SCALING METHOD FOR THE LIOUVILLE TYPE THEOREM
MINGFENG ZHAO
March 06, 2014
Contents
1.
Whole Space
1
2.
Upper Half Space with Neumann Boundary Reaction
8
References
14
1. Whole Space
Lemma 1.1. Let u satisfy

 −∆u ≥ 0, ∀x ∈ Rn ,
 u(x) > 0, ∀x ∈ Rn .
Let C1 = min u(x) > 0, then
|x|=1
u(x) ≥
C1
,
|x|n−2
Proof. Let C1 = min u(x) > 0, for any R > 1, let v(x) =
|x|=1
∀|x| ≥ 1.
C1
C1
+ u(x) − n−2 , then v satisfies
n−2
R
|x|



−∆v(x) = −∆u(x) ≥ 0,



v(x) ≥ 0, ∀|x| = 1




 v(x) > 0, ∀|x| = R.
∀1 < |x| < R,
By the Strong Maximum Principle, we have v(x) > 0 for all 1 < |x| < R, that is,
C1
C1
+ u(x) ≥
,
Rn−2
|x|n−2
∀1 < |x| < R.
By taking R → ∞, we get
u(x) ≥
C1
,
|x|n−2
∀|x| ≥ 1.
1
2
MINGFENG ZHAO
Theorem 1.1. Let n ≥ 3, 1 ≤ p <
(1)
n
and u satisfies
n−2

 −∆u ≥ up ,
 u(x) ≥ 0,
∀x ∈ Rn ,
∀x ∈ Rn .
Then u(x) ≡ 0 in Rn .
Proof. If u(x) 6≡ 0 in Rn , by the Strong Maximum Principle, we have u(x) > 0 in Rn . So we can define
C1 = min u(x) > 0.
|x|=1
By Lemma 1.1, we have
u(x) ≥
C1
,
|x|n−2
∀|x| ≥ 1.
ξr2
on the first equation of (1) and integrate over R2+ , use the integration by parts, we have
u
Z
up−1 dx =
ξr2 up−1 dx Since ξr (x) = 1 for all 2r < |x| < 3r dx
For any r 2, multiply
Z
2r<|x|<3r
2r<|x|<3r
Z
≤
Rn
ξr2 p
·u
u
ξr2
· ∆u dx
u
Z
≤
−
Rn
ξr2
· ∆u dx
u
Z
=
−
r<|x|<4r
Z
∇u · ∇
=
r<|x|<4r
Since ξr (x) = 0 for all |x| < r and |x| > 4r
ξr2
dx
u
2ξr u∇ξr − ξr2 ∇u
dx
u2
Rn
Z
Z
ξr
ξr2
2
∇u · ∇ξr dx −
|∇u|2 dx
2
u
u
n
r<|x|<4r
R
Z
|∇ξr |2 dx By Cauchy-Schwarz’s inequality
Z
=
=
≤
∇u ·
r<|x|<4r
≤
C2 n
·r
r2
=
C2 rn−2 .
On the other hand, we know that
Z
up−1 dx ≥
2r<|x|<3r
Z
2r<|x|<3r
Z
=
C3
2
p−1
dx
Since p ≥ 1
3r
C3
2r
Z 3
=
C1
|x|n−2
ρ−(n−2)(p−1)+n−1 dρ
r−(n−2)(p−1)+n−1+1 t−(n−2)(p−1)+n−1 dt Let ρ = rt
SCALING METHOD FOR THE LIOUVILLE TYPE THEOREM
=
3
C4 r−(n−2)(p−1)+n .
So we get C4 r−(n−2)(p−1)+n ≤ C2 rn−2 , that is, 0 < C5 ≤ rn−2+(n−2)(p−1)−n = rp(n−2)−n for all r 2. But if
n
p(n − 2) − n < 0, that is, p <
, by taking r → ∞, we get 0 < C5 ≤ 0, contradiction.
n−2
Remark 1.1. By the same argument as Theorem 1.1, we can show that Theorem 1.1 holds for 0 < p < 1 if u is
bounded. In fact, when 0 < p < 1and u ∈ L∞ (Rn ), then
Z
Z
up−1 dx ≥
2r<|x|<3r
n
So we get C4 r ≤ C2 r
n−2
kukL∞ (Rn )
dx = C4 rn .
2r<|x|<3r
, that is, 0 < C5 < r
−2
for all r 1, contradiction.
Theorem 1.2. Let n ≥ 3, M > 0, p, q > 0 be such that min {p, q} <
(2)
p−1
2
, and u, v satisfies
n−2



