ON FINITE MORSE INDEX SOLUTIONS OF HIGHER ORDER FRACTIONAL
LANE-EMDEN EQUATIONS
MOSTAFA FAZLY AND JUNCHENG WEI
Abstract. We classify finite Morse index solutions of the following nonlocal Lane-Emden equation
(−∆)s u = |u|p−1 u
Rn
for 1 < s < 2 via a novel monotonicity formula. For local cases s = 1 and s = 2 this classification is provided
by Farina in [10] and Davila, Dupaigne, Wang and Wei in [8], respectively. Moreover, for the nonlocal case
0 < s < 1 finite Morse index solutions are classified by Davila, Dupaigne and Wei in [7].
1. Introduction and Main Results
We study the classification of stable solutions of the following equation
(−∆)s u = |u|p−1 u Rn
(1.1)
where (−∆)s is the fractional Laplacian operator for 1 < s < 2. For various parameters s and p this equation
has been of attention of many experts in the field of partial differential equations.
1.1. The local case. For the case of s = 1, a celebrated result of Gidas and Spruck in [12] shows that the
only nonnegative solution solution of the Lane-Emden equation is u = 0 for 1 < p < pS where
∞
if n ≤ 2,
pS (n) =
n+2
if n > 2,
n−2
that is called the Sobolev exponent. In addition, for the critical case p = pS (n) it is shown by CaffarelliGidas-Spruck [1] that there is a unique (up to translation and rescaling) positive solution for the Lane-Emden
equation. For finite Mose index solutions (not necessarily positive), such classification is provided by Farina
in [10] and the critical exponent, called Joseph-Lundgren [16] exponent, is given by
(
∞
if n ≤ 10,
√
pc (n) =
(1.2)
(n−2)2 −4n+8 n−1
if
n ≥ 11,
(n−2)(n−10)
Note that pc (n) > pS (n) for n > 2.
For the case of s = 2, Wei and Xu [19] (see also Lin [15]) proved that the only nonnegative solution of
the fourth order Lane-Emden equation is u = 0 for 1 < p < pS where pS (n) is the Sobolev exponent, i.e.
∞
if n ≤ 4,
(1.3)
pS (n) =
n+4
if
n > 4.
n−4
Moreover, for the critical case p = pS (n) they showed that there is a unique (up to translation and rescaling)
positive solution for the fourth order Lane-Emden equation. For finite Mose index solutions (not necessarily
positive), Davila, Dupaigne, Wang and Wei in [8] gave a complete classification. The Joseph-Lundgren
exponent, computed by Gazzola and Grunau in [11], is the following


∞
if n ≤ 12,
√
√
(1.4)
pc (n) =
n+2− n2 +4−n n2 −8n+32
√
if n ≥ 13,

√
2
2
n−6−
n +4−n n −8n+32
The first author is pleased to acknowledge the support of a University of Alberta start-up grant. Both authors are supported by Natural Sciences and Engineering Research Council of Canada (NSERC) grants. We thank Pacific Institute for the
Mathematical Sciences (PIMS) for hospitality.
1
The key idea of the proof of Davila, Dupaigne, Wang and Wei in [8] is proving and applying a monotonicity
formula. Note that a monotonicity formula for the second order equation is established by F. Pacard in [17].
We also refer the interested readers to Wei-Xu in [19] for classification of solutions of higher order conformally invariant equations, i.e. s any positive integer.
1.2. The nonlocal case. Assume that u ∈ C 2σ (Rn ), σ > s > 0 and
Z
|u(y)|
dy < ∞
(1
+
|y|)n+2s
n
R
so the fractional Laplacian of u
Z
s
(1.5)
(−∆) u(x) := p.v.
Rn
u(x) − u(y)
dy
|x − y|n+2s
n
is well-defined for every x ∈ R .
For the case of 0 < s < 1, a counterpart of the classification results of Gidas-Spruck [12] and CaffarelliGidas-Spruck [1] holds for the fractional Lane-Emden equation, see the work of Li [14] and Chen-Li-Ou [5].
In this case, the Sobolev exponent is the following
∞
if n ≤ 2s,
(1.6)
pS (n, s) =
n+2s
if n > 2s.
n−2s
Very recently, for the case of 0 < s < 1, Davila, Dupaigne and Wei [7] gave a complete classification of finite
Morse index solutions of (1.1) via proving and applying a monotonicity formula. As a matter of fact, they
proved that for either 1 < p < pS (n, s) or p > pS (n, s) and
p
Γ( n2 −
s
s
p−1 )Γ(s + p−1 )
s
s
Γ( p−1
)Γ( n−2s
− p−1
)
2
>
2
Γ( n+2s
4 )
2
Γ( n−2s
4 )
the only finite More index solution is zero. In this work, we are interested in knowing whether such classification results hold for finite Morse index solutions of (1.1) when 1 < s < 2.
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 [18]. Recently, Yang in [21]
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. See also Case-Chang [3] and Chang-Gonzales [4].
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 ).
b
Theorem 1.1. [21] Let 1 < s < 2. For functions ue ∈ W 2,2 (Rn+1
+ , y ) satisfying the equation
∆2b ue = 0
on the upper half space for (x, y) ∈ Rn × R+ where y is the special direction, and the boundary conditions
ue (x, 0)
b
lim y ∂y ue (x, 0)
y→0
= f (x)
=
0
along {y = 0} where f (x) is some function defined on H s (Rn ) we have the result that
(−∆)s f (x) = Cn,s lim y b ∂y ∆b ue (x, y)
y→0
2
Moreover,
Z
ˆ 2 dξ = Cn,s
|ξ|2s |u(ξ)|
Z
Rn+1
+
Rn
Applying the above theorem to solutions of
x = (x1 , · · · , xn ) and y ∈ R+ satisfies

∆2b ue

b
(1.7)
limy→0 y ∂y ue

limy→0 y b ∂y ∆b ue
y b |∆b ue (x, y)|2 dxdy
(1.1) we conclude that the extended function ue (x, y) where
= 0 in Rn+1
+ ,
= 0 in ∂Rn+1
+ ,
= Cn,s |u|p−1 u in ∂Rn+1
+
Moreover,
Z
ˆ 2 dξ = Cn,s
|ξ| |u(ξ)|
2s
Z
Rn+1
+
Rn
y b |∆b ue (x, y)|2 dxdy
Then u(x) = ue (x, 0).
For 1 < s < 2, Chen et al in [6] have classified all positive solutions of (1.1) for 1 < p ≤ pS (n, s). The
main goal of this paper is to classify all (positive or sign-changing) solutions of (1.1) which are stable outside
a compact set. To this end, we first introduce the corresponding Joseph-Lungren’s exponent. As it is shown
by Herbst in [13] (and also [20]), for 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 )
Definition 1.1. We say that a solution u of (1.1) is stable outside a compact set if there exists R0 > 0 such
that
Z Z
Z
(φ(x) − φ(y))2
(1.8)
dxdy
−
p
|u|p−1 φ2 ≥ 0
|x − y|n+2s
Rn Rn
Rn
for any φ ∈ Cc∞ (Rn \ BR0 ).
