BREZIS TYPE THEOREM INVOLVING FRACTIONAL LAPLACIANS
MINGFENG ZHAO
March 05, 2014
Theorem 1 (Theorem 1 on Page 271 in Brezis[1]). Let n ≥ 1 and p > 1, the only nonnegative solution of ∆u = up in
Rn is zero.
Remark 1. Let u(x) = ex1 > 0 in Rn , then ∆u = u in Rn .
Theorem 2. Let 0 < s < 1, p > 1, and u be a solution of the following problem:



Ls (u)(z) := div [y 1−2s ∇u(x, y)] = 0, ∀z = (x, y) ∈ Rn+1

+ ,

 ∂u
(x, 0) := − lim y 1−2s uy (x, y) = −up (x, 0), ∀x ∈ Rn ,
(1)

y&0
∂ν
s



 u(x, y) ≥ 0, ∀(x, y) ∈ Rn+1 .
+
p
2s
+ b in
Then then either u(x, y) ≡ 0 in Rn+1
+ , or there exists some a, b > 0 such that 2sa = b , and u(x, y) = ay
Rn+1
+ .
Corollary 1. Let 0 < s < 1 and p > 1, any non-negative bounded solution of (−∆)s u = −up in Rn must be constant.
Remark 2. When s = 1/2, Theorem 2 is proved by Lou and Zhu[2]. Indeed, our proof of Theorem 2 follows the same
idea as Lou and Zhu[2].
n
Remark 3. When s = 1/2, let u(x, y) = ey cos(x1 ) in Rn+1
+ , then ∆u(x, y) = 0 and uy (x, 0) = cos(x1 ) = u(x, 0) in R .
Conjecture 1. Theorem 2 does not hold for p = 1.
n+1
n+1
If u(x, y) ≡ 0 in Rn+1
such
+ , we are done. If u(x, y) 6≡ 0 in R+ , and there exists some z0 = (x0 , y0 ) ∈ R+
that u(z0 ) = 0. Since u(x, y) ≥ 0 in Rn+1
+ , by the Strong Maximum Principle, then y0 = 0 and u(x, y) > u(x0 , 0) for
∂u
∂u
all (x, y) ∈ Rn+1
(x0 , 0) < 0, which contradicts with
(x0 , 0) =
+ . By the Hopf’s Lemma for Ls , we know that
∂νs
∂νs
−up (x0 , 0) = 0. So the rest case should be u(x, y) > 0 for all (x, y) ∈ Rn+1
+ .
Consider the s-Kelvin transformation of u as:
v(z) =
1
|z|n−2s
u
z
|z|2
,
∀z ∈ Rn+1
+ \{0}.
It’s easy to see that v(z) → 0 as |z| → ∞. For all i = 1, · · · , n + 1, we have
n+1
X z ∂ zj z
∂
2s−n
2s−n
vi (z) = u
|z|
+
|z|
uj
·
|z|2 ∂zi
|z|2
∂zi |z|2
j=1
1
2
MINGFENG ZHAO
=
=
u
z
|z|2
· (2s − n)|z|2s−n−2 zi + |z|2s−n
n+1
X
uj
j=1
2s−n−2
(2s − n)|z|
u
z
|z|2
z
|z|2
zi + |z|
2s−n−2
ui
z
|z|2
δij |z|2 − 2zi zj
|z|4
·
z
|z|2