−∆u ≥ v p , ∀x ∈ Rn ,



−∆v ≥ uq , ∀x ∈ Rn ,




 M ≥ u(x), v(x) ≥ 0, ∀x ∈ Rn .
Then u(x) ≡ v(x) ≡ 0 in Rn .
Proof. If Theorem 1.2 is not true, by the Strong Maximum Principle, we have u(x), v(x) > 0 in Rn . So we can define
C1 = min min u(x), min v(x) > 0.
|x|=1
|x|=1
By 1.1, we have
M ≥ u(x), v(x) ≥
C1
,
|x|n−2
∀|x| ≥ 1.
ξr2
ξ2
on the first equation of (2), multiply r on the second equation of (2), add them together,
u
v
and integrate over R2+ , use the integration by parts, we have
Z
Z
1
1
[v p + uq ] dx =
ξ 2 [v p + uq ] dx Since ξr (x) = 1 for all 2r < |x| < 3r dx
M 2r<|x|<3r
M 2r<|x|<3r r
Z 2
ξr p ξr2 q
≤
v + u
Since M ≥ u(x), v(x) in Rn
u
v
n
R
Z
Z
ξr2
ξr2
≤ −
· ∆u dx −
· ∆v dx
Rn u
Rn v
Z
Z
ξr2
ξr2
= −
· ∆u dx −
· ∆v dx Since ξr (x) = 0 for all |x| < r and |x| > 4r
r<|x|<4r u
Rn v
Z
Z
ξ2
ξ2
=
∇u · ∇ r dx +
∇v · ∇ r dx
u
v
r<|x|<4r
r<|x|<4r
Z
Z
2
2ξr u∇ξr − ξr ∇u
2ξr v∇ξr − ξr2 ∇v
=
∇u ·
dx
+
∇v
·
dx
u2
v2
r<|x|<4r
r<|x|<4r
For any r 2, multiply
4
MINGFENG ZHAO
Z
=
ξr
∇u · ∇ξr dx −
u
2
r<|x|<4r
Z
|∇ξr |2 dx
≤ 2
Z
Rn
ξr2
|∇u|2 dx + 2
u2
Z
r<|x|<4r
ξr
∇v · ∇ξr dx −
v
Z
Rn
ξr2
|∇v|2 dx
v2
By Cauchy-Schwarz’s inequality
r<|x|<4r
≤
C2 n
·r
r2
= C2 rn−2 .
On the other hand, we know that
Z
p
v dx
Z
≥
2r<|x|<3r
2r<|x|<3r
C1
|x|n−2
p
dx
Since p, q > 0
3r
Z
ρ−(n−2)p+n−1 dρ
= C3
2r
Z 3
= C3
r−(n−2)p+n−1+1 t−(n−2)p+n−1 dt Let ρ = rt
2
= C4 r−(n−2)p+n .
Similarly, we have
Z
uq dx ≥ C4 r−(n−2)q+n .
2r<|x|<3r
Without loss of generality, assume p ≥ q > 0, so we get
C2 rn−2 ≥
1
M
Z
[v p + uq ] dx ≥ C5 r−(n−2)q+n .
2r<|x|<3r
That is, 0 < C6 < rn−2+(n−2)q−n = r(n−2)q−2 . If (n − 2)q − 2 < 0, that is, q <
2
, by taking r → ∞, we get
n−2
0 < C6 ≤ 0, contradiction.
Lemma 1.2. Let n ≥ 3, pq > 1, and u, v satisfy
(3)



