On stable solutions of the fractional Henon-Lane-Emden equation
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 H´ enon-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 u (−∆) ˆ(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 ˆ 2 |ξ| |φ| 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 kuk2H˙ 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
Theorem 2.2. Assume that n > Furthermore,
p+4s+2a−1 p+2s+a−1
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, ¯ e , λ) := λ E(u
(2.6)
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 ¯ e , λ) = E(u ¯ λe , 1) = E(u
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 λ ¯ λe , 1) Z dE(u 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 Z λ ¯ λe , 1) Z dE(u duλ b λ dve (2.10) = y ve dxdy − lim y b ∂y veλ e dλ dλ dλ Rn+1 ∩B1 ∂Rn+1 ∩B1 y→0 + + 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 ¯ λe , 1) duλe dE(u 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 ¯ λ , 1) dE(u 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 ¯ λe , 1) dE(u dλ
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 ¯ λ , 1) dE(u d u due du d u e e e (2.25) e = 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)
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 ¯ λ , 1) dE(u β d 1 d d du β d e 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 .
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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)
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)
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
)(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 ∈ H˙ s (Rn ) ∩ Lp+1 (Rn ). Multiplying (1.1) with u and doing integration, we obtain Z (5.1) |x|a u|p+1 = ||u||2H˙ 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||2H˙ 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. References [1] L. Caffarelli, B. Gidas, J. Spruck, Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth, Comm. Pure Appl. Math. 42 (1989), no. 3, 271-297 [2] L. Caffarelli and L. Silvestre, An extension problem related to the fractional Laplacian, Comm. Partial Differential Equations 32 (2007), no. 7-9, 1245-1260. [3] J. Case, Sun-Yung Alice Chang, On fractional GJMS operators, preprint http://arxiv.org/abs/1406.1846 [4] Sun-Yung Alice Chang and Maria del Mar Gonzalez, Fractional Laplacian in conformal geometry, Advances in Mathematics 226 (2011), no. 2, 1410-1432. [5] W. Chen, C. Li, B. Ou, Classification of solutions for an integral equation, Comm. Pure Appl. Math. 59 (2006), no. 3, 330-343.
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
17
[6] C. Cowan, M. Fazly, On stable entire solutions of semilinear elliptic equations with weights, Proc. Amer. Math. Soc. 140 (2012), 2003-2012. [7] E.N. Dancer, Yihong Du, Z. Guo, Finite Morse index solutions of an elliptic equation with supercritical exponent, J. Differential Equations 250 (2011) 3281-3310. [8] J. Davila, L. Dupaigne, A. Farina, Partial regularity of finite Morse index solutions to the Lane-Emden equation, J. Funct. Anal. 261 (2011), no. 1, 218-232. [9] J. Davila, L. Dupaigne, J. Wei, On the fractional Lane-Emden equation, preprint. [10] J. Davila, L. Dupaigne, K. Wang and J. Wei, A Monotonicity Formula and a Liouville-type Theorem for a Fourth Order Supercritical Problem, Advances in Mathematics 258 (2014), 240-285. [11] Yihong Du, Zongming Guo and Kelei Wang, Monotonicity formula and -regularity of stable solutions to supercritical problems and applications to finite Morse index solutions. Calc. Var. Partial Differential Equations 50 (2014), no. 3-4, 615638. [12] E. Fabes, C. Kenig, R. Serapioni, The local regularity of solutions of degenerate elliptic equations, Comm. Partial Differential Equations 7 (1982) 77-116. [13] M. Fall, Semilinear elliptic equations for the fractional Laplacian with Hardy potential, preprint. http://arxiv.org/pdf/1109.5530v4.pdf [14] A. Farina; On the classification of solutions of the Lane-Emden equation on unbounded domains of RN , J. Math. Pures Appl. (9) 87 (2007), no. 5, 537-561. [15] M. Fazly, N. Ghossoub, On the Henon-Lane-Emden conjecture, Disc. Cont. Dyn. Syst. A 34 no 6 (2014) 2513-2533. [16] M. Fazly, J. Wei, On the higher order fractional Lane-Emden equation, preprint. [17] B. Gidas, J. Spruck; Global and local behavior of positive solutions of nonlinear elliptic equations, Commun. Pure Appl. Math. 34 (1981) 525-598. [18] B. Gidas, J. Spruck; A priori bounds for positive solutions of nonlinear elliptic equations, Comm. Partial Differential Equations 6 (1981), no. 8, 883-901. [19] Ira W. Herbst, Spectral theory of the operator (p2 + m2 )1/2 − Ze2 /r, Comm. Math. Phys. 53 (1977), no. 3, 285-294. [20] L. Hu, Liouville-type theorems for the fourth order nonlinear elliptic equation, J. Differential Equations 256 (2014) 