− 2|z|
2s−n−4
n+1
X
uj
j=1
z
|z|2
zi zj .
So for all i = 1, · · · , n + 1, we have
vii (z)
=
(2s − n)(2s − n − 2)|z|2s−n−4 u
2s−n−2
+(2s − n)|z|
u
z
|z|2
uj
n+1
X
−2|z|
uj
j=1
=
z
|z|2
zj − 2|z|
2s−n−2
+(2s − n)|z|
u
z
|z|2
−2(2s − n − 4)|z|2s−n−6
−2|z|2s−n−4 z · ∇u
=
z
|z|2
(2s − n)(2s − n − 2)|z|
+(2s − n)|z|2s−n−2 u
−2|z|2s−n−6
n+1
X
uj
j=1
n+1
X
z
|z|2
z
|z|2
u
z
|z|2
ui
uj
z
|z|2
zi + |z|
zi2
ui
n+1
X
zj
|z|2
uij
n+1
X
ujk
z
|z|2
·
∂
∂zi
zk
|z|2
z
|z|2
∂
·
∂zi
zj
|z|2
· zi zj
n+1
X
uj
z
|z|2
zi + |z|
z
|z|2
2s−n−2
·
δij |z|2 − 2zi zj
|z|4
n+1
X
uij
j=1
n+1
X
ujk
z
|z|2
·
z
|z|2
·
δij |z|2 − 2zi zj
|z|4
δik |z|2 − 2zi zk
· zi zj
|z|4
2s−n−4
+ (2s − n)|z|
− 2(2s − n − 4)|z|2s−n−6
z
|z|2
n+1
X
zi ui
z
|z|2
− 2(2s −
zi + |z|2s−n−4 uii
uj
j=1
∂
∂zi
j=1
j,k=1
z
|z|2
·
zi δij
zi2 zj − 2|z|2s−n−4
2s−n−2
j=1
2s−n−4
z
|z|2
+ (2s − n − 2)|z|2s−n−4 ui
z
|z|2
z
|z|2
zi2 + (2s − n)|z|2s−n−2 zi
− 2|z|2s−n−4 zi ui
2s−n−4
zi zj uij
n+1
X
+ (2s − n − 2)|z|
n+1
X
uj
j,k=1
2s−n−4
z
|z|2
zi2 zj − 2|z|2s−n−4
j=1
2s−n−4
j=1
(2s − n)(2s − n − 2)|z|2s−n−4 u
z
|z|2
n+1
X
j=1
+ (2s − n − 2)|z|
j=1
2s−n−4
zi2 + (2s − n)|z|2s−n−2 zi
n+1
X
−2(2s − n − 4)|z|2s−n−6
z
|z|2
n)|z|2s−n−6 zi2 z
z
|z|2
· ∇u
z
|z|2
zi2 zj − 2|z|2s−n−6
n+1
X
j=1
uij
z
|z|2
z
z
z
− 2|z|2s−n−4 z · ∇u
− 2|z|2s−n−4 zi ui
|z|2
|z|2
|z|2
j,k=1
z
z
z
2
2s−n−4
2s−n−6 2
= (2s − n)(2s − n − 2)|z|2s−n−4 u
z
+
(2s
−
n)|z|
z
u
−
2(2s
−
n)|z|
z
z
·
∇u
i i
i
i
|z|2
|z|2
|z|2
z
z
z
+(2s − n)|z|2s−n−2 u
+ (2s − n − 2)|z|2s−n−4 ui
zi + |z|2s−n−4 uii
2
2
|z|
|z|
|z|2
n+1
X
z
z
2s−n−6 2
−4|z|2s−n−6
zi zj uij
−
2(2s
−
n
−
4)|z|
z
z
·
∇u
i
|z|2
|z|2
j=1
+4|z|2s−n−8
zi2 zj zk ujk
BREZIS TYPE THEOREM INVOLVING FRACTIONAL LAPLACIANS
n+1
X
3
z
z
z
2s−n−4
2s−n−4
−
2|z|
−
2|z|
z
·
∇u
z
u
i i
|z|2
|z|2
|z|2
j,k=1
z
z
2s−n−4
2
2s−n−4
= (2s − n)(2s − n − 2)|z|
u
zi + (4s − 2n − 4)|z|
zi ui
|z|2
|z|2
z
z
z
2s−n−2
2s−n−4
−2(4s − 2n − 4)|z|2s−n−6 zi2 z · ∇u
+
(2s
−
n)|z|
+
|z|
u
u
ii
|z|2
|z|2