−∆u ≥ v p , ∀x ∈ Rn ,



−∆v ≥ uq , ∀x ∈ Rn ,




 u(x), v(x) ≥ 0, ∀x ∈ Rn .
Then
Z
v p dx
≤ Crn−2−
2(p+1)
pq−1
uq dx
≤ Crn−2−
2(q+1)
pq−1
,
|x|<r
Z
|x|<r
.
SCALING METHOD FOR THE LIOUVILLE TYPE THEOREM
5
Proof. Let ζ ∈ Cc∞ ([0, ∞)) be such that ζ(t) = 1 for all 0 ≤ t ≤ 1, ζ(t) = 0 for all t ≥ 2, and 0 ≤ ζ(t) ≤ 1 for all t ≥ 0.
For any r 2, let
ζr (x) = ζ
|x|
r
,
∀x ∈ Rn .
Let m > 2, notice that ζr is radially symmetric and
ζr0 (|x|)
=
ζr00 (|x|)
=
dζrm
d|x|
|x|
r
1 00 |x|
ζ
r2
r
1 0
ζ
r
= mζrm−1 ζr0
d2 ζrm
d|x|2
= m(m − 1)ζrm−2 |ζr0 |2 + mζrm−1 ζr00
∆ζrm
=
d2 ζrm
n − 1 dζrm
+
d|x|2
r d|x|
= m(m − 1)ζrm−2 |ζr0 |2 + mζrm−1 ζr00 +
≤
n−1 0
ζr (|x|)
r
C m−2
ζ
.
r2 r
Let m1 , m2 > 2. Multiply ζrm1 on the first equation of (3), and integration by parts, we get
Z
ζrm1 v p
Z
dx
≤
−
Rn
|x|<2r
ζrm1 ∆u dx
Z
u∆ζrm1 dx
= −
|x|<2r
Use integration by parts, and ζr (x) = 0 for all |x| ≥ 2r
Z
= −
u∆ζrm1 dx Since ζr (x) = 1 for all |x| < r
r<|x|<2r
≤
≤
Z
C1
r2
uζrm1 −2 dx
r<|x|<2r
! q1
Z
C1
r2
ζr(m1 −2)·q uq dx
r<|x|<2r
= C2 r
n(q−1)
−2+
q
! q−1
q
Z
dx
By Holder’s inequality
r<|x|<2r
Z
! q1
ζr(m1 −2)q uq dx
.
r<|x|<2r
Multiply ζrm2 on the second equation of (3), and integration by parts, we get
Z
ζrm2 uq dx
Z
≤
−
Rn
|x|<2r
ζrm2 ∆v dx
Z
= −
v∆ζrm2 dx
|x|<2r
Use integration by parts, and ζr (x) = 0 for all |x| ≥ 2r
6
MINGFENG ZHAO
Z
v∆ζrm2 dx
= −
Since ζr (x) = 1 for all |x| < r
r<|x|<2r
≤
≤
C1
r2
Z
! p1
Z
C1
r2
= C2 r
vζrm2 −2 dx
r<|x|<2r
ζr(m1 −2)·q v p
dx
dx
By Holder’s inequality
r<|x|<2r
r<|x|<2r
−2+
! p−1
p
Z
! p1
Z
n(p−1)
p
ζr(m2 −2)p v p
.
dx
r<|x|<2r
Since pq > 1, then we have