1817-1846. [21] D. D. Joseph and T. S. Lundgren, Quasilinear Dirichlet problems driven by positive sources, Arch. Rational Mech. Anal. 49 (1972/73), 241-269. [22] Y. Li, Remark on some conformally invariant integral equations: the method of moving spheres, J. Eur. Math. Soc. (JEMS) 6 (2004), no. 2, 153-180. [23] Q. H. Phan, Ph. Souplet; Liouville-type theorems and bounds of solutions of Hardy-H´enon equations, J. Diff. Equ.,252 (2012), 2544-2562. [24] P. Pol´ aˇcik, P. Quittner, Ph. Souplet; Singularity and decay estimates in superlinear problems via Liouvilletype theorems, Part I: Elliptic systems, Duke Math. J. 139 (2007) 555-579. [25] C. S. Lin, A classification of solutions of a conformally invariant fourth order equation in RN , Comment. Math. Helv. 73 (1998) 206-231. [26] Ros-Oton, J. Serra, Local integration by parts and Pohozaev identities for higher order fractional Laplacians, preprint http://arxiv.org/abs/1406.1107. [27] F. Spitzer, Some theorems concerning 2-dimensional Brownian motion, Trans. Amer. Math. Soc. 87 (1958), 187-197. [28] J. Wei, X. Xu; Classification of solutions of higher order conformally invariant equations, Math. Ann. 313 (1999), no. 2, 207-228. [29] R. Yang, On higher order extensions for the fractional Laplacian, preprint. http://arxiv.org/pdf/1302.4413v1.pdf [30] D. Yafaev, Sharp Constants in the Hardy-Rellich Inequalities, Journal of Functional Analysis 168, (1999) 121-144.
18
MOSTAFA FAZLY, JUNCHENG WEI
[31] C. Wang, D. Ye, Some Liouville theorems for H´enon type elliptic equations, Journal of Functional Analysis 262 (2012) 1705-1727. Department of Mathematical and Statistical Sciences, CAB 632, University of Alberta, Edmonton, Alberta, Canada T6G 2G1 E-mail address: [email protected] Department of Mathematics, University of British Columbia, Vancouver, B.C. Canada V6T 1Z2. E-mail address: [email protected]
Abstract. We derive a monotonicity formula for solutions of the fractional H´ enon-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 u (−∆) ˆ(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 ˆ 2 |ξ| |φ| 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 kuk2H˙ 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
Theorem 2.2. Assume that n > Furthermore,
p+4s+2a−1 p+2s+a−1
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, ¯ e , λ) := λ E(u
(2.6)
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 ¯ e , λ) = E(u ¯ λe , 1) = E(u
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 λ ¯ λe , 1) Z dE(u 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 Z λ ¯ λe , 1) Z dE(u duλ b λ dve (2.10) = y ve dxdy − lim y b ∂y veλ e dλ dλ dλ Rn+1 ∩B1 ∂Rn+1 ∩B1 y→0 + + 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 ¯ λe , 1) duλe dE(u 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 ¯ λ , 1) dE(u 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 ¯ λe , 1) dE(u dλ
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 ¯ λ , 1) dE(u d u due du d u e e e (2.25) e = 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)
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 ¯ λ , 1) dE(u β d 1 d d du β d e 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)
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)
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
)(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 ∈ H˙ s (Rn ) ∩ Lp+1 (Rn ). Multiplying (1.1) with u and doing integration, we obtain Z (5.1) |x|a u|p+1 = ||u||2H˙ 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||2H˙ 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. References [1] L. Caffarelli, B. Gidas, J. Spruck, Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth, Comm. Pure Appl. Math. 42 (1989), no. 3, 271-297 [2] L. Caffarelli and L. Silvestre, An extension problem related to the fractional Laplacian, Comm. Partial Differential Equations 32 (2007), no. 7-9, 1245-1260. [3] J. Case, Sun-Yung Alice Chang, On fractional GJMS operators, preprint http://arxiv.org/abs/1406.1846 [4] Sun-Yung Alice Chang and Maria del Mar Gonzalez, Fractional Laplacian in conformal geometry, Advances in Mathematics 226 (2011), no. 2, 1410-1432. [5] W. Chen, C. Li, B. Ou, Classification of solutions for an integral equation, Comm. Pure Appl. Math. 59 (2006), no. 3, 330-343.
ON STABLE SOLUTIONS OF THE FRACTIONAL HENON-LANE-EMDEN EQUATION
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MOSTAFA FAZLY, JUNCHENG WEI
[31] C. Wang, D. Ye, Some Liouville theorems for H´enon type elliptic equations, Journal of Functional Analysis 262 (2012) 1705-1727. Department of Mathematical and Statistical Sciences, CAB 632, University of Alberta, Edmonton, Alberta, Canada T6G 2G1 E-mail address: [email protected] Department of Mathematics, University of British Columbia, Vancouver, B.C. Canada V6T 1Z2. E-mail address: [email protected]
