LIOUVILLE TYPE THEOREM IN UPPER HALF SPACE
MINGFENG ZHAO
March 06, 2014
Theorem 1. Let 0 < s < 1






(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
1−2s
n
(x, 0) := − lim y
uy (x, y) = 0, ∀x ∈ R ,
y&0
∂νs
∀(x, y) ∈ Rn+1
+ .
u(x, y) ≥ 0,
Then u is a constant function in Rn+1
+ .
Corollary 1. Let 0 < s < 1 and p > 1, any nonnegative bounded solution of (−∆)s u = 0 in Rn must be constant.
Remark 1. When s = 1/2, Theorem 1 is proved by Li and Zhu[1]. Indeed, our proof of Theorem 1 follows the same
idea as Li and Zhu[1], or Lou and Zhu[2].
n+1
n+1
such that
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+
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 all
∂u
∂u
(x, y) ∈ Rn+1
(x0 , 0) < 0, which contradicts with
(x0 , 0) = 0. So
+ . By the Hopf’s Lemma for Ls , we know that
∂νs
∂νs
the rest case should be u(x, y) > 0 for all (x, y) ∈ Rn+1
+ .
Let C1 = max u(z), by Lemma 2.1 on Page 9 in Zhao, we know that
z∈S1
(2)
u(z) ≥
C1
,
|z|n−2s
+
∀z ∈ Rn+1
+ \B1 .
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| → ∞. Also we know that v : Rn+1
+ \{0} → R satisfies



Ls (v)(x, y) = 0, ∀(x, y) ∈ Rn+1

+ ,



∂v


(x, 0) = 0, ∀x ∈ Rn \{0},

∂ν
s
(3)


v(x, y) > 0, ∀(x, y) ∈ Rn+1

+ \{0},





 lim v(z) = 0.
|z|→∞
Since v(x, y) > 0 for all (x, y) ∈ Rn+1
+ \{0}, then min v(z) > 0.
z∈S1
1
2
MINGFENG ZHAO
Lemma 1. Let C1 = min v(z), we have
z∈S1
v(z) ≥ C1 ,
∀x ∈ B1+ \{0}.
Proof. By the definition of v, we know that u(z) = v(z) for all z ∈ S1 , which implies that
C1 = max v(z).
z∈S1
+
∈ Rn+1
+ \B1 . By (2), we get
z
1
C1
1
≥ n−2s · n−2s = C1 ,
v(z) = n−2s u
|z|
|z|2
|z|
1
1
For any z ∈ B+
\{0}, then
z
|z|2
∀z ∈ B1+ \{0}.
|z|
Corollary 2. Let Cr = max v(z), then
z∈Sr
v(z) ≥ Cr ,
∀z ∈ Br+ \{0}.
Proof. Let w(z) = v(rz) for all z ∈ Rn+1
+ , then w is a solution to (3). By Lemma 1, we have
w(z) = v(rz) ≥ max w(z) = max v(z) = Cr ,
z∈Sr
z∈S1
∀z ∈ B1+ \{0}.
Hence we have
v(z) ≥ Cr ,
∀z ∈ Br+ \{0}.
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
∀z ∈ ∂Σλ
∀z ∈ Σλ .
\
Rn+1
+ ,
LIOUVILLE TYPE THEOREM IN UPPER HALF SPACE
For all x ∈ ∂Rn+1
+
T
3
Σλ , it’s easy to see that |x| < |xλ |, and
∂wλ
∂v
∂v λ
(x, 0) =
(x, 0) −
(x , 0) = 0
∂νs
∂νs
∂νs
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
∂wλ
∂wλ
the Hopf’s Lemma for Ls , we know that
(z1 ) < 0, which contradicts with
(z1 ) = 0 . Therefore, we know that
∂νs
∂νs
wλ (z) ≥ 0 for all z ∈ Σλ .
Lemma 3. There exists some Λ −1 such that for all λ ≤ Λ, we have
wλ (x) ≥ 0,
∀x ∈ 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 .
4
MINGFENG ZHAO
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
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 in Σλ0 , by the Strong Maximum Principle, we have wλ0 (z) > 0 in Σλ0 . If there exists some
T
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
+
∂wλs
∂wλ0
for Ls , we have
(x0 , 0) < 0, which contradicts with
(x0 , 0) = 0. In summary, we know that wλ0 (z) > 0 for all
∂νs
∂νs
Σλ0 \Tλ0 .
|λ0 |
Claim I: For any 0 < r < min
, 1 , there exists some constant δ > 0 which only depends on λ0 and r such that
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) = 0, ∀x ∈ Γr \{0},

∂ν
s



 w (z) > 0, ∀z ∈ B + \{0}.
r
λ0
Since wλ0 (z) > 0 for all z ∈ Br+ \{0}, then min wλ0 (z) > 0. Let 0 < δ < min 1, min wλ0 (z) . By Corollary 2, we
z∈Sr
z∈Sr
have
wλ0 (z) ≥ δ,
∀z ∈ Br+ \{0}.
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+ .
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.
Proof of Theorem 1. By Lemma 4, we know that
v(x1 , · · · , xn , y) ≥ v(−x1 , · · · , xn , y),
∀z ∈ Rn+1
with x1 > 0.
+
Let v(z) = v(−x1 , · · · , xn , y) for all z ∈ Rn+1
+ \{0}, then v is also a solution of 3, 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
u
|z|n−2s
z
|z|2
,
∀z ∈ Rn+1
+ \{0}.
LIOUVILLE TYPE THEOREM IN UPPER HALF SPACE
5
Then
u(z) =
1
|z|
v
n−2s
z
|z|2
,
∀z ∈ Rn+1
+ \{0}.
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
in Rn , that is, u(x, y) = U (y) for some function U on [0, ∞). By (1), we have
[y
U (y)] = 0, then there
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 = 0, then a = 0. So u(x, y) ≡ b for all (x, y) ∈ Rn+1
+ .
y&0
∂νs
References
[1] Yanyan Li and Meijun Zhu. Uniqueness theorems through the method of moving spheres. Duke Mathematical Journal, 80:383–417, 1995.
[2] Yuan Lou and Meijun Zhu. Classifications of nonnegative solutions to some elliptic problems. Differential and Integral Equations, 12:601–
612, 1999.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
E-mail address: mingfeng.zhao@uconn.edu