where x ∈ Rn
0 in Rn+1
+ ,
0 in ∂Rn+1
+ ,
Cn,s |x|a |u|p−1 u in Rn
Moreover,
Z
ˆ 2 dξ = Cn,s
|ξ|2s |u(ξ)|
Z
Rn
Rn+1
+
y b |∆b ue (x, y)|2 dxdy
Note that u(x) = ue (x, 0) in Rn .
On the other hand, Herbst in [19] (see also [30]), shoed that when n > 2s the following Hardy inequality holds
Z
Z
2s
2
|ξ| |φ̂| dξ > Λn,s
|x|−2s φ2 dx
Rn
Rn
for any φ ∈ Cc∞ (Rn ) where the optimal constant given by
Λn,s = 22s
2
Γ( n+2s
4 )
.
2
Γ( n−2s
4 )
Here we fix a constant that plays an important role in the classification of solutions of (1.1)

+∞ if n ≤ 2s

(1.6)
pS (n, a) = n + 2s + 2a

if n > 2s
n − 2s
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
3
Remark 1.1. Note that for p > pS (n, a) the function
2s+a
us (x) = A|x|− p−1
(1.7)
where
Ap−1 = λ
n − 2s 2s + a
−
2
p−1
for constant
λ(α) = 22s
(1.8)
)Γ( n+2s−2α
)
Γ( n+2s+2α
4
4
n−2s−2α
n−2s+2α
Γ(
)Γ(
)
4
4
is a singular solution of (1.1) where 0 < s < 2. For details, we refer the interested readers to [13] for the case
of 0 < s < 1 and to [16] for the case of 1 < s < 2.
Here is our main result
Theorem 1.2. Assume that n ≥ 1 and 0 < s < σ < 2. Let u ∈ C 2σ (Rn ) ∩ L1 (Rn , (1 + |y|)n+2s dy) be a stable
solution to (1.1).
• If 1 < p < pS (n, a) or if pS (n, a) < p and
(1.9)
p
Γ( n2 −
s+ a
2
p−1
)Γ(s +
s+ a
2
Γ( p−1 )Γ( n−2s
−
2
then u ≡ 0;
• If p = pS (n, a), then u has finite energy i.e.
Z
kuk2Ḣ s (Rn ) =
s+ a
2
p−1 )
a
s+ 2
p−1 )
>
2
Γ( n+2s
4 )
,
2
Γ( n−2s
4 )
|x|a |u|p+1 < +∞.
Rn
If in addition u is stable, then in fact u ≡ 0.
Note that the classification of finite Morse index solutions of (1.1) when a = 0 is given by Davila-Dupaigne-Wei
in [9] when 0 < s < 1 and by Fazly-Wei in [16] 1 < s < 2.
Note also that in the absence of stability it is expected that the only nonnegative bounded solution of (1.1)
must be zero for the subcritical exponents 1 < p < pS (n, a) where a ≥ 0. To our knowledge not much is known
about the classification of solutions when a 6= 0 even for the standard case s = 1. For the case of s = 1,
Phan-Souplet in [23] proved that the only nonnegative bounded solution of (1.1) in three dimensions must be
zero for the case of 1 < p < pS (n, a) and a > −2. Some partial results are given in [17].
2. The monotonicity formula
Here is the monotonicity formula for the case of 0 < s < 1.
n+1
Theorem 2.1. Suppose that 0 < s < 1. Let ue ∈ C 2 (Rn+1
+ ) ∩ C(R+ ) be a solution of (1.1) such that
n+1
y 1−2s ∂y ue ∈ C(Rn+1
+ ). For x0 ∈ ∂R+ , λ > 0, let
!
Z
Z
2s(p+1)+2a
1
κ
s
−n
1−2s
2
a
p+1
E(ue , λ) := λ p−1
y
|∇ue | dx dy −
|x| |ue |
dx
2 Rn+1
p + 1 ∂Rn+1
∩Bλ
∩Bλ
+
+
a Z
2s(p+1)+2a
−n−1 s + 2
p−1
+λ
y 1−2s u2e dσ.
p + 1 ∂Bλ ∩Rn+1
+
Then, E is a nondecreasing function of λ. Furthermore,
2
Z
2s(p+1)+a
dE
∂ue
2s + a ue
= λ p−1 −n+1
y 1−2s
+
dσ
dλ
∂r
p−1 r
∂B(x0 ,λ)∩Rn+1
+
4
MOSTAFA FAZLY, JUNCHENG WEI
Proof. Let
(2.1)
p+1
2s p−1
−n
I(ue , λ) = λ
Z
y
2
1−2s |∇ue |
Rn+1
∩Bλ
+
2
κs
dx dy −
p+1
!
Z
a
∂Rn+1
∩Bλ
+
p+1
|x| |ue |
dx
Now for X ∈ Rn+1
+ , define
2s+a
uλe (X) = λ p−1 ue (λX).
(2.2)
Then, uλe solves (1.5) and in addition
I(ue , λ) = I(uλe , 1).
(2.3)
Taking partial derivatives we get
λ∂λ uλe =
(2.4)
2s + a λ
u + r∂r uλe .
p−1 e
Differentiating the operator (2.1) w.r.t. λ, we find
Z
Z
∂λ I(ue , λ) =
y 1−2s ∇uλe · ∇∂λ uλe dx dy − κs
Rn+1
∩B1
+
∩B1
∂Rn+1
+
|x|a |uλe |p−1 ∂λ uλe dx.
Integrating by parts and then using (2.4),
Z
y 1−2s ∂r uλe ∂λ uλe dσ
∂λ I(ue , λ) =
∂B1 ∩Rn+1
+
Z
=λ
∂B1 ∩Rn+1
+
y 1−2s (∂λ uλe )2 dσ −
Z
=λ
y
∂B1 ∩Rn+1
+
1−2s
(∂λ uλe )2 dσ
2s + a
p−1
Z
s + a2
−
∂λ
p−1
∂B1 ∩Rn+1
+
y 1−2s uλe ∂λ uλe dσ
!
Z
y
∂B1 ∩Rn+1
+
1−2s
(uλe )2
dσ
Scaling finishes the proof.
We now consider the case of 1 < s < 2 and a > 0. Note that a monotonicity formula is given for the case of
a = 0 and s = 2 and 1 < s < 2 by Davila-Dupaigne-Wang-Wei in [10] and Fazly-Wei in [16], respectively. We
define the energy functional
!