 (m1 − 2)q = m2
2(pq + p)
2(pq + q)
=⇒ m1 =
> 2, m2 =
> 2.
 (m − 2)p = m
pq − 1
pq − 1
2
1
Since 0 ≤ ζr (x) ≤ 1 for all x ∈ Rn , then we have
Z
ζrm1 v p dx
≤
C2 r
! q1
Z
n(q−1)
−2+
q
|x|<2r
ζr(m1 −2)q uq dx
r<|x|<2r
≤
C2 r
−2+
! q1
Z
n(q−1)
q
ζrm2 uq dx
r<|x|<2r
Z
ζrm2 uq
dx
≤
C2 r
−2+
! p1
Z
n(p−1)
p
|x|<2r
ζr(m2 −2)p v p
dx
r<|x|<2r
≤
C2 r
−2+
! p1
Z
n(p−1)
p
ζrm1 v p
dx
.
r<|x|<2r
So we get
Z
ζrm1 v p
dx
≤ C2 r
−2+
n(q−1)
q
! q1
Z
|x|<2r
ζrm2 uq
dx
r<|x|<2r
! p1  q1

≤ C2 r−2+
n(q−1)
q
C2 r−2+
Z
n(p−1)
p
ζrm1 v p dx
r<|x|<2r
= C3 r
−2+
n(q−1)
+ q1
q
1
! pq
Z
(−2+ n(p−1)
)
p
ζrm1 v p
dx
r<|x|<2r
= C3 r
−2pq+np(q−1)−2p+n(p−1)
pq
1
! pq
Z
ζrm1 v p
dx
r<|x|<2r
= C3 r
−2pq+npq−np−2p+np−n
pq
1
! pq
Z
ζrm1 v p dx
r<|x|<2r
= C3 r
−2pq+npq−2p−n
pq
Z
r<|x|<2r
1
! pq
ζrm1 v p dx
.

SCALING METHOD FOR THE LIOUVILLE TYPE THEOREM
7
Then
Z
v p dx
Z
ζrm1 v p dx
≤
|x|<r
Since ζr (x) = 1 for all |x| < r
|x|<2r
pq
−2pq+npq−2p−n
· pq−1
pq
≤
C4 r
=
C4 r
=
C4 rn−2−
(n−2)pq−n−2p
pq−1
2(p+1)
pq−1
.
And also we have
Z
ζrm2 uq
dx
≤ C2 r
−2+
n(p−1)
p
! p1
Z
|x|<2r
ζrm1 v p
dx
r<|x|<2r
! q1  p1

≤ C2 r−2+
n(p−1)
p
C2 r−2+
Z
n(q−1)
q
ζrm2 uq dx

r<|x|<2r
= C3 r
−2+
n(p−1)
1
+p
p
1
! pq
Z
(−2+ n(q−1)
)
q
ζrm2 uq
dx
r<|x|<2r
= C3 r
1
! pq
Z
−2pq+nq(p−1)−2p+n(q−1)
pq
ζrm2 uq
dx
r<|x|<2r
= C3 r
1
! pq
Z
−2pq+npq−nq−2q+nq−n
pq
ζrm2 uq dx
r<|x|<2r
= C3 r
−2pq+npq−2q−n
pq
1
! pq
Z
ζrm2 uq dx
.
r<|x|<2r
So we get
Z
uq dx
Z
ζrm2 uq dx
≤
|x|<r
Since ζr (x) = 1 for all |x| < r
|x|<2r
−2pq+npq−2q−n
pq
· pq−1
pq
≤
C4 r
=
C4 r
=
C4 rn−2−
(n−2)pq−n−2q
pq−1
2(q+1)
pq−1
.
Corollary 1.1. Let n ≥ 3, p > 1 and u satisfy

 −∆u ≥ up , ∀x ∈ Rn ,
 u(x) ≥ 0, ∀x ∈ Rn .
Then
Z
|x|<r
2
up dx ≤ Crn−2− p+1 ,
∀r > 1.
8
MINGFENG ZHAO
Corollary 1.2. Let n ≥ 3, pq > 1 be such that max
2(p + 1) 2(q + 1)
,
pq − 1 pq − 1
≥ n − 2, and u, v satisfy