In the following lemma we provide an explicit singular solution for (1.1).
Lemma 1.1. Suppose that 1 < s < 2 and p > pS (n, s) then
2s
us (x) = A|x|− p−1
(1.9)
where
Ap−1 =
Γ( n2 −
s
s
p−1 )Γ(s + p−1 )
s
s
Γ( p−1
)Γ( n−2s
− p−1
)
2
solves (1.1).
Proof. From Lemma 3.1 in [9], we conclude that when 0 < t < 1, for any − n+2t
<β<
2
(−∆)t |x|
(1.10)
2t−n
+β
2
= γt (β)|x|
n−2t
2
2t−n
+β−2t
2
where
γt (β) = 22t
(1.11)
Γ( n+2t+2β
)Γ( n+2t−2β
)
4
4
Γ( n−2t−2β
)Γ( n−2t+2β
)
4
4
From the fact that (−∆)s = (−∆)o(−∆)t for 0 < t = s − 1 < 1 we have
(1.12)
where ηt =
(1.13)
(1.14)
(−∆)s |x|
2t−n
2
2t−n
+β
2
= γt (β)(−∆)|x|
2t−n
+β−2t
2
= −γt (β)ηt (n + ηt − 2)|x|
2t−n
+β−2t−2
2
+ β − 2t. Now using the change of variable t = s − 1 we get
(−∆)s |x|
2s−n
+β−1
2
=
=
2s−n
−γs−1 (β)ηs−1 (n + ηs−1 − 2)|x| 2 +β−2s−1
2s−n
p
−γs−1 (β)ηs−1 (n + ηs−1 − 2) |x| 2 +β−1
3
where
2s−n
+β−2s−1
2
2s−n
+β−1
2
= p. From this we conclude that β =
−2s
p−1
+
n−2s
2
+ 1. This implies
2s
us (x) = A|x|− p−1
(1.15)
where
Ap−1 = λ
n − 2s
2s
−
+1
2
p−1
is a solution of (1.1) for
λ(β) = −γs−1 (β)ηs−1 (n + ηs−1 − 2)
(1.16)
Elementary calculations show that
(1.17)
γs−1 (β)
=
22s−2
Γ( n2 −
s
s
p−1 )Γ(s + p−1 − 1)
s
s
Γ( p−1
)Γ( n−2s
− p−1
+ 1)
2
and
(1.18)
−ηs−1 (n + ηs−1 − 2) = 4 s +
s
−1
p−1
n − 2s
s
−
2
p−1
From (1.17) and (1.18) and using the property aΓ(a) = Γ(a + 1) we conclude the desired result.
Here is our main result.
Theorem 1.2. Suppose that n ≥ 1 and 1 < s < δ < 2. Let u ∈ C 2δ (Rn ) ∩ L1 (Rn , (1 + |y|)n+2s dy) be a
solution of (1.1) that is stable outside a compact set. Then either for 1 < p < pS (n, s) or for p > pS (n, s)
and
s
s
2
Γ( n2 − p−1
)Γ(s + p−1
)
Γ( n+2s
4 )
(1.19)
p
>
s
n−2s 2
n−2s
s
Γ( p−1 )Γ( 2 − p−1 )
Γ( 4 )
solution u must be zero. Moreover for the case p = pc (n), solution u has finite energy that is
Z
Z Z
(u(x) − u(y))2
|u|p+1 =
<∞
|x − y|n+2s
Rn
Rn Rn
If in addition u has finite energy then u must be zero.
Note that when s = 1 and s = 2 assumption (1.19) is equivalent to 1 < p < pc (n) where pc (n) is given
by (1.2) and (1.4), respectively. Here is the computation for the case of s = 1. Note that when s = 1 the
assumption (1.19) is
(1.20)
p
Γ( n2 −
1
p−1 )Γ(1
1
Γ( p−1
)Γ( n2 − 1
+
−
1
p−1 )
1
p−1 )
>
2
Γ( n+2
4 )
n−2 2 .
Γ( 4 )
We now use properties of the gamma function, i.g. Γ(1 + a) = aΓ(a) for a > 0, to get
n
1
n
1
n
1
(1.21)
Γ
−
=
−1−
Γ
−1−
2
p−1
2
p−1
2
p−1
1
1
1
(1.22)
Γ 1+
=
Γ
p−1
p−1
p−1
n+2
n−2
n−2
(1.23)
Γ
=
Γ
.
4
4
4
Substituting this in (1.20) we get
p
n
1
−1−
2
p−1
1
p−1
>
n−2
4
2
.
Straightforward calculations show that this is equivalent to 1 < p < pc (n) where pc (n) is given by (1.2).
Some remarks are in order. Even though the proof of Theorem 1.2 follows from the general procedure
used in [8] and [7], there are a few new ingredients in our proofs. First (in Section 2) we have derived the
4
monotonicity formula involving higher order fractional operators. Second (in Section 3) we have developed
a new and direct method to prove the non-existence of stable homogeneous solutions. This method avoids
multiplication or integration by parts and works for any fractional operator.
The monotonicity formula we derived in Section 2 implicitly used the Pohozaev’s type identity. For higher
order factional operator the Pohozaev identity has been derived recently by Ros-Oton and Serra [18].
2. Monotonicity Formula
The key technique of our proof is a monotonicity formula that is developed in this section. Define
!
Z
Z
p+1
1 3−2s
Cn,s
−n
2s p−1
2
p+1
y
|∆b ue | −
ue
E(r, x, ue ) := r
p + 1 ∂Rn+1
Rn+1
∩Br (x0 ) 2
∩Br (x0 )
+
+
Z
4s
s
p + 2s − 1
−
− n r−3+2s+ p−1 −n
y 3−2s u2e
p−1
p−1
Rn+1
∩∂B
(x
)
r
0
+
#
"
Z
4s
s
p + 2s − 1
d
+2s−2−n
3−2s
2
r p−1
y
ue
−
−n
p−1
p−1
dr
Rn+1
∩∂Br (x0 )
+
"
2 #
Z
4s
1 3 d
∂ue
2s −1
+2s−3−n
3−2s
p−1
+ r
r
r u+
y
2 dr
p−1
∂r
∩∂Br (x0 )
Rn+1
+
"
2 !#
Z
∂ue p+1
1 d
+
r2s p−1 −n
y 3−2s |∇ue |2 − 2 dr
∂r ∩∂Br (x0 )
Rn+1
+
!