Z
Z
p+1
C
1 3−2s
n,s
2s p−1
−n
E(ue , r) := r
y
|∆b ue |2 −
|x|a up+1
e
p + 1 ∂Rn+1
∩Br
Rn+1
∩Br 2
+
+
Z
4s+2a
s + a2 p + 2s + a − 1
−
− n − b r−3+2s+ p−1 −n
y 3−2s u2e
n+1
p−1
p−1
R+ ∩∂Br
"
#
Z
a 4s+2a
s + 2 p + 2s + a − 1
d
−
−n−b
r p−1 +2s−2−n
y 3−2s u2e
p−1
p−1
dr
Rn+1
∩∂B
r
+
"
2 #
Z
4s+2a
1 3 d
2s
+
a
∂u
e
+2s−3−n
−1
3−2s
+ r
r p−1
y
r u+
2 dr
p−1
∂r
Rn+1
∩∂Br
+
"
2 !#
Z
∂ue 2s(p+1)+2a
1 d
+
y 3−2s |∇ue |2 − r p−1 −n
n+1
2 dr
∂r R+ ∩∂Br
!
Z
∂ue 2
1 2s(p+1)+2a
−n−1
3−2s
2
y
|∇ue | − + r p−1
2
∂r
Rn+1
∩∂B
r
+
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
p+4s+2a−1
p+2s+a−1
Theorem 2.2. Assume that n >
Furthermore,
2s+a
p−1
+
− b. Then, E(ue , λ) is a nondecreasing function of λ > 0.
4s+2a
dE(λ, ue )
≥ C(n, s, p) λ p−1 +2s−2−n
dλ
(2.5)
5
Z
y
3−2s
Rn+1
∩∂Bλ
+
2s + a −1
∂ue
r u+
p−1
∂r
2
where C(n, s, p) is independent from λ.
Proof: Set,
(2.6)
Ē(ue , λ) := λ
Z
2s(p+1)+2a
−n
p−1
Rn+1
∩Bλ
+
2s+a
!
Z
1 b
Cn,s
y |∆b ue |2 dxdy −
2
p+1
∂Rn+1
∩Bλ
+
|x|a up+1
e
2s+a
Define ve := ∆b ue , uλe (X) := λ p−1 ue (λX), and veλ (X) := λ p−1 +2 ve (λX) where X = (x, y) ∈ Rn+1
+ . Therefore,
∆b uλe (X) = veλ (X) and

∆b veλ = 0 in Rn+1

+ ,
limy→0 y b ∂y uλe = 0 in ∂Rn+1
(2.7)
+ ,

a λ p
b
λ
limy→0 y ∂y ve = Cn,s |x| (ue ) in Rn
In addition, differentiating with respect to λ we have
(2.8)
∆b
Note that
Ē(ue , λ) =
Ē(uλe , 1)
Z
=
∩B1
Rn+1
+
duλe
dv λ
= e.
dλ
dλ
1 b λ 2
Cn,s
y (ve ) dxdy −
2
p+1
Z
∩B1
∂Rn+1
+
|x|a |uλe |p+1
Taking derivate of the energy with respect to λ, we have
Z
Z
λ
dĒ(uλe , 1)
duλ
b λ dve
(2.9)
=
y ve
dxdy − Cn,s
|x|a |uλe |p e
dλ
dλ
dλ
Rn+1
∩B1
∂Rn+1
∩B1
+
+
Using (2.7) we end up with
dĒ(uλe , 1)
=
dλ
(2.10)
Z
Rn+1
∩B1
+
dv λ
y b veλ e
dλ
Z
dxdy −
lim y b ∂y veλ
∂Rn+1
∩B1 y→0
+
duλe
dλ
From (2.8) and by integration by parts we have
Z
Z
λ
duλ
b λ dve
y ve
=
y b ∆b uλe ∆b e
dλ
dλ
Rn+1
∩B1
Rn+1
∩B1
+
+
λ
λ
Z
Z
due
due
λ
b
λ b
= −
∇∆b ue · ∇
y +
∆b ue y ∂ν
n+1
n+1
dλ
dλ
R+ ∩B1
∂(R+ ∩B1 )
Note that
duλ
−
∇∆b ue · ∇ e y b
dλ
Rn+1
∩B1
+
Z
Z
=
Rn+1
∩B1
+
Z
=
Rn+1
∩B1
+
duλ
div(∇∆b uλe y b ) e
y b ∆2b uλe
Z
= −
∂(Rn+1
∩B1 )
+
duλe
−
dλ
Z
−
dλ
Z
∂(Rn+1
∩B1 )
+
∂(Rn+1
∩B1 )
+
y b ∂ν (∆b uλe )
y b ∂ν (∆b uλe )
y b ∂ν (∆b uλe )
duλe
dλ
duλe
dλ
Therefore,
Z
Rn+1
∩B1
+
y b veλ
dveλ
dλ
Z
=
∂(Rn+1
∩B1 )
+
∆b uλe y b ∂ν
duλe
dλ
Z
−
∂(Rn+1
∩B1 )
+
y b ∂ν (∆b uλe )
duλe
dλ
duλe
dλ
6
MOSTAFA FAZLY, JUNCHENG WEI
Boundary of Rn+1
∩ B1 consists of ∂Rn+1
∩ B1 and Rn+1
∩ ∂B1 . Therefore,
+
+
+
λ
Z
Z
λ
due
duλ
b λ dve
λ
b
y ve
=
+ lim y b ∂y veλ e
−ve lim y ∂y
y→0
y→0
dλ
dλ
dλ
Rn+1
∩B1
∂Rn+1
∩B1
+
+
λ
Z
λ
due
du
+
y b veλ ∂r
− y b ∂r veλ e
n+1
dλ
dλ
R+ ∩∂B1
where r = |X|, X = (x, y) ∈ Rn+1
and ∂r = ∇· Xr is
+
the corresponding radial derivative. Note that the first
duλ
e
integral in the right-hand side vanishes since ∂y dλ = 0 on ∂Rn+1
+ . From (2.10) we obtain
Z
duλe
dĒ(uλe , 1)
duλe
(2.11)
=
y b veλ ∂r
− ∂r veλ
dλ
dλ
dλ
Rn+1
∩∂B1
+
Now note that from the definition of uλe and veλ and by differentiating in λ we get the following for X ∈ Rn+1
+
λ
due (X)