−∆u ≥ v p , ∀x ∈ Rn ,



−∆v ≥ uq , ∀x ∈ Rn ,




 u(x), v(x) ≥ 0, ∀x ∈ Rn .
(4)
Then u(x) ≡ v(x) ≡ 0 in Rn .
Proof. Without loss of generality, we assume p ≥ q > 0, by the assumption, we have
2(p + 1)
≥ n − 2. By Lemma 1.2,
pq − 1
we have
Z
v p (x) dx ≤ C,
∀r > 1,
|x|<r
which implies that v ∈ Lp (Rn ), in particular,
Z
v p (x) dx = 0.
lim
r→∞
r<|x|<2r
On the other hand, by the proof of Lemma 1.2, we know that
Z
p
v (x) dx ≤ C3 r
−2pq+npq−2p−n
pq
|x|<r
Since
1
! pq
Z
p
v dx
.
r<|x|<2r
2(p + 1)
≥ n − 2, then 2p + 2 ≥ (n − 2)(pq − 1), that is, 2p + 2 ≥ npq − n − 2pq + 2, which implies that
pq − 1
−2pq + npq − 2p − n ≤ 0.
So we get
Z
Z
p
v (x) dx ≤ C3
|x|<r
1
! pq
p
v dx
.
r<|x|<2r
By taking r → ∞, we get
Z
v p (x) dx = 0.
Rn
So v(x) ≡ 0 in Rn . Since 0 ≡ −∆v(x) ≥ up (x) in Rn , then u(x) ≡ 0 in Rn .
2. Upper Half Space with Neumann Boundary Reaction
Let 0 < s < 1, n ≥ 1 and p ≥ 1, we will consider 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 ,
(5)

y&0
∂νs



 u(x, y) ≥ 0, ∀(x, y) ∈ Rn+1 .
SCALING METHOD FOR THE LIOUVILLE TYPE THEOREM
9
n+1
n+1
Remark 2.1. If u(x, y) ≡ 0 in Rn+1
+ , we are done. If u(x, y) 6≡ 0 in R+ , and there exists some z0 = (x0 , y0 ) ∈ R+
such 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)
∂u
∂u
(x0 , 0) < 0, which contradicts with
(x0 , 0) =
for all (x, y) ∈ Rn+1
+ . 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
+ .
Lemma 2.1. Let 0 < s < 1, n ≥ 1 and u satisfy



−Ls (u)(z) ≥ 0, ∀z = (x, y) ∈ Rn+1

+ ,

 ∂u
(x, 0) ≥ 0, ∀x ∈ Rn ,

∂νs



 u(x, y) > 0, ∀(x, y) ∈ Rn+1 .
+
Let C1 = min u(z) > 0, then
z∈S1
u(z) ≥
C1
,
|z|n−2s
+
∀z ∈ Rn+1
+ \B1 .
+
Proof. Let C1 = min u(z) > 0 and v(z) = C1 |z|2s−n for all z ∈ Rn+1
+ \B1 , then v satisfies
z∈S1

+


Ls (v)(z) = 0, ∀z ∈ Rn+1

+ \B1 ,

 ∂v
(x, 0) = 0, ∀x ∈ Rn \Γ1 ,

∂ν
s



 v(z) ≤ u(z), ∀z ∈ S .
1
+
\B1+ , then wR satisfies
For any R > 1, let wR (z) = u(z) + v(R) − v(z) for all z ∈ BR