Z
∂ue 2
p+1
1 2s p−1
−n−1
3−2s
2
y
|∇ue | − + r
2
∂r
∩∂B
(x
)
Rn+1
r
0
+
Theorem 2.1. Assume that n >
Furthermore,
(2.1)
p+4s−1
p+2s−1
+
2s
p−1
− b. Then, E(λ, x, ue ) is a nondecreasing function of λ > 0.
4s
dE(λ, x, ue )
≥ C(n, s, p) λ p−1 +2s−2−n
dλ
Z
y 3−2s
∩∂Bλ (x0 )
Rn+1
+
2s −1
∂ue
r u+
p−1
∂r
2
where C(n, s, p) is independent from λ.
Proof: Suppose that x0 = 0 and the balls Bλ are centred at zero. Set,
!
Z
Z
p+1
1 b
C(n,
s)
2s p−1
−n
(2.2)
Ē(ue , λ) := λ
y |∆b ue |2 dxdy −
up+1
e
p + 1 ∂Rn+1
∩Bλ 2
Rn+1
∩Bλ
+
+
2s
2s
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

+ ,
b
limy→0 y ∂y uλe = 0 in ∂Rn+1
(2.3)
+ ,

p
limy→0 y b ∂y veλ = Cn,s (uλe ) in ∂Rn+1
+
In addition, differentiating with respect to λ we have
(2.4)
∆b
Note that
Ē(ue , λ) = Ē(uλe , 1) =
Z
Rn+1
∩B1
+
dv λ
duλe
= e.
dλ
dλ
1 b λ 2
Cn,s
y (ve ) dxdy −
2
p+1
Z
∂Rn+1
∩B1
+
|uλe |p+1
Taking derivate of the energy with respect to λ, we have
Z
Z
dv λ
duλ
dĒ(uλe , 1)
(2.5)
=
y b veλ e dxdy − Cn,s
|uλe |p e
dλ
dλ
dλ
Rn+1
∩B1
∂Rn+1
∩B1
+
+
5
Using (2.3) we end up with
dĒ(uλe , 1)
=
dλ
(2.6)
Z
Rn+1
∩B1
+
y b veλ
dveλ
dxdy −
dλ
Z
lim y b ∂y veλ
∂Rn+1
∩B1 y→0
+
duλe
dλ
From (2.4) and by integration by parts we have
Z
y b veλ
∩B1
Rn+1
+
Note that
Z
−
Rn+1
∩B1
+
dveλ
dλ
duλe
dλ
Rn+1
∩B1
+
λ
λ
Z
Z
due
due
yb +
∆b uλe y b ∂ν
= −
∇∆b uλe · ∇
dλ
dλ
∂(Rn+1
∩B1 )
Rn+1
∩B1
+
+
Z
y b ∆b uλe ∆b
=
∇∆b ue · ∇
duλe b
y
dλ
Z
=
Rn+1
∩B1
+
Z
=
∩B1
Rn+1
+
div(∇∆b uλe y b )
y b ∆2b uλe
Z
= −
∂(Rn+1
∩B1 )
+
duλe
−
dλ
duλe
−
dλ
Z
Z
∂(Rn+1
∩B1 )
+
y b ∂ν (∆b uλe )
∩B1 )
∂(Rn+1
+
y b ∂ν (∆b uλe )
y b ∂ν (∆b uλe )
duλe
dλ
duλe
dλ
duλe
dλ
Therefore,
Z
Rn+1
∩B1
+
dv λ
y b veλ e
dλ
Z
=
∩B1 )
∂(Rn+1
+
∆b uλe y b ∂ν
duλe
dλ
Z
−
∂(Rn+1
∩B1 )
+
y b ∂ν (∆b uλe )
duλe
dλ
Boundary of Rn+1
∩ B1 consists of ∂Rn+1
∩ B1 and Rn+1
∩ ∂B1 . Therefore,
+
+
+
Z
Rn+1
∩B1
+
y b veλ
dveλ
dλ
duλe
duλ
+ lim y b ∂y veλ e
y→0
y→0
dλ
dλ
∂Rn+1
∩B1
+
λ
Z
due
duλ
+
y b veλ ∂r
− y b ∂r veλ e
dλ
dλ
Rn+1
∩∂B1
+
Z
=
−veλ lim y b ∂y
and ∂r = ∇ · X
where r = |X|, X = (x, y) ∈ Rn+1
+
r isλ the
corresponding radial derivative. Note that the first
due
integral in the right-hand side vanishes since ∂y dλ = 0 on ∂Rn+1
+ . From (2.6) we obtain
(2.7)
dĒ(uλe , 1)
=
dλ
Z
y
Rn+1
∩∂B1
+
b
veλ ∂r
duλe
dλ
−
∂r veλ
duλe
dλ
Now note that from the definition of uλe and veλ and by differentiating in λ we get the following for X ∈ Rn+1
+
duλe (X)
dλ
=
1
λ
(2.8)
2s λ
u (X) + r∂r uλe (X)
p−1 e
dveλ (X)
dλ
=
1
λ
(2.9)
2(p + s − 1) λ
ve (X) + r∂r veλ (X)
p−1
Therefore, differentiating with respect to λ we get
λ
d2 uλe (X) duλe (X)
2s duλe (X)
duλ (X)
+
=
+ r∂r e
2
dλ
dλ
p − 1 dλ
dλ
6
So, for all X ∈ Rn+1
∩ ∂B1
+
2s λ
duλ (X)
−
u (X)
∂r uλe (X) = λ e
dλ
p−1 e
λ
due (X)
d2 uλe (X) p − 1 − 2s duλe (X)
+
∂r
= λ
dλ
dλ2
p−1
dλ
λ
dv (X) 2(p + s − 1) λ
−
ve (X)
∂r veλ (X) = λ e
dλ
p−1
(2.10)
(2.11)
(2.12)
Substituting (2.11) and (2.12) in (2.7) we get
dĒ(uλe , 1)
(2.13)
dλ
Z
=
Rn+1
∩∂B1
+
y b veλ
p − 1 − 2s duλe
d2 uλ
λ 2e +
dλ
p−1
dλ
−y
b
2(p + s − 1) λ
dv λ
λ e −
ve
dλ
p−1
duλe
dλ
2 λ
λ
dveλ duλe
λ d ue
λ due
y λve
−λ
+ 3ve
dλ2
dλ
dλ dλ
∩∂B1
Rn+1
+
Z
=
b
Taking derivative of (2.8) in r we get
r
∂ 2 uλe
∂uλe
∂
+
=λ
2
∂r
∂r
∂r
duλe
dλ
−
2s ∂uλe
p − 1 ∂r
So, from (2.11) for all X ∈ Rn+1
∩ ∂B1 we have
+
∂ 2 uλe
∂r2
(2.14)
∂ duλe
p + 2s − 1 ∂uλe
= λ
−
∂r dλ
p−1
∂r
2 λ
d ue
p − 2s − 1 duλe
p + 2s − 1
duλ
2s λ
= λ λ 2 +
−
λ e −
ue
dλ
p−1
dλ
p−1
dλ
p−1
=
λ2
d2 uλe
4s duλe
2s(p + 2s − 1) λ
−
λ
+
ue
2
dλ
p − 1 dλ
(p − 1)2
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, (2.10) and (2.14) we get
veλ
=
λ2
d2 uλe
duλ
4s
2s
p + 2s − 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
7
From this and (2.13) we get
dĒ(uλe , 1)
dλ
(2.15)
=
(2.16)
(2.17)
(2.18)
(2.19)
(2.20)
4s
p−1
where α := n + b −
dĒ(uλe , 1)
(2.21)
dλ
2s
p−1
and β :=
Z
=
∩∂B1
Rn+1
+
θ1b
Z
+
Rn+1
∩∂B1
+
2λ
θ1b
Z
+
λ
Rn+1
∩∂B1
+
3
θ1b λ
p+2s−1
p−1
λ
2d
− n − b . Simplifying the integrals we get
λ 2 !