1 2s + a λ
=
u (X) + r∂r uλe (X)
(2.12)
dλ
λ p−1 e
1 2(p + s − 1) + a λ
dveλ (X)
(2.13)
=
ve (X) + r∂r veλ (X)
dλ
λ
p−1
Therefore, differentiating with respect to λ we get
λ
d2 uλe (X) duλe (X)
2s + a duλe (X)
duλe (X)
+
=
+
r∂
r
dλ2
dλ
p−1
dλ
dλ
So, for all X ∈ Rn+1
∩ ∂B1
+
(2.14)
(2.15)
(2.16)
∂r uλe (X)
λ
due (X)
∂r
dλ
∂r veλ (X)
duλe (X) 2s + a λ
−
u (X)
dλ
p−1 e
d2 uλe (X) p − 1 − 2s − a duλe (X)
= λ
+
dλ2
p−1
dλ
= λ
= λ
dveλ (X) 2(p + s − 1) + a λ
−
ve (X)
dλ
p−1
Substituting (2.15) and (2.16) in (2.11) we get
2 λ
Z
dĒ(uλ , 1)
p − 1 − 2s − a duλe
dv λ
2(p + s − 1) + a λ duλe
d u
(2.17) e
=
y b veλ λ 2e +
− yb λ e −
ve
dλ
dλ
p−1
dλ
dλ
p−1
dλ
Rn+1
∩∂B1
+
Z
d2 uλe
duλ
dv λ duλe
=
y b λveλ
+ 3veλ e − λ e
2
dλ
dλ
dλ dλ
Rn+1
∩∂B1
+
Taking derivative of (2.12) in r we get
r
∂ 2 uλe
∂uλe
∂
+
=λ
2
∂r
∂r
∂r
duλe
dλ
−
2s + a ∂uλe
p − 1 ∂r
So, from (2.15) for all X ∈ Rn+1
∩ ∂B1 we have
+
∂ 2 uλe
p + 2s + a − 1 ∂uλe
∂ duλe
(2.18)
−
=
λ
∂r2
∂r dλ
p−1
∂r
2 λ
λ
d ue
p − 2s − 1 − a due
p + 2s + a − 1
duλe
2s + a λ
= λ λ 2 +
−
λ
−
u
dλ
p−1
dλ
p−1
dλ
p−1 e
= λ2
d2 uλe
4s + 2a duλe
(2s + a)(p + 2s + a − 1) λ
−
λ
+
ue
2
dλ
p−1
dλ
(p − 1)2
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
7
Note that
veλ = ∆b uλe = y −b div(y b ∇uλe )
and on Rn+1
∩ ∂B1 , we have
+
div(y b ∇uλe ) = (urr + (n + b)ur )θ1b + divS n (θ1b ∇S n uλe )
where θ1 = yr . From the above, (2.14) and (2.18) we get
veλ
= λ2
duλ
d2 uλe
4s + 2a
2s + a p + 2s + a − 1
+ λ e (n + b −
) + uλe (
)(
− n − b) + θ1−b divS n (θ1b ∇S n uλe )
2
dλ
dλ
p−1
p−1
p−1
From this and (2.17) we get
2 λ
duλe
d ue
d2 uλe
λ
+
αλ
+
βu
θ1b λ λ2
e
2
n+1
dλ
dλ
dλ2
R+ ∩∂B1
λ
Z
d2 uλe
duλe
due
λ
θ1b 3 λ2
+
(2.20)
+
αλ
+
βu
e
2
dλ
dλ
dλ
∩∂B1
Rn+1
+
Z
λ
2 λ
λ
due
b due d
2 d ue
λ
θ1 λ
−
λ
+ αλ
+ βue
(2.21)
dλ dλ
dλ2
dλ
Rn+1
∩∂B1
+
Z
d2 uλ
(2.22)
θ1b λ 2e θ1−b divS n (θ1b ∇S n uλe )
+
dλ
∩∂B1
Rn+1
+
Z
duλe −b
(2.23)
+
3θ1b
θ divS n (θ1b ∇S n uλe )
dλ 1
Rn+1
∩∂B1
+
Z
duλe
d
θ1b λ
−
(2.24)
θ1−b divS n (θ1b ∇S n uλe )
dλ
dλ
∩∂B1
Rn+1
+
2s+a p+2s+a−1
− n − b . Simplifying the integrals we get
where α := n + b − 4s+2a
p−1 and β := p−1
p−1
2 λ 2
λ 2 !
Z
λ
2 λ
dĒ(uλe , 1)
d
u
due
du
d
u
e
e
e
(2.25)
=
θ1b 2λ3
+ 2(α − β)λ
+ 4λ2
2
2 dλ
n+1
dλ
dλ
dλ
dλ
R+ ∩∂B1
!
!
λ 2
Z
1 d
β d2
due
β d λ 2
b
λ 2
3 d
θ1
+
λ(ue ) −
λ
+
(u )
2 dλ2
2 dλ
dλ dλ
2 dλ e
Rn+1
∩∂B1
+
Z
duλe
d2 uλ
duλ
d
divS n (θ1b ∇S n uλe )
+
λ 2e divS n (θ1b ∇S n uλe ) + 3 divS n (θ1b ∇S n uλe ) e − λ
dλ
dλ
dλ
dλ
Rn+1
∩∂B1
+
(2.19)
dĒ(uλe , 1)
dλ
Z
=
Note that from the assumptions we have α − β − 1 > 0, therefore the first term in the RHS of (2.25) is positive
that is
2 λ 2
λ 2
2 λ
2
λ 2
2 λ
λ
d ue
due
d ue
duλe
due
3
2 d ue due
2λ
+ 4λ
+ 2(α − β)λ
= 2λ λ 2 +
+ 2(α − β − 1)λ
>0
dλ2
dλ2 dλ
dλ
dλ
dλ
dλ
From this we have
!
λ 2 !
Z
2
dĒ(uλe , 1)
β
d
1
d
d
du
β
d
e
≥
θ1b
λ(uλe )2 −
λ3
+
(uλ )2
dλ
2 dλ2
2 dλ
dλ dλ
2 dλ e
Rn+1
∩∂B1
+
Z
duλe
d2 uλ
duλ
d
+
λ 2e divS n (θ1b ∇S n uλe ) + 3 divS n (θ1b ∇S n uλe ) e − λ
divS n (θ1b ∇S n uλe )
dλ
dλ
dλ
dλ
Rn+1
∩∂B1
+
=: R1 + R2 .