|z|2
n+1
n+1
X
X
z
z
2s−n−6
2s−n−8
2
−4|z|
zi zj uij
+ 4|z|
zi zj zk ujk
|z|2
|z|2
j=1
j,k=1
z
−2|z|2s−n−4 z · ∇u
.
|z|2
+4|z|2s−n−8
zi2 zj zk ujk
So we get
vy (z)
=
=
∆v(z)
=
n+1
X
z
|z|2
y + |z|2s−n−2 uy
z
|z|2
n+1
X
z
yzj
|z|2
j=1
z
z
z
2s−n−2
2s−n−4
2s−n−2
y + |z|
uy
− 2|z|
z · ∇u
y
(2s − n)|z|
u
|z|2
|z|2
|z|2
(2s − n)|z|2s−n−2 u
− 2|z|2s−n−4
uj
vii (z)
i=1
=
=
Ls (v)(x, y)
z
z
2s−n−4
+
(4s
−
2n
−
4)|z|
z
·
∇
|z|2
|z|2
z
z
z
2s−n−2
2s−n−4
2s−n−4
+ (2s − n)(n + 1)|z|
u
+ |z|
∆u
−2(4s − 2n − 4)|z|
z · ∇u
|z|2
|z|2
|z|2
n+1
n+1
X
X
z
z
2s−n−6
−4|z|2s−n−6
zi zj uij
+
4|z|
z
z
u
j k jk
|z|2
|z|2
i,j=1
j,k=1
z
2s−n−4
−2(n + 1)|z|
z · ∇u
|z|2
z
z
z
2s−n−4
2s−n−4
(2s − n)(2s − 1)|z|2s−n−2 u
−
2(2s
−
1)|z|
z
·
∇u
+
|z|
∆u
|z|2
|z|2
|z|2
(2s − n)(2s − n − 2)|z|2s−n−2 u
:=
div [y 1−2s ∇v(x, y)]
=
y 1−2s ∆v(z) + (1 − 2s)y −2s vy (z)
=
y −2s [y∆v(z) + (1 − 2s)vy (z)]
z
z
z
2s−n−4
2s−n−4
−
y2(2s
−
1)|z|
z
·
∇u
+
y|z|
∆u
y −2s y(2s − n)(2s − 1)|z|2s−n−2 u
|z|2
|z|2
|z|2
z
z
z
2s−n−2
2s−n−2
2s−n−4
(1 − 2s)(2s − n)|z|
u
y + (1 − 2s)|z|
uy
− 2(1 − 2s)|z|
z · ∇u
y
|z|2
|z|2
|z|2
z
z
y −2s y|z|2s−n−4 ∆u
+ (1 − 2s)|z|2s−n−2 uy
2
|z|
|z|2
y
z
z
y −2s |z|2s−n−2
· ∆u
+ (1 − 2s)uy
|z|2
|z|2
|z|2
=
=
=
4
MINGFENG ZHAO
=
=
∂v
(x, 0)
∂νs
−2s y
y
z
z
|z|
·
· ∆u
+ (1 − 2s)uy
|z|2
|z|2
|z|2
|z|2
z
|z|−n−2s−2 Ls (u)
|z|2
−2s−n−2
:= − lim y 1−2s vy (x, y)
y&0
=
=
=
=
=
z
z
z
2s−n−2
2s−n−2
2s−n−4
y + |z|
− 2|z|
y
− lim y
(2s − n)|z|
u
uy
z · ∇u
y&0
|z|2
|z|2
|z|2
z
z
2s−n−4
− lim (2s − n)y 2−2s (2s − n)|z|2s−n−2 u
−
2|z|
z
·
∇u
y&0
|z|2
|z|2
z
− lim y 1−2s |z|2s−n−2 uy
y&0
|z|2
z
− lim y 1−2s |z|2s−n−2 uy
Since 0 < s < 1
y&0
|z|2
1−2s y
z
− lim |z|−n−2s
u
y
y&0
|z|2
|z|2
x
∂u
, 0 , ∀x ∈ Rn \{0}.
|x|−n−2s
∂νs |x|2
1−2s
So we get
Ls (v)(z)
=
∂v
(x, 0)
∂vs
=
=
=
=
∀z ∈ Rn+1
+ ,
x
−n−2s ∂u
−|x|
,0
∂νs |x|2
p
x
,
0
−|x|−n−2s u
|x|2
p
1
x
p(n−2s)−n−2s
−|x|
u
,0
|x|n−2s
|x|2
0,
−|x|α v p (x, 0),
∀x ∈ Rn .
Then v : Rn+1
+ \{0} → R satisfies
(2)