duλe
due
+ 4λ
+ 2(α − β)λ
dλ
dλ
!
λ 2 !
1
d
d
d
du
β
e
λ(uλe )2 −
λ3
(uλ )2
+
2 dλ
dλ dλ
2 dλ e
d2 uλe
dλ2
β d2
2 dλ2
2 λ
ue
dλ2
duλ
+ αλ e + βuλe
dλ
d2 uλe
dλ2
Rn+1
∩∂B1
+
Z
d2 uλe
duλe
duλe
λ
+
θ1b 3 λ2
+
αλ
+
βu
e
dλ2
dλ
dλ
Rn+1
∩∂B1
+
Z
λ
2 λ
λ
due
2 d ue
λ
b due d
+ αλ
λ
+ βue
−
θ1 λ
dλ dλ
dλ2
dλ
Rn+1
∩∂B1
+
Z
d2 uλ
+
θ1b λ 2e θ1−b divS n (θ1b ∇S n uλe )
dλ
Rn+1
∩∂B1
+
Z
duλe −b
+
3θ1b
θ1 divS n (θ1b ∇S n uλe )
n+1
dλ
R+ ∩∂B1
Z
duλe
d
θ1−b divS n (θ1b ∇S n uλe )
−
θ1b λ
dλ
dλ
Rn+1
∩∂B1
+
Z
2
2 λ
ue
dλ2
2d
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
Note that from the assumptions we have α − β − 1 > 0, therefore the first term in the RHS of (2.21) is
positive that is
2λ3
d2 uλe
dλ2
2
+ 4λ2
d2 uλe duλe
+ 2(α − β)λ
dλ2 dλ
duλe
dλ
2
2 λ
2
λ 2
d u
duλ
due
= 2λ λ 2e + e
+ 2(α − β − 1)λ
>0
dλ
dλ
dλ
From this we have
!
λ 2 !
Z
1 d
dĒ(uλe , 1)
β d2
due
β d λ 2
b
λ 2
3 d
≥
θ1
λ(ue ) −
λ
+
(u )
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 .
Note that the terms appeared in R1 are of the following form
d2
λ(uλe )2
2
dλ
Rn+1
∩∂B1
+
"
λ 2 #
Z
due
b d
3 d
θ1
λ
n+1
dλ
dλ dλ
R+ ∩∂B1
Z
d
y b (uλe )2
n+1
dλ
R+ ∩∂B1
Z
θ1b
=
=
=
!
Z
4s
d2
+2(s−1)−n
b
2
λ p−1
y ue
dλ2
Rn+1
∩∂Bλ
+
"
2 !#
Z
4s
d
2s −1
∂ue
+2s−3−n
b
3 d
p−1
λ
λ
y
λ ue +
dλ
dλ
p−1
∂r
Rn+1
∩∂Bλ
+
!
Z
4s
d
λ2s−3+ p−1 −n
y b u2e
n+1
dλ
R+ ∩∂Bλ
8
We now apply integration by parts to simplify the terms appeared in R2 .
Z
duλe
duλ
d
d2 uλ
R2 =
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
+
Z
2
duλe duλe
d2 uλe
b
λ
b n u · ∇S n
n
−
3θ
∇
∇
=
−θ1b λ∇S n uλe · ∇S n
+
θ
λ
S
S
1
e
1 dλ2
dλ
dλ Rn+1
∩∂B1
+
!
!
Z
Z
Z
duλe 2
λ d2
3 d
b
λ 2
b
λ 2
b
= −
∇
θ
|∇
u
|
−
θ
|∇
u
|
+
2λ
θ
θ e
θ e
1
1
1 θ
2 dλ2
2 dλ
dλ Rn+1
∩∂B1
Rn+1
∩∂B1
Rn+1
∩∂B1
+
+
+
!
!
Z
Z
Z
duλe 2
1 d2
1 d
b
b
λ 2
λ 2
b
= −
λ
θ1 |∇θ ue | + 2λ
θ1 |∇θ ue | −
θ1 ∇θ
n+1
n+1
2 dλ2
2
dλ
dλ
R
∩∂B
Rn+1
∩∂B
R
∩∂B
1
1
1
+
+
+
!
!
Z
Z
1 d2
1 d
b
λ 2
≥ −
λ
θ
θ1b |∇θ uλe |2
|∇
u
|
−
θ e
1
n+1
n+1
2 dλ2
2
dλ
R+ ∩∂B1
R+ ∩∂B1
Note that the two terms that appear as lower bound for R3 are of the form
!
"
2 !#
Z
Z
∂u p+1
d2
d2
−n
2s p−1
b
2
b
λ 2
y |∇u| − θ1 |∇θ ue |
λ
=
λ
2
2
n+1
n+1
dλ
dλ
∂r
R+ ∩∂Bλ
R+ ∩∂B1
!
"
2 !#
Z
Z
∂u p+1
d
d
−n−1
2s p−1
b
λ 2
b
2
θ1 |∇θ ue |
=
y |∇u| − λ
dλ
dλ
∂r
∩∂B1
∩∂Bλ
Rn+1
Rn+1
+
+
2
Remark 2.1. It is straightforward to show that n >
p+1
p−1 2s
implies n >
p+4s−1
p+2s−1
+
2s
p−1
− b.
3. Homogeneous Solutions
2s
In this section, we examine homogenous solutions of the form u = r− p−1 ψ(θ). Note that the methods
and ideas that we apply here are different from the ones used in [7].
2s
Theorem 3.1. Suppose that u = r− p−1 ψ(θ) is a stable solution of (1.1) then ψ = 0 provided p >
p
Γ( n2
s
s
− p−1
)Γ(s + p−1
)
s
n−2s
s
Γ( p−1 )Γ( 2 − p−1 )
>
2
Γ( n+2s
4 )
n−2s 2
Γ( 4 )
Proof. Since u satisfies (1.1), the function ψ satisfies (we omit the P.V.)