8
MOSTAFA FAZLY, JUNCHENG WEI
Note that the terms appeared in R1 are of the following form
d2
θ1b 2 λ(uλe )2
n+1
dλ
R+ ∩∂B1
"
λ 2 #
Z
due
3 d
b d
λ
θ1
dλ
dλ
dλ
Rn+1
∩∂B
1
+
Z
d
y b (uλe )2
n+1
dλ
R+ ∩∂B1
!
Z
4s+2a
d2
λ p−1 +2(s−1)−n
y b u2e
dλ2
Rn+1
∩∂B
λ
+
"
2 !#
Z
4s+2a
d
∂u
2s
+
a
e
+2s−3−n
3 d
−1
b
λ
λ p−1
λ ue +
y
dλ
dλ
p−1
∂r
Rn+1
∩∂Bλ
+
!
Z
4s+2a
d
λ2s−3+ p−1 −n
y b u2e
dλ
Rn+1
∩∂B
λ
+
Z
=
=
=
We now apply integration by parts to simplify the terms appeared in R2 .
λ
duλe
d2 uλe
d
b
λ
b
λ due
b
λ
n (θ ∇S n u ) + 3 divS n (θ ∇S n u )
n (θ ∇S n u )
div
−
λ
div
S
S
1
e
1
e
1
e
dλ2
dλ
dλ
dλ
∩∂B1
Rn+1
+
Z
2
duλ d2 uλe
duλ
−θ1b λ∇S n uλe · ∇S n
− 3θ1b ∇S n uλe · ∇S n e + θ1b λ ∇S n e 2
dλ
dλ
dλ
∩∂B1
Rn+1
+
!
!
Z
Z
Z
duλe 2
3 d
λ d2
b
λ 2
b
λ 2
b
θ1 |∇θ ue | −
θ1 |∇θ ue | + 2λ
θ 1 ∇θ
−
n+1
n+1
2 dλ2
2
dλ
dλ
Rn+1
∩∂B
R
∩∂B
R
∩∂B
1
1
1
+
+
+
!
!
Z
Z
Z
duλe 2
1 d
1 d2
b
λ 2
b
λ 2
b
θ
|∇
u
|
−
θ
|∇
u
|
+
2λ
θ
∇
λ
−
θ e
θ e
1
1
1 θ
2 dλ2
2 dλ
dλ ∩∂B1
∩∂B1
Rn+1
∩∂B1
Rn+1
Rn+1
+
+
+
!
!
Z
Z
1
d
1 d2
θ1b |∇θ uλe |2
λ
θ1b |∇θ uλe |2 −
−
2 dλ2
2 dλ
∩∂B1
Rn+1
Rn+1
∩∂B1
+
+
Z
R2
=
=
=
=
≥
λ
Note that the two terms that appear as lower bound for R3 are of the form
d2
dλ2
d
dλ
Z
λ
Rn+1
∩∂B1
+
Z
Rn+1
∩∂B1
+
!
θ1b |∇θ uλe |2
=
!
θ1b |∇θ uλe |2
=
"
2 !#
Z
∂u 2s(p+1)+2a
d2
−n
b
2
y |∇u| − λ p−1
2
dλ
∂r
∩∂Bλ
Rn+1
+
"
2 !#
Z
∂u 2s(p+1)+2a
d
−n−1
b
2
p−1
λ
y |∇u| − n+1
dλ
∂r
R+ ∩∂Bλ
2
Remark 2.1. It is straightforward to show that n >
2s(p+1)+2a
p−1
implies n >
p+4s+2a−1
p+2s+a−1
+
2s+a
p−1
− b.
3. Homogeneous Solutions
Theorem 3.1. Suppose that u = r
(3.1)
− 2s+a
p−1
p
ψ(θ) is a stable solution of (1.1) then ψ = 0 provided p >
Γ( n2 −
s+ a
2
s+ a
2
p−1
)Γ(s +
Γ( p−1 )Γ( n−2s
−
2
s+ a
2
p−1 )
a
s+ 2
p−1 )
>
2
Γ( n+2s
4 )
,
2
Γ( n−2s
4 )
n+2s+2a
n−2s
and
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
9
Proof. Since u satisfies (1.1), the function ψ satisfies
2s+a
2s+a
Z
|x|− p−1 ψ(θ) − |y|− p−1 ψ(σ)
− 2ps+ap
ap
p
p−1
dy
|x| |x|
ψ (θ) =
|x − y|n+2s
2s+a
2s+a
2s+a
Z
|x|− p−1 ψ(θ) − r− p−1 t− p−1 ψ(σ) n n−1
=
|x| t
dtdσ where |y| = rt
n+2s
(t2 + 1 − 2t < θ, σ >) 2 |x|n+2s
2s+a
Z
ψ(θ) − t− p−1 ψ(θ)
− 2ps+a
n−1
p−1
= |x|
[
dtdσ
n+2s t
(t2 + 1 − 2t < θ, σ >) 2
2s+a
Z
t− p−1 (ψ(θ) − ψ(σ)
n−1
dtdσ]
+
n+2s t
(t2 + 1 − 2t < θ, σ >) 2
We now drop |x|−
2ps+a
p−1
and get
Z
(3.2)
K 2s+a (< θ, σ >)(ψ(θ) − ψ(σ))dσ = ψ p (θ)
ψ(θ)An,s,a (θ) +
p−1
Sn−1
where
Z
∞
2s+a
1 − t− p−1
Z
An,s,a :=
(t2 + 1 − 2t < θ, σ >)
Sn−1
0
n+2s
2
tn−1 dσdt
and
p−1
2s
∞
Z
K 2s+a (< θ, σ >) :=
0
tn−1− p−1
(t2 + 1 − 2t < θ, σ >)
n+2s
2
dt
Note that
Z
K 2s+a (< θ, σ >)
p−1
=
0
We now set Kα (< θ, σ >) =
R1
0
Z
(t2 + 1 − 2t < θ, σ >)
0
Z
2s+a
tn−1− p−1
1
=
1
2s+a
n+2s dt +
(t2 + 1 − 2t < θ, σ >)
tn−1+α +t2s−1+α
n+2s
2
tn−1− p−1
(t2 + 1 − 2t < θ, σ >)
n+2s
2
dt
2s+a
tn−1− p−1 + t2s−1+ p−1
(t2 +1−2t<θ,σ>)
1
2
2s+a
∞
n+2s
2
dt
dt. The most important property of the Kα is that Kα is
decreasing in α. This can be seen by the following elementary calculations
Z 1 n−1−α
−t
ln t + t2s−1+α ln t
∂α Kα =
n+2s dt
0 (t2 + 1 − 2t < θ, σ >) 2
Z 1
ln t(−tn−1−α + t2s−1+α )
=
n+2s dt < 0
0 (t2 + 1 − 2t < θ, σ >) 2
For the last part we have used the fact that for p > n+2s+2a
we have 2s − 1 + α < n − 1 − α.
n−2s
From (3.2) we get the following
Z
Z
Z
2
2
(3.3)
ψ (θ)An,s,a +
K 2s+a (< θ, σ >)(ψ(θ) − ψ(σ)) dθdσ =
ψ p+1 (θ)dθ
Sn−1
Sn−1
p−1
Sn−1
We set a standard cut-off function η ∈
at the origin and at infinity that is η = 1 for < r < −1 and
n−2s
η = 0 for either r < /2 or r > 2/. We test the stability (1.2) on the function φ(x) = r− 2 ψ(θ)η (r).