div [y 1−2s ∇v(x, y)] = 0, ∀(x, y) ∈ Rn+1
+ ,




∂v


(x, 0) := − lim y 1−2s vy (x, y) = −|x|α v p (x, 0),


y&0
∂ν

s

v(x, y) > 0, ∀(x, y) ∈ Rn+1
+ \{0},





lim v(z) = 0,



|z|→∞



 α = p(n − 2s) − (n + 2s).
∀x ∈ Rn \{0},
T
n + 2s
+
+T
, that is, α > 0. For any R > 0, let BR
= BR (0) Rn+1
∂Rn+1
Firstly, we will assume p >
+ , and
+ , ΓR = ∂BR
n − 2s
T n+1
n+1
SR = ∂BR (0) R+ . Since v(x, y) > 0 for all (x, y) ∈ R+ \{0}, then min v(z) > 0.
z∈S1
BREZIS TYPE THEOREM INVOLVING FRACTIONAL LAPLACIANS
Lemma 1. For any 0 < < min 1, min v(z) , we have
z∈S1
v(z) ≥
2s
,
1 + 2s
∀x ∈ B1+ \{0}.
Proof. For any 0 < r < 1, consider the function
ψr (z) =
y 2s
2s
rn−2s − n−2s +
,
1 + 2s |z|
1 + 2s
∀z ∈ Rn+1
+ \{0}.
By Remark 4, we know that
Ls (ψr )(z)
=
=
=
=
=
∂ψr
(x, 0)
∂νs
div [y 1−2s ∇ψr (x, y)]
Ls (y 2s )
1 + 2s
div [y 1−2s ∇y 2s ]
1 + 2s
div [y 1−2s (2s)y 2s−1 ]
1 + 2s
div [2s]
1 + 2s
∀z ∈ Rn+1
+ ,
=
0,
=
− lim y 1−2s Dy ψr (x, y)
=
=
=
y&0
lim y 1−2s Dy (y 2s )
1 + 2s y&0
−
lim y 1−2s (2s)y 2s−1
1 + 2s y&0
2s
−
, ∀x ∈ Rn \{0}.
1 + 2s
−
For all z ∈ Sr , we have
ψr (z)
=
≤
=
y 2s
2s
−+
1 + 2s
1 + 2s
2s
−+
1 + 2s
1 + 2s
Since 0 < y ≤ r < 1
0
≤ v(z).
For all z ∈ S1 , we have
ψr (z)
=
<
2s
y 2s
− rn−2s +
1 + 2s
1 + 2s
2s
+
Since 0 < y ≤ 1
1 + 2s 1 + 2s
<
≤
v(z).
5
6
MINGFENG ZHAO
Let Pr (z) = v(z) − ψr (z) in B1+ \Br+ , then Pr satisfies