2s
2s
Z
2ps
|x|− p−1 ψ(θ) − |y|− p−1 ψ(σ)
|x|− p−1 ψ p (θ) =
dy
|x − y|n+2s
2s
2s
2s
Z
|x|− p−1 ψ(θ) − r− p−1 t− p−1 ψ(σ)
=
|x|n tn−1 dtdσ where |y| = rt
n+2s
(t2 + 1 − 2t < θ, σ >) 2 |x|n+2s
2s
Z
2ps
ψ(θ) − t− p−1 ψ(θ)
n−1
dtdσ
= |x|− p−1 [
n+2s t
(t2 + 1 − 2t < θ, σ >) 2
2s
Z
t− p−1 (ψ(θ) − ψ(σ)
n−1
+
dtdσ]
n+2s t
(t2 + 1 − 2t < θ, σ >) 2
2ps
We now drop |x|− p−1 and get
Z
(3.1)
ψ(θ)An,s (θ) +
Sn−1
p
2s (< θ, σ >)(ψ(θ) − ψ(σ))dσ = ψ (θ)
K p−1
where
Z
∞
Z
An,s :=
0
Sn−1
2s
1 − t− p−1
(t2 + 1 − 2t < θ, σ >)
9
n+2s
2
tn−1 dσdt
n+2s
n−2s
and
and
2s (< θ, σ >) :=
K p−1
2s
∞
Z
tn−1− p−1
(t2 + 1 − 2t < θ, σ >)
0
n+2s
2
dt
Note that
Z
2s (< θ, σ >)
K p−1
=
1
=
R1
0
2s
n+2s
2
dt +
1
2s
∞
tn−1− p−1
(t2 + 1 − 2t < θ, σ >)
n+2s
2
dt
2s
tn−1− p−1 + t2s−1+ p−1
(t2 + 1 − 2t < θ, σ >)
0
We now set Kα (< θ, σ >) =
Z
(t2 + 1 − 2t < θ, σ >)
0
Z
2s
tn−1− p−1
1
tn−1+α +t2s−1+α
(t2 +1−2t<θ,σ>)
n+2s
2
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
n+2s
For the last part we have used the fact that for p > n−2s
we have 2s − 1 + α < n − 1 − α.
From (3.1) we get the following
Z
Z
Z
2
2
2s (< θ, σ >)(ψ(θ) − ψ(σ)) dθdσ =
(3.2)
ψ (θ)An,s +
K p−1
ψ p+1 (θ)dθ
Sn−1
Sn−1
Sn−1
at the origin and at infinity that is η = 1 for < r < −1
We set a standard cut-off function η ∈
n−2s
and η = 0 for either r < /2 or r > 2/. We test the stability (1.8) 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|)
dσd(|y|)
dy
=
n+2s
n+2s
(r2 + |y|2 − 2r|y| < θ, σ >) 2
Rn |x − y|
Sn−1
Cc1 (R+ )
Now set |y| = rt then
Z
Z ∞Z
n−2s
φ(x) − φ(y)
ψ(θ)η(r) − t− 2 ψ(σ)η(rt) n−1
−s
−n
2
dy = r
dtdσ
n+2s t
n+2s
Rn |x − y|
0
Sn−1 (t2 + 1 − 2t < θ, σ >) 2
Z Z
n−2s
n−2s
ψ(θ)η(r) − t− 2 ψ(σ)η(r) + t− 2 (η(r)ψ(θ) − η(rt)ψ(σ)) n−1
−s
−n
2
= r
t
dtdσ
n+2s
(t2 + 1 − 2t < θ, σ >) 2
Sn−1
Z ∞Z
n−2s
1−t 2
−n
−s
n−1
2
= r
η(r)ψ(θ)
dtdσ
n+2s t
0
Sn−1 (t2 + 1 − 2t < θ, σ >) 2
Z ∞Z
n−2s
tn−1− 2 (ψ(θ) − ψ(σ))
−s
−n
2
+r
η(r)
n+2s dtdσ
0
Sn−1 (t2 + 1 − 2t < θ, σ >) 2
Z ∞Z
n−2s
tn−1− 2 (η(r) − η(rt))ψ(σ)
−n
−s
2
+r
dtdσ
n+2s
0
Sn−1 (t2 + 1 − 2t < θ, σ >) 2
n−2s
R∞R
1−t 2
n−1
Define Λn,s := 0 Sn−1
dσdt. Therefore,
n+2s t
(t2 +1−2t<θ,σ>)
Z
Rn
φ(x) − φ(y)
dy
|x − y|n+2s
=
2
n
r− 2 −s η(r)ψ(θ)Λn,s
Z
n
+r− 2 −s η(r)
K n−2s (< θ, σ >)(ψ(θ) − ψ(σ))dσ
2
Sn−1
n
+r− 2 −s
Z
∞
Z
Sn−1
0
10
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 (1.8),
Z Z
Z Z
(φ(x) − φ(y))φ(x)
(φ(x) − φ(y))2
dxdy
=
2
dxdy
n+2s
|x − y|
|x − y|n+2s
n
n
Rn Rn
ZR∞ R
Z
= 2
r−1 η 2 (r)dr
ψ 2 Λn,s dθ
0
Sn−1
Z
Z ∞
−1 2
r η (r)dr
K n−2s (< θ, σ >)(ψ(θ) − ψ(σ))2 dσdθ
+2
Sn−1
0
∞
Z
(3.3)
2
∞
Z
r−1 η(r)(η(r) − η(rt))dr
+2
tn−1−
Z
Sn−1
0
0
Z
Sn−1
n−2s
2
(t2 + 1 − 2t < θ, σ >)
We now compute the second term in the stability inequality (1.8) for the test function φ(x) = r−
2s
and u = r− p−1 ψ(θ),
Z ∞
Z ∞
r−2s r−(n−2s) ψ p+1 η 2 (r)dr
|u|p−1 φ2 = p
p
0
Z0 ∞
Z
r−1 η 2 (r)dr
(3.4)
= p
ψ p+1 (θ)dθ
n−2s
2
n+2s
2
ψ(θ)η(r)
Sn−1
0
R∞
ψ(σ)ψ(θ)
r−1 η2 (r)dr
Due to the definition of the η , we have 0
= ln(2/) + O(1). Note that this term appears in
both terms of the stability inequality that we computed in (3.3) and (3.5). We now claim that
Z ∞
f (t) :=
r−1 η (r)(η (r) − η (rt))dr = O(ln t)
0
t
1
t
2
Note that η (rt) = 1 for < r <
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
f (t) =
r−1 η (r)(η (r) − η (rt))dr
2
In what follows we consider a few cases to explain the claim. For example when <
Z t
Z t2
f (t) ≈
r−1 dr +
r−1 dr ≈ ln t
2
Now consider the case
1
<
t
<
1
then
1
1
then t ≈ 2 . So,
Z t
Z
f (t) ≈
r−1 dr +
<
t
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
(3.5)
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
(3.6) ≈
n+2s ψ(σ)ψ(θ)dtdσdθ
(t2 + 1 − 2t < θ, σ >) 2
Sn−1 Sn−1 0
(3.7) = O(1)
Collecting higher order terms of the stability inequality we get
Z
Z
Z
2
2
(3.8)
Λn,s
ψ +
K n−2s (< θ, σ >)(ψ(θ) − ψ(σ)) dσ ≥ p
Sn−1
Sn−1
From this and (3.2) we obtain
Z
Z
(Λn,s − pAn,s )
ψ2 +
Sn−1
Sn−1
2
ψ p+1
Sn−1
2
2s )(< θ, σ >)(ψ(θ) − ψ(σ)) dσ ≥ 0
(K n−2s − pK p−1
2
11
dσdθdt
n+2s
2s
2s
for p > n−2s
< 0. On
Note that Kα is decreasing in α. This implies K n−2s < K p−1
. So, K n−2s − pK p−1
2
2
the other hand the assumption of the theorem implies that Λn,s − pAn,s < 0. Therefore, ψ = 0.