Note that
Z
Z Z
n−2s
n−2s
φ(x) − φ(y)
r− 2 ψ(θ)η (r) − |y|− 2 ψ(σ)η (|y|)
dy
=
dσd(|y|)
n+2s
n+2s
(r2 + |y|2 − 2r|y| < θ, σ >) 2
Rn |x − y|
Sn−1
Cc1 (R+ )
10
MOSTAFA FAZLY, JUNCHENG WEI
Now set |y| = rt then
Z
Rn
φ(x) − φ(y)
dy
|x − y|n+2s
=
n
r− 2 −s
∞
Z
n
r− 2 −s
ψ(θ)η (r) − t−
Z Z
n−2s
2
n
r− 2 −s η (r)ψ(θ)
∞
Z
n
∞
Z
n
Z
∞
Define Λn,s :=
R∞R
1−t
Sn−1
0
Z
Rn
n−2s
2
n+2s
2
n−2s
2
n−2s
2
n−2s
2
n+2s
2
(ψ(θ) − ψ(σ))
(t2 + 1 − 2t < θ, σ >)
tn−1−
n−2s
2
φ(x) − φ(y)
dy
|x − y|n+2s
tn−1−
n+2s
2
(η (r) − η (rt))ψ(σ)
(t2 + 1 − 2t < θ, σ >)
Sn−1
(t2 +1−2t<θ,σ>)
tn−1 dtdσ
(t2 + 1 − 2t < θ, σ >)
Sn−1
Sn−1
Z
0
1−t
Z
0
+r− 2 −s
n+2s
2
ψ(σ)η (r) + t−
Z
0
+r− 2 −s η (r)
ψ(σ)η (rt)
(η(r)ψ(θ) − η (rt)ψ(σ))
(t2 + 1 − 2t < θ, σ >)
Sn−1
=
n−2s
2
(t2 + 1 − 2t < θ, σ >)
Sn−1
0
=
ψ(θ)η (r) − t−
Z
n+2s
2
n+2s
2
tn−1 dtdσ
tn−1 dtdσ
dtdσ
dtdσ
tn−1 dσdt. Therefore,
n
= r− 2 −s η (r)ψ(θ)Λn,s
Z
n
K n−2s (< θ, σ >)(ψ(θ) − ψ(σ))dσ
+r− 2 −s η (r)
2
Sn−1
n
+r− 2 −s
Z
0
∞Z
Sn−1
t
− n−2s
2
(t2
(η (r) − η (rt))ψ(σ)
+ 1 − 2t < θ, σ >)
n+2s
2
dtdσ
Applying the above, we compute the left-hand side of the stability inequality (1.2),
Z Z
Z Z
(φ(x) − φ(y))2
(φ(x) − φ(y))φ(x)
dxdy
=
2
dxdy
n+2s
|x
−
y|
|x − y|n+2s
n
n
n
n
R
R
R
R
Z ∞
Z
= 2
r−1 η2 (r)dr
ψ 2 Λn,s dθ
n−1
0
S
Z ∞
Z
−1 2
+2
r η (r)dr
K n−2s (< θ, σ >)(ψ(θ) − ψ(σ))2 dσdθ
Sn−1
0
Z
(3.4)
∞ Z
+2
∞
r
0
2
−1
Z
η (r)(η (r) − η (rt))dr
Sn−1
0
tn−1−
Z
Sn−1
(t2
n−2s
2
+ 1 − 2t < θ, σ >)
We now compute the second term in the stability inequality (1.2) for the test function φ(x) = r−
2s
and u = r− p−1 ψ(θ),
Z ∞
Z ∞
a
p−1 2
p
r |u| φ = p
ra r−(2s+a) r−(n−2s) ψ p+1 η2 (r)dr
0
Z0 ∞
Z
(3.5)
= p
r−1 η2 (r)dr
ψ p+1 (θ)dθ
0
R∞
ψ(σ)ψ(θ)
n−2s
2
n+2s
2
ψ(θ)η (r)
Sn−1
Due to the definition of the η , we have 0 r−1 η2 (r)dr = ln(2/) + O(1). Note that this term appears in both
terms of the stability inequality that we computed in (3.4) and (3.6). We now claim that
Z ∞
f (t) :=
r−1 η (r)(η (r) − η (rt))dr = O(ln t)
0
dσdθdt
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
11
1
2
Note that η (rt) = 1 for t < r < t
and η (rt) = 0 for either r < 2t
or r > t
. Now consider various ranges of
value of t ∈ (0, ∞) to compare the support of η (r) and η (rt). From the definition of η , we have
Z 2
r−1 η (r)(η (r) − η (rt))dr
f (t) =
2
In what follows we consider a few cases to explain the claim. For example when <
Z t
Z t2
−1
f (t) ≈
r dr +
r−1 dr ≈ ln t
2
Now consider the case
1
<
t
<
t
<
1
then
1
1
then t ≈ 2 . So,
Z t
Z
−1
f (t) ≈
r dr +
2
2
r−1 dr ≈ ln t + ln ≈ ln t
t
Other cases can be treated similarly. From this one can see that
Z
Z ∞ Z ∞
Z
n−2s
tn−1− 2
(3.6)
r−1 η(r)(η(r) − η(rt))dr
n+2s ψ(σ)ψ(θ)dσdθdt
0
0
Sn−1 Sn−1 (t2 + 1 − 2t < θ, σ >) 2
Z
Z
Z ∞
n−2s
tn−1− 2 ln t
(3.7)
≈
n+2s ψ(σ)ψ(θ)dtdσdθ
(t2 + 1 − 2t < θ, σ >) 2
Sn−1 Sn−1 0
(3.8)
= O(1)
Collecting higher order terms of the stability inequality we get
Z
Z
Z
(3.9)
Λn,s
ψ2 +
K n−2s (< θ, σ >)(ψ(θ) − ψ(σ))2 dσ ≥ p
Sn−1
From this and (3.3) we obtain
Z
(Λn,s − pAn,s,a )
2
Sn−1
ψ2 +
Z
Sn−1
ψ p+1
Sn−1
(K n−2s − pK 2s+a )(< θ, σ >)(ψ(θ) − ψ(σ))2 dσ ≥ 0
2
Sn−1
p−1
n−2s − pK 2s+a < 0. On
Note that Kα is decreasing in α. This implies K n−2s < K 2s+a for p > n+2s+2a
n−2s . So, K 2
2
p−1
p−1
the other hand the assumption of the theorem implies that Λn,s − pAn,s,a < 0. Therefore, ψ = 0.