div [y 1−2s ∇Pr (x, y)] = 0, ∀z ∈ B1+ \Br+ ,


 ∂P
2s
r
(x, 0) = −|x|α v p (x, 0) +
, ∀x ∈ Γ1 \Γr ,

∂ν
1
+ 2s
s


S

 P (z) ≥ 0, ∀z ∈ S
S .
r
r
1
Claim I: Pr (z) ≥ 0 for all B1+ \Br+ .
If the Claim I is not true, by the Strong Maximum Principle, there exists some x0 ∈ Γ1 \Γr such that
Pr (x0 , 0) =
inf
Pr (z) < 0,
and
Pr (z) > Pr (x0 , 0),
z∈B1+ \Br+
∀z ∈ B1+ \Br+ .
2s
∂Pr
(x0 , 0) < 0, that is, −|x0 |α v p (x0 , 0) +
< 0, which implies that
∂νs
1 + 2s
1
α
2s p
· |x0 |− p
1 + 2s
By the Hopf’s Lemma for Ls , we know that
v(x0 , 0) >
≥
2s
,
1 + 2s
Since p > 1 > , α > 0 and |x0 | < 1.
On the other hand, we have
Pr (x0 , 0)
=
v(x0 , 0) −
>
v(x0 , 0) −
rn−2s 2s
−
1 + 2s |x0 |n−2s
2s
1 + 2s
> 0,
which contradicts with Pr (x0 , 0) < 0.
By the Claim I, we know that
v(z) ≥
rn−2s y 2s
2s
rn−2s 2s
− n−2s +
≥
− n−2s ,
1 + 2s |z|
1 + 2s
1 + 2s |z|
∀z ∈ B1+ \Br+ .
By taking r & 0 in the above inequality, we know that
v(z) ≥
2s
,
1 + 2s
∀z ∈ B1+ \{0}.
Remark 4. Let n ≥ 2, define
u(x, y) =
1
1
=
n−2s ,
2
|(x, y)|n−2s
[|x| + y 2 ] 2
∀(x, y) ∈ Rn+1
+ .
Let r = |(x, y)|, then for all (x, y) 6= 0, we have
div [y 1−2s ∇u(x, y)]
= y 1−2s ∆u(x, y) + (1 − 2s)y −2s uy (x, y)
h
i
n
y
= y 1−2s · u00 (r) + · u0 (r) + (1 − 2s)y −2s u0 (r) ·
r
r
BREZIS TYPE THEOREM INVOLVING FRACTIONAL LAPLACIANS
= y 1−2s [(2s − n)(2s − n − 1)r2s−n−2 + n(2s − n)r2s−n−2 ] + (1 − 2s)y 1−2s (2s − n)r2s−n−2
= y 1−2s (2s − n)(2s − 1)r2s−n−2 + (1 − 2s)y 1−2s (2s − n)r2s−n−2
=
uy (x, y)
= u0 (r) ·
=
−y 1−2s uy (x, y)
0
y
r
(2s − n)yr2s−n−2
= −y 2−2s (2s − n)r2s−n−2
= y 2−2s (n − 2s)|(x, y)|2s−n−2
≥ 0,
Since n ≥ 2 ≥ 2s.
Now for any λ < 0, we define
Σλ
=
z ∈ Rn+1
: x1 > λ ,
+
Tλ
=
z ∈ Rn+1
: x1 = λ ,
+
Σλ
= Σλ \{0},
zλ
=
(2λ − x1 , x2 , · · · , xn , y),
vλ (z)
= v(z λ ),
wλ (z)
= v(z) − vλ (z).
wλ (z)
=
0,
Ls (wλ )(z)
=
Ls (v)(2λ − x1 , x2 , · · · , xn , y)
=
0,
Notice that
For all x ∈ ∂Rn+1
+
T
∀z ∈ ∂Σλ
\
Rn+1
+ ,
∀z ∈ Σλ .
Σλ , it’s easy to see that |x| < |xλ |, which implies that
∂wλ
(x, 0)
∂νs
=
∂v
∂v λ
(x, 0) −
(x , 0)
∂νs
∂νs
=
−|x|α v p (x, 0) + |xλ |α v p (xλ , 0)
=
−|x|α [v p (x, 0) − vλp (xλ , 0)] + [|xλ |α − |x|α ]v p (xλ , 0)
≥
−|x|α [v p (x, 0) − vλp (xλ , 0)]
= −c1 (x, λ)wλ (x, 0),
By the Mean Value Theorem.
where c1 (x, λ) = p|x|α ξ1p−1 (x, λ) for some positive function ξ1 (x, λ) which is between v(x, 0) and vλ (x, 0).
7
8
MINGFENG ZHAO
Lemma 2. For any fixed λ < 0, if there exists some small 0 < r < |λ| such that wλ (z) > 0 for all z ∈ Br+ \{0}, then
wλ (z) ≥ 0 for all z ∈ Σλ .
Proof. If the Lemma 2 is not true, then there exists some z0 ∈ Σλ such that wλ (z0 ) < 0. Since lim