Remark 3.1. Note that in this section we never used the fact that 1 < s < 2. So this proof holds for a
larger range of the parameter s.
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 [7, 8].
Lemma 4.1. The following identities hold for any functions ζ and η,
(4.1)
∆b ζ∆b (ζη 2 ) − |∆b (ζη)|2
(4.2)
∆b (ζη)
= −ζ 2 |∆b η|2 + 2ζ∆b ζ|∇η|2 − 4|∇ζ · ∇η|2 − 4ζ∆b η∇ζ · ∇η
= η∆b ζ + ζ∆b η + 2∇ζ · ∇η
Proof. We omit the proof, since it is elementary.
We apply the given identities to get some energy estimates.
Lemma 4.2. Let u be a solution of (1.1) that is stable outside a ball BR0 and ue satisfies (1.7). Then there
exists a positive constant C such that
Z
Z
Z
p+1 2
b
2 2
(4.3)
|ue | η +
y |∆b ue | η ≤ C
y b u2e |∆b η|2 + |∆b |∇η|2 | + |∇η · ∇∆b η|
∂Rn+1
+
Rn+1
+
Rn+1
+
Z
(4.4)
+C
Rn+1
+
y b |ue ||∆b ue ||∇η|2
Proof. Multiply the equation with y b uη 2 where η is a test function to get
Z
Z
0 =
y b (ue η 2 )∆2b ue =
ue η 2 div(y b ∇∆b ue )
Rn+1
+
Rn+1
+
Z
=
b
−
Rn+1
+
Z
=
−
Rn+1
+
Z
2
y ∇(ue η ) · ∇∆b ue +
lim y b ∂y (∆b ue )(ue η 2 )
y→0
∂Rn+1
+
y b ∇(ue η 2 ) · ∇∆b ue + Cn,s
Z
∂Rn+1
+
|ue |p+1 η 2
From this we get
Z
(4.5)
Cn,s
∂Rn+1
+
Apply Lemma 4.1 for ζ = ue we get
Z
Z
Cn,s
|ue |p+1 η 2 =
∂Rn+1
+
Rn+1
+
|ue |p+1 η 2 =
b
−4
n+1
R+
Note that the last integral is
Z
−4
y b ue ∆b η∇ue · ∇η
Rn+1
+
y |∇ue · ∇η| − 4
Z
n+1
R+
y b ∆b ue ∆b (ue η 2 )
Z
2
= −2
Rn+1
+
Z
=
Rn+1
+
y b |∆b (ue η)|2 −
Z
(4.6)
Z
2
Rn+1
+
y b u2e |∆b η|2 + 2
Z
Rn+1
+
Z
Rn+1
+
y b ue ∆b ue |∇η|2
y b ue ∆b η∇ue · ∇η
y b ∆b η∇(u2e ) · ∇η
u2e div(y b ∆b η∇η) = 2
12
Z
Rn+1
+
y b u2e (|∆b η|2 + ∇η · ∇∆b η)
From this and (4.6) we get
Z
Cn,s
(4.7)
|ue |p+1 η 2
∂Rn+1
+
Z
b
y |∆b (ue η)| + 2
=
Rn+1
+
Z
−4
(4.8)
Z
2
Rn+1
+
Rn+1
+
y b |∇ue · ∇η|2 +
y b ue ∆b ue |∇η|2
Z
y b u2e (|∆b η|2 + 2∇η · ∇∆b η)
Rn+1
+
We now apply the stability inequality (1.8) for φ = uη to get
Z
Z
(4.9)
p
|u|p+1 η 2 ≤
y b |∆b (ue η)|2
Rn+1
+
Rn
From (4.9) and (4.7) we obtain
Z
Z
|ue |p+1 η 2 +
y b |∆b (ue η)|2
∂Rn+1
+
Z
≤ C
Rn+1
+
Rn+1
+
y b |ue ||∆b ue ||∇η|2 + C
Z
+C
(4.10)
Rn+1
+
Z
Rn+1
+
y b |∇ue |2 |∇η|2
y b u2e (|∆b η|2 + |∇η · ∇∆b η|)
Note that from Lemma 4.1 we have ∆b (ue η) = η∆b ue + ue ∆b η + 2∇ue · ∇η. So from (4.10) we get
Z
Z
Z
Z
p+1 2
b
2 2
b
2
(4.11)
|ue | η +
y |∆b ue | η ≤ C
y |ue ||∆b ue ||∇η| + C
y b |∇ue |2 |∇η|2
∂Rn+1
+
Rn+1
+
Rn+1
+
Rn+1
+
Z
(4.12)
+C
Rn+1
+
y b u2e (|∆b η|2 + |∇η · ∇∆b η|)
Note also that 2|∇ue |2 = ∆b (u2e ) − 2ue ∆b ue . Therefore,
Z
Z
Z
(4.13)
2
y b |∇ue |2 |∇η|2 =
y b |∇η|2 ∆b (u2e ) − 2
Rn+1
+
Rn+1
+
Z
(4.14)
=
Rn+1
+
From this and (4.11) we get
Z
Z
|ue |p+1 η 2 +
Rn+1
+
∂Rn+1
+
y b |∆b ue |2 η 2
Rn+1
+
y b u2e ∆b |∇η|2
Z
≤
C
Rn+1
+
−2
Rn+1
+
y b ue ∆b ue |∇η|2
y b |ue ||∆b ue ||∇η|2
Z
(4.15)
Z
y b ue ∆b ue |∇η|2
+C
Rn+1
+
y b u2e (|∆b η|2 + |∇η · ∇∆b η| + |∆b |∇η|2 |)
This finishes the proof.