4. Energy Estimates
In this section, we provide some estimates for solutions of (1.1). These estimates are needed in the next
section when we perform a blow-down analysis argument. The methods and ideas provided in this section are
strongly motivated by [9, 10].
Lemma 4.1. Let u be a stable solution to (1.1). Let also η ∈ Cc∞ (Rn ) and for x ∈ Rn , define
Z
(η(x) − η(y))2
dy
(4.1)
ρ(x) =
|x − y|n+2s
Rn
Then,
Z
(4.2)
Rn
|x|a |u|p+1 η 2 dx +
Z
Rn
Z
Rn
|u(x)η(x) − u(y)η(y)|2
dxdy ≤ C
|x − y|n+2s
Proof. Proof is quite similar to Lemma 2.1 in [9] and we omit it here.
Z
u2 ρdx
Rn
12
MOSTAFA FAZLY, JUNCHENG WEI
Lemma 4.2. Let m > n/2 and x ∈ Rn . Set
Z
(η(x) − η(y))2
(4.3)
ρ(x) =
dy where η(x) = (1 + |x|2 )−m/2
|x − y|n+2s
Rn
Then there is a constant C = C(n, s, m) > 0 such that
C −1 (1 + |x|2 )−n/2−s ≤ ρ(x) ≤ C(1 + |x|2 )−n/2−s
(4.4)
Proof. Proof is quite similar to Lemma 2.2 in [9] and we omit it here.
Corollary 4.1. Suppose that m > n/2, η given by (4.3) and R > 1. Define
Z
(ηR (x) − ηR (y))2
(4.5)
ρR (x) =
dy where ηR (x) = η(x/R)φ(x)
|x − y|n+2s
Rn
where φ ∈ C ∞ (Rn ) ∩ [0, 1] is a cut-off function. Then there exists a constant C > 0 such that
x 2
x
ρR (x) ≤ Cη
|x|−n−2s + R−2s ρ
R
R
Lemma 4.3. Suppose that u is a stable solution of (1.1). Consider ρR that is defined in Corollary 4.1 for
n/2 < m < n/2 + s(p + 1)/2. Then there exists a constant C > 0 such that
Z
2s(p+1)+2a
u2 ρR ≤ CRn− p−1
Rn
for any R > 1
Proof. Note that
Z
2
Z
a
u ρR dx ≤
Rn
|x|
Rn
2
p+1
Z
2
||u|p+1 ηR
dx
|x|
2a
− p−1
Rn
p+1
p−1
ρR
p−1
p+1
− 4
ηR p−1 dx
From Lemma 4.1 we get
Z
u2 ρR dx ≤
Rn
Z
Rn
2a
p+1
−
4
|x|− p−1 ρRp−1 ηR p−1 dx
Now applying Corollary 4.1 for two different cases |x| > R and |x| < R one can get ρR (x) ≤ C(|x|−n−2s + R−2s )
−n2/−s
2
and ρ(x) ≤ CR−2s 1 + |x|
. This finishes the proof.
2
R
Note that
We are now ready to state the essential estimate on stable solutions. Since the proofs are similar to the ones
given in [9], for the case of 0 < s < 1, and in [16], for the case of 1 < s < 2, we omit them here.
Lemma 4.4. Suppose that p 6= n+2s+2a
n−2s . Let u be a stable solution of (1.1) and ue satisfies (1.5). Then there
exists a constant C > 0 such that
(i) for 0 < s < 1
Z
2s(p+1)+2a
y 1−2s u2e ≤ CRn+2− p−1
BR
and
(ii) for 1 < s < 2
Z
y 3−2s u2e ≤ CRn+4−
2s(p+1)+2a
p−1
BR
Lemma 4.5. Let u be a stable solution of (1.1) and ue satisfies (1.5). Then there exists a positive constant C
such that
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
(i) for 0 < s < 1
Z
(4.6)
|x|a |ue |p+1 dx +
BR ∩∂Rn+1
+
and
(ii) for 1 < s < 2
Z
(4.7)
a
BR ∩∂Rn+1
+
p+1
|x| |ue |
Z
BR ∩Rn+1
+
Z
dx +
BR ∩Rn+1
+
y 1−2s |∇ue |2 dxdy ≤ CRn−
13
2s(p+1)+2a
p−1
y 3−2s |∆b ue |2 dxdy ≤ CRn−
2s(p+1)+2a
p−1
5. Blow-down analysis
This section is devoted to the proof of Theorem 1.2. The methods and ideas are strongly motivated by the
ones given in [9, 10].
Proof of Theorem 1.2: Let u be a stable solution of (1.1) and let ue be its extension solving (1.5). For the case
1 < p ≤ pS (n, a) the conclusion follows from the Pohozaev identity. Note that for the subcritical case Lemma
4.5 implies that u ∈ Ḣ s (Rn ) ∩ Lp+1 (Rn ). Multiplying (1.1) with u and doing integration, we obtain
Z
(5.1)
|x|a u|p+1 = ||u||2Ḣ s (Rn )
Rn
in addition multiplying (1.1) with uλ (x) = u(λx) yields
Z
Z
Z
|x|a |u|p−1 uλ =
(−∆)s/2 u(−∆)s/2 uλ = λs
Rn
Rn
wwλ
Rn
√
where w = (−∆)s/2 u. Following ideas provided in [10, 26] and using the change of variable z = λx one can
get the following Pohozaev identity
Z
Z
Z
√
√
n+a
2s − n
2s − n
d
a
p+1
2
−
|x| |u|
=
w +
|λ=1
||u||2Ḣ s (Rn )
w λ w1/ λ dz =
p + 1 Rn
2
dλ
2
n
n
R
R
This equality together and (5.1) proves the theorem for the subcritical case.