|z|→∞
v(z) = 0,
then lim wλ (z) = 0. Since wλ (z) > 0 for all z ∈ Br+ \{0} and wλ (z) = 0 for all z ∈ Tλ , then there exists some
|z|→∞
S z1 ∈ Σλ \ Br+ Tλ such that
wλ (z1 ) = inf wλ (z) < 0.
z∈Σλ
T
By the Strong Maximum Principle, we know that z1 ∈ ∂Σλ ∂Rn+1
+ \{0} and wλ (z) > wλ (z1 ) for all z ∈ Σλ . By the
∂wλ
Hopf’s Lemma for Ls , we know that
(z1 ) < 0. On the other hand, we know that
∂νs
∂wλ
(z1 ) ≥ −c1 (z0 , λ)wλ (z0 ) > 0.
∂νs
We get a contradiction. Therefore, we know that wλ (z) ≥ 0 for all z ∈ Σλ .
Lemma 3. There exists some Λ −1 such that for all λ ≤ Λ, we have
wλ (z) ≥ 0,
∀z ∈ Σλ .
Proof. By Lemma 1, there exists some 1 > 0 such that
v(z) ≥ 2,
Since lim
|z|→∞
∀z ∈ B1+ \{0}.
v(z) = 0, then there exists some R0 1 such that
|v(z)| ≤ ,
∀|z| ≥ R0 .
Let Λ = −R0 − 1 −1, for all λ ≤ Λ, then B1+ \{0} ⊂ Σλ . For all z ∈ B1+ \{0}, we know that |z λ | = |(2λ −
x1 , · · · , xn , y)| ≥ 2|λ| − |z| ≥ 2|λ| − 1 ≥ |λ| > R0 , which implies that |vλ (z)| = |v(z λ )| ≤ . Hence we get
wλ (z) = v(z) − vλ (z) ≥ 2 − = > 0,
∀z ∈ B1+ \{0}.
By Lemma 2, we know that wλ (z) ≥ 0 for all z ∈ Σλ .
By the Lemma 3, we can define
λ0 = sup
Lemma 4. λ0 = 0.
λ < 0 : wµ (z) ≥ 0 in Σµ , ∀µ ≤ λ .
BREZIS TYPE THEOREM INVOLVING FRACTIONAL LAPLACIANS
9
Proof. If the Lemma 4 is not true, that is, λ0 < 0. By the Definition of λ0 , we know that wλ0 (z) ≥ 0 in Σλ0 . If
wλ0 (z) ≡ 0 in Σλ0 , that is, v(z) = v(z λ ) in Σλ0 , in particular, we get
0
=
=
∂wλ0
(x, 0)
∂νs
∂v
∂v λ0
(x, 0) −
(x , 0)
∂νs
∂νs
=
−|x|α v p (x, 0) + |xλ0 |α v p (xλ , 0)
=
[|xλ0 |α − |x|α ]v p (x, 0),
On the other hand, for all x ∈ ∂Rn+1
+
T
∀x ∈ ∂Rn+1
+
\
Σλ 0
Σλ0 , since λ0 < 0, then we know that |x| < |xλ0 |. Since v(x, 0) > 0, then the
above inequality will give us 0 > 0, contradiction. So the only possibility is that wλ0 (z) 6≡ 0 in Σλ0 . Since wλ0 (z) ≥ 0
T
in Σλ0 , by the Strong Maximum Principle, we have wλ0 (z) > 0 in Σλ0 . If there exists some x0 ∈ Σλ0 ∂Rn+1
such
+
that wλ0 (x0 , 0) = 0, then wλ0 (x, y) > 0 = wλ0 (x0 , 0) for all (x, y) ∈ Σλ0 , by the Hopf’s Lemma for Ls , we have
T
∂wλs
(x0 , 0) < 0. On the other hand, since wλ0 (x0 , 0) = 0, then v(x0 , 0) = v(xλ0 0 , 0) > 0. Since (x0 , 0) ∈ Σλ0 ∂Rn+1
+ ,
∂νs
∂wλs
then |xλ0 0 | > |x0 |. Hence we get
(x0 , 0) = −|x0 |α v p (x0 , 0) + |xλ0 0 |α v p (xλ0 0 , 0) > 0, contradiction. In summary, we
∂νs
know that wλ0 (z) > 0 for all Σλ0 \Tλ0 .
|λ0 |
, 1 , there exists some constant δ > 0 which only depends on λ0 and r such that
Claim I: For any 0 < r < min
2
wλ0 (z) > δ for all z ∈ Br+ \{0}.
Notice that wλ0 satisfies the following PDEs:



Ls (wλ0 )(z) = 0, ∀z ∈ Br+ ,


 ∂w
λ0
(x, 0) = −|x|α v p (x, 0) + |xλ0 |α vλp0 (x, 0),

∂ν
s



 w (z) > 0, ∀z ∈ B + \{0}.
∀x ∈ Γr \{0},
r
λ0
Since wλ0 (z) > 0 for all z ∈ Br+ \{0}, then min wλ0 (z) > 0. Let 0 < < min 1, min wλ0 (z) . For all z ∈ Br+ \{0},
z∈Sr
z∈Sr
|λ0 |
, then z λ0 ∈
/ Br+ , which implies that
since 0 < r <
2
C1 :=
sup
vλ0 (z) ∈ (0, ∞).
z∈Br+ \{0}
For any 0 < t < r and 0 < µ < 1, let
ψt,µ (z) =
2sµ
tn−2s (1 − µ)y 2s
− n−2s +
,
1 + 2s |z|
1 + 2s
∀z ∈ Br+ \Bt+ .
By the proof of Lemma 1, we know that


 Ls (ψt,µ ) = 0, ∀z ∈ Br+ \Bt+ ,
∂ψt,µ
2s(1 − µ)


(x, 0) = −
, ∀x ∈ Γr \Γt .
∂νs
1 + 2s
10
MINGFENG ZHAO
For all z ∈ Sr , we have
ψt,µ (z)
=
<
≤
≤
2sµ
1 + 2s
2sµ
1 + 2s
2sµ
1 + 2s
2s
1 + 2s
tn−2s (1 − µ)y 2s
+
rn−2s
1 + 2s
2s
(1 − µ)y
+
1 + 2s
(1 − µ)
+
Since 0 < y < r < 1
1 + 2s
+
Since 0 < µ < 1
1 + 2s
−
= ≤ wλ0 (z).
For all z ∈ St , we have
ψt,µ (z)
=
≤
=
(1 − µ)y 2s
2sµ
−+
1 + 2s
1 + 2s
2s
−+
Since 0 < y, µ < 1
1 + 2s
1 + 2s
0
≤ wλ0 (z).
Let Pt,µ (z) = wλ0 (z) − ψt,µ (z) in Br+ \Bt+ , then Pt,µ satisfies



Ls (Pt,µ )(z) = 0, ∀z ∈ Br+ \Bt+ ,



∂Pt,µ
2s(1 − µ)
− |x|α v p (x, 0) + |xλ0 |α vλp0 (x, 0),
(x, 0) =

∂ν
1
+
2s
s



 P (z) ≥ 0, ∀z ∈ S S S .
t,µ
r
∀x ∈ Γr \{0},
t
If there exists some point z1 ∈ Br+ \Bt+ such that Pt,µ (z1 ) < 0, by the Strong Maximum Principle, there exists some
x0 ∈ Γr \Γt such that
Pt,µ (x0 , 0) =
inf
Pt,µ (z) < 0,
and
Pt,µ (z) > Pt,µ (x0 , 0),
z∈Bt+ \Bt+
Since Pt,µ (x0 , 0) < 0, then
wλ0 (x0 , 0) <
=
<
ψt,µ (x0 , 0)
2sµ
tn−2s −
1 + 2s |x0 |n−2s
2sµ
.
1 + 2s
Since x0 ∈ Γr \Γt , then t < |x0 | < r, which implies that
ψt,µ (x0 , 0) =
2sµ
tn−2s 2sµ
2s
−
≤
<
< 1.
1 + 2s |x0 |n−2s
1 + 2s
1 + 2s
∀z ∈ Br+ \Bt+ .
BREZIS TYPE THEOREM INVOLVING FRACTIONAL LAPLACIANS
11
Since Pt,µ (x0 , 0) < 0, then
v(x0 , 0) <
≤
vλ0 (x0 , 0) + ψt,µ (x0 , 0)
C1 + 1
= C2 .
By the Hopf’s Lemma for Ls , we know that
2s(1 − µ)
− |x0 |α v p (x0 , 0) + |xλ0 0 |α vλp0 (x0 , 0) < 0, which implies that
1 + 2s
|xλ0 0 |α v p (x0 , 0) − |xλ0 0 |α vλp0 (x0 , 0)
> |xλ0 0 |α v p (x0 , 0) +
≥
2s(1 − µ)
,
1 + 2s
2s(1 − µ)
− |x0 |α v p (x0 , 0)
1 + 2s
Since |xλ0 0 | > |x0 |.
By the Mean Value Theorem, there exists some positive number ξ2 (λ0 , x0 ) which is between v(x0 , 0) and vλ0 (x0 , 0)
such that
2s(1 − µ)
1 + 2s
≤
|xλ0 0 |α v p (x0 , 0) − |xλ0 0 |α vλp0 (x0 , 0)
= |xλ0 0 |α pξ2p−1 (λ0 , x0 )wλ0 (x0 , 0)
Since wλ0 (z) > 0 in Σλ0 , then 0 ≤ v(x0 , 0) ≤ vλ0 (x0 , 0) ≤ C1 , which implies that
2s(1 − µ)
1 + 2s
≤
|xλ0 0 |α pC1p−1 wλ0 (x0 , 0)
≤ C3 wλ0 (x0 , 0).
In summary, we get
2s(1 − µ) 1
2sµ
> wλ0 (x0 , 0) ≥
·
.
1 + 2s
1 + 2s
C3
In particular, we have
C3 µ > (1 − µ).
That is, µ >
1
1
. But if we take some fixed µ <
, we must have Pt,µ (z) ≥ 0 for all z ∈ Br+ \Bt+ , that is,
1 + C3
1 + C3
wλ0 (z) ≥ ψt,µ (z) =
2sµ
tn−2s (1 − µ)y 2s
2sµ
tn−2s − n−2s +
≥
− n−2s ,
1 + 2s |z|
1 + 2s
1 + 2s |z|
∀z ∈ Br+ \Bt+ .
2sµ
, for all z ∈ Br+ \{0}.
1 + 2s