+


−Ls (wR )(z) ≥ 0, ∀z ∈ BR
\B1+ ,


 ∂w
R
(x, 0) ≥ 0, ∀x ∈ ΓR \Γ1 ,

∂ν
s



 w (z) ≥ 0, ∀z ∈ S S S .
R
+
\B1+ be such that wR (zR ) =
Let zR ∈ BR
1
min
+
z∈BR
\B1+
R
+
\B1+ . If wR (zR ) < 0,
wR (z). If wR (zR ) ≥ 0, then wR (z) ≥ 0 in BR
+
by the Strong Maximum Priniciple, then zR ∈ ΓR \Γ1 and wR (z) > wR (zR ) for all z ∈ BR
\B1+ . By the Hopf’s Lemma
∂wR
+
for Ls , we have
(x, 0) < 0, contradiction. Hence, we must have wR (z) ≥ 0 in BR
\B1+ , that is,
∂νs
C1
C1
− n−2s ≤ u(z),
n−2s
|z|
R
+
∀z ∈ BR
\B1+ .
By taking R → ∞, we get
u(z) ≥
C1
,
|z|n−2s
+
∀z ∈ Rn+1
+ \B1 .
10
MINGFENG ZHAO
Theorem 2.1. Let 0 < s < 1, n > 2s and 1 ≤ p <
n
and u satisfy
n − 2s
(6)



−Ls (u)(z) ≥ 0, ∀z = (x, y) ∈ Rn+1

+ ,

 ∂u
p
n
(x, 0) ≥ u (x, 0), ∀x ∈ R ,

 ∂νs


 u(x, y) ≥ 0, ∀(x, y) ∈ Rn+1 .
Then u(x, y) ≡ 0 in Rn+1
+ .
Remark 2.2. We follow the proof of Lemma 6.2 on Page 133 in Hu and Yin[2] with s = 1/2.
n+1
Proof. Assume u(x, y) 6≡ 0 in Rn+1
+ , by Remark 2.1, we have u(x, y) > 0 in R+ . Let C1 = min u(z) > 0, by Lemma
z∈S1
2.1, we have
u(z) ≥
(7)
C1
,
|z|n−2s
+
∀z ∈ Rn+1
+ \B1 .
Now consider the s-Kevin transform of u:
u(z) =
1
u
|z|n−2s
z
|z|2
,
∀z ∈ Rn+1
+ \{0}.
By (7), then for all z ∈ B1+ \{0}, we have
n−2s
|z|
u(z) = u
z
|z|2
C1
≥
= C1 |z|n−2s .
z n−2s
|z|2 That is, we have
u(z) ≥ C1 > 0,
∀z ∈ B1+ \{0}.
Then u satisfies
(8)


−Ls (u)(z) ≥ 0, ∀z ∈ Rn+1

+ ,




∂u


(x, 0) ≥ |x|−α up (x, 0), ∀x ∈ Rn ,


∂ν

s



 u(z) > 0, ∀z ∈ Rn+1 \{0},


u(z) ≥ C1 > 0, ∀z ∈ B1+ \{0},







lim u(z) = 0,


|z|→∞




α = (n + 2s) − p(n − 2s).
Let ξ ∈ Cc∞ ([0, ∞) such that



ξ(t) = 0, ∀|z| ≤ 1 and |z| ≥ 4,



ξ(t) = 1, ∀2 ≤ |z| ≤ 3,




 0 ≤ ξ(t) ≤ 1, ∀t ≥ 0.
SCALING METHOD FOR THE LIOUVILLE TYPE THEOREM
11
For any 1/5 r > 0, let
ξr (z) = ξ
z r
,
∀z ∈ Rn+1 .
Then ξr satisfies