Now suppose that p > pS (n, a).
Case 1: 0 < s < 1. We perform the proof in a few steps.
Step 1. limλ→+∞ E(ue , λ) < +∞. From the fact that E is nondecreasing in λ, it suffices to show that E(ue , λ)
is bounded. Write E = I + J, where I is given by (2.1) and
Z
2s(p+1)+2a
s+a
J(ue , λ) = λ p−1 −n−1
y 1−2s u2e dσ
p + 1 ∂Bλ ∩Rn+1
+
Note that Lemma 4.5 implies that I is bounded. To show that E is bounded we state the following argument.
The nondecreasing property of E yields
Z
Z
2s(p+1)+2a
1 2λ
E(ue , λ) ≤
E(u, t)dt ≤ C + λ p−1 −n−1
y 1−2s u2e .
λ λ
B2λ ∩Rn+1
+
From Lemma 4.4 we conclude that E is bounded.
1
1−2s
Step 2. There exists a sequence λi → +∞ such that (uλe i ) converges weakly in Hloc
(Rn+1
dydx) to a
+ ;y
∞
function ue .
1
1−2s
This follows from the fact that (uλe i ) is bounded in Hloc
(Rn+1
dxdy) by Lemma 4.5.
+ ;y
∞
Step 3. ue is homogeneous.
14
MOSTAFA FAZLY, JUNCHENG WEI
To see this, apply the scale invariance of E, its finiteness and the monotonicity formula: given R2 > R1 > 0,
0
lim E(ue , λi R2 ) − E(ue , λi R1 )
=
n→+∞
lim E(uλe i , R2 ) − E(uλe i , R1 )
=
n→+∞
Z
≥
n→+∞
2
2s + a uλe i
∂uλe i
+
dxdy
p−1 r
∂r
2
∂u∞
2s + a u∞
e
e
dxdy
+
p−1 r
∂r
y 1−2s r2−n+
lim inf
(BR2 \BR1 )∩Rn+1
+
Z
y 1−2s r2−n+
≥
4s+2a
p−1
(BR2 \BR1 )∩Rn+1
+
4s+2a
p−1
n+1 1−2s
1
Note that in the last inequality we only used the weak convergence of (uλe i ) to u∞
dxdy).
e in Hloc (R+ ; y
So,
2s + a u∞
∂u∞
e
e
+
= 0 a.e. in Rn+1
+ .
p−1 r
∂r
And so, u∞
e is homogeneous.
Step 4. u∞
e ≡ 0. This is a direct consequence of Theorem 3.1.
Step 5. (uλe i ) converges strongly to zero in H 1 (BR \ Bε ; y 1−2s dxdy) and (uλi ) converges strongly to zero in
Lp+1 (BR \ Bε ) for all R > > 0.
1−2s
1
(Rn+1
dxdy) and converges weakly to 0.
From Step 2 and Step 3, we have (uλe i ) is bounded in Hloc
+ ;y
n+1 1−2s
2
λi
dxdy). By the standard Rellich-Kondrachov
Therefore, (ue ) converges strongly to zero in Lloc (R+ ; y
theorem and a diagonal argument, passing to a subsequence, for any BR = BR (0) ⊂ Rn+1 and A of the form
A = {(x, t) ∈ Rn+1
: 0 < t < r/2}, where R, r > 0 we obtain
+
Z
y 1−2s |uλe i |2 dxdy → 0.
lim
i→∞
∩(BR \A)
Rn+1
+
By [12, Theorem 1.2],
Z
Rn+1
∩Br (x)
+
y 1−2s |uλe i |2 dxdy ≤ Cr2
Z
∩Br (x)
Rn+1
+
y 1−2s |∇uλe i |2 dxdy
λi
for any x ∈ ∂Rn+1
+ , |x| ≤ R, with a uniform constant C. Applying similar arguments as [9] one can get (ue )
p+1
n+1
1
1−2s
n
converges strongly to 0 in Hloc (R+ \ {0}; y
dxdy) and the convergence also holds in Lloc (R \ {0}).
Step 6.
ue ≡ 0.
I(ue , λ)
= I(uλe , 1)
Z
Z
κs
1
=
y 1−2s |∇uλe |2 dxdy −
|x|a |uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B1
∂R
∩B1
Z +
Z +
1
κ
s
=
y 1−2s |∇uλe |2 dxdy −
|x|a |uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B
∂R
∩B
+
Z+
Z
κs
1
1−2s
λ 2
+
y
|∇ue | dxdy −
|x|a |uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B
\B
∂R
∩B
\B
1
1
+
+
Z
Z
2s(p+1)+2a
1
κs
n−
1−2s
λ 2
p−1
= ε
I(ue , 0, λε) +
y
|∇ue | dxdy −
|x|a |uλe |p+1 dx
2 Rn+1
p + 1 ∂Rn+1
∩B1 \B
∩B1 \B
+
+
Z
Z
2s(p+1)+2a
1
κs
≤ Cεn− p−1
+
y 1−2s |∇uλe |2 dxdy −
|x|a |uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B
\B
∩B
\B
∂R
1
1
+
+
Letting λ → +∞ and then ε → 0, we deduce that limλ→+∞ I(ue , λ) = 0. Using the monotonicity of E,
Z
Z
2s(p+1)+2s
1 2λ
(5.2)
E(ue , λ) ≤
E(t) dt ≤ sup I + Cλ−n−1+ p−1
u2e
λ λ
[λ,2λ]
B2λ \Bλ
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
15
and so limλ→+∞ E(ue , λ) = 0. Since u is smooth, we also have E(ue , 0) = 0. Since E is monotone, E ≡ 0 and
so ue must be homogeneous, a contradiction unless ue ≡ 0.
Case 2: 1 < s < 2. Proof of this case is very similar to Case 1. We perform the proof in a few steps.
Step 1. limλ→∞ E(ue , λ) < ∞.
From Theorem 2.2, E is nondecreasing. So, we only need to show that E(ue , λ) is bounded. Note that
E(ue , λ) ≤
1
λ
Z
2λ
E(ue , t)dt ≤
λ
1
λ2
Z
2λ
Z
t+λ
E(ue , γ)dγdt
t
λ
From Lemma 4.5 we conclude that
1
λ2
Z
2λ
t+λ
Z
γ
λ
p+1
2s p−1
−n
Z
∩Bγ
Rn+1
+
t
1 3−2s
Cn,s
y
|∆b ue |2 dydx −
2
p+1
!