By the Definition of λ0 , there exists a strictly decreasing sequence 0 > λk > λ0 such that lim λk = λ0 and
k→∞
|λ0 |
, 1 , by the Claim I, there exists some small δ > 0 such that
inf wλk (z) < 0. For any fixed 0 < r < min
2
z∈Σλ
wλ0 (z) ≥ 2δ for all z ∈ Br+ . Since lim λk = λ0 , then there exists some K 1 such that wλk (z) ≥ δ > 0 for all z ∈ Br+ .
By taking t & 0, we get wλ0 (z) ≥ δ with δ =
k→∞
By Lemma 2, we know that wλk (z) ≥ 0 for all z ∈ Σλk , which contradicts with inf wλk (z) < 0.
z∈Σλ
In summary, we must have λ0 = 0.
12
MINGFENG ZHAO
Proof of Theorem 2. Case I: p >
n + 2s
. By Lemma 4, we know that
n − 2s
∀z ∈ Rn+1
with x1 > 0.
+
v(x1 , · · · , xn , y) ≥ v(−x1 , · · · , xn , y),
Let v(z) = v(−x1 , · · · , xn , y) for all z ∈ Rn+1
+ \{0}, then v is also a solution of 2, then
v(−x1 , · · · , xn , y) = v(x1 , · · · , xn , y) ≥ v(−x1 , · · · , xn , y) = v(x1 , · · · , xn , y),
∀z ∈ Rn+1
with x1 > 0.
+
So v is symmetric with respect to x1 = 0. By the rotation, we know that v is radial symmetric with respect to x ∈ Rn .
Since we know that
v(z) =
1
|z|n−2s
u
z
|z|2
z
|z|2
,
∀z ∈ Rn+1
+ \{0}.
,
∀z ∈ Rn+1
+ \{0}.
Then
u(z) =
1
|z|
v
n−2s
Since v is radially symmetric with respect to x ∈ Rn , then u is also radially symmetric with respect to x ∈ Rn .
On the other hand, we know that any translation with respect to x ∈ Rn of u is also a solution of 1, then for each
y > 0, u(·, y) is radial symmetric with respect to each x ∈ Rn , which implies that u(·, y) is a constant function
d 1−2s 0
[y
U (y)] = 0, then there
in Rn , that is, u(x, y) = U (y) for some function U on [0, ∞). By (1), we have
dy
exists some a, b ∈ R such that U (y) = ay 2s + b for all y > 0. Since U (y) > 0 for all y ≥ 0, then a, b > 0. Since
∂U
(x, 0) = − lim y 1−2s U 0 (y) = −a · 2s = −bp , then 2sa = bp .
y&0
∂νs
n + 2s
n + m + 2s
Case II: 1 < p ≤
. In this case, there exists some large m ∈ N such that p >
. Let u(x, t, y) = u(x, y)
n − 2s
n − m − 2s
n+m+1
, then u is a solution of (2) in Rn+m+1
. By the result of Case I, we know that, u(x, t, y) =
for all (x, t, y) ∈ R+
+
u(x, y) = ay 2s + b in Rn+m+1
.
+
Conjecture 2. Let 0 < s < 1 and p, q > 1, then the only solutions of the following problem:



Ls (u)(z) = Lz (v)(z) = 0, ∀z = (x, y) ∈ Rn+1

+ ,

 ∂u
∂v
(3)
(x, 0) = −v p (x, 0),
(x, 0) = −uq (x, 0) ∀x ∈ Rn ,

∂ν
∂ν
s
s



 u(x, y), v(x, y) > 0, ∀(x, y) ∈ Rn+1 .
+
must be the form of u(x, y) = ay 2s + b and v(x, y) = cy 2s + d in Rn+1
for some a, b, c, d > 0 such that 2sa = dp and
+
2sc = bq .
References
[1] H. Brezis. Semilinear equations in RN without condition at infinity. Appl. Math. Optim., 12:271–282, 1984.
[2] Yuan Lou and Meijun Zhu. Classifications of nonnegative solutions to some elliptic problems. Differential and Integral Equations, 12:601–
612, 1999.
BREZIS TYPE THEOREM INVOLVING FRACTIONAL LAPLACIANS
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
E-mail address: mingfeng.zhao@uconn.edu
13