Corollary 4.1. With the same assumption as Lemma 4.2. Then there exists a positive constant C such that
Z
Z
Z
(4.16)
|ue |p+1 +
y b |∆b ue |2 ≤ CR−4
y b u2e
BR ∩∂Rn+1
+
BR ∩Rn+1
+
BR ∩Rn+1
+
Proof. This is a direct consequence of the estimate (4.3). Substitute η with η m in (4.3) for a number
3 < m ∈ N. Therefore
Z
Z
Z
(4.17) m2
y b |ue ||∆b ue ||∇η|2 η 2m−2 ≤ y b |∆b ue |2 wη 2m + C()
y b u2e η 2m−4 |∇η|4
Rn+1
+
Rn+1
+
Rn+1
+
for a small enough > 0. One can apply the standard test function to finish the proof.
Lemma 4.3. Suppose that u is a solution of (1.1) that is stable outside some ball BR0 ⊂ Rn . For η ∈
Cc∞ (Rn \ BR0 ) and x ∈ Rn define
Z
(η(x) − η(y))2
(4.18)
ρ(x) =
dy.
|x − y|n+2s
Rn
13
Then
Z
(4.19)
|u|p+1 η 2 dx +
Rn
Z
Z
Rn
Rn
|u(x)η(x) − u(y)η(y)|2
dxdy ≤ C
|x − y|n+2s
Z
u2 ρdx
Rn
Proof. Proof is quite similar to Lemma 2.1 in [7] and we omit it here.
Lemma 4.4. Let m > n/2 and x ∈ Rn . Set
Z
(η(x) − η(y))2
dy where η(x) = (1 + |x|2 )−m/2
(4.20)
ρ(x) =
|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
(4.21)
Proof. Proof is quite similar to Lemma 2.2 in [7] and we omit it here.
Corollary 4.2. Suppose that m > n/2, η given by (4.20) and R > R0 > 1. Define
Z
(ηR (x) − ηR (y))2
(4.22)
ρR (x) =
dy where ηR (x) = η(x/R)ψ(x/R0 )
|x − y|n+2s
Rn
for the standard test function ψ that is ψ ∈ C ∞ (Rn ) and 0 ≤ ψ ≤ 1, ψ = 0 on B1 and ψ = 1 on Rn \ B2 .
Then there exists a constant C > 0 such that
ρR (x) ≤ Cη 2 (x/R)|x|−(n+2s) + R−2s ρ(x/R).
Lemma 4.5. Suppose that u is a solution of (1.1) that is stable outside a ball BR0 . Consider ρR that is
defined in Corollary 5.2 for n/2 < m < n/2 + s(p + 1)/2. Then there exists a constant C > 0 such that
!
Z
Z
p+1
u2 ρR ≤ C
u2 ρR + Rn−2s p−1
Rn
B3R0
for any R > 3R0
Proof. Proof is quite similar to Lemma 2.4 in [7] and we omit it here.
n+2s
. Let u be a solution of (1.1) that is stable outside a ball BR0 and ue
Lemma 4.6. Suppose that p 6= n−2s
satisfies (1.7). Then there exists a constant C > 0 such that
Z
p+1
y b u2e ≤ CRn+4−2s p−1
BR
for any R > 3R0 .
Proof. The extension ue satisfies
Z
u2 (z)
ū(x, y) ≤ Cn,s
Rn
From this we have
Z
y 3−2s u2e dxdy ≤
Z
Cn,s
|x|≤R,z∈Rn
BR
Z
≤
Cn,s
|x|≤R,z∈Rn
Z
=
u2e (z)
"Z
(|x − z|2 + y 2 )
R
Z
0
u2e (z)
y 2s
n+2s
2
!
y3
(|x − z|2 + y 2 )
R2
2
n+2s
2
dy dzdx
2 1− n
2 −s
(|x − z| + α )
0
dz
2
Z
dα − |x − z|
R2
#
2
2 −n
2 −s
(|x − z| + α )
dα
0
n
n
n
+ s)−1 (|x − z|2 )2− 2 −s − (|x − z|2 + R2 )2− 2 −s dα
2
|x|≤R,z∈Rn
Z
n
n
n
+Cn,s
u2e (z)|x − z|2 ( + s − 1)−1 (|x − z|2 + R2 )1− 2 −s − (|x − z|2 )2− 2 −s dα
2
|x|≤R,z∈Rn
Cn,s
u2e (z)(−2 +
We now split the integral to |x − z| < 2R and |x − z| > 2R. For the case of |x − z| < 2R we get
14
Z
≤
n
n
n
+ s)−1 (|x − z|2 )2− 2 −s − (|x − z|2 + R2 )2− 2 −s
2
|x|≤R,|x−z|<2R
Z
n
n
n
+
u2e (z)|x − z|2 ( + s − 1)−1 (|x − z|2 + R2 )1− 2 −s − (|x − z|2 )2− 2 −s
2
|x|≤R,|x−z|<2R
Z
n
C
u2e (z)(|x − z|2 )2− 2 −s
u2e (z)(−2 +
|x|≤R,|x−z|<2R
≤
R4−2s
Z
u2e (z)dz ≤ CR4−2s
Z
B3R
≤
CR
4−2s+n p−1
p+1
Z
2
|u|p+1 ηR
2/(p+1) Z
B3R
2/(p+1)
−4/(p−1)
B3R
(p−1)/(p+1)
ηR
u2 (z)ρR (z)dz
B3R
≤
CR
p+1
n+4−2s p−1
Here we have used Lemma 4.3 and Lemma 4.5. For the case of |x − z| > 2R we apply the mean value
inequality to get
Z
n
n
n
u2e (z)(−2 + + s)−1 (|x − z|2 )2− 2 −s − (|x − z|2 + R2 )2− 2 −s
2
|x|≤R,|x−z|≥2R
Z
n
n
n
+
u2e (z)|x − z|2 ( + s − 1)−1 (|x − z|2 + R2 )1− 2 −s − (|x − z|2 )2− 2 −s
2
|x|≤R,|x−z|≥2R
Z
n
≤ CR4
u2e (z)(|x − z|2 )− 2 −s
|x|≤R,|x−z|≥2R
Z
≤ CR4
u2e (z)ρdz
|z|≥R
≤
CR
p+1
n+4−2s p−1
.
Here we have used Corollary 4.2 and Lemma 4.5. This finishes the proof.
Lemma 4.7. Let u be a solution of (1.1) that is stable outside a ball BR0 and ue satisfies (1.7). Then there
exists a positive constant C such that
Z
Z
p+1
|ue |p+1 +
y b |∆b ue |2 ≤ CRn−2s p−1
(4.23)
BR ∩Rn+1
+
BR ∩∂Rn+1
+
Proof. This is a direct consequence of Corollary 4.1 and Lemma 4.6.