Z
∩Bγ
∂Rn+1
+
|x|a up+1
dx dγdt ≤ C
e
where C > 0 is independent from λ. For the next term in the energy we have
1
λ2
Z
2λ
Z
t+λ
γ
λ
−3+2s+ 4s+2a
p−1 −n
Z
∩∂Bγ
Rn+1
+
t
!
y 3−2s u2e dydx
dγdt ≤
≤
≤
≤
1
λ2
Z
1
λ2
Z
2λ
t−3+2s+
Z
y 3−2s u2e dydxdt
Bt+λ \Bt
λ
2λ
t−3+2s+
4s+2a
p−1 −n
λ
2s(p+1)+2a
n+4−
p−1
λ
4s+2a
p−1 −n
Z
y 3−2s u2e dydx dt
B3λ
1
λ2
Z
2λ
t−3+2s+
4s+2a
p−1 −n
dt
λ
C
where C > 0 is independent from λ. In the above estimates we have applied Lemma 4.4. For the next term we
have
"
2 #
Z
4s+2a
2s
+
a
γ3 d
∂u
e
y 3−2s
γ 2s−3−n+ p−1
γ −1 ue +
dγdt
2 dγ
p−1
∂r
∂Bγ
λ
t
2
Z 2λ
Z
2s + a
∂ue
1
2s−n+ 4s+2a
3−2s
−1
p−1
y
(t + λ) ue +
[(t + λ)
=
2λ2 λ
p−1
∂r
∂Bt+λ
2
Z
4s+2a
2s + a −1
∂ue
−t2s−n+ p−1
y 3−2s
γ ue +
]dt
p
−
1
∂r
∂Bλ
"
2 #
Z 2λ Z t+λ
Z
4s+2a
3
2s
+
a
∂u
e
− 2
y 3−2s
γ −1 ue +
dγdt
γ 2s−1−n+ p−1
2λ λ
p−1
∂r
∂Bγ
t
2
Z
2s + a −1
∂ue
−2+2s−n+ 4s+2a
3−2s
p−1
≤ λ
y
λ ue +
≤C
p−1
∂r
B3λ \Bλ
1
λ2
Z
2λ
Z
t+λ
where C > 0 is independent from λ. The rest of the terms can be treated similarly.
1
Step 2. There exists a sequence λi → ∞ such that (uλe i ) converges weakly in Hloc
(Rn , y 3−2s dxdy) to a function
∞
ue .
Note that this is a direct consequence of Lemma 4.5.
∞
Step 3. u∞
e is homogeneous and therefore ue = 0.
16
MOSTAFA FAZLY, JUNCHENG WEI
To prove this claim, apply the scale invariance of E, its finiteness and the monotonicity formula; given
R2 > R1 > 0,
0
=
=
≥
≥
lim (E(ue , R2 λi ) − E(ue , R1 λi ))
lim E(uλe i , R2 ) − E(uλe i , R1 )
i→∞
2
Z
4s+2a
2s + a −1 λi ∂uλe i
lim inf
y 3−2s r p−1 +2s−2−n
r ue +
dydx
i→∞
p−1
∂r
(BR2 \BR1 )∩Rn+1
+
2
Z
4s+2a
2s + a −1 ∞ ∂u∞
e
dydx
r ue +
y 3−2s r p−1 +2s−2−n
p−1
∂r
(BR2 \BR1 )∩Rn+1
+
i→∞
1
n 3−2s
In the last inequality we have used the weak convergence of (uλe i ) to u∞
dydx). This implies
e in Hloc (R , y
2s + a −1 ∞ ∂u∞
e
r ue +
= 0 a.e. in Rn+1
+ .
p−1
∂r
∞
Therefore, u∞
e is homogeneous. Apply Theorem 3.1 we get ue = 0.
λi
1
3−2s
Step 5. (ue ) converges strongly to zero in H (BR \ B , y
dydx) and (uλe i ) converges strongly to zero in
p+1
L (BR \ B ) for all R > > 0.
Step 6. ue ≡ 0.
I(ue , λ)
=
=
=
=
≤
I(uλe , 1)
Z
Z
1
κs
y 3−2s |∆b uλe |2 dxdy −
|x|a |uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B
∂R
∩B
1
1
Z +
Z +
1
κs
3−2s
λ 2
y
|∆b ue | dxdy −
|x|a |uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B
∂R+ ∩B
Z+
Z
κ
1
s
3−2s
λ 2
y
|∆b ue | dxdy −
|x|a |uλe |p+1 dx
+
n+1
2 Rn+1
p
+
1
∩B
\B
∩B
\B
∂R
1
1
+
+
Z
Z
2s(p+1)+2a
κs
1
n−
3−2s
p−1
ε
y
|∆b uλe |2 dxdy −
|x|a |uλe |p+1 dx
I(ue , λε) +
n+1
2 Rn+1
p
+
1
∩B1 \B
∂R+ ∩B1 \B
+
Z
Z
2s(p+1)+2a
κ
1
s
y 3−2s |∆uλe |2 dxdy −
|x|a |uλe |p+1 dx
Cεn− p−1
+
n+1
2 Rn+1
p
+
1
∩B
\B
∩B
\B
∂R
1
1
+
+
Letting λ → +∞ and then ε → 0, we deduce that limλ→+∞ I(ue , λ) = 0. Using the monotonicity of E,
Z
Z
2s(p+1)+2a
1 2λ
(5.3)
E(ue , λ) ≤
E(t) dt ≤ sup I + Cλ−n−1+ p−1
u2e
λ λ
[λ,2λ]
B2λ \Bλ
and so limλ→+∞ E(ue , λ) = 0. Since u is smooth, we also have E(ue , 0) = 0. Since E is monotone, E ≡ 0 and
so ue must be homogeneous, a contradiction unless ue ≡ 0.
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MOSTAFA FAZLY, JUNCHENG WEI
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Department of Mathematical and Statistical Sciences, CAB 632, University of Alberta, Edmonton, Alberta,
Canada T6G 2G1
E-mail address: fazly@ualberta.ca
Department of Mathematics, University of British Columbia, Vancouver, B.C. Canada V6T 1Z2.
E-mail address: jcwei@math.ubc.ca