ξr (t) = 0,







ξr (t) = 1,



∀|z| ≤ r and |z| ≥ 4r,
∀2r ≤ |z| ≤ 3r,
0 ≤ ξr (t) ≤ 1, ∀t ≥ 0,





∇ξr (z) = 0, ∀|z| ≤ r and |z| ≥ 4r,





 |∇ξ (z)| ≤ C2 , ∀2r < |z| < 3r.
r
r
S
ξr2
on the first equation of (8) and integrate over B1+ , since ξr (z) = 0 for all z ∈ S1 Γr , use the integration
u
by parts, we have
Z
ξr2
· div [y 1−2s ∇u] dz
0 ≥
B1+ u
Z
Z
ξr2 ∂u
ξ2
=
y 1−2s ∇u · ∇ r dz
·
dx −
∂νs
u
Γ1 u
B1+
Z
Z
2
2ξr u∇ξr − ξr2 ∇u
ξr
· |x|−α up dx −
y 1−2s ∇u ·
≥
dz By (8)
u2
Γ1 u
B1+
2
Z
Z
Z
ξr
ξr
y 1−2s · ∇ξr · ∇u dz +
y 1−2s
=
ξr2 |x|−α up−1 dx − 2
|∇u|2 dz.
+
+
u
u
B1
Γ1
B1
Multiply
So we have
Z
ξr2 |x|−α up−1 dx
Γ1
2
Z
ξr
ξr
y 1−2s
∇ξr · ∇u dz −
|∇u|2 dz
u
u
B1+
B1+
2
2
Z
Z
Z
ξr
ξr
1−2s
1−2s
2
1−2s
2
y
y
|∇ξr | dz +
≤
y
|∇u| dz −
|∇u|2 dz
+
+
+
u
u
B1
B1
B1
Z
≤ 2
y 1−2s ·
By the Cauchy-Schwartz inequality
Z
=
B1+
Z
=
+
B4r
Z
≤
y 1−2s |∇ξr |2 dz
y 1−2s |∇ξr |2 dz
y 1−2s ·
+
B4r
C22
dz
r2
≤ C3 rn+2−2s−2
= C3 rn−2s .
On the other hand, we know that
Z
Z
ξr2 |x|−α up−1 dx =
Γ1
Γ1 \Γr
ξr2 |x|−α up−1 dx
12
MINGFENG ZHAO
Z
≥
Γ1 \Γr
ξr2 |x|−α C1p−1 dx
Z
Since u(z) ≥ C1 in B1+ \{0} and p ≥ 1
ξr2 |x|−α C1p−1 dx
=
r<|x|<4r
Z
ξr2 |x|−α C1p−1 dx
≥
2r<|x|<3r
Z
|x|−α C1p−1 dx
=
Since ξr (z) = 1 for all 2r < |z| < 3r
2r<|x|<3r
Z
3r
= C4
ρn−1−α dρ
2r
= C5 rn−1−α+1
= C5 rn−α .
So we get C5 rn−α ≤ C4 rn−2s , that is, 0 < C6 ≤ rα−2s for all 1/5 r > 0. If α > 2s, that is, α = (n + 2s) − p(n − 2s),
n
, by taking r & 0, we get 0 < C6 ≤ 0, contradiction.
which is the same as 1 ≤ p <
n − 2s
n
In summary, if n > 2s and 1 ≤ p <
, then u(z) ≡ 0 in Rn+1
+ .
n − 2s
Remark 2.3. By the same argument as Theorem 2.1, we can show that Theorem 2.1 holds for 0 < p < 1 if u is
bounded.
Theorem 2.2. Assume that 1/2 ≤ s < 1 and p ≥ 1, or 0 < s < 1/2 and 1 ≤ p <
1
. Let u satisfy
1 − 2s