5. Blow-Down Analysis
In this section we provide the proof of Theorem 1.2.
Proof of Theorem 1.2. Suppose that u is a solution of (1.1) that is stable outside the ball of radius R0
and suppose that ue is its extension satisfying (1.7).
Let’s first consider the subcritical case, i.e. 1 < p ≤ pS (n). Note that for the subcritical case Lemma
implies that u ∈ Ḣ s (Rn ) ∩ Lp+1 (Rn ). Multiplying (1.1) with u and doing integration, we obtain
Z
(5.1)
|u|p+1 = ||u||2Ḣ s (Rn )
Rn
λ
in addition multiplying (1.1) with u (x) = u(λx) yields
Z
Z
Z
|u|p−1 uλ =
(−∆)s/2 u(−∆)s/2 uλ = λs
Rn
Rn
Rn
15
wwλ
where w = (−∆)s/2 u. Following ideas provided in [8, 18] and the using the change of variable z =
can get the following Pohozaev identity
−
n
p+1
Z
|u|p+1 =
Rn
2s − n
2
Z
w2 +
Rn
d
|λ=1
dλ
√
Z
w
λ
w1/
√
λ
dz =
Rn
√
λx one
2s − n
||u||2Ḣ s (Rn )
2
This equality together and (5.1) proves the theorem for the subcritical case.
We now focus on the supercritical case, i.e. p > pS (n). We perform the proof in a few steps.
Step 1. limλ→∞ E(ue , 0, λ) < ∞.
From Theorem 2.1 E is nondecreasing. So, we only need to show that E(ue , 0, λ) is bounded. Note that
E(ue , 0, λ) ≤
1
λ
Z
2λ
E(ue , 0, t)dt ≤
λ
1
λ2
Z
2λ
t+λ
Z
E(ue , 0, γ)dγdt
t
λ
From Lemma 4.7 we conclude that
1
λ2
Z
2λ
t+λ
Z
γ
λ
Z
p+1
2s p−1
−n
∩Bγ
Rn+1
+
t
1 3−2s
Cn,s
y
|∆b ue |2 dydx −
2
p+1
!
Z
∩Bγ
∂Rn+1
+
dx
up+1
e
dγdt ≤ C
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
t
4s
−n
−3+2s+ p−1
Z
∩∂Bγ
Rn+1
+
!
y 3−2s u2e dydx
dγdt ≤
≤
1
λ2
Z
1
λ2
Z
2λ
4s
−n
−3+2s+ p−1
Z
y 3−2s u2e dydxdt
t
Bt+λ \Bt
λ
2λ
4s
−n
−3+2s+ p−1
Z
y 3−2s u2e dydx
t
λ
B3λ
p+1
≤
λn+4−2s p−1
≤
C
1
λ2
Z
2λ
4s
t−3+2s+ p−1 −n dt
λ
where C > 0 is independent from λ. In the above estimates we have applied Lemma 4.6. For the next term
we have
"
2 #
Z 2λ Z t+λ 3
Z
4s
2s −1
1
∂ue
γ d
2s−3−n+ p−1
3−2s
γ
y
γ ue +
dγdt
λ2 λ
2 dγ
p−1
∂r
∂Bγ
t
2
Z 2λ
Z
4s
1
2s
∂ue
2s−n+ p−1
3−2s
−1
=
[(t + λ)
y
(t + λ) ue +
2λ2 λ
p−1
∂r
∂Bt+λ
2
Z
4s
2s −1
∂ue
−t2s−n+ p−1
y 3−2s
γ ue +
]dt
p
−
1
∂r
∂Bλ
"
2 #
Z
Z 2λ Z t+λ
4s
3
2s −1
∂ue
2s−1−n+ p−1
3−2s
− 2
γ
y
γ ue +
dγdt
2λ λ
p−1
∂r
t
∂Bγ
2
Z
4s
2s −1
∂ue
−2+2s−n+ p−1
3−2s
≤ λ
y
λ ue +
≤C
p−1
∂r
B3λ \Bλ
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 4.7.
Step 3. u∞
e is homogeneous.
16
dt
To prove this claim, apply the scale invariance of E, its finiteness and the monotonicity formula; given
R2 > R1 > 0,
0
=
=
≥
≥
lim (E(ue , 0, R2 λi ) − E(ue , 0, R1 λi ))
lim E(uλe i , 0, R2 ) − E(uλe i , 0, R1 )
i→∞
2
Z
4s
2s −1 λi ∂uλe i
r ue +
dydx
lim inf
y 3−2s r p−1 +2s−2−n
i→∞
p−1
∂r
(BR2 \BR1 )∩Rn+1
+
2
Z
4s
2s −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 −1 ∞ ∂ue
r ue +
= 0 a.e. in Rn+1
+ .
p−1
∂r
Therefore, 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 \ B , y 3−2s dydx) and (uλe i ) converges strongly to zero in
Lp+1 (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 −
|uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B1
∂R
∩B1
Z +
Z +
1
κ
s
y 3−2s |∆b uλe |2 dxdy −
|uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B
∂R
∩B
+
Z+
Z
κs
1
3−2s
λ 2
y
|∆b ue | dxdy −
|uλe |p+1 dx
+
n+1
2 Rn+1
p
+
1
∩B1 \B
∂R+ ∩B1 \B
+
Z
Z
2s(p+1)
1
κs
n− p−1
3−2s
λ 2
ε
I(ue , λε) +
y
|∆b ue | dxdy −
|uλe |p+1 dx
n+1
2 Rn+1
p
+
1
∩B
\B
∩B
\B
∂R
1
1
+
+
Z
Z
2s(p+1)
κs
1
n− p−1
3−2s
λ 2
y
|∆ue | dxdy −
|uλe |p+1 dx
Cε
+
n+1
2 Rn+1
p
+
1
∩B
\B
∂R
∩B
\B
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)
1 2λ
(5.2)
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 ū must be homogeneous, a contradiction unless ue ≡ 0.
Remark 5.1. Note that we expect that when (1.19) does not hold that is when
(5.3)
p
Γ( n2 −
s
s
p−1 )Γ(s + p−1 )
s
s
Γ( p−1
)Γ( n−2s
− p−1
)
2
≤
2
Γ( n+2s
4 )
n−2s 2
Γ( 4 )
there exist radial entire stable solutions. The method of construction of such solutions is the one that is applied
in [7] and references therein. More precisely, one needs to mimic the standard proof for the existence of a
minimal solution that is axially symmetric for the associated problem on bounded domains. Then applying
the truncation method and the moving plane method one can show that the minimal solution is bounded and
radially decreasing. From elliptic estimates and some classical convexity arguments the minimal solution
would converge to the singular solution that is stable. This implies that (5.3) should hold. Finally using
the singular solution and the minimal solution one can construct a radial, bounded and smooth solution via
rescaling arguments.
17
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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
18