−Ls (u)(z) ≥ 0, ∀z = (x, y) ∈ R2+ ,


 ∂u
(x, 0) ≥ up (x, 0), ∀x ∈ Rn ,

∂ν
s



 u(x, y) ≥ 0, ∀(x, y) ∈ R2 .
(9)
+
Then u(x, y) ≡ 0 in R2+ .
Remark 2.4. We follow the proof of Theorem on Page 302 in Hu[1] with s = 1/2.
Proof. Assume u(x, y) ≡ 0 in R2+ , by Remark 2.1, we have u(x, y) > 0 in R2+ . If 1/2 ≤ s < 1, Lemma 2.2 follows from
Lemma 4.7.6 on Page 173 in Zhao[3]. Now assume 0 < s < 1/2, let C1 = min u(z) > 0, by the proof of Theorem 2.1,
z∈S1
we have
u(z) ≥
(10)
For any r 2, multiply
Z
0
≥
R2+
C1
,
|z|1−2s
+
∀z ∈ Rn+1
+ \B1 .
ξr2
on the first equation of (5) and integrate over R2+ , use the integration by parts, we have
u
ξr2
· div [y 1−2s ∇u] dz
u
SCALING METHOD FOR THE LIOUVILLE TYPE THEOREM
13
Z
ξr2 ∂u
ξ2
=
·
dx −
y 1−2s ∇u · ∇ r dz
+
∂νs
u
Γ4r \Γr u
B4r
Z
Z
ξr2 p
2ξr u∇ξr − ξr2 ∇u
dz By (8)
≥
· u dx −
y 1−2s ∇u ·
+
u2
Γ4r \Γr u
B4r
2
Z
Z
Z
ξr
2 p−1
1−2s ξr
1−2s
=
|∇u|2 dz.
ξr u
dx − 2
y
y
· ∇ξr · ∇u dz +
+
+
u
u
Γ4r \Γr
B4r
B4r
Z
So we have
Z
ξr2 up−1 dx
Γ4r \Γr
2
Z
ξr
ξr
|∇u|2 dz
∇ξr · ∇u dz −
y 1−2s
+
+
u
u
B4r
B4r
2
2
Z
Z
Z
ξr
ξr
2
1−2s
2
1−2s
1−2s
|∇u| dz −
|∇u|2 dz
≤
y
y
|∇ξr | dz +
y
+
+
+
u
u
B4r
B4r
B4r
Z
y 1−2s ·
≤ 2
By the Cauchy-Schwartz inequality
Z
=
+
B4r
Z
y 1−2s |∇ξr |2 dz
y 1−2s ·
≤
+
B4r
C22
dz
r2
≤ C3 r1+2−2s−2
= C3 r1−2s .
On the other hand, we know that
Z
Γ4r \Γr
ξr2 up−1 dx
p−1
C1
dx By (10) and p ≥ 1
|x|1−2s
Γ4r \Γr
p−1
Z
C1
ξr2
≥
dx
|x|1−2s
Γ3r \Γ2r
p−1
Z
C1
=
dx Since ξr (z) = 1 for all 2r < |z| < 3r
|x|1−2s
Γ3r \Γ2r
Z
|x|(2s−1)(p−1) dx
= C1p−1
Z
=
ξr2
2r<|x|<3r
=
2C1p−1
Z
3r
x−(1−2s)(p−1) dx
2r
= C4 r1−(1−2s)(p−1)
So we get C4 r1−(1−2s)(p−1) ≤ C3 r1−2s , that is, 0 < C5 ≤ r(1−2s)(p−1)−2s for all r 2. If (1 − 2s)(p − 1) − 2s < 0,
2s
1
that is, (1 − 2s)(p − 1) < 2s, which implies that 1 ≤ p < 1 +
=
. In this case, by taking r → ∞, we get
1 − 2s
1 − 2s
0 < C5 ≤ 0, contradiction.
Conjecture 2.1. Let n = 1, 0 < s < 1/2, p ≥ 1 and u be a solution of (5), then u(x, y) ≡ 0 in R2+ .
14
MINGFENG ZHAO
Conjecture 2.2. Let 0 < s < 1, p > 1, and u be a solution of the following problem:

 (−∆)s u(x) = up (x), ∀x ∈ Rn ,
(11)
 u(x) > 0, ∀x ∈ Rn .
Then for any 0 < s0 < s < 1, we have
0
(−∆)s u(x) > 0,
∀x ∈ Rn .
References
[1] Bei Hu. Nonexistence of a positive solution of the laplace equation with a nonlinear boundary condition. Differetial and Integral Equation,
7:301–313, 1994.
[2] Bei Hu and Hongming Yin. The profile near blowup time for solution of the heat equation with a nonlinear boundary condition. Transactions of the American Mathematical Society, 346:117–135, 1994.
[3] M. Zhao. The Fractional Sobolev Spaces and the Fractional Laplacians. http://www.math.uconn.edu/ mingfeng/notes.html, 2012.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
E-mail address: mingfeng.zhao@uconn.edu